id string | source string | formal_statement string | header string | lean4_code string | has_proof bool | proof_body string | natural_language null | lean_version string | split string | tags list | category null | metadata string | verification string | v4210_is_valid bool | v4210_compiles bool | v4210_has_sorry bool | v4210_latency_s float64 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
compfiles_Bulgaria1998P1 | compfiles | Copyright (c) 2023 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
# Bulgarian Mathematical Olympiad 1998, Problem 1
We will be considering colorings in 2 colors of n (distinct) points
A₁, A₂, ..., Aₙ. Call such a coloring "good" if there exist three points
Aᵢ, Aⱼ, A₂ⱼ₋ᵢ, 1 ≤ i < 2j - i ≤ n, which are colored the same color.
Find the least natural number n (n ≥ 3) such that all colorings
of n points are good.
-/
namespace Bulgaria1998P1
abbrev coloring_is_good {m : ℕ} (color : Set.Icc 1 m → Fin 2) : Prop :=
∃ i j : Set.Icc 1 m,
i < j ∧
∃ h3 : 2 * j.val - i ∈ Set.Icc 1 m,
color i = color j ∧ color i = color ⟨2 * j - i, h3⟩
abbrev all_colorings_are_good (m : ℕ) : Prop :=
3 ≤ m ∧ ∀ color : Set.Icc 1 m → Fin 2, coloring_is_good color
lemma lemma1 {m n : ℕ} (hmn : m ≤ n) (hm : all_colorings_are_good m) :
all_colorings_are_good n := by
constructor
· exact hm.1.trans hmn
· intro c
let c' : Set.Icc 1 m → Fin 2 :=
fun x ↦ c ⟨x.val, by rw [Set.mem_Icc]; exact ⟨x.prop.1, x.prop.2.trans hmn⟩⟩
obtain ⟨⟨i, hi1, hi2⟩, ⟨j, hj1, hj2⟩, hij1, hij2, hc1, hc2⟩ := hm.2 c'
use ⟨i, hi1, hi2.trans hmn⟩
use ⟨j, hj1, hj2.trans hmn⟩
simp only [Subtype.mk_lt_mk, Set.mem_Icc, tsub_le_iff_right, exists_and_left]
simp only [Subtype.mk_lt_mk] at hij1
refine ⟨hij1, ?_⟩
simp only [Set.mem_Icc, tsub_le_iff_right] at hij2
simp only [c'] at hc1
refine ⟨hc1, ?_⟩
have hij2' : 1 ≤ 2 * j - i ∧ 2 * j ≤ n + i :=
⟨hij2.1, le_add_of_le_add_right hij2.2 hmn⟩
use hij2'
def coloring_of_eight {n : ℕ} : Set.Icc 1 n → Fin 2
| ⟨1, _⟩ => 0
| ⟨2, _⟩ => 1
| ⟨3, _⟩ => 0
| ⟨4, _⟩ => 1
| ⟨5, _⟩ => 1
| ⟨6, _⟩ => 0
| ⟨7, _⟩ => 1
| ⟨8, _⟩ => 0
| _ => 0 -- unreachable
lemma lemma2 :
∃ f : Set.Icc 1 8 → Fin 2, ¬coloring_is_good f := by
use coloring_of_eight
intro h
obtain ⟨⟨i, hi1, hi2⟩, ⟨j, hj1, hj2⟩, hij1, hij2, hc1, hc2⟩ := h
dsimp [coloring_of_eight] at *
interval_cases i <;> interval_cases j <;> sorry --aesop (simp_config := {decide := true})
abbrev solution_value : ℕ := 9
theorem bulgaria1998_p1 : IsLeast { m | all_colorings_are_good m } solution_value := by
constructor
· rw [Set.mem_setOf_eq]
refine ⟨by norm_num, ?_⟩
intro color
sorry
· rw [mem_lowerBounds]
intro n hn
rw [Set.mem_setOf_eq] at hn
by_contra! H
have h1 : n ≤ solution_value - 1 := Nat.le_pred_of_lt H
have ⟨h2, h3⟩ := lemma1 h1 hn
obtain ⟨f, hf⟩ := lemma2
exact (hf (h3 f)).elim
end Bulgaria1998P1 | /- | /-
Copyright (c) 2023 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
# Bulgarian Mathematical Olympiad 1998, Problem 1
We will be considering colorings in 2 colors of n (distinct) points
A₁, A₂, ..., Aₙ. Call such a coloring "good" if there exist three points
Aᵢ, Aⱼ, A₂ⱼ₋ᵢ, 1 ≤ i < 2j - i ≤ n, which are colored the same color.
Find the least natural number n (n ≥ 3) such that all colorings
of n points are good.
-/
namespace Bulgaria1998P1
abbrev coloring_is_good {m : ℕ} (color : Set.Icc 1 m → Fin 2) : Prop :=
∃ i j : Set.Icc 1 m,
i < j ∧
∃ h3 : 2 * j.val - i ∈ Set.Icc 1 m,
color i = color j ∧ color i = color ⟨2 * j - i, h3⟩
abbrev all_colorings_are_good (m : ℕ) : Prop :=
3 ≤ m ∧ ∀ color : Set.Icc 1 m → Fin 2, coloring_is_good color
lemma lemma1 {m n : ℕ} (hmn : m ≤ n) (hm : all_colorings_are_good m) :
all_colorings_are_good n := by
constructor
· exact hm.1.trans hmn
· intro c
let c' : Set.Icc 1 m → Fin 2 :=
fun x ↦ c ⟨x.val, by rw [Set.mem_Icc]; exact ⟨x.prop.1, x.prop.2.trans hmn⟩⟩
obtain ⟨⟨i, hi1, hi2⟩, ⟨j, hj1, hj2⟩, hij1, hij2, hc1, hc2⟩ := hm.2 c'
use ⟨i, hi1, hi2.trans hmn⟩
use ⟨j, hj1, hj2.trans hmn⟩
simp only [Subtype.mk_lt_mk, Set.mem_Icc, tsub_le_iff_right, exists_and_left]
simp only [Subtype.mk_lt_mk] at hij1
refine ⟨hij1, ?_⟩
simp only [Set.mem_Icc, tsub_le_iff_right] at hij2
simp only [c'] at hc1
refine ⟨hc1, ?_⟩
have hij2' : 1 ≤ 2 * j - i ∧ 2 * j ≤ n + i :=
⟨hij2.1, le_add_of_le_add_right hij2.2 hmn⟩
use hij2'
def coloring_of_eight {n : ℕ} : Set.Icc 1 n → Fin 2
| ⟨1, _⟩ => 0
| ⟨2, _⟩ => 1
| ⟨3, _⟩ => 0
| ⟨4, _⟩ => 1
| ⟨5, _⟩ => 1
| ⟨6, _⟩ => 0
| ⟨7, _⟩ => 1
| ⟨8, _⟩ => 0
| _ => 0 -- unreachable
lemma lemma2 :
∃ f : Set.Icc 1 8 → Fin 2, ¬coloring_is_good f := by
use coloring_of_eight
intro h
obtain ⟨⟨i, hi1, hi2⟩, ⟨j, hj1, hj2⟩, hij1, hij2, hc1, hc2⟩ := h
dsimp [coloring_of_eight] at *
interval_cases i <;> interval_cases j <;> sorry --aesop (simp_config := {decide := true})
abbrev solution_value : ℕ := 9
theorem bulgaria1998_p1 : IsLeast { m | all_colorings_are_good m } solution_value := by
constructor
· rw [Set.mem_setOf_eq]
refine ⟨by norm_num, ?_⟩
intro color
sorry
· rw [mem_lowerBounds]
intro n hn
rw [Set.mem_setOf_eq] at hn
by_contra! H
have h1 : n ≤ solution_value - 1 := Nat.le_pred_of_lt H
have ⟨h2, h3⟩ := lemma1 h1 hn
obtain ⟨f, hf⟩ := lemma2
exact (hf (h3 f)).elim
end Bulgaria1998P1
| false | null | null | v4.29.0-rc6 | test | [
"competition"
] | null | {"filename": "Bulgaria1998P1.lean"} | {"v4.21.0": {"is_valid": false, "has_sorry": true, "errors": [], "timeout_s": 600.0, "latency_s": 34.5438, "verified_at": "2026-03-26T18:15:55.711664+00:00"}} | false | true | true | 34.5438 |
compfiles_Bulgaria1998P11 | compfiles | Copyright (c) 2023 The Compfiles Contributors. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw, Adam Kurkiewicz
-/
import Mathlib
/-!
Bulgarian Mathematical Olympiad 1998, Problem 11
Let m,n be natural numbers such that
A = ((m + 3)ⁿ + 1) / (3m)
is an integer. Prove that A is odd.
-/
namespace Bulgaria1998P11
lemma mod_3_add_3_under_exponent (m n : ℕ) : ((m + 3) ^ n) ≡ (m ^ n) [MOD 3] := by
change (m + 3)^n % 3 = m ^ n % 3
simp [Nat.pow_mod]
lemma zero_pow_mod_3 {m n : ℕ} (h1 : n > 0) (h2 : m ≡ 0 [MOD 3]) : m ^ n ≡ 0 [MOD 3]:= by
change _ % 3 = 0 at h2 ⊢
rw [←Nat.dvd_iff_mod_eq_zero] at h2 ⊢
exact Dvd.dvd.pow h2 (Nat.ne_zero_of_lt h1)
lemma one_pow_mod_3 {m n : ℕ} (h2 : m ≡ 1 [MOD 3]) : m ^ n ≡ 1 [MOD 3]:= by
change _ % _ = 1 at h2 ⊢
simp [Nat.pow_mod, h2]
lemma two_even_pow_mod_3 {m n : ℕ} (h1 : Even n) (h2 : m ≡ 2 [MOD 3]) : m ^ n ≡ 1 [MOD 3] := by
change _ % _ = _ at h2 ⊢
obtain ⟨k, hk⟩ := h1
simp (config := { arith := true, decide := true }) [Nat.pow_mod, h2, hk, pow_mul]
theorem n_odd_and_m_eq_2_mod_3 (m n A : ℕ) (h : 3 * m * A = (m + 3)^n + 1) :
Odd n ∧ m ≡ 2 [MOD 3] := by
by_cases n_gt_zero : n > 0
· have h_mod_3 : 0 ≡ (m ^ n + 1) [MOD 3] := by
calc 0 ≡ 3 * (m * A) [MOD 3] := (Nat.mul_mod_right 3 (m * A)).symm
_ ≡ (3 * m * A) [MOD 3] := by rw[show 3 * (m * A) = (3 * m * A) by ring]
_ ≡ ((m + 3) ^ n + 1) [MOD 3] := by rw[h]
_ ≡ (m ^ n + 1) [MOD 3] := Nat.ModEq.add (mod_3_add_3_under_exponent m n) rfl
mod_cases mod_case : m % 3
· have towards_contradiction : 0 ≡ 1 [MOD 3] := by
calc 0 ≡ m ^ n + 1 [MOD 3] := h_mod_3
_ ≡ 0 + 1 [MOD 3] := Nat.ModEq.add (zero_pow_mod_3 n_gt_zero mod_case) rfl
_ ≡ 1 [MOD 3] := by rfl
contradiction
· have towards_contradiction : 0 ≡ 2 [MOD 3] := by
rw[show 2 = 1 + 1 by ring]
calc 0 ≡ m ^ n + 1 [MOD 3] := h_mod_3
_ ≡ 1 + 1 [MOD 3] := Nat.ModEq.add (one_pow_mod_3 mod_case) rfl
contradiction
· by_cases n_even : Even n
· have towards_contradiction : 0 ≡ 2 [MOD 3] := by
calc 0 ≡ m ^ n + 1 [MOD 3] := h_mod_3
_ ≡ 1 + 1 [MOD 3] := Nat.ModEq.add (two_even_pow_mod_3 n_even mod_case) rfl
_ ≡ 2 [MOD 3] := by rfl
contradiction
· rw [Nat.not_even_iff_odd] at n_even
exact ⟨n_even, mod_case⟩
· rw [Nat.eq_zero_of_not_pos n_gt_zero] at h
omega
lemma not_one_le_k {k : ℕ} (h : ¬1 ≤ k) : k = 0 := by
simp_all only [not_le, Nat.lt_one_iff]
lemma two_le_pow_two (l : ℕ) : 2 ≤ 2 ^ (l + 1) := by
rw [pow_succ]
suffices 1 ≤ 2 ^ l from Nat.le_mul_of_pos_left 2 this
exact Nat.one_le_two_pow
lemma two_n_and_rest_factorisation (m : ℕ) (even_m : Even m) (h: 0 < m) :
∃ (l : ℕ) (m₁ : ℕ), 1 ≤ l ∧ Odd m₁ ∧ m = 2 ^ l * m₁ := by
have ⟨a, ha⟩ := (pow_padicValNat_dvd : 2 ^ padicValNat 2 m ∣ m)
refine ⟨padicValNat 2 m, a, ?_⟩
obtain ⟨k, hk⟩ := even_iff_two_dvd.mp even_m
have hk0 : 0 < k := by omega
refine ⟨?_, ?_, ha⟩
· rw [hk]
rw [show 2 * k = 2^1 * k from rfl]
rw [padicValNat_base_pow_mul one_lt_two hk0.ne']
exact Nat.le_add_left 1 (padicValNat 2 k)
· by_contra! Ha
rw [Nat.not_odd_iff_even] at Ha
obtain ⟨b, hb⟩ := Ha
have h3 : 2 ^ (padicValNat 2 m + 1) ∣ m := by
use b
rw [hb] at ha
rw [pow_succ]
linarith only [ha]
have h4 := (Nat.pow_dvd_iff_le_padicValNat (by omega) (Nat.ne_zero_of_lt h)).mp h3
exact (lt_self_iff_false _).mp (Nat.succ_le_iff.mp h4)
lemma m_mod_2_contradiction (m n A : ℕ)
(even_A : Even A)
(m_eq_2_mod_4 : m ≡ 2 [MOD 4])
(h : 3 * m * A = (m + 3)^n + 1) : False := by
rw [← ZMod.natCast_eq_natCast_iff] at m_eq_2_mod_4
apply_fun (fun x ↦ (x : ZMod 4)) at h
push_cast at h
rw [m_eq_2_mod_4] at h
obtain ⟨a, rfl⟩ := even_A
push_cast at h
rw [←two_mul] at h
ring_nf at h; reduce_mod_char at h
rw [one_pow n] at h
simp (config := { arith := true, decide := true })at h
lemma m_add_3_pow_n_mod_m (n m : ℕ) : (m + 3)^n ≡ 3^n [MOD m] := by
simp [Nat.ModEq, Nat.pow_mod]
lemma too_good_to_be_true (n l : ℕ)
(three_le_l : 3 ≤ l)
(two_pow_l_divides_expresion : 2^l ∣ (3^n + 1))
(expression_eq_4_mod_8 : 3^n + 1 ≡ 4 [MOD 8]) : False := by
have : 8 ∣ 3 ^ n + 1 := by
obtain ⟨a, Ha⟩ := two_pow_l_divides_expresion
obtain ⟨b, Hb⟩ := Nat.exists_eq_add_of_le' three_le_l
rw[Hb] at Ha
dsimp [Dvd.dvd]
use 2 ^ b * a
rw[Ha]
ring
obtain ⟨a, Ha⟩ := this
rw [Ha] at expression_eq_4_mod_8
dsimp[Nat.ModEq] at expression_eq_4_mod_8
simp at expression_eq_4_mod_8
lemma thue
(a n : ℕ) (hn : 1 < n) :
∃ (x y : ℤ), (a : ZMod n) * x + y = 0 ∧ 0 < |x| ∧ x ^ 2 ≤ n ∧ y ^ 2 ≤ n := by
-- https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvMy8yL2QzYjEzOGM0ODE3YzYwZGU4NGFmOTEwZDc0ZGNhODRjOGMyMzZlLnBkZg==&rn=dGh1ZS12NC5wZGY=
-- let r = ⌊√n⌋, i.e. r is the unique integer for which
-- r² ≤ n < (r + 1)².
let r := Nat.sqrt n
-- The number of pairs (x,y) so that 0 ≤ x,y ≤ r is (r + 1)² which
-- is greater than n.
--
-- Therefore there must be two different pairs (x₁, y₁) and (x₂, y₂)
-- such that
-- ax₁ + y₁ = ax₂ + ay₂ (mod n)
-- i.e.
-- a(x₁ - x₂) = y₂ - y₁ (mod n)
--
let D := Fin (r + 1) × Fin (r + 1)
let f : D → ZMod n := fun ⟨x,y⟩ ↦ a * (x.val : ZMod n) + (y.val : ZMod n)
have cardD : Fintype.card D = (r + 1)^2 := by
unfold D
rw [sq, Fintype.card_prod, Fintype.card_fin]
have : NeZero n := NeZero.of_gt hn
have cardZ : Fintype.card (ZMod n) = n := ZMod.card n
have hDZ : Fintype.card (ZMod n) < Fintype.card D := by
rw [cardD, cardZ]
exact Nat.lt_succ_sqrt' n
obtain ⟨⟨x₁, y₁⟩, ⟨x₂, y₂⟩, hxyn, hxy⟩ :=
Fintype.exists_ne_map_eq_of_card_lt f hDZ
dsimp [f] at hxy
replace hxy : (a : ZMod n) * (↑↑x₁ - ↑↑x₂) + (↑↑y₁ - ↑↑y₂) = 0 :=
by linear_combination hxy
use (x₁ : ℤ) - (x₂ : ℤ)
use (y₁ : ℤ) - (y₂ : ℤ)
have hx1 : ((x₁ : ℕ): ZMod n) = (((x₁ : ℕ): ℤ) : ZMod n) := by
norm_cast
have hx2 : ((x₂ : ℕ): ZMod n) = (((x₂ : ℕ): ℤ) : ZMod n) := by
norm_cast
have hx3 : (((x₁ : ℕ): ℤ) : ZMod n) - (((x₂ : ℕ): ℤ) : ZMod n) =
(((((x₁ : ℕ): ℤ) -((x₂ : ℕ): ℤ)) : ℤ) : ZMod n) := by
norm_cast
have hy1 : ((y₁ : ℕ): ZMod n) = (((y₁ : ℕ): ℤ) : ZMod n) := by
norm_cast
have hy2 : ((y₂ : ℕ): ZMod n) = (((y₂ : ℕ): ℤ) : ZMod n) := by
norm_cast
have hy3 : (((y₁ : ℕ): ℤ) : ZMod n) - (((y₂ : ℕ): ℤ) : ZMod n) =
(((((y₁ : ℕ): ℤ) -((y₂ : ℕ): ℤ)) : ℤ) : ZMod n) := by
norm_cast
refine ⟨?_, ?_, ?_, ?_⟩
· rw [hx1, hx2, hx3, hy1, hy2, hy3] at hxy
rw [hxy]
· by_contra! H
replace H : |((x₁:ℕ):ℤ) - ↑↑x₂| = 0 := by
have : (0 : ℤ) ≤ |((x₁:ℕ):ℤ) - ↑↑x₂| := abs_nonneg _
exact (Int.le_antisymm this H).symm
rw [abs_eq_zero] at H
replace H : ((x₁:ℕ):ℤ) = ↑↑x₂ := Int.eq_of_sub_eq_zero H
replace H : x₁.val = x₂.val := Int.ofNat_inj.mp H
replace H : x₁ = x₂ := Fin.eq_of_val_eq H
rw [H, sub_self, mul_zero, zero_add] at hxy
have h4 : ((y₁:ℕ): ZMod n) = ↑↑y₂ := by linear_combination hxy
rw [ZMod.natCast_eq_natCast_iff'] at h4
have p1 := y₁.prop
have p2 := y₂.prop
have : r + 1 ≤ n := by
have h31 := Nat.sqrt_lt_self hn
omega
have h10 : y₁.val < n := by omega
have h11 : y₂.val < n := by omega
rw [Nat.mod_eq_of_lt h10, Nat.mod_eq_of_lt h11] at h4
have h12 : y₁ = y₂ := Fin.eq_of_val_eq h4
rw [H,h12] at hxyn
simp at hxyn
· have hx₁ : x₁.val < r + 1 := x₁.prop
have hx₂ : x₂.val < r + 1 := x₂.prop
have hr1 : r^2 ≤ n := Nat.sqrt_le' n
nlinarith only [hx₁, hx₂, hr1]
· have hy₁ : y₁.val < r + 1 := y₁.prop
have hy₂ : y₂.val < r + 1 := y₂.prop
have hr1 : r^2 ≤ n := Nat.sqrt_le' n
nlinarith only [hy₁, hy₂, hr1]
/--
In this version of Thue's lemma, we get an `x` that is nonzero and
a `y` that might be zero. If we wanted to guarantee that `y` is nonzero,
we would need an extra hypothesis `Nat.Coprime a n`.
-/
lemma Thue's_lemma
(a n : ℕ)
(hn : 1 < n) :
∃ (x y : ℤ),
a * x + y ≡ 0 [ZMOD n] ∧
0 < |x| ∧
x ^ 2 ≤ n ∧
y ^ 2 ≤ n := by
obtain ⟨x, y, hxay, hx, hx1, hy1⟩ := thue a n hn
refine ⟨x, y, ?_, hx, hx1, hy1⟩
have h1 : (a : ZMod n) * (x : ZMod n) + (y : ZMod n) =
((((a : ℤ) * x + y):ℤ) : ZMod n) := by norm_cast
have h2 : ((0:ℤ):ZMod n) = 0 := by norm_cast
rw [h1, ←h2] at hxay
exact (ZMod.intCast_eq_intCast_iff (a * x + y) 0 n).mp hxay
example (a b n : ℕ) (ha : a < n + 1) (hb : b < n + 1) : ((a : ℤ) - (b : ℤ))^2 ≤ n^2 := by
nlinarith
lemma square_mod_4_zmod (x : ZMod 4) : x ^ 2 = 1 ∨ x ^ 2 = 0 := by
fin_cases x <;> simp (config := { arith := true, decide := true })lemma square_mod_3_zmod_0 : ∀ {x : ZMod 3} (_ : x ^ 2 = 0), x = 0 := by
decide
lemma leaf_contradiction {x y m₁ : ℤ} (h: 3 * x ^ 2 + y ^ 2 = m₁) (h2 : m₁ - 5 ≡ 0 [ZMOD 6]) :
False := by
replace h2 : m₁ - 5 ≡ 0 [ZMOD 3] := Int.ModEq.of_mul_left 2 h2
rw [show 3 = ((3:ℕ):ℤ) by rfl] at h2
rw [← ZMod.intCast_eq_intCast_iff] at h2
apply_fun (fun x ↦ (x : ZMod 3)) at h
push_cast at h h2
replace h2 : (m₁:ZMod 3) = 5 := by linear_combination h2
rw [h2] at h
reduce_mod_char at h
generalize hz : (y : ZMod 3) = z
fin_cases z <;> rw [hz] at h <;> grind
theorem bulgaria1998_p11
(m n A : ℕ)
(h : 3 * m * A = (m + 3)^n + 1) : Odd A := by
-- Follows the second solution in _Mathematical Olympiads 1998-1999_
-- (edited by Titu Andreescu and Zuming Feng)
have ⟨⟨k, Hk⟩, m_eq_2_mod_3⟩ : Odd n ∧ m ≡ 2 [MOD 3] := n_odd_and_m_eq_2_mod_3 m n A h
by_contra even_A
rw [Nat.not_odd_iff_even] at even_A
have zero_lt_m : 0 < m := by
by_contra m_eq_0
replace m_eq_0 := not_one_le_k m_eq_0
rw [m_eq_0] at h
ring_nf at h
have : 1 + 3 ^ n > 0 := by positivity
rw [← h] at this
contradiction
have even_m : Even m := by
have n_ne_zero : n ≠ 0 := by omega
replace evenA : (A : ZMod 2) = (0:ℕ) := by
rw [Nat.even_iff] at even_A
exact (ZMod.natCast_eq_natCast_iff' A 0 2).mpr even_A
suffices H : (m : ZMod 2) = (0:ℕ)
· rw [ZMod.natCast_eq_natCast_iff] at H
exact Nat.even_iff.mpr H
apply_fun (fun x ↦ (x : ZMod 2)) at h
push_cast at h
rw [evenA] at h
simp only [Nat.cast_zero, mul_zero] at h
generalize hm' : (m : ZMod 2) = m'
rw [hm'] at h
fin_cases m'
· rfl
· dsimp only [Nat.reduceAdd, Fin.mk_one, Fin.isValue] at h
reduce_mod_char at h
rw [zero_pow n_ne_zero, zero_add] at h
exact (zero_ne_one h).elim
obtain ⟨l, m₁, ⟨one_le_l, odd_m₁, m_factorisation⟩⟩ :=
two_n_and_rest_factorisation m even_m zero_lt_m
by_cases l_eq_one : (l = 1)
· rw [l_eq_one] at m_factorisation
ring_nf at m_factorisation
have : m₁ ≡ 1 [MOD 4] ∨ m₁ ≡ 3 [MOD 4] := by
obtain ⟨a, rfl⟩ := odd_m₁
obtain ⟨b, rfl⟩ | ⟨b, rfl⟩ := Nat.even_or_odd a
· left
dsimp [Nat.ModEq]
ring_nf
simp
· right
dsimp [Nat.ModEq]
ring_nf
simp
have m_eq_2_mod_4 : m ≡ 2 [MOD 4] := by
rw [m_factorisation]
obtain left | right := this
· nth_rw 2 [show 2 = 1 * 2 by rfl]
exact Nat.ModEq.mul_right _ left
· calc m₁ * 2 ≡ 3 * 2 [MOD 4] := Nat.ModEq.mul right rfl
_ ≡ 2 [MOD 4] := rfl
exact m_mod_2_contradiction m n A even_A m_eq_2_mod_4 h
· have eq_2 : 0 ≡ 3^n + 1 [MOD m] := by
calc 0 ≡ m * (3 * A) [MOD m] := by
rw[show 0 = 0 * (3 * A) by ring]
refine Nat.ModEq.mul_right _ ?_
simp [Nat.ModEq]
_ ≡ 3 * m * A [MOD m] := by rw[show m * (3 * A) = 3 * m * A by ring]
_ ≡ (m + 3)^n + 1 [MOD m] := by rw[h]
_ ≡ 3^n + 1 [MOD m] := by
refine Nat.ModEq.add_right _ ?_
exact m_add_3_pow_n_mod_m n m
have l_eq_2 : l = 2 := by
have two_le_l : 2 ≤ l := by omega
obtain left | right := lt_or_eq_of_le two_le_l
· have two_pow_l_divides_expresion : 2 ^ l ∣ (3^n + 1) := by
have m_divides_expression : m ∣ (3 ^ n) + 1 := by
exact (@Nat.modEq_zero_iff_dvd m (3^n + 1)).mp eq_2.symm
dsimp[Dvd.dvd]
obtain ⟨a, Ha⟩ := m_divides_expression
use m₁ * a
rw[show 2 ^ l * (m₁ * a) = (2 ^ l * m₁) * a by ring]
rw[← m_factorisation]
exact Ha
have expression_eq_4_mod_8 : 3^n + 1 ≡ 4 [MOD 8] := by
rw [←ZMod.natCast_eq_natCast_iff, Hk]
rw [pow_succ, pow_mul]
rw[show (3 ^ 2) = 9 by rfl]
push_cast
reduce_mod_char
rw [one_pow (M := ZMod 8) k]
simp (config := { arith := true, decide := true })exfalso
exact too_good_to_be_true
n l (show 3 ≤ l by exact left) two_pow_l_divides_expresion expression_eq_4_mod_8
· exact right.symm
have m_eq_4_m₁ : m = 4 * m₁ := by
rw[l_eq_2] at m_factorisation
exact m_factorisation
have m₁_divides_expresion : m₁ ∣ (3^n + 1) := by
have m_divides_expression : m ∣ (3 ^ n) + 1 := by
exact (@Nat.modEq_zero_iff_dvd m (3^n + 1)).mp eq_2.symm
dsimp[Dvd.dvd]
obtain ⟨a, Ha⟩ := m_divides_expression
use 2 ^ l * a
rw[show m₁ * (2 ^ l * a) = (2 ^ l * m₁) * a by ring]
rw[← m_factorisation]
exact Ha
have m₁_eq_2_mod_3 : m₁ ≡ 2 [MOD 3] := by
have step_1 : 4 * m₁ ≡ 2 [MOD 3] := by
rw[m_eq_4_m₁] at m_eq_2_mod_3
exact m_eq_2_mod_3
have step_2 : 4 * m₁ ≡ m₁ [MOD 3] := by
nth_rw 2 [show m₁ = 1 * m₁ by ring]
exact Nat.ModEq.mul rfl rfl
calc m₁ ≡ 4 * m₁ [MOD 3] := step_2.symm
_ ≡ 2 [MOD 3] := step_1
have m₁_eq_5_mod_6 : m₁ ≡ 5 [MOD 6] := by
mod_cases m_mod_six : m₁ % 6
· have step_1: m₁ * 2 ≡ 0 * 2 [MOD 6]:= Nat.ModEq.mul_right 2 m_mod_six
have step_2: m₁ * 2 ≡ 2 * 2 [MOD 3 * 2] := Nat.ModEq.mul_right' 2 m₁_eq_2_mod_3
simp at step_1
simp at step_2
exfalso
have := Nat.ModEq.trans step_2.symm step_1
contradiction
· have step_1: m₁ * 2 ≡ 1 * 2 [MOD 6]:= Nat.ModEq.mul_right 2 m_mod_six
have step_2: m₁ * 2 ≡ 2 * 2 [MOD 3 * 2] := Nat.ModEq.mul_right' 2 m₁_eq_2_mod_3
simp at step_1
simp at step_2
exfalso
have := Nat.ModEq.trans step_2.symm step_1
contradiction
· have : m₁ ≡ 1 [MOD 2] := by
obtain ⟨a, Ha⟩ := odd_m₁
rw[show 1 = 0 + 1 by norm_num]
rw[Ha]
refine Nat.ModEq.add_right _ ?_
simp[Nat.ModEq]
have step_1: m₁ * 2 ≡ 2 * 2 [MOD 3 * 2] := Nat.ModEq.mul_right' 2 m₁_eq_2_mod_3
have step_2: m₁ * 3 ≡ 1 * 3 [MOD 2 * 3] := Nat.ModEq.mul_right' 3 this
change m₁ * 2 + m₁ ≡ 2 * 2 + 5 [MOD 2 * 3] at step_2
exact Nat.ModEq.add_left_cancel step_1 step_2
· have step_1: m₁ * 2 ≡ 3 * 2 [MOD 6]:= Nat.ModEq.mul_right 2 m_mod_six
have step_2: m₁ * 2 ≡ 2 * 2 [MOD 3 * 2] := Nat.ModEq.mul_right' 2 m₁_eq_2_mod_3
simp at step_1
simp at step_2
exfalso
have := Nat.ModEq.trans step_2.symm step_1
contradiction
· have step_1: m₁ * 2 ≡ 4 * 2 [MOD 6]:= Nat.ModEq.mul_right 2 m_mod_six
have step_2: m₁ * 2 ≡ 2 * 2 [MOD 3 * 2] := Nat.ModEq.mul_right' 2 m₁_eq_2_mod_3
simp at step_1
simp at step_2
exfalso
have := Nat.ModEq.trans step_2.symm step_1
contradiction
· exact m_mod_six
have exists_a : ∃ a : ℕ, a = 3 ^ (k + 1) := by
use 3 ^ (k + 1)
obtain ⟨a, Ha⟩ := exists_a
have m₁_divides_for_thues_lemma : m₁ ∣ a^2 + 3 := by
apply Nat.modEq_zero_iff_dvd.mp
have step_1:= Nat.modEq_zero_iff_dvd.mpr m₁_divides_expresion
rw[Hk] at step_1
have step_2: 3 * (3 ^ (2 * k + 1) + 1) ≡ 3 * 0 [MOD m₁] := Nat.ModEq.mul_left 3 step_1
ring_nf at step_2
rw[show 3 ^ (k * 2) * 9 = 3 ^ ((k + 1) * 2 ) by ring] at step_2
have step_3 : 3 ^ ((k + 1) * 2) = (3 ^ (k + 1)) ^ (2) :=
pow_mul 3 (k + 1) 2
rw[step_3] at step_2
rw[← Ha] at step_2
ring_nf
exact step_2
have hn1 : 1 < m₁ := by
change _ % _ = _ % _ at m₁_eq_2_mod_3
omega
obtain ⟨x, y, x_y_props⟩ := Thue's_lemma a m₁ hn1
clear hn1
obtain ⟨mod_expression, x_lower, x_higher, y_higher⟩ := x_y_props
have lifted_m₁_result := Int.natCast_modEq_iff.mpr (Nat.modEq_zero_iff_dvd.mpr m₁_divides_for_thues_lemma)
norm_num at lifted_m₁_result
have expression : 3 * x ^ 2 + y ^ 2 ≡ 0 [ZMOD m₁] := by
have step_2 : a * x ≡ -y [ZMOD m₁] := by
rw [show -y = 0 - y by ring]
rw [show a * x = a * x + y - y by ring]
exact Int.ModEq.sub_right _ mod_expression
have step_3 : a ^ 2 * x ^ 2 ≡ y ^ 2 [ZMOD m₁] := by
rw [show a ^ 2 * x ^ 2 = (a * x) * (a * x) by ring]
rw [show y ^ 2 = (-y) * (-y) by ring]
exact Int.ModEq.mul step_2 step_2
have step_4: (-3) * x ^ 2 ≡ y ^ 2 [ZMOD m₁] := by
have step_1 : a ^ 2 ≡ -3 [ZMOD m₁] :=
Int.ModEq.add_right_cancel' 3 lifted_m₁_result
trans a ^ 2 * x ^ 2
· exact Int.ModEq.mul (step_1.symm) rfl
· exact step_3
have : ((-3) * x ^ 2) + (3 * x ^ 2) ≡ (y ^ 2) + (3 * x ^ 2) [ZMOD m₁] :=
Int.ModEq.add_right _ step_4
rw [show (-3) * x ^ 2 + (3 * x ^ 2) = 0 by ring] at this
rw [show y ^ 2 + 3 * x ^ 2 = 3 * x ^ 2 + y ^ 2 by ring] at this
exact this.symm
obtain ⟨s, Hs⟩ := Int.modEq_zero_iff_dvd.mp expression
rw [m_eq_4_m₁] at zero_lt_m
have zero_lt_m₁ : 0 < @Nat.cast ℤ _ m₁ := by omega
have upper_bound_s : s ≤ 4 := by
have upper_bound_expression: 3 * x ^ 2 + y ^ 2 ≤ 4 * m₁ := by omega
rw [Hs] at upper_bound_expression
rw [show m₁ * s = s * m₁ by ring] at upper_bound_expression
exact le_of_mul_le_mul_right upper_bound_expression zero_lt_m₁
have lower_bound_s : 0 < s := by
have lower_bound_expression : 0 < 3 * x ^ 2 + y ^ 2 := by
have h1 : 0 < 3 * x ^ 2 := by rw [←sq_abs]; positivity
exact Int.add_pos_of_pos_of_nonneg h1 (sq_nonneg y)
rw [Hs] at lower_bound_expression
exact Int.pos_of_mul_pos_right lower_bound_expression zero_lt_m₁
have m₁_sub_5_mod_6 : ↑m₁ - 5 ≡ 0 [ZMOD 6] := by
exact Int.ModEq.sub (Int.natCast_modEq_iff.mpr m₁_eq_5_mod_6) rfl
interval_cases s
· -- s = 1
rw [mul_one] at Hs
exact leaf_contradiction Hs m₁_sub_5_mod_6
· -- s = 2
have := Int.modEq_zero_iff_dvd.mp m₁_sub_5_mod_6
dsimp[Dvd.dvd] at this
obtain ⟨c, this⟩ := this
have expr_m₁_mod_6 : ↑m₁ = 6 * c + 5 := sub_eq_iff_eq_add.mp this
rw[expr_m₁_mod_6] at Hs
ring_nf at Hs
have expression_mod_4 : ((x : (ZMod 4)) ^ 2 * 3 + y ^ 2) = ((10 + c * 12) : (ZMod 4)) := by
norm_cast
rw[Hs]
reduce_mod_char at expression_mod_4
obtain left_x | right_x := square_mod_4_zmod x
· rw[left_x] at expression_mod_4
obtain left_y | right_y := square_mod_4_zmod y
· rw[left_y] at expression_mod_4
simp (config := { arith := true, decide := true })at expression_mod_4
· rw[right_y] at expression_mod_4
simp (config := { arith := true, decide := true })at expression_mod_4
· rw[right_x] at expression_mod_4
obtain left_y | right_y := square_mod_4_zmod y
· rw[left_y] at expression_mod_4
simp (config := { arith := true, decide := true })at expression_mod_4
· rw[right_y] at expression_mod_4
simp (config := { arith := true, decide := true })at expression_mod_4
· -- s = 3
have := Int.modEq_zero_iff_dvd.mp m₁_sub_5_mod_6
dsimp[Dvd.dvd] at this
obtain ⟨c, this⟩ := this
have expr_m₁_mod_6 : ↑m₁ = 6 * c + 5 := by omega
rw[expr_m₁_mod_6] at Hs
ring_nf at Hs
have expression_mod_3 :
((x : (ZMod 3)) ^ 2 * 3 + y ^ 2) = ((15 + c * 18) : (ZMod 3)) := by
norm_cast
rw[Hs]
reduce_mod_char at expression_mod_3
have : (y : ZMod 3) = 0 := square_mod_3_zmod_0 expression_mod_3
have := (ZMod.intCast_zmod_eq_zero_iff_dvd y 3).mp this
dsimp[Dvd.dvd] at this
obtain⟨y', Hy'⟩ := this
rw[Hy'] at Hs
ring_nf at Hs
rw[show x ^ 2 * 3 + y' ^ 2 * 9 = 3 * (x ^ 2 + 3 * y' ^ 2) by ring] at Hs
rw[show 15 + c * 18 = 3 * (5 + 6 * c) by ring] at Hs
have reduced_expression : (x ^ 2 + 3 * y' ^ 2) = (5 + 6 * c) := by
omega
rw[show 5 + 6 * c = 6 * c + 5 by ring] at reduced_expression
rw[← expr_m₁_mod_6] at reduced_expression
rw[show x ^ 2 + 3 * y' ^ 2 = 3 * y' ^ 2 + x ^ 2 by ring] at reduced_expression
exact leaf_contradiction reduced_expression m₁_sub_5_mod_6
· -- s = 4
have : y^2 < m₁ := by
by_contra! H
have h1 : y^2 = m₁ := (Int.le_antisymm H y_higher).symm
rw [←sq_abs, Int.abs_eq_natAbs] at h1
norm_cast at h1
generalize hyy : y.natAbs = yy
rw [hyy] at h1
rw [←h1] at m₁_eq_2_mod_3
mod_cases Hyy : yy % 3 <;>
change _ % _ = _ % _ at Hyy m₁_eq_2_mod_3 <;>
simp [Nat.pow_mod, Hyy] at m₁_eq_2_mod_3
omega
end Bulgaria1998P11 | /- | /-
Copyright (c) 2023 The Compfiles Contributors. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw, Adam Kurkiewicz
-/
import Mathlib
/-!
Bulgarian Mathematical Olympiad 1998, Problem 11
Let m,n be natural numbers such that
A = ((m + 3)ⁿ + 1) / (3m)
is an integer. Prove that A is odd.
-/
namespace Bulgaria1998P11
lemma mod_3_add_3_under_exponent (m n : ℕ) : ((m + 3) ^ n) ≡ (m ^ n) [MOD 3] := by
change (m + 3)^n % 3 = m ^ n % 3
simp [Nat.pow_mod]
lemma zero_pow_mod_3 {m n : ℕ} (h1 : n > 0) (h2 : m ≡ 0 [MOD 3]) : m ^ n ≡ 0 [MOD 3]:= by
change _ % 3 = 0 at h2 ⊢
rw [←Nat.dvd_iff_mod_eq_zero] at h2 ⊢
exact Dvd.dvd.pow h2 (Nat.ne_zero_of_lt h1)
lemma one_pow_mod_3 {m n : ℕ} (h2 : m ≡ 1 [MOD 3]) : m ^ n ≡ 1 [MOD 3]:= by
change _ % _ = 1 at h2 ⊢
simp [Nat.pow_mod, h2]
lemma two_even_pow_mod_3 {m n : ℕ} (h1 : Even n) (h2 : m ≡ 2 [MOD 3]) : m ^ n ≡ 1 [MOD 3] := by
change _ % _ = _ at h2 ⊢
obtain ⟨k, hk⟩ := h1
simp (config := { arith := true, decide := true }) [Nat.pow_mod, h2, hk, pow_mul]
theorem n_odd_and_m_eq_2_mod_3 (m n A : ℕ) (h : 3 * m * A = (m + 3)^n + 1) :
Odd n ∧ m ≡ 2 [MOD 3] := by
by_cases n_gt_zero : n > 0
· have h_mod_3 : 0 ≡ (m ^ n + 1) [MOD 3] := by
calc 0 ≡ 3 * (m * A) [MOD 3] := (Nat.mul_mod_right 3 (m * A)).symm
_ ≡ (3 * m * A) [MOD 3] := by rw[show 3 * (m * A) = (3 * m * A) by ring]
_ ≡ ((m + 3) ^ n + 1) [MOD 3] := by rw[h]
_ ≡ (m ^ n + 1) [MOD 3] := Nat.ModEq.add (mod_3_add_3_under_exponent m n) rfl
mod_cases mod_case : m % 3
· have towards_contradiction : 0 ≡ 1 [MOD 3] := by
calc 0 ≡ m ^ n + 1 [MOD 3] := h_mod_3
_ ≡ 0 + 1 [MOD 3] := Nat.ModEq.add (zero_pow_mod_3 n_gt_zero mod_case) rfl
_ ≡ 1 [MOD 3] := by rfl
contradiction
· have towards_contradiction : 0 ≡ 2 [MOD 3] := by
rw[show 2 = 1 + 1 by ring]
calc 0 ≡ m ^ n + 1 [MOD 3] := h_mod_3
_ ≡ 1 + 1 [MOD 3] := Nat.ModEq.add (one_pow_mod_3 mod_case) rfl
contradiction
· by_cases n_even : Even n
· have towards_contradiction : 0 ≡ 2 [MOD 3] := by
calc 0 ≡ m ^ n + 1 [MOD 3] := h_mod_3
_ ≡ 1 + 1 [MOD 3] := Nat.ModEq.add (two_even_pow_mod_3 n_even mod_case) rfl
_ ≡ 2 [MOD 3] := by rfl
contradiction
· rw [Nat.not_even_iff_odd] at n_even
exact ⟨n_even, mod_case⟩
· rw [Nat.eq_zero_of_not_pos n_gt_zero] at h
omega
lemma not_one_le_k {k : ℕ} (h : ¬1 ≤ k) : k = 0 := by
simp_all only [not_le, Nat.lt_one_iff]
lemma two_le_pow_two (l : ℕ) : 2 ≤ 2 ^ (l + 1) := by
rw [pow_succ]
suffices 1 ≤ 2 ^ l from Nat.le_mul_of_pos_left 2 this
exact Nat.one_le_two_pow
lemma two_n_and_rest_factorisation (m : ℕ) (even_m : Even m) (h: 0 < m) :
∃ (l : ℕ) (m₁ : ℕ), 1 ≤ l ∧ Odd m₁ ∧ m = 2 ^ l * m₁ := by
have ⟨a, ha⟩ := (pow_padicValNat_dvd : 2 ^ padicValNat 2 m ∣ m)
refine ⟨padicValNat 2 m, a, ?_⟩
obtain ⟨k, hk⟩ := even_iff_two_dvd.mp even_m
have hk0 : 0 < k := by omega
refine ⟨?_, ?_, ha⟩
· rw [hk]
rw [show 2 * k = 2^1 * k from rfl]
rw [padicValNat_base_pow_mul one_lt_two hk0.ne']
exact Nat.le_add_left 1 (padicValNat 2 k)
· by_contra! Ha
rw [Nat.not_odd_iff_even] at Ha
obtain ⟨b, hb⟩ := Ha
have h3 : 2 ^ (padicValNat 2 m + 1) ∣ m := by
use b
rw [hb] at ha
rw [pow_succ]
linarith only [ha]
have h4 := (Nat.pow_dvd_iff_le_padicValNat (by omega) (Nat.ne_zero_of_lt h)).mp h3
exact (lt_self_iff_false _).mp (Nat.succ_le_iff.mp h4)
lemma m_mod_2_contradiction (m n A : ℕ)
(even_A : Even A)
(m_eq_2_mod_4 : m ≡ 2 [MOD 4])
(h : 3 * m * A = (m + 3)^n + 1) : False := by
rw [← ZMod.natCast_eq_natCast_iff] at m_eq_2_mod_4
apply_fun (fun x ↦ (x : ZMod 4)) at h
push_cast at h
rw [m_eq_2_mod_4] at h
obtain ⟨a, rfl⟩ := even_A
push_cast at h
rw [←two_mul] at h
ring_nf at h; reduce_mod_char at h
rw [one_pow n] at h
simp (config := { arith := true, decide := true })at h
lemma m_add_3_pow_n_mod_m (n m : ℕ) : (m + 3)^n ≡ 3^n [MOD m] := by
simp [Nat.ModEq, Nat.pow_mod]
lemma too_good_to_be_true (n l : ℕ)
(three_le_l : 3 ≤ l)
(two_pow_l_divides_expresion : 2^l ∣ (3^n + 1))
(expression_eq_4_mod_8 : 3^n + 1 ≡ 4 [MOD 8]) : False := by
have : 8 ∣ 3 ^ n + 1 := by
obtain ⟨a, Ha⟩ := two_pow_l_divides_expresion
obtain ⟨b, Hb⟩ := Nat.exists_eq_add_of_le' three_le_l
rw[Hb] at Ha
dsimp [Dvd.dvd]
use 2 ^ b * a
rw[Ha]
ring
obtain ⟨a, Ha⟩ := this
rw [Ha] at expression_eq_4_mod_8
dsimp[Nat.ModEq] at expression_eq_4_mod_8
simp at expression_eq_4_mod_8
lemma thue
(a n : ℕ) (hn : 1 < n) :
∃ (x y : ℤ), (a : ZMod n) * x + y = 0 ∧ 0 < |x| ∧ x ^ 2 ≤ n ∧ y ^ 2 ≤ n := by
-- https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvMy8yL2QzYjEzOGM0ODE3YzYwZGU4NGFmOTEwZDc0ZGNhODRjOGMyMzZlLnBkZg==&rn=dGh1ZS12NC5wZGY=
-- let r = ⌊√n⌋, i.e. r is the unique integer for which
-- r² ≤ n < (r + 1)².
let r := Nat.sqrt n
-- The number of pairs (x,y) so that 0 ≤ x,y ≤ r is (r + 1)² which
-- is greater than n.
--
-- Therefore there must be two different pairs (x₁, y₁) and (x₂, y₂)
-- such that
-- ax₁ + y₁ = ax₂ + ay₂ (mod n)
-- i.e.
-- a(x₁ - x₂) = y₂ - y₁ (mod n)
--
let D := Fin (r + 1) × Fin (r + 1)
let f : D → ZMod n := fun ⟨x,y⟩ ↦ a * (x.val : ZMod n) + (y.val : ZMod n)
have cardD : Fintype.card D = (r + 1)^2 := by
unfold D
rw [sq, Fintype.card_prod, Fintype.card_fin]
have : NeZero n := NeZero.of_gt hn
have cardZ : Fintype.card (ZMod n) = n := ZMod.card n
have hDZ : Fintype.card (ZMod n) < Fintype.card D := by
rw [cardD, cardZ]
exact Nat.lt_succ_sqrt' n
obtain ⟨⟨x₁, y₁⟩, ⟨x₂, y₂⟩, hxyn, hxy⟩ :=
Fintype.exists_ne_map_eq_of_card_lt f hDZ
dsimp [f] at hxy
replace hxy : (a : ZMod n) * (↑↑x₁ - ↑↑x₂) + (↑↑y₁ - ↑↑y₂) = 0 :=
by linear_combination hxy
use (x₁ : ℤ) - (x₂ : ℤ)
use (y₁ : ℤ) - (y₂ : ℤ)
have hx1 : ((x₁ : ℕ): ZMod n) = (((x₁ : ℕ): ℤ) : ZMod n) := by
norm_cast
have hx2 : ((x₂ : ℕ): ZMod n) = (((x₂ : ℕ): ℤ) : ZMod n) := by
norm_cast
have hx3 : (((x₁ : ℕ): ℤ) : ZMod n) - (((x₂ : ℕ): ℤ) : ZMod n) =
(((((x₁ : ℕ): ℤ) -((x₂ : ℕ): ℤ)) : ℤ) : ZMod n) := by
norm_cast
have hy1 : ((y₁ : ℕ): ZMod n) = (((y₁ : ℕ): ℤ) : ZMod n) := by
norm_cast
have hy2 : ((y₂ : ℕ): ZMod n) = (((y₂ : ℕ): ℤ) : ZMod n) := by
norm_cast
have hy3 : (((y₁ : ℕ): ℤ) : ZMod n) - (((y₂ : ℕ): ℤ) : ZMod n) =
(((((y₁ : ℕ): ℤ) -((y₂ : ℕ): ℤ)) : ℤ) : ZMod n) := by
norm_cast
refine ⟨?_, ?_, ?_, ?_⟩
· rw [hx1, hx2, hx3, hy1, hy2, hy3] at hxy
rw [hxy]
· by_contra! H
replace H : |((x₁:ℕ):ℤ) - ↑↑x₂| = 0 := by
have : (0 : ℤ) ≤ |((x₁:ℕ):ℤ) - ↑↑x₂| := abs_nonneg _
exact (Int.le_antisymm this H).symm
rw [abs_eq_zero] at H
replace H : ((x₁:ℕ):ℤ) = ↑↑x₂ := Int.eq_of_sub_eq_zero H
replace H : x₁.val = x₂.val := Int.ofNat_inj.mp H
replace H : x₁ = x₂ := Fin.eq_of_val_eq H
rw [H, sub_self, mul_zero, zero_add] at hxy
have h4 : ((y₁:ℕ): ZMod n) = ↑↑y₂ := by linear_combination hxy
rw [ZMod.natCast_eq_natCast_iff'] at h4
have p1 := y₁.prop
have p2 := y₂.prop
have : r + 1 ≤ n := by
have h31 := Nat.sqrt_lt_self hn
omega
have h10 : y₁.val < n := by omega
have h11 : y₂.val < n := by omega
rw [Nat.mod_eq_of_lt h10, Nat.mod_eq_of_lt h11] at h4
have h12 : y₁ = y₂ := Fin.eq_of_val_eq h4
rw [H,h12] at hxyn
simp at hxyn
· have hx₁ : x₁.val < r + 1 := x₁.prop
have hx₂ : x₂.val < r + 1 := x₂.prop
have hr1 : r^2 ≤ n := Nat.sqrt_le' n
nlinarith only [hx₁, hx₂, hr1]
· have hy₁ : y₁.val < r + 1 := y₁.prop
have hy₂ : y₂.val < r + 1 := y₂.prop
have hr1 : r^2 ≤ n := Nat.sqrt_le' n
nlinarith only [hy₁, hy₂, hr1]
/--
In this version of Thue's lemma, we get an `x` that is nonzero and
a `y` that might be zero. If we wanted to guarantee that `y` is nonzero,
we would need an extra hypothesis `Nat.Coprime a n`.
-/
lemma Thue's_lemma
(a n : ℕ)
(hn : 1 < n) :
∃ (x y : ℤ),
a * x + y ≡ 0 [ZMOD n] ∧
0 < |x| ∧
x ^ 2 ≤ n ∧
y ^ 2 ≤ n := by
obtain ⟨x, y, hxay, hx, hx1, hy1⟩ := thue a n hn
refine ⟨x, y, ?_, hx, hx1, hy1⟩
have h1 : (a : ZMod n) * (x : ZMod n) + (y : ZMod n) =
((((a : ℤ) * x + y):ℤ) : ZMod n) := by norm_cast
have h2 : ((0:ℤ):ZMod n) = 0 := by norm_cast
rw [h1, ←h2] at hxay
exact (ZMod.intCast_eq_intCast_iff (a * x + y) 0 n).mp hxay
example (a b n : ℕ) (ha : a < n + 1) (hb : b < n + 1) : ((a : ℤ) - (b : ℤ))^2 ≤ n^2 := by
nlinarith
lemma square_mod_4_zmod (x : ZMod 4) : x ^ 2 = 1 ∨ x ^ 2 = 0 := by
fin_cases x <;> simp (config := { arith := true, decide := true })lemma square_mod_3_zmod_0 : ∀ {x : ZMod 3} (_ : x ^ 2 = 0), x = 0 := by
decide
lemma leaf_contradiction {x y m₁ : ℤ} (h: 3 * x ^ 2 + y ^ 2 = m₁) (h2 : m₁ - 5 ≡ 0 [ZMOD 6]) :
False := by
replace h2 : m₁ - 5 ≡ 0 [ZMOD 3] := Int.ModEq.of_mul_left 2 h2
rw [show 3 = ((3:ℕ):ℤ) by rfl] at h2
rw [← ZMod.intCast_eq_intCast_iff] at h2
apply_fun (fun x ↦ (x : ZMod 3)) at h
push_cast at h h2
replace h2 : (m₁:ZMod 3) = 5 := by linear_combination h2
rw [h2] at h
reduce_mod_char at h
generalize hz : (y : ZMod 3) = z
fin_cases z <;> rw [hz] at h <;> grind
theorem bulgaria1998_p11
(m n A : ℕ)
(h : 3 * m * A = (m + 3)^n + 1) : Odd A := by
-- Follows the second solution in _Mathematical Olympiads 1998-1999_
-- (edited by Titu Andreescu and Zuming Feng)
have ⟨⟨k, Hk⟩, m_eq_2_mod_3⟩ : Odd n ∧ m ≡ 2 [MOD 3] := n_odd_and_m_eq_2_mod_3 m n A h
by_contra even_A
rw [Nat.not_odd_iff_even] at even_A
have zero_lt_m : 0 < m := by
by_contra m_eq_0
replace m_eq_0 := not_one_le_k m_eq_0
rw [m_eq_0] at h
ring_nf at h
have : 1 + 3 ^ n > 0 := by positivity
rw [← h] at this
contradiction
have even_m : Even m := by
have n_ne_zero : n ≠ 0 := by omega
replace evenA : (A : ZMod 2) = (0:ℕ) := by
rw [Nat.even_iff] at even_A
exact (ZMod.natCast_eq_natCast_iff' A 0 2).mpr even_A
suffices H : (m : ZMod 2) = (0:ℕ)
· rw [ZMod.natCast_eq_natCast_iff] at H
exact Nat.even_iff.mpr H
apply_fun (fun x ↦ (x : ZMod 2)) at h
push_cast at h
rw [evenA] at h
simp only [Nat.cast_zero, mul_zero] at h
generalize hm' : (m : ZMod 2) = m'
rw [hm'] at h
fin_cases m'
· rfl
· dsimp only [Nat.reduceAdd, Fin.mk_one, Fin.isValue] at h
reduce_mod_char at h
rw [zero_pow n_ne_zero, zero_add] at h
exact (zero_ne_one h).elim
obtain ⟨l, m₁, ⟨one_le_l, odd_m₁, m_factorisation⟩⟩ :=
two_n_and_rest_factorisation m even_m zero_lt_m
by_cases l_eq_one : (l = 1)
· rw [l_eq_one] at m_factorisation
ring_nf at m_factorisation
have : m₁ ≡ 1 [MOD 4] ∨ m₁ ≡ 3 [MOD 4] := by
obtain ⟨a, rfl⟩ := odd_m₁
obtain ⟨b, rfl⟩ | ⟨b, rfl⟩ := Nat.even_or_odd a
· left
dsimp [Nat.ModEq]
ring_nf
simp
· right
dsimp [Nat.ModEq]
ring_nf
simp
have m_eq_2_mod_4 : m ≡ 2 [MOD 4] := by
rw [m_factorisation]
obtain left | right := this
· nth_rw 2 [show 2 = 1 * 2 by rfl]
exact Nat.ModEq.mul_right _ left
· calc m₁ * 2 ≡ 3 * 2 [MOD 4] := Nat.ModEq.mul right rfl
_ ≡ 2 [MOD 4] := rfl
exact m_mod_2_contradiction m n A even_A m_eq_2_mod_4 h
· have eq_2 : 0 ≡ 3^n + 1 [MOD m] := by
calc 0 ≡ m * (3 * A) [MOD m] := by
rw[show 0 = 0 * (3 * A) by ring]
refine Nat.ModEq.mul_right _ ?_
simp [Nat.ModEq]
_ ≡ 3 * m * A [MOD m] := by rw[show m * (3 * A) = 3 * m * A by ring]
_ ≡ (m + 3)^n + 1 [MOD m] := by rw[h]
_ ≡ 3^n + 1 [MOD m] := by
refine Nat.ModEq.add_right _ ?_
exact m_add_3_pow_n_mod_m n m
have l_eq_2 : l = 2 := by
have two_le_l : 2 ≤ l := by omega
obtain left | right := lt_or_eq_of_le two_le_l
· have two_pow_l_divides_expresion : 2 ^ l ∣ (3^n + 1) := by
have m_divides_expression : m ∣ (3 ^ n) + 1 := by
exact (@Nat.modEq_zero_iff_dvd m (3^n + 1)).mp eq_2.symm
dsimp[Dvd.dvd]
obtain ⟨a, Ha⟩ := m_divides_expression
use m₁ * a
rw[show 2 ^ l * (m₁ * a) = (2 ^ l * m₁) * a by ring]
rw[← m_factorisation]
exact Ha
have expression_eq_4_mod_8 : 3^n + 1 ≡ 4 [MOD 8] := by
rw [←ZMod.natCast_eq_natCast_iff, Hk]
rw [pow_succ, pow_mul]
rw[show (3 ^ 2) = 9 by rfl]
push_cast
reduce_mod_char
rw [one_pow (M := ZMod 8) k]
simp (config := { arith := true, decide := true })exfalso
exact too_good_to_be_true
n l (show 3 ≤ l by exact left) two_pow_l_divides_expresion expression_eq_4_mod_8
· exact right.symm
have m_eq_4_m₁ : m = 4 * m₁ := by
rw[l_eq_2] at m_factorisation
exact m_factorisation
have m₁_divides_expresion : m₁ ∣ (3^n + 1) := by
have m_divides_expression : m ∣ (3 ^ n) + 1 := by
exact (@Nat.modEq_zero_iff_dvd m (3^n + 1)).mp eq_2.symm
dsimp[Dvd.dvd]
obtain ⟨a, Ha⟩ := m_divides_expression
use 2 ^ l * a
rw[show m₁ * (2 ^ l * a) = (2 ^ l * m₁) * a by ring]
rw[← m_factorisation]
exact Ha
have m₁_eq_2_mod_3 : m₁ ≡ 2 [MOD 3] := by
have step_1 : 4 * m₁ ≡ 2 [MOD 3] := by
rw[m_eq_4_m₁] at m_eq_2_mod_3
exact m_eq_2_mod_3
have step_2 : 4 * m₁ ≡ m₁ [MOD 3] := by
nth_rw 2 [show m₁ = 1 * m₁ by ring]
exact Nat.ModEq.mul rfl rfl
calc m₁ ≡ 4 * m₁ [MOD 3] := step_2.symm
_ ≡ 2 [MOD 3] := step_1
have m₁_eq_5_mod_6 : m₁ ≡ 5 [MOD 6] := by
mod_cases m_mod_six : m₁ % 6
· have step_1: m₁ * 2 ≡ 0 * 2 [MOD 6]:= Nat.ModEq.mul_right 2 m_mod_six
have step_2: m₁ * 2 ≡ 2 * 2 [MOD 3 * 2] := Nat.ModEq.mul_right' 2 m₁_eq_2_mod_3
simp at step_1
simp at step_2
exfalso
have := Nat.ModEq.trans step_2.symm step_1
contradiction
· have step_1: m₁ * 2 ≡ 1 * 2 [MOD 6]:= Nat.ModEq.mul_right 2 m_mod_six
have step_2: m₁ * 2 ≡ 2 * 2 [MOD 3 * 2] := Nat.ModEq.mul_right' 2 m₁_eq_2_mod_3
simp at step_1
simp at step_2
exfalso
have := Nat.ModEq.trans step_2.symm step_1
contradiction
· have : m₁ ≡ 1 [MOD 2] := by
obtain ⟨a, Ha⟩ := odd_m₁
rw[show 1 = 0 + 1 by norm_num]
rw[Ha]
refine Nat.ModEq.add_right _ ?_
simp[Nat.ModEq]
have step_1: m₁ * 2 ≡ 2 * 2 [MOD 3 * 2] := Nat.ModEq.mul_right' 2 m₁_eq_2_mod_3
have step_2: m₁ * 3 ≡ 1 * 3 [MOD 2 * 3] := Nat.ModEq.mul_right' 3 this
change m₁ * 2 + m₁ ≡ 2 * 2 + 5 [MOD 2 * 3] at step_2
exact Nat.ModEq.add_left_cancel step_1 step_2
· have step_1: m₁ * 2 ≡ 3 * 2 [MOD 6]:= Nat.ModEq.mul_right 2 m_mod_six
have step_2: m₁ * 2 ≡ 2 * 2 [MOD 3 * 2] := Nat.ModEq.mul_right' 2 m₁_eq_2_mod_3
simp at step_1
simp at step_2
exfalso
have := Nat.ModEq.trans step_2.symm step_1
contradiction
· have step_1: m₁ * 2 ≡ 4 * 2 [MOD 6]:= Nat.ModEq.mul_right 2 m_mod_six
have step_2: m₁ * 2 ≡ 2 * 2 [MOD 3 * 2] := Nat.ModEq.mul_right' 2 m₁_eq_2_mod_3
simp at step_1
simp at step_2
exfalso
have := Nat.ModEq.trans step_2.symm step_1
contradiction
· exact m_mod_six
have exists_a : ∃ a : ℕ, a = 3 ^ (k + 1) := by
use 3 ^ (k + 1)
obtain ⟨a, Ha⟩ := exists_a
have m₁_divides_for_thues_lemma : m₁ ∣ a^2 + 3 := by
apply Nat.modEq_zero_iff_dvd.mp
have step_1:= Nat.modEq_zero_iff_dvd.mpr m₁_divides_expresion
rw[Hk] at step_1
have step_2: 3 * (3 ^ (2 * k + 1) + 1) ≡ 3 * 0 [MOD m₁] := Nat.ModEq.mul_left 3 step_1
ring_nf at step_2
rw[show 3 ^ (k * 2) * 9 = 3 ^ ((k + 1) * 2 ) by ring] at step_2
have step_3 : 3 ^ ((k + 1) * 2) = (3 ^ (k + 1)) ^ (2) :=
pow_mul 3 (k + 1) 2
rw[step_3] at step_2
rw[← Ha] at step_2
ring_nf
exact step_2
have hn1 : 1 < m₁ := by
change _ % _ = _ % _ at m₁_eq_2_mod_3
omega
obtain ⟨x, y, x_y_props⟩ := Thue's_lemma a m₁ hn1
clear hn1
obtain ⟨mod_expression, x_lower, x_higher, y_higher⟩ := x_y_props
have lifted_m₁_result := Int.natCast_modEq_iff.mpr (Nat.modEq_zero_iff_dvd.mpr m₁_divides_for_thues_lemma)
norm_num at lifted_m₁_result
have expression : 3 * x ^ 2 + y ^ 2 ≡ 0 [ZMOD m₁] := by
have step_2 : a * x ≡ -y [ZMOD m₁] := by
rw [show -y = 0 - y by ring]
rw [show a * x = a * x + y - y by ring]
exact Int.ModEq.sub_right _ mod_expression
have step_3 : a ^ 2 * x ^ 2 ≡ y ^ 2 [ZMOD m₁] := by
rw [show a ^ 2 * x ^ 2 = (a * x) * (a * x) by ring]
rw [show y ^ 2 = (-y) * (-y) by ring]
exact Int.ModEq.mul step_2 step_2
have step_4: (-3) * x ^ 2 ≡ y ^ 2 [ZMOD m₁] := by
have step_1 : a ^ 2 ≡ -3 [ZMOD m₁] :=
Int.ModEq.add_right_cancel' 3 lifted_m₁_result
trans a ^ 2 * x ^ 2
· exact Int.ModEq.mul (step_1.symm) rfl
· exact step_3
have : ((-3) * x ^ 2) + (3 * x ^ 2) ≡ (y ^ 2) + (3 * x ^ 2) [ZMOD m₁] :=
Int.ModEq.add_right _ step_4
rw [show (-3) * x ^ 2 + (3 * x ^ 2) = 0 by ring] at this
rw [show y ^ 2 + 3 * x ^ 2 = 3 * x ^ 2 + y ^ 2 by ring] at this
exact this.symm
obtain ⟨s, Hs⟩ := Int.modEq_zero_iff_dvd.mp expression
rw [m_eq_4_m₁] at zero_lt_m
have zero_lt_m₁ : 0 < @Nat.cast ℤ _ m₁ := by omega
have upper_bound_s : s ≤ 4 := by
have upper_bound_expression: 3 * x ^ 2 + y ^ 2 ≤ 4 * m₁ := by omega
rw [Hs] at upper_bound_expression
rw [show m₁ * s = s * m₁ by ring] at upper_bound_expression
exact le_of_mul_le_mul_right upper_bound_expression zero_lt_m₁
have lower_bound_s : 0 < s := by
have lower_bound_expression : 0 < 3 * x ^ 2 + y ^ 2 := by
have h1 : 0 < 3 * x ^ 2 := by rw [←sq_abs]; positivity
exact Int.add_pos_of_pos_of_nonneg h1 (sq_nonneg y)
rw [Hs] at lower_bound_expression
exact Int.pos_of_mul_pos_right lower_bound_expression zero_lt_m₁
have m₁_sub_5_mod_6 : ↑m₁ - 5 ≡ 0 [ZMOD 6] := by
exact Int.ModEq.sub (Int.natCast_modEq_iff.mpr m₁_eq_5_mod_6) rfl
interval_cases s
· -- s = 1
rw [mul_one] at Hs
exact leaf_contradiction Hs m₁_sub_5_mod_6
· -- s = 2
have := Int.modEq_zero_iff_dvd.mp m₁_sub_5_mod_6
dsimp[Dvd.dvd] at this
obtain ⟨c, this⟩ := this
have expr_m₁_mod_6 : ↑m₁ = 6 * c + 5 := sub_eq_iff_eq_add.mp this
rw[expr_m₁_mod_6] at Hs
ring_nf at Hs
have expression_mod_4 : ((x : (ZMod 4)) ^ 2 * 3 + y ^ 2) = ((10 + c * 12) : (ZMod 4)) := by
norm_cast
rw[Hs]
reduce_mod_char at expression_mod_4
obtain left_x | right_x := square_mod_4_zmod x
· rw[left_x] at expression_mod_4
obtain left_y | right_y := square_mod_4_zmod y
· rw[left_y] at expression_mod_4
simp (config := { arith := true, decide := true })at expression_mod_4
· rw[right_y] at expression_mod_4
simp (config := { arith := true, decide := true })at expression_mod_4
· rw[right_x] at expression_mod_4
obtain left_y | right_y := square_mod_4_zmod y
· rw[left_y] at expression_mod_4
simp (config := { arith := true, decide := true })at expression_mod_4
· rw[right_y] at expression_mod_4
simp (config := { arith := true, decide := true })at expression_mod_4
· -- s = 3
have := Int.modEq_zero_iff_dvd.mp m₁_sub_5_mod_6
dsimp[Dvd.dvd] at this
obtain ⟨c, this⟩ := this
have expr_m₁_mod_6 : ↑m₁ = 6 * c + 5 := by omega
rw[expr_m₁_mod_6] at Hs
ring_nf at Hs
have expression_mod_3 :
((x : (ZMod 3)) ^ 2 * 3 + y ^ 2) = ((15 + c * 18) : (ZMod 3)) := by
norm_cast
rw[Hs]
reduce_mod_char at expression_mod_3
have : (y : ZMod 3) = 0 := square_mod_3_zmod_0 expression_mod_3
have := (ZMod.intCast_zmod_eq_zero_iff_dvd y 3).mp this
dsimp[Dvd.dvd] at this
obtain⟨y', Hy'⟩ := this
rw[Hy'] at Hs
ring_nf at Hs
rw[show x ^ 2 * 3 + y' ^ 2 * 9 = 3 * (x ^ 2 + 3 * y' ^ 2) by ring] at Hs
rw[show 15 + c * 18 = 3 * (5 + 6 * c) by ring] at Hs
have reduced_expression : (x ^ 2 + 3 * y' ^ 2) = (5 + 6 * c) := by
omega
rw[show 5 + 6 * c = 6 * c + 5 by ring] at reduced_expression
rw[← expr_m₁_mod_6] at reduced_expression
rw[show x ^ 2 + 3 * y' ^ 2 = 3 * y' ^ 2 + x ^ 2 by ring] at reduced_expression
exact leaf_contradiction reduced_expression m₁_sub_5_mod_6
· -- s = 4
have : y^2 < m₁ := by
by_contra! H
have h1 : y^2 = m₁ := (Int.le_antisymm H y_higher).symm
rw [←sq_abs, Int.abs_eq_natAbs] at h1
norm_cast at h1
generalize hyy : y.natAbs = yy
rw [hyy] at h1
rw [←h1] at m₁_eq_2_mod_3
mod_cases Hyy : yy % 3 <;>
change _ % _ = _ % _ at Hyy m₁_eq_2_mod_3 <;>
simp [Nat.pow_mod, Hyy] at m₁_eq_2_mod_3
omega
end Bulgaria1998P11
| true | Copyright (c) 2023 The Compfiles Contributors. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw, Adam Kurkiewicz
-/
import Mathlib
/-!
Bulgarian Mathematical Olympiad 1998, Problem 11
Let m,n be natural numbers such that
A = ((m + 3)ⁿ + 1) / (3m)
is an integer. Prove that A is odd.
-/
namespace Bulgaria1998P11
lemma mod_3_add_3_under_exponent (m n : ℕ) : ((m + 3) ^ n) ≡ (m ^ n) [MOD 3] := by
change (m + 3)^n % 3 = m ^ n % 3
simp [Nat.pow_mod]
lemma zero_pow_mod_3 {m n : ℕ} (h1 : n > 0) (h2 : m ≡ 0 [MOD 3]) : m ^ n ≡ 0 [MOD 3]:= by
change _ % 3 = 0 at h2 ⊢
rw [←Nat.dvd_iff_mod_eq_zero] at h2 ⊢
exact Dvd.dvd.pow h2 (Nat.ne_zero_of_lt h1)
lemma one_pow_mod_3 {m n : ℕ} (h2 : m ≡ 1 [MOD 3]) : m ^ n ≡ 1 [MOD 3]:= by
change _ % _ = 1 at h2 ⊢
simp [Nat.pow_mod, h2]
lemma two_even_pow_mod_3 {m n : ℕ} (h1 : Even n) (h2 : m ≡ 2 [MOD 3]) : m ^ n ≡ 1 [MOD 3] := by
change _ % _ = _ at h2 ⊢
obtain ⟨k, hk⟩ := h1
simp (config := { arith := true, decide := true }) [Nat.pow_mod, h2, hk, pow_mul]
theorem n_odd_and_m_eq_2_mod_3 (m n A : ℕ) (h : 3 * m * A = (m + 3)^n + 1) :
Odd n ∧ m ≡ 2 [MOD 3] := by
by_cases n_gt_zero : n > 0
· have h_mod_3 : 0 ≡ (m ^ n + 1) [MOD 3] := by
calc 0 ≡ 3 * (m * A) [MOD 3] := (Nat.mul_mod_right 3 (m * A)).symm
_ ≡ (3 * m * A) [MOD 3] := by rw[show 3 * (m * A) = (3 * m * A) by ring]
_ ≡ ((m + 3) ^ n + 1) [MOD 3] := by rw[h]
_ ≡ (m ^ n + 1) [MOD 3] := Nat.ModEq.add (mod_3_add_3_under_exponent m n) rfl
mod_cases mod_case : m % 3
· have towards_contradiction : 0 ≡ 1 [MOD 3] := by
calc 0 ≡ m ^ n + 1 [MOD 3] := h_mod_3
_ ≡ 0 + 1 [MOD 3] := Nat.ModEq.add (zero_pow_mod_3 n_gt_zero mod_case) rfl
_ ≡ 1 [MOD 3] := by rfl
contradiction
· have towards_contradiction : 0 ≡ 2 [MOD 3] := by
rw[show 2 = 1 + 1 by ring]
calc 0 ≡ m ^ n + 1 [MOD 3] := h_mod_3
_ ≡ 1 + 1 [MOD 3] := Nat.ModEq.add (one_pow_mod_3 mod_case) rfl
contradiction
· by_cases n_even : Even n
· have towards_contradiction : 0 ≡ 2 [MOD 3] := by
calc 0 ≡ m ^ n + 1 [MOD 3] := h_mod_3
_ ≡ 1 + 1 [MOD 3] := Nat.ModEq.add (two_even_pow_mod_3 n_even mod_case) rfl
_ ≡ 2 [MOD 3] := by rfl
contradiction
· rw [Nat.not_even_iff_odd] at n_even
exact ⟨n_even, mod_case⟩
· rw [Nat.eq_zero_of_not_pos n_gt_zero] at h
omega
lemma not_one_le_k {k : ℕ} (h : ¬1 ≤ k) : k = 0 := by
simp_all only [not_le, Nat.lt_one_iff]
lemma two_le_pow_two (l : ℕ) : 2 ≤ 2 ^ (l + 1) := by
rw [pow_succ]
suffices 1 ≤ 2 ^ l from Nat.le_mul_of_pos_left 2 this
exact Nat.one_le_two_pow
lemma two_n_and_rest_factorisation (m : ℕ) (even_m : Even m) (h: 0 < m) :
∃ (l : ℕ) (m₁ : ℕ), 1 ≤ l ∧ Odd m₁ ∧ m = 2 ^ l * m₁ := by
have ⟨a, ha⟩ := (pow_padicValNat_dvd : 2 ^ padicValNat 2 m ∣ m)
refine ⟨padicValNat 2 m, a, ?_⟩
obtain ⟨k, hk⟩ := even_iff_two_dvd.mp even_m
have hk0 : 0 < k := by omega
refine ⟨?_, ?_, ha⟩
· rw [hk]
rw [show 2 * k = 2^1 * k from rfl]
rw [padicValNat_base_pow_mul one_lt_two hk0.ne']
exact Nat.le_add_left 1 (padicValNat 2 k)
· by_contra! Ha
rw [Nat.not_odd_iff_even] at Ha
obtain ⟨b, hb⟩ := Ha
have h3 : 2 ^ (padicValNat 2 m + 1) ∣ m := by
use b
rw [hb] at ha
rw [pow_succ]
linarith only [ha]
have h4 := (Nat.pow_dvd_iff_le_padicValNat (by omega) (Nat.ne_zero_of_lt h)).mp h3
exact (lt_self_iff_false _).mp (Nat.succ_le_iff.mp h4)
lemma m_mod_2_contradiction (m n A : ℕ)
(even_A : Even A)
(m_eq_2_mod_4 : m ≡ 2 [MOD 4])
(h : 3 * m * A = (m + 3)^n + 1) : False := by
rw [← ZMod.natCast_eq_natCast_iff] at m_eq_2_mod_4
apply_fun (fun x ↦ (x : ZMod 4)) at h
push_cast at h
rw [m_eq_2_mod_4] at h
obtain ⟨a, rfl⟩ := even_A
push_cast at h
rw [←two_mul] at h
ring_nf at h; reduce_mod_char at h
rw [one_pow n] at h
simp (config := { arith := true, decide := true })at h
lemma m_add_3_pow_n_mod_m (n m : ℕ) : (m + 3)^n ≡ 3^n [MOD m] := by
simp [Nat.ModEq, Nat.pow_mod]
lemma too_good_to_be_true (n l : ℕ)
(three_le_l : 3 ≤ l)
(two_pow_l_divides_expresion : 2^l ∣ (3^n + 1))
(expression_eq_4_mod_8 : 3^n + 1 ≡ 4 [MOD 8]) : False := by
have : 8 ∣ 3 ^ n + 1 := by
obtain ⟨a, Ha⟩ := two_pow_l_divides_expresion
obtain ⟨b, Hb⟩ := Nat.exists_eq_add_of_le' three_le_l
rw[Hb] at Ha
dsimp [Dvd.dvd]
use 2 ^ b * a
rw[Ha]
ring
obtain ⟨a, Ha⟩ := this
rw [Ha] at expression_eq_4_mod_8
dsimp[Nat.ModEq] at expression_eq_4_mod_8
simp at expression_eq_4_mod_8
lemma thue
(a n : ℕ) (hn : 1 < n) :
∃ (x y : ℤ), (a : ZMod n) * x + y = 0 ∧ 0 < |x| ∧ x ^ 2 ≤ n ∧ y ^ 2 ≤ n := by
-- https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvMy8yL2QzYjEzOGM0ODE3YzYwZGU4NGFmOTEwZDc0ZGNhODRjOGMyMzZlLnBkZg==&rn=dGh1ZS12NC5wZGY=
-- let r = ⌊√n⌋, i.e. r is the unique integer for which
-- r² ≤ n < (r + 1)².
let r := Nat.sqrt n
-- The number of pairs (x,y) so that 0 ≤ x,y ≤ r is (r + 1)² which
-- is greater than n.
--
-- Therefore there must be two different pairs (x₁, y₁) and (x₂, y₂)
-- such that
-- ax₁ + y₁ = ax₂ + ay₂ (mod n)
-- i.e.
-- a(x₁ - x₂) = y₂ - y₁ (mod n)
--
let D := Fin (r + 1) × Fin (r + 1)
let f : D → ZMod n := fun ⟨x,y⟩ ↦ a * (x.val : ZMod n) + (y.val : ZMod n)
have cardD : Fintype.card D = (r + 1)^2 := by
unfold D
rw [sq, Fintype.card_prod, Fintype.card_fin]
have : NeZero n := NeZero.of_gt hn
have cardZ : Fintype.card (ZMod n) = n := ZMod.card n
have hDZ : Fintype.card (ZMod n) < Fintype.card D := by
rw [cardD, cardZ]
exact Nat.lt_succ_sqrt' n
obtain ⟨⟨x₁, y₁⟩, ⟨x₂, y₂⟩, hxyn, hxy⟩ :=
Fintype.exists_ne_map_eq_of_card_lt f hDZ
dsimp [f] at hxy
replace hxy : (a : ZMod n) * (↑↑x₁ - ↑↑x₂) + (↑↑y₁ - ↑↑y₂) = 0 :=
by linear_combination hxy
use (x₁ : ℤ) - (x₂ : ℤ)
use (y₁ : ℤ) - (y₂ : ℤ)
have hx1 : ((x₁ : ℕ): ZMod n) = (((x₁ : ℕ): ℤ) : ZMod n) := by
norm_cast
have hx2 : ((x₂ : ℕ): ZMod n) = (((x₂ : ℕ): ℤ) : ZMod n) := by
norm_cast
have hx3 : (((x₁ : ℕ): ℤ) : ZMod n) - (((x₂ : ℕ): ℤ) : ZMod n) =
(((((x₁ : ℕ): ℤ) -((x₂ : ℕ): ℤ)) : ℤ) : ZMod n) := by
norm_cast
have hy1 : ((y₁ : ℕ): ZMod n) = (((y₁ : ℕ): ℤ) : ZMod n) := by
norm_cast
have hy2 : ((y₂ : ℕ): ZMod n) = (((y₂ : ℕ): ℤ) : ZMod n) := by
norm_cast
have hy3 : (((y₁ : ℕ): ℤ) : ZMod n) - (((y₂ : ℕ): ℤ) : ZMod n) =
(((((y₁ : ℕ): ℤ) -((y₂ : ℕ): ℤ)) : ℤ) : ZMod n) := by
norm_cast
refine ⟨?_, ?_, ?_, ?_⟩
· rw [hx1, hx2, hx3, hy1, hy2, hy3] at hxy
rw [hxy]
· by_contra! H
replace H : |((x₁:ℕ):ℤ) - ↑↑x₂| = 0 := by
have : (0 : ℤ) ≤ |((x₁:ℕ):ℤ) - ↑↑x₂| := abs_nonneg _
exact (Int.le_antisymm this H).symm
rw [abs_eq_zero] at H
replace H : ((x₁:ℕ):ℤ) = ↑↑x₂ := Int.eq_of_sub_eq_zero H
replace H : x₁.val = x₂.val := Int.ofNat_inj.mp H
replace H : x₁ = x₂ := Fin.eq_of_val_eq H
rw [H, sub_self, mul_zero, zero_add] at hxy
have h4 : ((y₁:ℕ): ZMod n) = ↑↑y₂ := by linear_combination hxy
rw [ZMod.natCast_eq_natCast_iff'] at h4
have p1 := y₁.prop
have p2 := y₂.prop
have : r + 1 ≤ n := by
have h31 := Nat.sqrt_lt_self hn
omega
have h10 : y₁.val < n := by omega
have h11 : y₂.val < n := by omega
rw [Nat.mod_eq_of_lt h10, Nat.mod_eq_of_lt h11] at h4
have h12 : y₁ = y₂ := Fin.eq_of_val_eq h4
rw [H,h12] at hxyn
simp at hxyn
· have hx₁ : x₁.val < r + 1 := x₁.prop
have hx₂ : x₂.val < r + 1 := x₂.prop
have hr1 : r^2 ≤ n := Nat.sqrt_le' n
nlinarith only [hx₁, hx₂, hr1]
· have hy₁ : y₁.val < r + 1 := y₁.prop
have hy₂ : y₂.val < r + 1 := y₂.prop
have hr1 : r^2 ≤ n := Nat.sqrt_le' n
nlinarith only [hy₁, hy₂, hr1]
/--
In this version of Thue's lemma, we get an `x` that is nonzero and
a `y` that might be zero. If we wanted to guarantee that `y` is nonzero,
we would need an extra hypothesis `Nat.Coprime a n`.
-/
lemma Thue's_lemma
(a n : ℕ)
(hn : 1 < n) :
∃ (x y : ℤ),
a * x + y ≡ 0 [ZMOD n] ∧
0 < |x| ∧
x ^ 2 ≤ n ∧
y ^ 2 ≤ n := by
obtain ⟨x, y, hxay, hx, hx1, hy1⟩ := thue a n hn
refine ⟨x, y, ?_, hx, hx1, hy1⟩
have h1 : (a : ZMod n) * (x : ZMod n) + (y : ZMod n) =
((((a : ℤ) * x + y):ℤ) : ZMod n) := by norm_cast
have h2 : ((0:ℤ):ZMod n) = 0 := by norm_cast
rw [h1, ←h2] at hxay
exact (ZMod.intCast_eq_intCast_iff (a * x + y) 0 n).mp hxay
example (a b n : ℕ) (ha : a < n + 1) (hb : b < n + 1) : ((a : ℤ) - (b : ℤ))^2 ≤ n^2 := by
nlinarith
lemma square_mod_4_zmod (x : ZMod 4) : x ^ 2 = 1 ∨ x ^ 2 = 0 := by
fin_cases x <;> simp (config := { arith := true, decide := true })lemma square_mod_3_zmod_0 : ∀ {x : ZMod 3} (_ : x ^ 2 = 0), x = 0 := by
decide
lemma leaf_contradiction {x y m₁ : ℤ} (h: 3 * x ^ 2 + y ^ 2 = m₁) (h2 : m₁ - 5 ≡ 0 [ZMOD 6]) :
False := by
replace h2 : m₁ - 5 ≡ 0 [ZMOD 3] := Int.ModEq.of_mul_left 2 h2
rw [show 3 = ((3:ℕ):ℤ) by rfl] at h2
rw [← ZMod.intCast_eq_intCast_iff] at h2
apply_fun (fun x ↦ (x : ZMod 3)) at h
push_cast at h h2
replace h2 : (m₁:ZMod 3) = 5 := by linear_combination h2
rw [h2] at h
reduce_mod_char at h
generalize hz : (y : ZMod 3) = z
fin_cases z <;> rw [hz] at h <;> grind
theorem bulgaria1998_p11
(m n A : ℕ)
(h : 3 * m * A = (m + 3)^n + 1) : Odd A := by
-- Follows the second solution in _Mathematical Olympiads 1998-1999_
-- (edited by Titu Andreescu and Zuming Feng)
have ⟨⟨k, Hk⟩, m_eq_2_mod_3⟩ : Odd n ∧ m ≡ 2 [MOD 3] := n_odd_and_m_eq_2_mod_3 m n A h
by_contra even_A
rw [Nat.not_odd_iff_even] at even_A
have zero_lt_m : 0 < m := by
by_contra m_eq_0
replace m_eq_0 := not_one_le_k m_eq_0
rw [m_eq_0] at h
ring_nf at h
have : 1 + 3 ^ n > 0 := by positivity
rw [← h] at this
contradiction
have even_m : Even m := by
have n_ne_zero : n ≠ 0 := by omega
replace evenA : (A : ZMod 2) = (0:ℕ) := by
rw [Nat.even_iff] at even_A
exact (ZMod.natCast_eq_natCast_iff' A 0 2).mpr even_A
suffices H : (m : ZMod 2) = (0:ℕ)
· rw [ZMod.natCast_eq_natCast_iff] at H
exact Nat.even_iff.mpr H
apply_fun (fun x ↦ (x : ZMod 2)) at h
push_cast at h
rw [evenA] at h
simp only [Nat.cast_zero, mul_zero] at h
generalize hm' : (m : ZMod 2) = m'
rw [hm'] at h
fin_cases m'
· rfl
· dsimp only [Nat.reduceAdd, Fin.mk_one, Fin.isValue] at h
reduce_mod_char at h
rw [zero_pow n_ne_zero, zero_add] at h
exact (zero_ne_one h).elim
obtain ⟨l, m₁, ⟨one_le_l, odd_m₁, m_factorisation⟩⟩ :=
two_n_and_rest_factorisation m even_m zero_lt_m
by_cases l_eq_one : (l = 1)
· rw [l_eq_one] at m_factorisation
ring_nf at m_factorisation
have : m₁ ≡ 1 [MOD 4] ∨ m₁ ≡ 3 [MOD 4] := by
obtain ⟨a, rfl⟩ := odd_m₁
obtain ⟨b, rfl⟩ | ⟨b, rfl⟩ := Nat.even_or_odd a
· left
dsimp [Nat.ModEq]
ring_nf
simp
· right
dsimp [Nat.ModEq]
ring_nf
simp
have m_eq_2_mod_4 : m ≡ 2 [MOD 4] := by
rw [m_factorisation]
obtain left | right := this
· nth_rw 2 [show 2 = 1 * 2 by rfl]
exact Nat.ModEq.mul_right _ left
· calc m₁ * 2 ≡ 3 * 2 [MOD 4] := Nat.ModEq.mul right rfl
_ ≡ 2 [MOD 4] := rfl
exact m_mod_2_contradiction m n A even_A m_eq_2_mod_4 h
· have eq_2 : 0 ≡ 3^n + 1 [MOD m] := by
calc 0 ≡ m * (3 * A) [MOD m] := by
rw[show 0 = 0 * (3 * A) by ring]
refine Nat.ModEq.mul_right _ ?_
simp [Nat.ModEq]
_ ≡ 3 * m * A [MOD m] := by rw[show m * (3 * A) = 3 * m * A by ring]
_ ≡ (m + 3)^n + 1 [MOD m] := by rw[h]
_ ≡ 3^n + 1 [MOD m] := by
refine Nat.ModEq.add_right _ ?_
exact m_add_3_pow_n_mod_m n m
have l_eq_2 : l = 2 := by
have two_le_l : 2 ≤ l := by omega
obtain left | right := lt_or_eq_of_le two_le_l
· have two_pow_l_divides_expresion : 2 ^ l ∣ (3^n + 1) := by
have m_divides_expression : m ∣ (3 ^ n) + 1 := by
exact (@Nat.modEq_zero_iff_dvd m (3^n + 1)).mp eq_2.symm
dsimp[Dvd.dvd]
obtain ⟨a, Ha⟩ := m_divides_expression
use m₁ * a
rw[show 2 ^ l * (m₁ * a) = (2 ^ l * m₁) * a by ring]
rw[← m_factorisation]
exact Ha
have expression_eq_4_mod_8 : 3^n + 1 ≡ 4 [MOD 8] := by
rw [←ZMod.natCast_eq_natCast_iff, Hk]
rw [pow_succ, pow_mul]
rw[show (3 ^ 2) = 9 by rfl]
push_cast
reduce_mod_char
rw [one_pow (M := ZMod 8) k]
simp (config := { arith := true, decide := true })exfalso
exact too_good_to_be_true
n l (show 3 ≤ l by exact left) two_pow_l_divides_expresion expression_eq_4_mod_8
· exact right.symm
have m_eq_4_m₁ : m = 4 * m₁ := by
rw[l_eq_2] at m_factorisation
exact m_factorisation
have m₁_divides_expresion : m₁ ∣ (3^n + 1) := by
have m_divides_expression : m ∣ (3 ^ n) + 1 := by
exact (@Nat.modEq_zero_iff_dvd m (3^n + 1)).mp eq_2.symm
dsimp[Dvd.dvd]
obtain ⟨a, Ha⟩ := m_divides_expression
use 2 ^ l * a
rw[show m₁ * (2 ^ l * a) = (2 ^ l * m₁) * a by ring]
rw[← m_factorisation]
exact Ha
have m₁_eq_2_mod_3 : m₁ ≡ 2 [MOD 3] := by
have step_1 : 4 * m₁ ≡ 2 [MOD 3] := by
rw[m_eq_4_m₁] at m_eq_2_mod_3
exact m_eq_2_mod_3
have step_2 : 4 * m₁ ≡ m₁ [MOD 3] := by
nth_rw 2 [show m₁ = 1 * m₁ by ring]
exact Nat.ModEq.mul rfl rfl
calc m₁ ≡ 4 * m₁ [MOD 3] := step_2.symm
_ ≡ 2 [MOD 3] := step_1
have m₁_eq_5_mod_6 : m₁ ≡ 5 [MOD 6] := by
mod_cases m_mod_six : m₁ % 6
· have step_1: m₁ * 2 ≡ 0 * 2 [MOD 6]:= Nat.ModEq.mul_right 2 m_mod_six
have step_2: m₁ * 2 ≡ 2 * 2 [MOD 3 * 2] := Nat.ModEq.mul_right' 2 m₁_eq_2_mod_3
simp at step_1
simp at step_2
exfalso
have := Nat.ModEq.trans step_2.symm step_1
contradiction
· have step_1: m₁ * 2 ≡ 1 * 2 [MOD 6]:= Nat.ModEq.mul_right 2 m_mod_six
have step_2: m₁ * 2 ≡ 2 * 2 [MOD 3 * 2] := Nat.ModEq.mul_right' 2 m₁_eq_2_mod_3
simp at step_1
simp at step_2
exfalso
have := Nat.ModEq.trans step_2.symm step_1
contradiction
· have : m₁ ≡ 1 [MOD 2] := by
obtain ⟨a, Ha⟩ := odd_m₁
rw[show 1 = 0 + 1 by norm_num]
rw[Ha]
refine Nat.ModEq.add_right _ ?_
simp[Nat.ModEq]
have step_1: m₁ * 2 ≡ 2 * 2 [MOD 3 * 2] := Nat.ModEq.mul_right' 2 m₁_eq_2_mod_3
have step_2: m₁ * 3 ≡ 1 * 3 [MOD 2 * 3] := Nat.ModEq.mul_right' 3 this
change m₁ * 2 + m₁ ≡ 2 * 2 + 5 [MOD 2 * 3] at step_2
exact Nat.ModEq.add_left_cancel step_1 step_2
· have step_1: m₁ * 2 ≡ 3 * 2 [MOD 6]:= Nat.ModEq.mul_right 2 m_mod_six
have step_2: m₁ * 2 ≡ 2 * 2 [MOD 3 * 2] := Nat.ModEq.mul_right' 2 m₁_eq_2_mod_3
simp at step_1
simp at step_2
exfalso
have := Nat.ModEq.trans step_2.symm step_1
contradiction
· have step_1: m₁ * 2 ≡ 4 * 2 [MOD 6]:= Nat.ModEq.mul_right 2 m_mod_six
have step_2: m₁ * 2 ≡ 2 * 2 [MOD 3 * 2] := Nat.ModEq.mul_right' 2 m₁_eq_2_mod_3
simp at step_1
simp at step_2
exfalso
have := Nat.ModEq.trans step_2.symm step_1
contradiction
· exact m_mod_six
have exists_a : ∃ a : ℕ, a = 3 ^ (k + 1) := by
use 3 ^ (k + 1)
obtain ⟨a, Ha⟩ := exists_a
have m₁_divides_for_thues_lemma : m₁ ∣ a^2 + 3 := by
apply Nat.modEq_zero_iff_dvd.mp
have step_1:= Nat.modEq_zero_iff_dvd.mpr m₁_divides_expresion
rw[Hk] at step_1
have step_2: 3 * (3 ^ (2 * k + 1) + 1) ≡ 3 * 0 [MOD m₁] := Nat.ModEq.mul_left 3 step_1
ring_nf at step_2
rw[show 3 ^ (k * 2) * 9 = 3 ^ ((k + 1) * 2 ) by ring] at step_2
have step_3 : 3 ^ ((k + 1) * 2) = (3 ^ (k + 1)) ^ (2) :=
pow_mul 3 (k + 1) 2
rw[step_3] at step_2
rw[← Ha] at step_2
ring_nf
exact step_2
have hn1 : 1 < m₁ := by
change _ % _ = _ % _ at m₁_eq_2_mod_3
omega
obtain ⟨x, y, x_y_props⟩ := Thue's_lemma a m₁ hn1
clear hn1
obtain ⟨mod_expression, x_lower, x_higher, y_higher⟩ := x_y_props
have lifted_m₁_result := Int.natCast_modEq_iff.mpr (Nat.modEq_zero_iff_dvd.mpr m₁_divides_for_thues_lemma)
norm_num at lifted_m₁_result
have expression : 3 * x ^ 2 + y ^ 2 ≡ 0 [ZMOD m₁] := by
have step_2 : a * x ≡ -y [ZMOD m₁] := by
rw [show -y = 0 - y by ring]
rw [show a * x = a * x + y - y by ring]
exact Int.ModEq.sub_right _ mod_expression
have step_3 : a ^ 2 * x ^ 2 ≡ y ^ 2 [ZMOD m₁] := by
rw [show a ^ 2 * x ^ 2 = (a * x) * (a * x) by ring]
rw [show y ^ 2 = (-y) * (-y) by ring]
exact Int.ModEq.mul step_2 step_2
have step_4: (-3) * x ^ 2 ≡ y ^ 2 [ZMOD m₁] := by
have step_1 : a ^ 2 ≡ -3 [ZMOD m₁] :=
Int.ModEq.add_right_cancel' 3 lifted_m₁_result
trans a ^ 2 * x ^ 2
· exact Int.ModEq.mul (step_1.symm) rfl
· exact step_3
have : ((-3) * x ^ 2) + (3 * x ^ 2) ≡ (y ^ 2) + (3 * x ^ 2) [ZMOD m₁] :=
Int.ModEq.add_right _ step_4
rw [show (-3) * x ^ 2 + (3 * x ^ 2) = 0 by ring] at this
rw [show y ^ 2 + 3 * x ^ 2 = 3 * x ^ 2 + y ^ 2 by ring] at this
exact this.symm
obtain ⟨s, Hs⟩ := Int.modEq_zero_iff_dvd.mp expression
rw [m_eq_4_m₁] at zero_lt_m
have zero_lt_m₁ : 0 < @Nat.cast ℤ _ m₁ := by omega
have upper_bound_s : s ≤ 4 := by
have upper_bound_expression: 3 * x ^ 2 + y ^ 2 ≤ 4 * m₁ := by omega
rw [Hs] at upper_bound_expression
rw [show m₁ * s = s * m₁ by ring] at upper_bound_expression
exact le_of_mul_le_mul_right upper_bound_expression zero_lt_m₁
have lower_bound_s : 0 < s := by
have lower_bound_expression : 0 < 3 * x ^ 2 + y ^ 2 := by
have h1 : 0 < 3 * x ^ 2 := by rw [←sq_abs]; positivity
exact Int.add_pos_of_pos_of_nonneg h1 (sq_nonneg y)
rw [Hs] at lower_bound_expression
exact Int.pos_of_mul_pos_right lower_bound_expression zero_lt_m₁
have m₁_sub_5_mod_6 : ↑m₁ - 5 ≡ 0 [ZMOD 6] := by
exact Int.ModEq.sub (Int.natCast_modEq_iff.mpr m₁_eq_5_mod_6) rfl
interval_cases s
· -- s = 1
rw [mul_one] at Hs
exact leaf_contradiction Hs m₁_sub_5_mod_6
· -- s = 2
have := Int.modEq_zero_iff_dvd.mp m₁_sub_5_mod_6
dsimp[Dvd.dvd] at this
obtain ⟨c, this⟩ := this
have expr_m₁_mod_6 : ↑m₁ = 6 * c + 5 := sub_eq_iff_eq_add.mp this
rw[expr_m₁_mod_6] at Hs
ring_nf at Hs
have expression_mod_4 : ((x : (ZMod 4)) ^ 2 * 3 + y ^ 2) = ((10 + c * 12) : (ZMod 4)) := by
norm_cast
rw[Hs]
reduce_mod_char at expression_mod_4
obtain left_x | right_x := square_mod_4_zmod x
· rw[left_x] at expression_mod_4
obtain left_y | right_y := square_mod_4_zmod y
· rw[left_y] at expression_mod_4
simp (config := { arith := true, decide := true })at expression_mod_4
· rw[right_y] at expression_mod_4
simp (config := { arith := true, decide := true })at expression_mod_4
· rw[right_x] at expression_mod_4
obtain left_y | right_y := square_mod_4_zmod y
· rw[left_y] at expression_mod_4
simp (config := { arith := true, decide := true })at expression_mod_4
· rw[right_y] at expression_mod_4
simp (config := { arith := true, decide := true })at expression_mod_4
· -- s = 3
have := Int.modEq_zero_iff_dvd.mp m₁_sub_5_mod_6
dsimp[Dvd.dvd] at this
obtain ⟨c, this⟩ := this
have expr_m₁_mod_6 : ↑m₁ = 6 * c + 5 := by omega
rw[expr_m₁_mod_6] at Hs
ring_nf at Hs
have expression_mod_3 :
((x : (ZMod 3)) ^ 2 * 3 + y ^ 2) = ((15 + c * 18) : (ZMod 3)) := by
norm_cast
rw[Hs]
reduce_mod_char at expression_mod_3
have : (y : ZMod 3) = 0 := square_mod_3_zmod_0 expression_mod_3
have := (ZMod.intCast_zmod_eq_zero_iff_dvd y 3).mp this
dsimp[Dvd.dvd] at this
obtain⟨y', Hy'⟩ := this
rw[Hy'] at Hs
ring_nf at Hs
rw[show x ^ 2 * 3 + y' ^ 2 * 9 = 3 * (x ^ 2 + 3 * y' ^ 2) by ring] at Hs
rw[show 15 + c * 18 = 3 * (5 + 6 * c) by ring] at Hs
have reduced_expression : (x ^ 2 + 3 * y' ^ 2) = (5 + 6 * c) := by
omega
rw[show 5 + 6 * c = 6 * c + 5 by ring] at reduced_expression
rw[← expr_m₁_mod_6] at reduced_expression
rw[show x ^ 2 + 3 * y' ^ 2 = 3 * y' ^ 2 + x ^ 2 by ring] at reduced_expression
exact leaf_contradiction reduced_expression m₁_sub_5_mod_6
· -- s = 4
have : y^2 < m₁ := by
by_contra! H
have h1 : y^2 = m₁ := (Int.le_antisymm H y_higher).symm
rw [←sq_abs, Int.abs_eq_natAbs] at h1
norm_cast at h1
generalize hyy : y.natAbs = yy
rw [hyy] at h1
rw [←h1] at m₁_eq_2_mod_3
mod_cases Hyy : yy % 3 <;>
change _ % _ = _ % _ at Hyy m₁_eq_2_mod_3 <;>
simp [Nat.pow_mod, Hyy] at m₁_eq_2_mod_3
omega
end Bulgaria1998P11 | null | v4.29.0-rc6 | test | [
"competition"
] | null | {"filename": "Bulgaria1998P11.lean"} | {"v4.21.0": {"is_valid": false, "has_sorry": false, "errors": ["unknown tactic", "unsolved goals\ncase neg\nm n A : \u2115\nh : 3 * m * A = (m + 3) ^ 0 + 1\nn_gt_zero : \u00acn > 0\n\u22a2 Odd n \u2227 m \u2261 2 [MOD 3]", "unknown tactic", "unsolved goals\nm : \u2115\neven_m : Even m\nh : 0 < m\na : \u2115\nha : m = 2 ^ padicValNat 2 m * a\nk : \u2115\nhk : m = 2 * k\n\u22a2 0 < k", "unsolved goals\ncase intro\nm : \u2115\neven_m : Even m\nh : 0 < m\na : \u2115\nha : m = 2 ^ padicValNat 2 m * a\nk : \u2115\nhk : m = 2 * k\nhk0 : 0 < k\n\u22a2 1 \u2264 padicValNat 2 m \u2227 Odd a \u2227 m = 2 ^ padicValNat 2 m * a", "unknown tactic", "unsolved goals\na n : \u2115\nhn : 1 < n\nr : \u2115 := n.sqrt\nD : Type := Fin (r + 1) \u00d7 Fin (r + 1)\nf : D \u2192 ZMod n :=\n fun x =>\n match x with\n | (x, y) => \u2191a * \u2191\u2191x + \u2191\u2191y\ncardD : Fintype.card D = (r + 1) ^ 2\nthis : NeZero n\ncardZ : Fintype.card (ZMod n) = n\nhDZ : Fintype.card (ZMod n) < Fintype.card D\nx\u2081 y\u2081 x\u2082 y\u2082 : Fin (r + 1)\nhxyn : (x\u2081, y\u2081) \u2260 (x\u2082, y\u2082)\nhxy : \u2191\u2191y\u2081 - \u2191\u2191y\u2082 = 0\nhx1 : \u2191\u2191x\u2081 = \u2191\u2191\u2191x\u2081\nhx2 : \u2191\u2191x\u2082 = \u2191\u2191\u2191x\u2082\nhx3 : \u2191\u2191\u2191x\u2081 - \u2191\u2191\u2191x\u2082 = \u2191(\u2191\u2191x\u2081 - \u2191\u2191x\u2082)\nhy1 : \u2191\u2191y\u2081 = \u2191\u2191\u2191y\u2081\nhy2 : \u2191\u2191y\u2082 = \u2191\u2191\u2191y\u2082\nhy3 : \u2191\u2191\u2191y\u2081 - \u2191\u2191\u2191y\u2082 = \u2191(\u2191\u2191y\u2081 - \u2191\u2191y\u2082)\nH : x\u2081 = x\u2082\nh4 : \u2191y\u2081 % n = \u2191y\u2082 % n\np1 : \u2191y\u2081 < r + 1\np2 : \u2191y\u2082 < r + 1\nh31 : n.sqrt < n\n\u22a2 r + 1 \u2264 n", "unsolved goals\ncase h.refine_2\na n : \u2115\nhn : 1 < n\nr : \u2115 := n.sqrt\nD : Type := Fin (r + 1) \u00d7 Fin (r + 1)\nf : D \u2192 ZMod n :=\n fun x =>\n match x with\n | (x, y) => \u2191a * \u2191\u2191x + \u2191\u2191y\ncardD : Fintype.card D = (r + 1) ^ 2\nthis\u271d : NeZero n\ncardZ : Fintype.card (ZMod n) = n\nhDZ : Fintype.card (ZMod n) < Fintype.card D\nx\u2081 y\u2081 x\u2082 y\u2082 : Fin (r + 1)\nhxyn : (x\u2081, y\u2081) \u2260 (x\u2082, y\u2082)\nhxy : \u2191\u2191y\u2081 - \u2191\u2191y\u2082 = 0\nhx1 : \u2191\u2191x\u2081 = \u2191\u2191\u2191x\u2081\nhx2 : \u2191\u2191x\u2082 = \u2191\u2191\u2191x\u2082\nhx3 : \u2191\u2191\u2191x\u2081 - \u2191\u2191\u2191x\u2082 = \u2191(\u2191\u2191x\u2081 - \u2191\u2191x\u2082)\nhy1 : \u2191\u2191y\u2081 = \u2191\u2191\u2191y\u2081\nhy2 : \u2191\u2191y\u2082 = \u2191\u2191\u2191y\u2082\nhy3 : \u2191\u2191\u2191y\u2081 - \u2191\u2191\u2191y\u2082 = \u2191(\u2191\u2191y\u2081 - \u2191\u2191y\u2082)\nH : x\u2081 = x\u2082\nh4 : \u2191y\u2081 % n = \u2191y\u2082 % n\np1 : \u2191y\u2081 < r + 1\np2 : \u2191y\u2082 < r + 1\nthis : r + 1 \u2264 n\n\u22a2 False", "unsolved goals\ncase h.refine_3\na n : \u2115\nhn : 1 < n\nr : \u2115 := n.sqrt\nD : Type := Fin (r + 1) \u00d7 Fin (r + 1)\nf : D \u2192 ZMod n :=\n fun x =>\n match x with\n | (x, y) => \u2191a * \u2191\u2191x + \u2191\u2191y\ncardD : Fintype.card D = (r + 1) ^ 2\nthis : NeZero n\ncardZ : Fintype.card (ZMod n) = n\nhDZ : Fintype.card (ZMod n) < Fintype.card D\nx\u2081 y\u2081 x\u2082 y\u2082 : Fin (r + 1)\nhxyn : (x\u2081, y\u2081) \u2260 (x\u2082, y\u2082)\nhxy : \u2191a * (\u2191\u2191x\u2081 - \u2191\u2191x\u2082) + (\u2191\u2191y\u2081 - \u2191\u2191y\u2082) = 0\nhx1 : \u2191\u2191x\u2081 = \u2191\u2191\u2191x\u2081\nhx2 : \u2191\u2191x\u2082 = \u2191\u2191\u2191x\u2082\nhx3 : \u2191\u2191\u2191x\u2081 - \u2191\u2191\u2191x\u2082 = \u2191(\u2191\u2191x\u2081 - \u2191\u2191x\u2082)\nhy1 : \u2191\u2191y\u2081 = \u2191\u2191\u2191y\u2081\nhy2 : \u2191\u2191y\u2082 = \u2191\u2191\u2191y\u2082\nhy3 : \u2191\u2191\u2191y\u2081 - \u2191\u2191\u2191y\u2082 = \u2191(\u2191\u2191y\u2081 - \u2191\u2191y\u2082)\n\u22a2 (\u2191\u2191x\u2081 - \u2191\u2191x\u2082) ^ 2 \u2264 \u2191n\n\ncase h.refine_4\na n : \u2115\nhn : 1 < n\nr : \u2115 := n.sqrt\nD : Type := Fin (r + 1) \u00d7 Fin (r + 1)\nf : D \u2192 ZMod n :=\n fun x =>\n match x with\n | (x, y) => \u2191a * \u2191\u2191x + \u2191\u2191y\ncardD : Fintype.card D = (r + 1) ^ 2\nthis : NeZero n\ncardZ : Fintype.card (ZMod n) = n\nhDZ : Fintype.card (ZMod n) < Fintype.card D\nx\u2081 y\u2081 x\u2082 y\u2082 : Fin (r + 1)\nhxyn : (x\u2081, y\u2081) \u2260 (x\u2082, y\u2082)\nhxy : \u2191a * (\u2191\u2191x\u2081 - \u2191\u2191x\u2082) + (\u2191\u2191y\u2081 - \u2191\u2191y\u2082) = 0\nhx1 : \u2191\u2191x\u2081 = \u2191\u2191\u2191x\u2081\nhx2 : \u2191\u2191x\u2082 = \u2191\u2191\u2191x\u2082\nhx3 : \u2191\u2191\u2191x\u2081 - \u2191\u2191\u2191x\u2082 = \u2191(\u2191\u2191x\u2081 - \u2191\u2191x\u2082)\nhy1 : \u2191\u2191y\u2081 = \u2191\u2191\u2191y\u2081\nhy2 : \u2191\u2191y\u2082 = \u2191\u2191\u2191y\u2082\nhy3 : \u2191\u2191\u2191y\u2081 - \u2191\u2191\u2191y\u2082 = \u2191(\u2191\u2191y\u2081 - \u2191\u2191y\u2082)\n\u22a2 (\u2191\u2191y\u2081 - \u2191\u2191y\u2082) ^ 2 \u2264 \u2191n", "unknown tactic", "unsolved goals\nm n A : \u2115\nh : 3 * m * A = (m + 3) ^ n + 1\nk : \u2115\nHk : n = 2 * k + 1\nm_eq_2_mod_3 : m \u2261 2 [MOD 3]\neven_A : Even A\nzero_lt_m : 0 < m\n\u22a2 n \u2260 0", "unsolved goals\nm n A : \u2115\nh : 3 * m * A = (m + 3) ^ n + 1\nk : \u2115\nHk : n = 2 * k + 1\nm_eq_2_mod_3 : m \u2261 2 [MOD 3]\neven_A : Even A\nzero_lt_m : 0 < m\nn_ne_zero : n \u2260 0\n\u22a2 Even m", "unsolved goals\nm n A : \u2115\nh : 3 * m * A = (m + 3) ^ n + 1\nk : \u2115\nHk : n = 2 * k + 1\nm_eq_2_mod_3 : m \u2261 2 [MOD 3]\neven_A : Even A\nzero_lt_m : 0 < m\neven_m : Even m\n\u22a2 False", "unexpected identifier; expected command"], "timeout_s": 600.0, "latency_s": 40.3366, "verified_at": "2026-03-26T18:16:01.505278+00:00"}} | false | true | false | 40.3366 |
compfiles_Bulgaria1998P2 | compfiles | Copyright (c) 2023 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
# Bulgarian Mathematical Olympiad 1998, Problem 2
A convex quadrilateral ABCD has AD = CD and ∠DAB = ∠ABC < 90°.
The line through D and the midpoint of BC intersects line AB
in point E. Prove that ∠BEC = ∠DAC. (Note: The problem is valid
without the assumption ∠ABC < 90°.)
-/
namespace Bulgaria1998P2
open EuclideanGeometry
theorem bulgaria1998_p2
(A B C D E M : EuclideanSpace ℝ (Fin 2))
(H1 : dist D A = dist D C)
(H2 : ∠ D A B = ∠ A B C)
(H3 : M = midpoint ℝ B C) :
∠ B E C = ∠ D A C := by
let x := ∠ D A C
have : ∠ D A C = ∠ D C A := EuclideanGeometry.angle_eq_angle_of_dist_eq H1
let y := ∠ C A B
have : ∠ A B C = x + y := by rw [←H2]; sorry -- need to switch to oangles?
sorry
end Bulgaria1998P2 | /- | /-
Copyright (c) 2023 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
# Bulgarian Mathematical Olympiad 1998, Problem 2
A convex quadrilateral ABCD has AD = CD and ∠DAB = ∠ABC < 90°.
The line through D and the midpoint of BC intersects line AB
in point E. Prove that ∠BEC = ∠DAC. (Note: The problem is valid
without the assumption ∠ABC < 90°.)
-/
namespace Bulgaria1998P2
open EuclideanGeometry
theorem bulgaria1998_p2
(A B C D E M : EuclideanSpace ℝ (Fin 2))
(H1 : dist D A = dist D C)
(H2 : ∠ D A B = ∠ A B C)
(H3 : M = midpoint ℝ B C) :
∠ B E C = ∠ D A C := by
let x := ∠ D A C
have : ∠ D A C = ∠ D C A := EuclideanGeometry.angle_eq_angle_of_dist_eq H1
let y := ∠ C A B
have : ∠ A B C = x + y := by rw [←H2]; sorry -- need to switch to oangles?
sorry
end Bulgaria1998P2
| false | null | null | v4.29.0-rc6 | test | [
"competition"
] | null | {"filename": "Bulgaria1998P2.lean"} | {"v4.21.0": {"is_valid": false, "has_sorry": true, "errors": [], "timeout_s": 600.0, "latency_s": 33.2751, "verified_at": "2026-03-26T18:15:54.443933+00:00"}} | false | true | true | 33.2751 |
compfiles_Bulgaria1998P3 | compfiles | Copyright (c) 2023 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
# Bulgarian Mathematical Olympiad 1998, Problem 3
Let ℝ⁺ be the set of positive real numbers. Prove that there does not exist a function
f: ℝ⁺ → ℝ⁺ such that
(f(x))² ≥ f(x + y) * (f(x) + y)
for every x,y ∈ ℝ⁺.
-/
namespace Bulgaria1998P3
lemma geom_sum_bound (n : ℕ) : ∑ i ∈ Finset.range n, (1:ℝ) / (2^i) < 3 :=
calc ∑ i ∈ Finset.range n, (1:ℝ) / ((2:ℝ)^i)
= ∑ i ∈ Finset.range n, ((1:ℝ) / 2)^i := by {congr; simp [div_eq_mul_inv]}
_ ≤ 2 := sum_geometric_two_le n
_ < 3 := by norm_num
theorem bulgaria1998_p3
(f : ℝ → ℝ)
(hpos : ∀ x : ℝ, 0 < x → 0 < f x)
(hf : (∀ x y : ℝ, 0 < x → 0 < y → (f (x + y)) * (f x + y) ≤ (f x)^2)) :
False := by
-- Follows the "second solution" of _Mathematical Olympiads 1998-1999_
-- (edited by Titu Andreescu and Zuming Feng)
have f_decr : ∀ x y : ℝ, 0 < x → 0 < y → f (x + y) < f x := by
intro x y hx hy
have h1 := hf x y hx hy
have h2 : 0 < f x + y := add_pos (hpos x hx) hy
have h4 : f x < f x + y := lt_add_of_pos_right (f x) hy
have h5 : f x / (f x + y) < 1 := (div_lt_one h2).mpr h4
calc f (x + y)
= f (x + y) * 1 := (mul_one (f (x + y))).symm
_ = f (x + y) * ((f x + y) / (f x + y)) := by rw [div_self (Ne.symm (ne_of_lt h2))]
_ = (f (x + y) * (f x + y)) / (f x + y) := mul_div_assoc' _ _ _
_ ≤ (f x)^2 / (f x + y) := (div_le_div_iff_of_pos_right h2).mpr h1
_ = (f x) * (f x / (f x + y)) := by field_simp [pow_two]
_ < f x := (mul_lt_iff_lt_one_right (hpos x hx)).mpr h5
have f_half : ∀ x : ℝ, 0 < x → f (x + f x) ≤ f x / 2 := by
intro x hx
have h0 := hpos x hx
have h1 := hf x (f x) hx h0
have h2 : 0 < f x + f x := add_pos h0 h0
have h3 : 0 ≠ f x + f x := ne_of_lt h2
have h6: 2 * f x ≠ 0 := by positivity
have h7 : (f x / (2 * f x)) = 1 / 2 := by { rw [div_eq_iff h6]; ring }
calc f (x + f x)
= f (x + f x) * 1 := (mul_one _).symm
_ = f (x + f x) * ((f x + f x) / (f x + f x)) := by rw [div_self (Ne.symm h3)]
_ = (f (x + f x) * (f x + f x)) / (f x + f x) := mul_div_assoc' _ _ _
_ ≤ (f x)^2 / (f x + f x) := (div_le_div_iff_of_pos_right h2).mpr h1
_ = (f x) * (f x / (f x + f x)) := by field_simp [pow_two]
_ = (f x) * (f x / (2 * f x)) := by rw [two_mul]
_ = f x / 2 := by field_simp [h7]
let x_seq : ℕ → ℝ := λ n : ℕ ↦ 1 + ∑ i ∈ Finset.range n, (f 1) / (2^i)
have hz : x_seq 0 = 1 := by
simp only [x_seq, add_eq_left, Finset.sum_empty, Finset.range_zero]
have hf1 := hpos 1 zero_lt_one
have x_seq_pos : ∀ n: ℕ, 0 < x_seq n := by
intro n
positivity
have f_x_seq: ∀ n : ℕ, f (x_seq n) ≤ f 1 / 2^n := by
intro n
induction n with
| zero => rw [hz]; simp only [pow_zero, div_one, le_refl]
| succ pn hpn =>
have hpp : x_seq pn.succ = x_seq pn + f 1 / 2^pn := by
rw [show x_seq _ = 1 + ∑ i ∈ Finset.range _, (f 1) / (2^i) by rfl]
have : ∑ i ∈ Finset.range pn.succ, f 1 / 2 ^ i =
f 1 / 2 ^ pn + ∑ i ∈ Finset.range pn, f 1 / 2 ^ i :=
Finset.sum_range_succ_comm (λ (x : ℕ) ↦ f 1 / 2 ^ x) pn
rw [this]
ring
have h1 : f (x_seq pn.succ) ≤ f (x_seq pn + f (x_seq pn)) := by
rw [hpp]
obtain heq | hlt := eq_or_lt_of_le hpn
· rw [heq]
· have := le_of_lt (f_decr (x_seq pn + f (x_seq pn)) (f 1 / 2 ^ pn - f (x_seq pn))
(add_pos (x_seq_pos pn) (hpos (x_seq pn) (x_seq_pos pn)))
(sub_pos.mpr hlt))
rw [add_add_sub_cancel] at this
exact this
calc f (x_seq pn.succ)
≤ f (x_seq pn + f (x_seq pn)) := h1
_ ≤ f (x_seq pn) / 2 := f_half (x_seq pn) (x_seq_pos pn)
_ ≤ (f 1 / 2 ^ pn) / 2 := (div_le_div_iff_of_pos_right two_pos).mpr hpn
_ = f 1 / 2 ^ pn.succ := by field_simp [ne_of_gt hf1]; exact pow_succ 2 pn
have h1: ∀ n: ℕ, x_seq n < 1 + 3 * f 1 := by
intro n
norm_num [x_seq]
have : MulPosStrictMono ℝ := MulPosReflectLT.toMulPosStrictMono ℝ
calc ∑ i ∈ Finset.range n, f 1 / (2:ℝ) ^ i
= (∑ i ∈ Finset.range n, 1 / (2:ℝ) ^ i) * f 1 := by rw [Finset.sum_mul]; field_simp
_ < 3 * f 1 := (mul_lt_mul_iff_left₀ hf1).mpr (geom_sum_bound n)
have h2 : ∀ n : ℕ, 0 < 1 + 3 * f 1 - x_seq n := by intro n; linarith [h1 n]
have h3 : ∀ n : ℕ, f (1 + 3 * f 1) < f 1 / 2 ^ n := by
intro n
calc f (1 + 3 * f 1)
= f (x_seq n + (1 + 3 * f 1 - x_seq n)) := by
simp only [x_seq, add_sub_add_left_eq_sub, add_add_sub_cancel]
_ < f (x_seq n) := f_decr (x_seq n) _ (x_seq_pos n) (h2 n)
_ ≤ f 1 / 2^n := f_x_seq n
have he : ∃n : ℕ, f 1 / 2^n < f (1 + 3 * f 1) := by
obtain ⟨N, hN⟩ := pow_unbounded_of_one_lt (f 1 / f (1 + 3 * f 1)) one_lt_two
use N
have hp : 0 < f (1 + 3 * f 1) :=
hpos (1 + 3 * f 1) (lt_trans (x_seq_pos 0) (h1 0))
have h2N : (0:ℝ) < 2^N := pow_pos two_pos N
exact (div_lt_iff₀ h2N).mpr ((div_lt_iff₀' hp).mp hN)
obtain ⟨N, hN⟩ := he
exact lt_irrefl _ (lt_trans (h3 N) hN)
end Bulgaria1998P3 | /- | /-
Copyright (c) 2023 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
# Bulgarian Mathematical Olympiad 1998, Problem 3
Let ℝ⁺ be the set of positive real numbers. Prove that there does not exist a function
f: ℝ⁺ → ℝ⁺ such that
(f(x))² ≥ f(x + y) * (f(x) + y)
for every x,y ∈ ℝ⁺.
-/
namespace Bulgaria1998P3
lemma geom_sum_bound (n : ℕ) : ∑ i ∈ Finset.range n, (1:ℝ) / (2^i) < 3 :=
calc ∑ i ∈ Finset.range n, (1:ℝ) / ((2:ℝ)^i)
= ∑ i ∈ Finset.range n, ((1:ℝ) / 2)^i := by {congr; simp [div_eq_mul_inv]}
_ ≤ 2 := sum_geometric_two_le n
_ < 3 := by norm_num
theorem bulgaria1998_p3
(f : ℝ → ℝ)
(hpos : ∀ x : ℝ, 0 < x → 0 < f x)
(hf : (∀ x y : ℝ, 0 < x → 0 < y → (f (x + y)) * (f x + y) ≤ (f x)^2)) :
False := by
-- Follows the "second solution" of _Mathematical Olympiads 1998-1999_
-- (edited by Titu Andreescu and Zuming Feng)
have f_decr : ∀ x y : ℝ, 0 < x → 0 < y → f (x + y) < f x := by
intro x y hx hy
have h1 := hf x y hx hy
have h2 : 0 < f x + y := add_pos (hpos x hx) hy
have h4 : f x < f x + y := lt_add_of_pos_right (f x) hy
have h5 : f x / (f x + y) < 1 := (div_lt_one h2).mpr h4
calc f (x + y)
= f (x + y) * 1 := (mul_one (f (x + y))).symm
_ = f (x + y) * ((f x + y) / (f x + y)) := by rw [div_self (Ne.symm (ne_of_lt h2))]
_ = (f (x + y) * (f x + y)) / (f x + y) := mul_div_assoc' _ _ _
_ ≤ (f x)^2 / (f x + y) := (div_le_div_iff_of_pos_right h2).mpr h1
_ = (f x) * (f x / (f x + y)) := by field_simp [pow_two]
_ < f x := (mul_lt_iff_lt_one_right (hpos x hx)).mpr h5
have f_half : ∀ x : ℝ, 0 < x → f (x + f x) ≤ f x / 2 := by
intro x hx
have h0 := hpos x hx
have h1 := hf x (f x) hx h0
have h2 : 0 < f x + f x := add_pos h0 h0
have h3 : 0 ≠ f x + f x := ne_of_lt h2
have h6: 2 * f x ≠ 0 := by positivity
have h7 : (f x / (2 * f x)) = 1 / 2 := by { rw [div_eq_iff h6]; ring }
calc f (x + f x)
= f (x + f x) * 1 := (mul_one _).symm
_ = f (x + f x) * ((f x + f x) / (f x + f x)) := by rw [div_self (Ne.symm h3)]
_ = (f (x + f x) * (f x + f x)) / (f x + f x) := mul_div_assoc' _ _ _
_ ≤ (f x)^2 / (f x + f x) := (div_le_div_iff_of_pos_right h2).mpr h1
_ = (f x) * (f x / (f x + f x)) := by field_simp [pow_two]
_ = (f x) * (f x / (2 * f x)) := by rw [two_mul]
_ = f x / 2 := by field_simp [h7]
let x_seq : ℕ → ℝ := λ n : ℕ ↦ 1 + ∑ i ∈ Finset.range n, (f 1) / (2^i)
have hz : x_seq 0 = 1 := by
simp only [x_seq, add_eq_left, Finset.sum_empty, Finset.range_zero]
have hf1 := hpos 1 zero_lt_one
have x_seq_pos : ∀ n: ℕ, 0 < x_seq n := by
intro n
positivity
have f_x_seq: ∀ n : ℕ, f (x_seq n) ≤ f 1 / 2^n := by
intro n
induction n with
| zero => rw [hz]; simp only [pow_zero, div_one, le_refl]
| succ pn hpn =>
have hpp : x_seq pn.succ = x_seq pn + f 1 / 2^pn := by
rw [show x_seq _ = 1 + ∑ i ∈ Finset.range _, (f 1) / (2^i) by rfl]
have : ∑ i ∈ Finset.range pn.succ, f 1 / 2 ^ i =
f 1 / 2 ^ pn + ∑ i ∈ Finset.range pn, f 1 / 2 ^ i :=
Finset.sum_range_succ_comm (λ (x : ℕ) ↦ f 1 / 2 ^ x) pn
rw [this]
ring
have h1 : f (x_seq pn.succ) ≤ f (x_seq pn + f (x_seq pn)) := by
rw [hpp]
obtain heq | hlt := eq_or_lt_of_le hpn
· rw [heq]
· have := le_of_lt (f_decr (x_seq pn + f (x_seq pn)) (f 1 / 2 ^ pn - f (x_seq pn))
(add_pos (x_seq_pos pn) (hpos (x_seq pn) (x_seq_pos pn)))
(sub_pos.mpr hlt))
rw [add_add_sub_cancel] at this
exact this
calc f (x_seq pn.succ)
≤ f (x_seq pn + f (x_seq pn)) := h1
_ ≤ f (x_seq pn) / 2 := f_half (x_seq pn) (x_seq_pos pn)
_ ≤ (f 1 / 2 ^ pn) / 2 := (div_le_div_iff_of_pos_right two_pos).mpr hpn
_ = f 1 / 2 ^ pn.succ := by field_simp [ne_of_gt hf1]; exact pow_succ 2 pn
have h1: ∀ n: ℕ, x_seq n < 1 + 3 * f 1 := by
intro n
norm_num [x_seq]
have : MulPosStrictMono ℝ := MulPosReflectLT.toMulPosStrictMono ℝ
calc ∑ i ∈ Finset.range n, f 1 / (2:ℝ) ^ i
= (∑ i ∈ Finset.range n, 1 / (2:ℝ) ^ i) * f 1 := by rw [Finset.sum_mul]; field_simp
_ < 3 * f 1 := (mul_lt_mul_iff_left₀ hf1).mpr (geom_sum_bound n)
have h2 : ∀ n : ℕ, 0 < 1 + 3 * f 1 - x_seq n := by intro n; linarith [h1 n]
have h3 : ∀ n : ℕ, f (1 + 3 * f 1) < f 1 / 2 ^ n := by
intro n
calc f (1 + 3 * f 1)
= f (x_seq n + (1 + 3 * f 1 - x_seq n)) := by
simp only [x_seq, add_sub_add_left_eq_sub, add_add_sub_cancel]
_ < f (x_seq n) := f_decr (x_seq n) _ (x_seq_pos n) (h2 n)
_ ≤ f 1 / 2^n := f_x_seq n
have he : ∃n : ℕ, f 1 / 2^n < f (1 + 3 * f 1) := by
obtain ⟨N, hN⟩ := pow_unbounded_of_one_lt (f 1 / f (1 + 3 * f 1)) one_lt_two
use N
have hp : 0 < f (1 + 3 * f 1) :=
hpos (1 + 3 * f 1) (lt_trans (x_seq_pos 0) (h1 0))
have h2N : (0:ℝ) < 2^N := pow_pos two_pos N
exact (div_lt_iff₀ h2N).mpr ((div_lt_iff₀' hp).mp hN)
obtain ⟨N, hN⟩ := he
exact lt_irrefl _ (lt_trans (h3 N) hN)
end Bulgaria1998P3
| true | Copyright (c) 2023 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
# Bulgarian Mathematical Olympiad 1998, Problem 3
Let ℝ⁺ be the set of positive real numbers. Prove that there does not exist a function
f: ℝ⁺ → ℝ⁺ such that
(f(x))² ≥ f(x + y) * (f(x) + y)
for every x,y ∈ ℝ⁺.
-/
namespace Bulgaria1998P3
lemma geom_sum_bound (n : ℕ) : ∑ i ∈ Finset.range n, (1:ℝ) / (2^i) < 3 :=
calc ∑ i ∈ Finset.range n, (1:ℝ) / ((2:ℝ)^i)
= ∑ i ∈ Finset.range n, ((1:ℝ) / 2)^i := by {congr; simp [div_eq_mul_inv]}
_ ≤ 2 := sum_geometric_two_le n
_ < 3 := by norm_num
theorem bulgaria1998_p3
(f : ℝ → ℝ)
(hpos : ∀ x : ℝ, 0 < x → 0 < f x)
(hf : (∀ x y : ℝ, 0 < x → 0 < y → (f (x + y)) * (f x + y) ≤ (f x)^2)) :
False := by
-- Follows the "second solution" of _Mathematical Olympiads 1998-1999_
-- (edited by Titu Andreescu and Zuming Feng)
have f_decr : ∀ x y : ℝ, 0 < x → 0 < y → f (x + y) < f x := by
intro x y hx hy
have h1 := hf x y hx hy
have h2 : 0 < f x + y := add_pos (hpos x hx) hy
have h4 : f x < f x + y := lt_add_of_pos_right (f x) hy
have h5 : f x / (f x + y) < 1 := (div_lt_one h2).mpr h4
calc f (x + y)
= f (x + y) * 1 := (mul_one (f (x + y))).symm
_ = f (x + y) * ((f x + y) / (f x + y)) := by rw [div_self (Ne.symm (ne_of_lt h2))]
_ = (f (x + y) * (f x + y)) / (f x + y) := mul_div_assoc' _ _ _
_ ≤ (f x)^2 / (f x + y) := (div_le_div_iff_of_pos_right h2).mpr h1
_ = (f x) * (f x / (f x + y)) := by field_simp [pow_two]
_ < f x := (mul_lt_iff_lt_one_right (hpos x hx)).mpr h5
have f_half : ∀ x : ℝ, 0 < x → f (x + f x) ≤ f x / 2 := by
intro x hx
have h0 := hpos x hx
have h1 := hf x (f x) hx h0
have h2 : 0 < f x + f x := add_pos h0 h0
have h3 : 0 ≠ f x + f x := ne_of_lt h2
have h6: 2 * f x ≠ 0 := by positivity
have h7 : (f x / (2 * f x)) = 1 / 2 := by { rw [div_eq_iff h6]; ring }
calc f (x + f x)
= f (x + f x) * 1 := (mul_one _).symm
_ = f (x + f x) * ((f x + f x) / (f x + f x)) := by rw [div_self (Ne.symm h3)]
_ = (f (x + f x) * (f x + f x)) / (f x + f x) := mul_div_assoc' _ _ _
_ ≤ (f x)^2 / (f x + f x) := (div_le_div_iff_of_pos_right h2).mpr h1
_ = (f x) * (f x / (f x + f x)) := by field_simp [pow_two]
_ = (f x) * (f x / (2 * f x)) := by rw [two_mul]
_ = f x / 2 := by field_simp [h7]
let x_seq : ℕ → ℝ := λ n : ℕ ↦ 1 + ∑ i ∈ Finset.range n, (f 1) / (2^i)
have hz : x_seq 0 = 1 := by
simp only [x_seq, add_eq_left, Finset.sum_empty, Finset.range_zero]
have hf1 := hpos 1 zero_lt_one
have x_seq_pos : ∀ n: ℕ, 0 < x_seq n := by
intro n
positivity
have f_x_seq: ∀ n : ℕ, f (x_seq n) ≤ f 1 / 2^n := by
intro n
induction n with
| zero => rw [hz]; simp only [pow_zero, div_one, le_refl]
| succ pn hpn =>
have hpp : x_seq pn.succ = x_seq pn + f 1 / 2^pn := by
rw [show x_seq _ = 1 + ∑ i ∈ Finset.range _, (f 1) / (2^i) by rfl]
have : ∑ i ∈ Finset.range pn.succ, f 1 / 2 ^ i =
f 1 / 2 ^ pn + ∑ i ∈ Finset.range pn, f 1 / 2 ^ i :=
Finset.sum_range_succ_comm (λ (x : ℕ) ↦ f 1 / 2 ^ x) pn
rw [this]
ring
have h1 : f (x_seq pn.succ) ≤ f (x_seq pn + f (x_seq pn)) := by
rw [hpp]
obtain heq | hlt := eq_or_lt_of_le hpn
· rw [heq]
· have := le_of_lt (f_decr (x_seq pn + f (x_seq pn)) (f 1 / 2 ^ pn - f (x_seq pn))
(add_pos (x_seq_pos pn) (hpos (x_seq pn) (x_seq_pos pn)))
(sub_pos.mpr hlt))
rw [add_add_sub_cancel] at this
exact this
calc f (x_seq pn.succ)
≤ f (x_seq pn + f (x_seq pn)) := h1
_ ≤ f (x_seq pn) / 2 := f_half (x_seq pn) (x_seq_pos pn)
_ ≤ (f 1 / 2 ^ pn) / 2 := (div_le_div_iff_of_pos_right two_pos).mpr hpn
_ = f 1 / 2 ^ pn.succ := by field_simp [ne_of_gt hf1]; exact pow_succ 2 pn
have h1: ∀ n: ℕ, x_seq n < 1 + 3 * f 1 := by
intro n
norm_num [x_seq]
have : MulPosStrictMono ℝ := MulPosReflectLT.toMulPosStrictMono ℝ
calc ∑ i ∈ Finset.range n, f 1 / (2:ℝ) ^ i
= (∑ i ∈ Finset.range n, 1 / (2:ℝ) ^ i) * f 1 := by rw [Finset.sum_mul]; field_simp
_ < 3 * f 1 := (mul_lt_mul_iff_left₀ hf1).mpr (geom_sum_bound n)
have h2 : ∀ n : ℕ, 0 < 1 + 3 * f 1 - x_seq n := by intro n; linarith [h1 n]
have h3 : ∀ n : ℕ, f (1 + 3 * f 1) < f 1 / 2 ^ n := by
intro n
calc f (1 + 3 * f 1)
= f (x_seq n + (1 + 3 * f 1 - x_seq n)) := by
simp only [x_seq, add_sub_add_left_eq_sub, add_add_sub_cancel]
_ < f (x_seq n) := f_decr (x_seq n) _ (x_seq_pos n) (h2 n)
_ ≤ f 1 / 2^n := f_x_seq n
have he : ∃n : ℕ, f 1 / 2^n < f (1 + 3 * f 1) := by
obtain ⟨N, hN⟩ := pow_unbounded_of_one_lt (f 1 / f (1 + 3 * f 1)) one_lt_two
use N
have hp : 0 < f (1 + 3 * f 1) :=
hpos (1 + 3 * f 1) (lt_trans (x_seq_pos 0) (h1 0))
have h2N : (0:ℝ) < 2^N := pow_pos two_pos N
exact (div_lt_iff₀ h2N).mpr ((div_lt_iff₀' hp).mp hN)
obtain ⟨N, hN⟩ := he
exact lt_irrefl _ (lt_trans (h3 N) hN)
end Bulgaria1998P3 | null | v4.29.0-rc6 | test | [
"competition"
] | null | {"filename": "Bulgaria1998P3.lean"} | {"v4.21.0": {"is_valid": false, "has_sorry": false, "errors": ["unknown identifier 'mul_lt_mul_iff_left\u2080'"], "timeout_s": 600.0, "latency_s": 35.1685, "verified_at": "2026-03-26T18:15:56.337288+00:00"}} | false | true | false | 35.1685 |
compfiles_Bulgaria1998P6 | compfiles | Copyright (c) 2023 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
# Bulgarian Mathematical Olympiad 1998, Problem 6
Prove that the equation
x²y² = z²(z² - x² - y²)
has no solutions in positive integers.
-/
namespace Bulgaria1998P6
lemma lemma_1'
(a b c : ℕ)
(ha : 0 < a)
(hb : 0 < b)
(hc : 0 < c)
(h : a^4 = b^4 + c^2) : False := by
induction' a using Nat.strongRecOn with a ih
have hab : Nat.gcd a b = 1 := sorry
obtain c_even | c_odd := Nat.even_or_odd c
· have a_odd : Odd a := sorry
have b_odd : Odd b := sorry
sorry
· sorry
lemma lemma_1
{s t u : ℤ}
(hs : 0 < s)
(ht : 0 < t)
(hu : 0 < u)
(h : s^4 - t^4 = u^2) : False := by
replace h : s^4 = t^4 + u^2 := eq_add_of_sub_eq' h
lift s to ℕ using Int.le_of_lt hs
lift t to ℕ using Int.le_of_lt ht
lift u to ℕ using Int.le_of_lt hu
replace hs : 0 < s := Int.natCast_pos.mp hs
replace ht : 0 < t := Int.natCast_pos.mp ht
replace hy : 0 < u := Int.natCast_pos.mp hu
have h' : s ^ 4 = t ^ 4 + u ^ 2 := by exact Int.ofNat_inj.mp h
exact lemma_1' s t u hs ht hy h'
theorem bulgaria1998_p6
(x y z : ℤ)
(hx : 0 < x)
(hy : 0 < y)
(_hz : 0 < z)
(h : x^2 * y^2 = z^2 * (z^2 - x^2 - y^2)) :
False := by
-- Follows the informal solution in _Mathematical Olympiads 1998-1999_
-- (edited by Titu Andreescu and Zuming Feng)
have h1 : 1 * (z^2 * z^2) + (- (x^2 + y^2)) * z^2 + -(x^2 * y^2) = 0 := by
rw[h]; ring
have : NeZero (2 : ℤ) := CharZero.NeZero.two
have h2 := (quadratic_eq_zero_iff_discrim_eq_sq one_ne_zero (z^2)).mp h1
dsimp [discrim] at h2
let a := x^2 + y^2
let b := 2 * x * y
have h3 : a^2 + b^2 = (2 * z ^ 2 - (x ^ 2 + y ^ 2)) ^ 2 :=
by linear_combination h2
have h4 : IsSquare (a^2 + b^2) := by use 2 * z ^ 2 - (x ^ 2 + y ^ 2); rwa [←sq]
have h5 : IsSquare (a^2 - b^2) := by use (x^2 - y^2); ring
have h6 : IsSquare ((a^2 + b^2) * (a^2 - b^2)) := IsSquare.mul h4 h5
rw [show (a^2 + b^2) * (a^2 - b^2) = a^4 - b^4 by ring] at h6
obtain ⟨c, hc⟩ := h6
rw [←sq, ←sq_abs] at hc
have ha' : 0 < a := by positivity
have hb' : 0 < b := by positivity
have hc' : 0 < |c| := by
obtain hc1 | hc2 | hc3 := lt_trichotomy 0 |c|
· exact hc1
· have hab : a^2 = b^2 := by
rw [← hc2, zero_pow (by norm_num)] at hc
have hc3 : (a^2)^2 = (b^2)^2 := by linear_combination hc
have hap : 0 ≤ a^2 := by positivity
have hbp : 0 ≤ b^2 := by positivity
exact (pow_left_inj₀ hap hbp two_ne_zero).mp hc3
rw [hab] at h4
obtain ⟨r, hr⟩ := h4
rw [←two_mul, ←sq] at hr
have h10 : b^2 ∣ r^2 := Dvd.intro_left _ hr
rw [Int.pow_dvd_pow_iff two_ne_zero] at h10
obtain ⟨e, rfl⟩ := h10
rw [show (b * e)^2 = e^2 * b^2 by ring] at hr
have h11 : b^2 ≠ 0 := by positivity
have h12 : 2 = e^2 := (Int.mul_eq_mul_right_iff h11).mp hr
clear h h1 h3 h5 hab
have h13 : e < 2 := by
by_contra! H
have h20 : 2^2 ≤ e^2 := by gcongr
rw [←h12] at h20
norm_num at h20
have h14 : -2 < e := by
by_contra! H
replace H : 2 ≤ -e := Int.le_neg_of_le_neg H
have h20 : 2^2 ≤ (-e)^2 := by gcongr
rw [neg_sq, ←h12] at h20
norm_num at h20
interval_cases e <;> linarith
· exact (Int.not_lt.mpr (abs_nonneg c) hc3).elim
exact lemma_1 ha' hb' hc' hc
end Bulgaria1998P6 | /- | /-
Copyright (c) 2023 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
# Bulgarian Mathematical Olympiad 1998, Problem 6
Prove that the equation
x²y² = z²(z² - x² - y²)
has no solutions in positive integers.
-/
namespace Bulgaria1998P6
lemma lemma_1'
(a b c : ℕ)
(ha : 0 < a)
(hb : 0 < b)
(hc : 0 < c)
(h : a^4 = b^4 + c^2) : False := by
induction' a using Nat.strongRecOn with a ih
have hab : Nat.gcd a b = 1 := sorry
obtain c_even | c_odd := Nat.even_or_odd c
· have a_odd : Odd a := sorry
have b_odd : Odd b := sorry
sorry
· sorry
lemma lemma_1
{s t u : ℤ}
(hs : 0 < s)
(ht : 0 < t)
(hu : 0 < u)
(h : s^4 - t^4 = u^2) : False := by
replace h : s^4 = t^4 + u^2 := eq_add_of_sub_eq' h
lift s to ℕ using Int.le_of_lt hs
lift t to ℕ using Int.le_of_lt ht
lift u to ℕ using Int.le_of_lt hu
replace hs : 0 < s := Int.natCast_pos.mp hs
replace ht : 0 < t := Int.natCast_pos.mp ht
replace hy : 0 < u := Int.natCast_pos.mp hu
have h' : s ^ 4 = t ^ 4 + u ^ 2 := by exact Int.ofNat_inj.mp h
exact lemma_1' s t u hs ht hy h'
theorem bulgaria1998_p6
(x y z : ℤ)
(hx : 0 < x)
(hy : 0 < y)
(_hz : 0 < z)
(h : x^2 * y^2 = z^2 * (z^2 - x^2 - y^2)) :
False := by
-- Follows the informal solution in _Mathematical Olympiads 1998-1999_
-- (edited by Titu Andreescu and Zuming Feng)
have h1 : 1 * (z^2 * z^2) + (- (x^2 + y^2)) * z^2 + -(x^2 * y^2) = 0 := by
rw[h]; ring
have : NeZero (2 : ℤ) := CharZero.NeZero.two
have h2 := (quadratic_eq_zero_iff_discrim_eq_sq one_ne_zero (z^2)).mp h1
dsimp [discrim] at h2
let a := x^2 + y^2
let b := 2 * x * y
have h3 : a^2 + b^2 = (2 * z ^ 2 - (x ^ 2 + y ^ 2)) ^ 2 :=
by linear_combination h2
have h4 : IsSquare (a^2 + b^2) := by use 2 * z ^ 2 - (x ^ 2 + y ^ 2); rwa [←sq]
have h5 : IsSquare (a^2 - b^2) := by use (x^2 - y^2); ring
have h6 : IsSquare ((a^2 + b^2) * (a^2 - b^2)) := IsSquare.mul h4 h5
rw [show (a^2 + b^2) * (a^2 - b^2) = a^4 - b^4 by ring] at h6
obtain ⟨c, hc⟩ := h6
rw [←sq, ←sq_abs] at hc
have ha' : 0 < a := by positivity
have hb' : 0 < b := by positivity
have hc' : 0 < |c| := by
obtain hc1 | hc2 | hc3 := lt_trichotomy 0 |c|
· exact hc1
· have hab : a^2 = b^2 := by
rw [← hc2, zero_pow (by norm_num)] at hc
have hc3 : (a^2)^2 = (b^2)^2 := by linear_combination hc
have hap : 0 ≤ a^2 := by positivity
have hbp : 0 ≤ b^2 := by positivity
exact (pow_left_inj₀ hap hbp two_ne_zero).mp hc3
rw [hab] at h4
obtain ⟨r, hr⟩ := h4
rw [←two_mul, ←sq] at hr
have h10 : b^2 ∣ r^2 := Dvd.intro_left _ hr
rw [Int.pow_dvd_pow_iff two_ne_zero] at h10
obtain ⟨e, rfl⟩ := h10
rw [show (b * e)^2 = e^2 * b^2 by ring] at hr
have h11 : b^2 ≠ 0 := by positivity
have h12 : 2 = e^2 := (Int.mul_eq_mul_right_iff h11).mp hr
clear h h1 h3 h5 hab
have h13 : e < 2 := by
by_contra! H
have h20 : 2^2 ≤ e^2 := by gcongr
rw [←h12] at h20
norm_num at h20
have h14 : -2 < e := by
by_contra! H
replace H : 2 ≤ -e := Int.le_neg_of_le_neg H
have h20 : 2^2 ≤ (-e)^2 := by gcongr
rw [neg_sq, ←h12] at h20
norm_num at h20
interval_cases e <;> linarith
· exact (Int.not_lt.mpr (abs_nonneg c) hc3).elim
exact lemma_1 ha' hb' hc' hc
end Bulgaria1998P6
| false | null | null | v4.29.0-rc6 | test | [
"competition"
] | null | {"filename": "Bulgaria1998P6.lean"} | {"v4.21.0": {"is_valid": false, "has_sorry": true, "errors": [], "timeout_s": 600.0, "latency_s": 1.6128, "verified_at": "2026-03-26T18:15:56.056907+00:00"}} | false | true | true | 1.6128 |
compfiles_Bulgaria1998P8 | compfiles | Copyright (c) 2023 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
# Bulgarian Mathematical Olympiad 1998, Problem 8
The polynomials Pₙ(x,y) for n = 1, 2, ... are defined by P₁(x,y) = 1 and
Pₙ₊₁(x,y) = (x + y - 1)(y + 1)Pₙ(x,y+2) + (y - y²)Pₙ(x,y)
Prove that Pₙ(x,y) = Pₙ(y,x) for all x,y,n.
-/
namespace Bulgaria1998P8
variable {R : Type} [CommRing R]
def P : ℕ → R → R → R
| 0, _, _ => 1
| n+1, x, y => (x + y - 1) * (y + 1) * P n x (y + 2) + (y - y^2) * P n x y
/- helper function
Sₙ₋₁(x,y) = [(x + y)² - 1](y + 1)(x + 1)Pₙ₋₁(y+2, x+2).
-/
def S : ℕ → R → R → R
| n, x, y => ((x + y)^2 - 1) * (y + 1) * (x + 1) * P n (y + 2) (x + 2)
/- helper function
Tₙ₋₁(x,y) = (y - y²)(x - x²)Pₙ₋₁(y, x).
-/
def T : ℕ → R → R → R
| n, x, y => (y - y^2) * (x - x^2) * P n y x
/- helper function
Uₙ₋₁(x,y) = (x + y - 1) [(y + 1)(x - x²)Pₙ₋₁(y + 2, x)
+ (x + 1)(y - y²) Pₙ₋₁(y, x + 2)]
-/
def U : ℕ → R → R → R
| n, x, y => (x + y - 1) *((y + 1)*(x - x^2) * P n (y + 2) x
+ (x + 1) * (y - y^2) * P n y (x + 2))
theorem bulgaria1998_p8 (n : ℕ) (x y : R) : P n x y = P n y x := by
-- Follows the proof in _Mathematical Olympiads 1998-1999_
-- by Titu Andreescu and Zuming Feng
-- We induct on n. For n = 1,2 the result is evident.
-- So we take n > 1 and suppose that the result is true for
-- Pₙ₋₁(x,y) and Pₙ₋₂(x,y).
revert x y
induction n using Nat.strong_induction_on with | h n ih =>
cases n with | zero => intros; rfl | succ n =>
cases n with | zero => intros; dsimp only [P]; ring | succ n =>
-- in the informal proof at this point, we're trying to
-- prove the end result at n+1.
-- In our formal version, that corresponds to proving the result for
-- n.succ.succ
/- We have
Pₙ₊₁(x,y) = (x + y - 1)(y + 1)Pₙ(x,y+2) + (y - y²)Pₙ(x,y)
= (x + y - 1)(y + 1)Pₙ(y+2,x) + (y - y²)Pₙ(y,x)
-/
have ih1 := ih n.succ (lt_add_one n.succ)
have h1 : ∀ x y : R, P n.succ.succ x y =
(x + y - 1) * (y + 1) * (P n.succ (y + 2) x) +
(y - y^2) * (P n.succ y x) := by
intro x y
calc P (n.succ.succ) x y
= (x + y - 1) * (y + 1) * (P n.succ x (y + 2)) +
(y - y^2) * (P n.succ x y) := rfl
_ = (x + y - 1) * (y + 1) * (P n.succ (y + 2) x) +
(y - y^2) * (P n.succ y x) := by rw [ih1 x y, ih1 x (y+2)]
have h2 : ∀ x y : R, (x + y - 1) * (y + 1) * P n.succ (y + 2) x
= S n x y + (x + y - 1)* (y + 1) * (x - x^2)* P n (y+2) x := by
intro x y; dsimp only [S, P]; ring
have h_s_symm : ∀ m : ℕ, m < n.succ.succ → ∀ x y : R, S m x y = S m y x := by
intro m hm x y; dsimp only [S]; rw [ih m hm (x + 2) (y + 2)]; ring
have h4 : ∀ x y : R, (y - y^2) * P n.succ y x =
(y - y^2) * (x + y -1) * (x + 1) * P n y (x + 2) + T n x y := by
intro x y; dsimp only [T, P]; ring
have h_t_symm : ∀ m : ℕ, m < n.succ.succ → ∀ x y : R, T m x y = T m y x := by
intro m hm x y; dsimp only[T]; rw [ih m hm x y]; ring
have h_u_symm : ∀ m : ℕ, m < n.succ.succ → ∀ x y : R, U m x y = U m y x := by
intro m hm x y; dsimp only [U]; rw [ih m hm (y+2) x, ih m hm (x+2) y]; ring
have h7 : ∀ x y : R, P n.succ.succ x y = S n x y + T n x y + U n x y := by
intro x y; rw [h1 x y, h2 x y, h4 x y]; dsimp only [U]; ring
have h8 : n < n.succ := lt_add_one n
have h9 : n < n.succ.succ := Nat.lt_succ_of_lt h8
intro x y
calc P n.succ.succ x y
= S n x y + T n x y + U n x y := h7 x y
_ = S n y x + T n x y + U n x y := by rw [h_s_symm n h9 x y]
_ = S n y x + T n y x + U n x y := by rw [h_t_symm n h9 x y]
_ = S n y x + T n y x + U n y x := by rw [h_u_symm n h9 x y]
_ = P n.succ.succ y x := (h7 y x).symm
end Bulgaria1998P8 | /- | /-
Copyright (c) 2023 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
# Bulgarian Mathematical Olympiad 1998, Problem 8
The polynomials Pₙ(x,y) for n = 1, 2, ... are defined by P₁(x,y) = 1 and
Pₙ₊₁(x,y) = (x + y - 1)(y + 1)Pₙ(x,y+2) + (y - y²)Pₙ(x,y)
Prove that Pₙ(x,y) = Pₙ(y,x) for all x,y,n.
-/
namespace Bulgaria1998P8
variable {R : Type} [CommRing R]
def P : ℕ → R → R → R
| 0, _, _ => 1
| n+1, x, y => (x + y - 1) * (y + 1) * P n x (y + 2) + (y - y^2) * P n x y
/- helper function
Sₙ₋₁(x,y) = [(x + y)² - 1](y + 1)(x + 1)Pₙ₋₁(y+2, x+2).
-/
def S : ℕ → R → R → R
| n, x, y => ((x + y)^2 - 1) * (y + 1) * (x + 1) * P n (y + 2) (x + 2)
/- helper function
Tₙ₋₁(x,y) = (y - y²)(x - x²)Pₙ₋₁(y, x).
-/
def T : ℕ → R → R → R
| n, x, y => (y - y^2) * (x - x^2) * P n y x
/- helper function
Uₙ₋₁(x,y) = (x + y - 1) [(y + 1)(x - x²)Pₙ₋₁(y + 2, x)
+ (x + 1)(y - y²) Pₙ₋₁(y, x + 2)]
-/
def U : ℕ → R → R → R
| n, x, y => (x + y - 1) *((y + 1)*(x - x^2) * P n (y + 2) x
+ (x + 1) * (y - y^2) * P n y (x + 2))
theorem bulgaria1998_p8 (n : ℕ) (x y : R) : P n x y = P n y x := by
-- Follows the proof in _Mathematical Olympiads 1998-1999_
-- by Titu Andreescu and Zuming Feng
-- We induct on n. For n = 1,2 the result is evident.
-- So we take n > 1 and suppose that the result is true for
-- Pₙ₋₁(x,y) and Pₙ₋₂(x,y).
revert x y
induction n using Nat.strong_induction_on with | h n ih =>
cases n with | zero => intros; rfl | succ n =>
cases n with | zero => intros; dsimp only [P]; ring | succ n =>
-- in the informal proof at this point, we're trying to
-- prove the end result at n+1.
-- In our formal version, that corresponds to proving the result for
-- n.succ.succ
/- We have
Pₙ₊₁(x,y) = (x + y - 1)(y + 1)Pₙ(x,y+2) + (y - y²)Pₙ(x,y)
= (x + y - 1)(y + 1)Pₙ(y+2,x) + (y - y²)Pₙ(y,x)
-/
have ih1 := ih n.succ (lt_add_one n.succ)
have h1 : ∀ x y : R, P n.succ.succ x y =
(x + y - 1) * (y + 1) * (P n.succ (y + 2) x) +
(y - y^2) * (P n.succ y x) := by
intro x y
calc P (n.succ.succ) x y
= (x + y - 1) * (y + 1) * (P n.succ x (y + 2)) +
(y - y^2) * (P n.succ x y) := rfl
_ = (x + y - 1) * (y + 1) * (P n.succ (y + 2) x) +
(y - y^2) * (P n.succ y x) := by rw [ih1 x y, ih1 x (y+2)]
have h2 : ∀ x y : R, (x + y - 1) * (y + 1) * P n.succ (y + 2) x
= S n x y + (x + y - 1)* (y + 1) * (x - x^2)* P n (y+2) x := by
intro x y; dsimp only [S, P]; ring
have h_s_symm : ∀ m : ℕ, m < n.succ.succ → ∀ x y : R, S m x y = S m y x := by
intro m hm x y; dsimp only [S]; rw [ih m hm (x + 2) (y + 2)]; ring
have h4 : ∀ x y : R, (y - y^2) * P n.succ y x =
(y - y^2) * (x + y -1) * (x + 1) * P n y (x + 2) + T n x y := by
intro x y; dsimp only [T, P]; ring
have h_t_symm : ∀ m : ℕ, m < n.succ.succ → ∀ x y : R, T m x y = T m y x := by
intro m hm x y; dsimp only[T]; rw [ih m hm x y]; ring
have h_u_symm : ∀ m : ℕ, m < n.succ.succ → ∀ x y : R, U m x y = U m y x := by
intro m hm x y; dsimp only [U]; rw [ih m hm (y+2) x, ih m hm (x+2) y]; ring
have h7 : ∀ x y : R, P n.succ.succ x y = S n x y + T n x y + U n x y := by
intro x y; rw [h1 x y, h2 x y, h4 x y]; dsimp only [U]; ring
have h8 : n < n.succ := lt_add_one n
have h9 : n < n.succ.succ := Nat.lt_succ_of_lt h8
intro x y
calc P n.succ.succ x y
= S n x y + T n x y + U n x y := h7 x y
_ = S n y x + T n x y + U n x y := by rw [h_s_symm n h9 x y]
_ = S n y x + T n y x + U n x y := by rw [h_t_symm n h9 x y]
_ = S n y x + T n y x + U n y x := by rw [h_u_symm n h9 x y]
_ = P n.succ.succ y x := (h7 y x).symm
end Bulgaria1998P8
| true | Copyright (c) 2023 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
# Bulgarian Mathematical Olympiad 1998, Problem 8
The polynomials Pₙ(x,y) for n = 1, 2, ... are defined by P₁(x,y) = 1 and
Pₙ₊₁(x,y) = (x + y - 1)(y + 1)Pₙ(x,y+2) + (y - y²)Pₙ(x,y)
Prove that Pₙ(x,y) = Pₙ(y,x) for all x,y,n.
-/
namespace Bulgaria1998P8
variable {R : Type} [CommRing R]
def P : ℕ → R → R → R
| 0, _, _ => 1
| n+1, x, y => (x + y - 1) * (y + 1) * P n x (y + 2) + (y - y^2) * P n x y
/- helper function
Sₙ₋₁(x,y) = [(x + y)² - 1](y + 1)(x + 1)Pₙ₋₁(y+2, x+2).
-/
def S : ℕ → R → R → R
| n, x, y => ((x + y)^2 - 1) * (y + 1) * (x + 1) * P n (y + 2) (x + 2)
/- helper function
Tₙ₋₁(x,y) = (y - y²)(x - x²)Pₙ₋₁(y, x).
-/
def T : ℕ → R → R → R
| n, x, y => (y - y^2) * (x - x^2) * P n y x
/- helper function
Uₙ₋₁(x,y) = (x + y - 1) [(y + 1)(x - x²)Pₙ₋₁(y + 2, x)
+ (x + 1)(y - y²) Pₙ₋₁(y, x + 2)]
-/
def U : ℕ → R → R → R
| n, x, y => (x + y - 1) *((y + 1)*(x - x^2) * P n (y + 2) x
+ (x + 1) * (y - y^2) * P n y (x + 2))
theorem bulgaria1998_p8 (n : ℕ) (x y : R) : P n x y = P n y x := by
-- Follows the proof in _Mathematical Olympiads 1998-1999_
-- by Titu Andreescu and Zuming Feng
-- We induct on n. For n = 1,2 the result is evident.
-- So we take n > 1 and suppose that the result is true for
-- Pₙ₋₁(x,y) and Pₙ₋₂(x,y).
revert x y
induction n using Nat.strong_induction_on with | h n ih =>
cases n with | zero => intros; rfl | succ n =>
cases n with | zero => intros; dsimp only [P]; ring | succ n =>
-- in the informal proof at this point, we're trying to
-- prove the end result at n+1.
-- In our formal version, that corresponds to proving the result for
-- n.succ.succ
/- We have
Pₙ₊₁(x,y) = (x + y - 1)(y + 1)Pₙ(x,y+2) + (y - y²)Pₙ(x,y)
= (x + y - 1)(y + 1)Pₙ(y+2,x) + (y - y²)Pₙ(y,x)
-/
have ih1 := ih n.succ (lt_add_one n.succ)
have h1 : ∀ x y : R, P n.succ.succ x y =
(x + y - 1) * (y + 1) * (P n.succ (y + 2) x) +
(y - y^2) * (P n.succ y x) := by
intro x y
calc P (n.succ.succ) x y
= (x + y - 1) * (y + 1) * (P n.succ x (y + 2)) +
(y - y^2) * (P n.succ x y) := rfl
_ = (x + y - 1) * (y + 1) * (P n.succ (y + 2) x) +
(y - y^2) * (P n.succ y x) := by rw [ih1 x y, ih1 x (y+2)]
have h2 : ∀ x y : R, (x + y - 1) * (y + 1) * P n.succ (y + 2) x
= S n x y + (x + y - 1)* (y + 1) * (x - x^2)* P n (y+2) x := by
intro x y; dsimp only [S, P]; ring
have h_s_symm : ∀ m : ℕ, m < n.succ.succ → ∀ x y : R, S m x y = S m y x := by
intro m hm x y; dsimp only [S]; rw [ih m hm (x + 2) (y + 2)]; ring
have h4 : ∀ x y : R, (y - y^2) * P n.succ y x =
(y - y^2) * (x + y -1) * (x + 1) * P n y (x + 2) + T n x y := by
intro x y; dsimp only [T, P]; ring
have h_t_symm : ∀ m : ℕ, m < n.succ.succ → ∀ x y : R, T m x y = T m y x := by
intro m hm x y; dsimp only[T]; rw [ih m hm x y]; ring
have h_u_symm : ∀ m : ℕ, m < n.succ.succ → ∀ x y : R, U m x y = U m y x := by
intro m hm x y; dsimp only [U]; rw [ih m hm (y+2) x, ih m hm (x+2) y]; ring
have h7 : ∀ x y : R, P n.succ.succ x y = S n x y + T n x y + U n x y := by
intro x y; rw [h1 x y, h2 x y, h4 x y]; dsimp only [U]; ring
have h8 : n < n.succ := lt_add_one n
have h9 : n < n.succ.succ := Nat.lt_succ_of_lt h8
intro x y
calc P n.succ.succ x y
= S n x y + T n x y + U n x y := h7 x y
_ = S n y x + T n x y + U n x y := by rw [h_s_symm n h9 x y]
_ = S n y x + T n y x + U n x y := by rw [h_t_symm n h9 x y]
_ = S n y x + T n y x + U n y x := by rw [h_u_symm n h9 x y]
_ = P n.succ.succ y x := (h7 y x).symm
end Bulgaria1998P8 | null | v4.29.0-rc6 | test | [
"competition"
] | null | {"filename": "Bulgaria1998P8.lean"} | {"v4.21.0": {"is_valid": true, "has_sorry": false, "errors": [], "timeout_s": 600.0, "latency_s": 1.7539, "verified_at": "2026-03-26T18:15:57.465676+00:00"}} | true | true | false | 1.7539 |
compfiles_CIIM2022P6 | compfiles | Copyright (c) 2024 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
# Iberoamerican Interuniversity Mathematics Competition 2022, Problem 6
Given a positive integer m, let d(m) be the number of postive
divisors of m. Show that for every positive integer n, one
has
d((n + 1)!) ≤ 2d(n!).
-/
namespace CIIM2022P6
def d : ℕ → ℕ
| m => (Nat.divisors m).card
theorem ciim2022_p6 (n : ℕ) (hn : 0 < n) :
d (Nat.factorial (n + 1)) ≤ 2 * d (Nat.factorial n) := by
sorry
end CIIM2022P6 | /- | /-
Copyright (c) 2024 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
# Iberoamerican Interuniversity Mathematics Competition 2022, Problem 6
Given a positive integer m, let d(m) be the number of postive
divisors of m. Show that for every positive integer n, one
has
d((n + 1)!) ≤ 2d(n!).
-/
namespace CIIM2022P6
def d : ℕ → ℕ
| m => (Nat.divisors m).card
theorem ciim2022_p6 (n : ℕ) (hn : 0 < n) :
d (Nat.factorial (n + 1)) ≤ 2 * d (Nat.factorial n) := by
sorry
end CIIM2022P6
| false | null | null | v4.29.0-rc6 | test | [
"competition"
] | null | {"filename": "CIIM2022P6.lean"} | {"v4.21.0": {"is_valid": false, "has_sorry": true, "errors": [], "timeout_s": 600.0, "latency_s": 0.0288, "verified_at": "2026-03-26T18:15:56.085817+00:00"}} | false | true | true | 0.0288 |
compfiles_Canada1998P3 | compfiles | Copyright (c) 2023 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
Canadian Mathematical Olympiad 1998, Problem 3
Let n be a natural number such that n ≥ 2. Show that
(1/(n + 1))(1 + 1/3 + ... + 1/(2n - 1)) > (1/n)(1/2 + 1/4 + ... + 1/2n).
-/
namespace Canada1998P3
theorem canada1998_p3 (n : ℕ) (hn : 2 ≤ n) :
(1/(n:ℝ)) * ∑ i ∈ Finset.range n, (1/(2 * (i:ℝ) + 2)) <
(1/((n:ℝ) + 1)) * ∑ i ∈ Finset.range n, (1/(2 * (i:ℝ) + 1)) := by
-- Follows the proof in _Mathematical Olympiads 1998-1999_
-- by Titu Andreescu and Zuming Feng
cases n with | zero => norm_num at hn | succ n =>
cases n with | zero => norm_num at hn | succ n =>
revert hn
intro h2; clear h2
-- We prove
-- (n + 1)(1/2 + 1/4 + ... + 1/2n) < n(1 + 1/3 + ... + 1/(2n - 1))
-- by induction.
suffices
((n.succ.succ:ℝ) + 1) * ∑ i ∈ Finset.range n.succ.succ, (1/(2 * (i:ℝ) + 2)) <
(n.succ.succ:ℝ) * ∑ i ∈ Finset.range n.succ.succ, (1/(2 * (i:ℝ) + 1))
by rw [div_mul_eq_mul_div₀, one_mul, div_mul_eq_mul_div₀, one_mul]
apply (div_lt_div_iff₀ (by positivity) (by positivity)).mpr
linarith
induction n with
| zero =>
-- Base case: when n = 2, we have 8/3 > 9/4.
norm_num
| succ m ih =>
let k := m.succ.succ
-- Inductive case: suppose claim is true for k ≥ 2. Then we have
-- k (1 + 1/3 + ... 1/(2k - 1)) > (k + 1)(1/2 + 1/4 + ... + 1/2k).
-- Note that
-- (1 + 1/3 + ... + 1/(2k-1)) + (k+1)/(2k+1)
-- = (1/2 + 1/3 + ... + 1/(2k-1)) + 1/2 + (k+1)/(2k+1)
-- > (1/2 + 1/4 + ... + 1/2k) + 1/2 + (k+1)/(2k+1)
-- > (1/2 + 1/4 + ... + 1/2k) + (k + 1)/(2k + 2) + 1/(2k+1)
-- > (1/2 + 1/4 + ... + 1/2k) + (k + 2)/(2k + 2).
have h1 : (1:ℝ) / (2 * ↑(0:ℕ) + 1) = 1 := by norm_num
have h2 : ∀ k' ∈ Finset.range (m + 1), (1:ℝ) / (2 * ↑(k' + 1) + 1 + 1) <
(1:ℝ) / (2 * ↑(k' + 1) + 1) := by
intro k' _
apply div_lt_div_of_pos_left zero_lt_one
· positivity
· exact lt_add_one _
have h3 : (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1)) + (k+1)/(2 * k + 1)
= _ := rfl
nth_rewrite 1 [Finset.sum_range_succ'] at h3
rw [h1, add_assoc] at h3
have h4 : Finset.Nonempty (Finset.range (m + 1)) := Finset.nonempty_range_add_one
have h5 := Finset.sum_lt_sum_of_nonempty h4 h2
norm_cast at h5
have h6 : (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1 + 1)) = _ := rfl
nth_rewrite 1 [Finset.sum_range_succ'] at h6
have h7' : (2:ℝ) * (k:ℝ) ≥ 4 := by
have hh2' : k ≥ 2 := Nat.succ_le_succ (Nat.succ_le_succ (Nat.zero_le m))
have hh2 : (k:ℝ) ≥ 2 := by exact_mod_cast hh2'
calc
(2:ℝ) * (k:ℝ) ≥ (2:ℝ) * 2 := mul_le_mul_of_nonneg_left hh2 zero_le_two
_ = 4 := by norm_num
have h7 : 1 / (2 * (k:ℝ) + 2) < 1 / 2 := by apply div_lt_div₀' <;> linarith
have h8 : ((k:ℝ)+1)/(2 * (k:ℝ) + 2) < ((k:ℝ)+1)/(2 * (k:ℝ) + 1) :=
by apply div_lt_div₀' <;> linarith
have h9 :=
-- (1 + 1/3 + ... + 1/(2k-1)) + (k+1)/(2k+1)
calc (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1)) + (k+1)/(2 * k + 1)
-- = (1/3 + ... + 1/(2k-1)) + (1 + (k+1)/(2k+1))
= (∑i ∈ Finset.range (m+1), 1 / (2 * ((i + 1):ℝ) + 1))
+ (1 + (k+1)/(2 * k + 1)) := by rw [←h3]; norm_cast
-- TODO shouldn't need casting?
-- > (1/4 + ... + 1/2k) + (1 + (k+1)/(2k+1))
_ > (∑i ∈ Finset.range (m+1), 1 / (2 * ((i + 1):ℝ) + 1 + 1))
+ (1 + (k+1)/(2 * k + 1)) := by norm_cast; linarith
_ = (∑i ∈ Finset.range (m+1), 1 / (2 * ((i + 1):ℝ) + 1 + 1))
+ (1/2 + 1/2 + (k+1)/(2 * k + 1)) := by ring
_ = (∑i ∈ Finset.range (m+1), 1 / (2 * ((i + 1):ℝ) + 1 + 1))
+ 1/2 + (1/2 + (k+1)/(2 * k + 1)) := by ring
_ = (∑i ∈ Finset.range (m+1), 1 / (2 * ((i + 1):ℝ) + 1 + 1))
+ 1/(2 *(((0:ℕ)):ℝ) + 1 + 1) + (1/2 + (k+1)/(2 * k + 1)) :=
by norm_num
_ = (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1 + 1)) +
(1/2 + (k+1)/(2 * k + 1)) := by rw [←h6]; norm_cast
-- > (1/2 + 1/4 + ... + 1/2k) + (k + 1)/(2k + 2) + 1/(2k+1)
_ > (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1 + 1)) +
((k+1)/(2 * k + 2) + 1/(2 * k + 2)) := by linarith
-- = (1/2 + 1/4 + ... + 1/2k) + (k + 2)/(2k + 2).
_ = (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1 + 1)) +
(k+2)/(2 * k + 2) := by field
--
-- Then we are done because
-- (k + 1)(1 + 1/3 + ... + 1/(2k - 1) + 1/(2k + 1))
-- = k (1 + 1/3 + ... + 1/(2k - 1))
-- + (1 + 1/3 + ... + 1/(2k - 1)) + (k + 1)/(2k + 1)
-- > k (1 + 1/3 + ... + 1/(2k - 1))
-- + (1/2 + 1/4 + ... + 1/2k)) + (k + 2)/(2k + 2)
-- (by h9, proved above)
-- > (k + 2)(1/2 + 1/4 + ... + 1/(2k + 2)).
have :=
calc (k + 1) * (∑i ∈ Finset.range k.succ, 1 / (2 * (i:ℝ) + 1))
= k * (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1)) +
((∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1)) +
(k + 1) / (2 * k + 1)) := by
rw [Finset.sum_range_succ]; ring
_ > k * (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1)) +
((∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1 + 1)) +
(k + 2) / (2 * k + 2)) := add_lt_add_right h9 _
_ > (k + 1) * (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 2)) +
((∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1 + 1)) +
(k + 2) / (2 * k + 2)) := add_lt_add_left ih _
_ = (k + 1) * (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 2)) +
((∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 2)) +
(k + 2) / (2 * k + 2)) := by
congr; funext x; rw [add_assoc, show (1:ℝ) + 1 = 2 by norm_num]
_ = (k + 2) * ((∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 2)) +
1 / (2 * k + 2)) := by ring
_ = (k + 2) * (∑i ∈ Finset.range k.succ, 1 / (2 * (i:ℝ) + 2))
:= by rw [←Finset.sum_range_succ]
norm_cast at this
norm_cast
end Canada1998P3 | /- | /-
Copyright (c) 2023 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
Canadian Mathematical Olympiad 1998, Problem 3
Let n be a natural number such that n ≥ 2. Show that
(1/(n + 1))(1 + 1/3 + ... + 1/(2n - 1)) > (1/n)(1/2 + 1/4 + ... + 1/2n).
-/
namespace Canada1998P3
theorem canada1998_p3 (n : ℕ) (hn : 2 ≤ n) :
(1/(n:ℝ)) * ∑ i ∈ Finset.range n, (1/(2 * (i:ℝ) + 2)) <
(1/((n:ℝ) + 1)) * ∑ i ∈ Finset.range n, (1/(2 * (i:ℝ) + 1)) := by
-- Follows the proof in _Mathematical Olympiads 1998-1999_
-- by Titu Andreescu and Zuming Feng
cases n with | zero => norm_num at hn | succ n =>
cases n with | zero => norm_num at hn | succ n =>
revert hn
intro h2; clear h2
-- We prove
-- (n + 1)(1/2 + 1/4 + ... + 1/2n) < n(1 + 1/3 + ... + 1/(2n - 1))
-- by induction.
suffices
((n.succ.succ:ℝ) + 1) * ∑ i ∈ Finset.range n.succ.succ, (1/(2 * (i:ℝ) + 2)) <
(n.succ.succ:ℝ) * ∑ i ∈ Finset.range n.succ.succ, (1/(2 * (i:ℝ) + 1))
by rw [div_mul_eq_mul_div₀, one_mul, div_mul_eq_mul_div₀, one_mul]
apply (div_lt_div_iff₀ (by positivity) (by positivity)).mpr
linarith
induction n with
| zero =>
-- Base case: when n = 2, we have 8/3 > 9/4.
norm_num
| succ m ih =>
let k := m.succ.succ
-- Inductive case: suppose claim is true for k ≥ 2. Then we have
-- k (1 + 1/3 + ... 1/(2k - 1)) > (k + 1)(1/2 + 1/4 + ... + 1/2k).
-- Note that
-- (1 + 1/3 + ... + 1/(2k-1)) + (k+1)/(2k+1)
-- = (1/2 + 1/3 + ... + 1/(2k-1)) + 1/2 + (k+1)/(2k+1)
-- > (1/2 + 1/4 + ... + 1/2k) + 1/2 + (k+1)/(2k+1)
-- > (1/2 + 1/4 + ... + 1/2k) + (k + 1)/(2k + 2) + 1/(2k+1)
-- > (1/2 + 1/4 + ... + 1/2k) + (k + 2)/(2k + 2).
have h1 : (1:ℝ) / (2 * ↑(0:ℕ) + 1) = 1 := by norm_num
have h2 : ∀ k' ∈ Finset.range (m + 1), (1:ℝ) / (2 * ↑(k' + 1) + 1 + 1) <
(1:ℝ) / (2 * ↑(k' + 1) + 1) := by
intro k' _
apply div_lt_div_of_pos_left zero_lt_one
· positivity
· exact lt_add_one _
have h3 : (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1)) + (k+1)/(2 * k + 1)
= _ := rfl
nth_rewrite 1 [Finset.sum_range_succ'] at h3
rw [h1, add_assoc] at h3
have h4 : Finset.Nonempty (Finset.range (m + 1)) := Finset.nonempty_range_add_one
have h5 := Finset.sum_lt_sum_of_nonempty h4 h2
norm_cast at h5
have h6 : (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1 + 1)) = _ := rfl
nth_rewrite 1 [Finset.sum_range_succ'] at h6
have h7' : (2:ℝ) * (k:ℝ) ≥ 4 := by
have hh2' : k ≥ 2 := Nat.succ_le_succ (Nat.succ_le_succ (Nat.zero_le m))
have hh2 : (k:ℝ) ≥ 2 := by exact_mod_cast hh2'
calc
(2:ℝ) * (k:ℝ) ≥ (2:ℝ) * 2 := mul_le_mul_of_nonneg_left hh2 zero_le_two
_ = 4 := by norm_num
have h7 : 1 / (2 * (k:ℝ) + 2) < 1 / 2 := by apply div_lt_div₀' <;> linarith
have h8 : ((k:ℝ)+1)/(2 * (k:ℝ) + 2) < ((k:ℝ)+1)/(2 * (k:ℝ) + 1) :=
by apply div_lt_div₀' <;> linarith
have h9 :=
-- (1 + 1/3 + ... + 1/(2k-1)) + (k+1)/(2k+1)
calc (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1)) + (k+1)/(2 * k + 1)
-- = (1/3 + ... + 1/(2k-1)) + (1 + (k+1)/(2k+1))
= (∑i ∈ Finset.range (m+1), 1 / (2 * ((i + 1):ℝ) + 1))
+ (1 + (k+1)/(2 * k + 1)) := by rw [←h3]; norm_cast
-- TODO shouldn't need casting?
-- > (1/4 + ... + 1/2k) + (1 + (k+1)/(2k+1))
_ > (∑i ∈ Finset.range (m+1), 1 / (2 * ((i + 1):ℝ) + 1 + 1))
+ (1 + (k+1)/(2 * k + 1)) := by norm_cast; linarith
_ = (∑i ∈ Finset.range (m+1), 1 / (2 * ((i + 1):ℝ) + 1 + 1))
+ (1/2 + 1/2 + (k+1)/(2 * k + 1)) := by ring
_ = (∑i ∈ Finset.range (m+1), 1 / (2 * ((i + 1):ℝ) + 1 + 1))
+ 1/2 + (1/2 + (k+1)/(2 * k + 1)) := by ring
_ = (∑i ∈ Finset.range (m+1), 1 / (2 * ((i + 1):ℝ) + 1 + 1))
+ 1/(2 *(((0:ℕ)):ℝ) + 1 + 1) + (1/2 + (k+1)/(2 * k + 1)) :=
by norm_num
_ = (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1 + 1)) +
(1/2 + (k+1)/(2 * k + 1)) := by rw [←h6]; norm_cast
-- > (1/2 + 1/4 + ... + 1/2k) + (k + 1)/(2k + 2) + 1/(2k+1)
_ > (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1 + 1)) +
((k+1)/(2 * k + 2) + 1/(2 * k + 2)) := by linarith
-- = (1/2 + 1/4 + ... + 1/2k) + (k + 2)/(2k + 2).
_ = (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1 + 1)) +
(k+2)/(2 * k + 2) := by field
--
-- Then we are done because
-- (k + 1)(1 + 1/3 + ... + 1/(2k - 1) + 1/(2k + 1))
-- = k (1 + 1/3 + ... + 1/(2k - 1))
-- + (1 + 1/3 + ... + 1/(2k - 1)) + (k + 1)/(2k + 1)
-- > k (1 + 1/3 + ... + 1/(2k - 1))
-- + (1/2 + 1/4 + ... + 1/2k)) + (k + 2)/(2k + 2)
-- (by h9, proved above)
-- > (k + 2)(1/2 + 1/4 + ... + 1/(2k + 2)).
have :=
calc (k + 1) * (∑i ∈ Finset.range k.succ, 1 / (2 * (i:ℝ) + 1))
= k * (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1)) +
((∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1)) +
(k + 1) / (2 * k + 1)) := by
rw [Finset.sum_range_succ]; ring
_ > k * (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1)) +
((∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1 + 1)) +
(k + 2) / (2 * k + 2)) := add_lt_add_right h9 _
_ > (k + 1) * (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 2)) +
((∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1 + 1)) +
(k + 2) / (2 * k + 2)) := add_lt_add_left ih _
_ = (k + 1) * (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 2)) +
((∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 2)) +
(k + 2) / (2 * k + 2)) := by
congr; funext x; rw [add_assoc, show (1:ℝ) + 1 = 2 by norm_num]
_ = (k + 2) * ((∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 2)) +
1 / (2 * k + 2)) := by ring
_ = (k + 2) * (∑i ∈ Finset.range k.succ, 1 / (2 * (i:ℝ) + 2))
:= by rw [←Finset.sum_range_succ]
norm_cast at this
norm_cast
end Canada1998P3
| true | Copyright (c) 2023 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
Canadian Mathematical Olympiad 1998, Problem 3
Let n be a natural number such that n ≥ 2. Show that
(1/(n + 1))(1 + 1/3 + ... + 1/(2n - 1)) > (1/n)(1/2 + 1/4 + ... + 1/2n).
-/
namespace Canada1998P3
theorem canada1998_p3 (n : ℕ) (hn : 2 ≤ n) :
(1/(n:ℝ)) * ∑ i ∈ Finset.range n, (1/(2 * (i:ℝ) + 2)) <
(1/((n:ℝ) + 1)) * ∑ i ∈ Finset.range n, (1/(2 * (i:ℝ) + 1)) := by
-- Follows the proof in _Mathematical Olympiads 1998-1999_
-- by Titu Andreescu and Zuming Feng
cases n with | zero => norm_num at hn | succ n =>
cases n with | zero => norm_num at hn | succ n =>
revert hn
intro h2; clear h2
-- We prove
-- (n + 1)(1/2 + 1/4 + ... + 1/2n) < n(1 + 1/3 + ... + 1/(2n - 1))
-- by induction.
suffices
((n.succ.succ:ℝ) + 1) * ∑ i ∈ Finset.range n.succ.succ, (1/(2 * (i:ℝ) + 2)) <
(n.succ.succ:ℝ) * ∑ i ∈ Finset.range n.succ.succ, (1/(2 * (i:ℝ) + 1))
by rw [div_mul_eq_mul_div₀, one_mul, div_mul_eq_mul_div₀, one_mul]
apply (div_lt_div_iff₀ (by positivity) (by positivity)).mpr
linarith
induction n with
| zero =>
-- Base case: when n = 2, we have 8/3 > 9/4.
norm_num
| succ m ih =>
let k := m.succ.succ
-- Inductive case: suppose claim is true for k ≥ 2. Then we have
-- k (1 + 1/3 + ... 1/(2k - 1)) > (k + 1)(1/2 + 1/4 + ... + 1/2k).
-- Note that
-- (1 + 1/3 + ... + 1/(2k-1)) + (k+1)/(2k+1)
-- = (1/2 + 1/3 + ... + 1/(2k-1)) + 1/2 + (k+1)/(2k+1)
-- > (1/2 + 1/4 + ... + 1/2k) + 1/2 + (k+1)/(2k+1)
-- > (1/2 + 1/4 + ... + 1/2k) + (k + 1)/(2k + 2) + 1/(2k+1)
-- > (1/2 + 1/4 + ... + 1/2k) + (k + 2)/(2k + 2).
have h1 : (1:ℝ) / (2 * ↑(0:ℕ) + 1) = 1 := by norm_num
have h2 : ∀ k' ∈ Finset.range (m + 1), (1:ℝ) / (2 * ↑(k' + 1) + 1 + 1) <
(1:ℝ) / (2 * ↑(k' + 1) + 1) := by
intro k' _
apply div_lt_div_of_pos_left zero_lt_one
· positivity
· exact lt_add_one _
have h3 : (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1)) + (k+1)/(2 * k + 1)
= _ := rfl
nth_rewrite 1 [Finset.sum_range_succ'] at h3
rw [h1, add_assoc] at h3
have h4 : Finset.Nonempty (Finset.range (m + 1)) := Finset.nonempty_range_add_one
have h5 := Finset.sum_lt_sum_of_nonempty h4 h2
norm_cast at h5
have h6 : (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1 + 1)) = _ := rfl
nth_rewrite 1 [Finset.sum_range_succ'] at h6
have h7' : (2:ℝ) * (k:ℝ) ≥ 4 := by
have hh2' : k ≥ 2 := Nat.succ_le_succ (Nat.succ_le_succ (Nat.zero_le m))
have hh2 : (k:ℝ) ≥ 2 := by exact_mod_cast hh2'
calc
(2:ℝ) * (k:ℝ) ≥ (2:ℝ) * 2 := mul_le_mul_of_nonneg_left hh2 zero_le_two
_ = 4 := by norm_num
have h7 : 1 / (2 * (k:ℝ) + 2) < 1 / 2 := by apply div_lt_div₀' <;> linarith
have h8 : ((k:ℝ)+1)/(2 * (k:ℝ) + 2) < ((k:ℝ)+1)/(2 * (k:ℝ) + 1) :=
by apply div_lt_div₀' <;> linarith
have h9 :=
-- (1 + 1/3 + ... + 1/(2k-1)) + (k+1)/(2k+1)
calc (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1)) + (k+1)/(2 * k + 1)
-- = (1/3 + ... + 1/(2k-1)) + (1 + (k+1)/(2k+1))
= (∑i ∈ Finset.range (m+1), 1 / (2 * ((i + 1):ℝ) + 1))
+ (1 + (k+1)/(2 * k + 1)) := by rw [←h3]; norm_cast
-- TODO shouldn't need casting?
-- > (1/4 + ... + 1/2k) + (1 + (k+1)/(2k+1))
_ > (∑i ∈ Finset.range (m+1), 1 / (2 * ((i + 1):ℝ) + 1 + 1))
+ (1 + (k+1)/(2 * k + 1)) := by norm_cast; linarith
_ = (∑i ∈ Finset.range (m+1), 1 / (2 * ((i + 1):ℝ) + 1 + 1))
+ (1/2 + 1/2 + (k+1)/(2 * k + 1)) := by ring
_ = (∑i ∈ Finset.range (m+1), 1 / (2 * ((i + 1):ℝ) + 1 + 1))
+ 1/2 + (1/2 + (k+1)/(2 * k + 1)) := by ring
_ = (∑i ∈ Finset.range (m+1), 1 / (2 * ((i + 1):ℝ) + 1 + 1))
+ 1/(2 *(((0:ℕ)):ℝ) + 1 + 1) + (1/2 + (k+1)/(2 * k + 1)) :=
by norm_num
_ = (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1 + 1)) +
(1/2 + (k+1)/(2 * k + 1)) := by rw [←h6]; norm_cast
-- > (1/2 + 1/4 + ... + 1/2k) + (k + 1)/(2k + 2) + 1/(2k+1)
_ > (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1 + 1)) +
((k+1)/(2 * k + 2) + 1/(2 * k + 2)) := by linarith
-- = (1/2 + 1/4 + ... + 1/2k) + (k + 2)/(2k + 2).
_ = (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1 + 1)) +
(k+2)/(2 * k + 2) := by field
--
-- Then we are done because
-- (k + 1)(1 + 1/3 + ... + 1/(2k - 1) + 1/(2k + 1))
-- = k (1 + 1/3 + ... + 1/(2k - 1))
-- + (1 + 1/3 + ... + 1/(2k - 1)) + (k + 1)/(2k + 1)
-- > k (1 + 1/3 + ... + 1/(2k - 1))
-- + (1/2 + 1/4 + ... + 1/2k)) + (k + 2)/(2k + 2)
-- (by h9, proved above)
-- > (k + 2)(1/2 + 1/4 + ... + 1/(2k + 2)).
have :=
calc (k + 1) * (∑i ∈ Finset.range k.succ, 1 / (2 * (i:ℝ) + 1))
= k * (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1)) +
((∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1)) +
(k + 1) / (2 * k + 1)) := by
rw [Finset.sum_range_succ]; ring
_ > k * (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1)) +
((∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1 + 1)) +
(k + 2) / (2 * k + 2)) := add_lt_add_right h9 _
_ > (k + 1) * (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 2)) +
((∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 1 + 1)) +
(k + 2) / (2 * k + 2)) := add_lt_add_left ih _
_ = (k + 1) * (∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 2)) +
((∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 2)) +
(k + 2) / (2 * k + 2)) := by
congr; funext x; rw [add_assoc, show (1:ℝ) + 1 = 2 by norm_num]
_ = (k + 2) * ((∑i ∈ Finset.range k, 1 / (2 * (i:ℝ) + 2)) +
1 / (2 * k + 2)) := by ring
_ = (k + 2) * (∑i ∈ Finset.range k.succ, 1 / (2 * (i:ℝ) + 2))
:= by rw [←Finset.sum_range_succ]
norm_cast at this
norm_cast
end Canada1998P3 | null | v4.29.0-rc6 | test | [
"competition"
] | null | {"filename": "Canada1998P3.lean"} | {"v4.21.0": {"is_valid": false, "has_sorry": false, "errors": ["unknown tactic", "unknown constant 'Finset.nonempty_range_add_one'", "unsolved goals\nm : \u2115\nih :\n (\u2191m.succ.succ + 1) * \u2211 i \u2208 Finset.range m.succ.succ, 1 / (2 * \u2191i + 2) <\n \u2191m.succ.succ * \u2211 i \u2208 Finset.range m.succ.succ, 1 / (2 * \u2191i + 1)\nk : \u2115 := m.succ.succ\nh1 : 1 / (2 * \u21910 + 1) = 1\nh2 : \u2200 k' \u2208 Finset.range (m + 1), 1 / (2 * \u2191(k' + 1) + 1 + 1) < 1 / (2 * \u2191(k' + 1) + 1)\nh3 :\n \u2211 k \u2208 Finset.range (m + 1), 1 / (2 * \u2191(k + 1) + 1) + (1 + (\u2191k + 1) / (2 * \u2191k + 1)) =\n \u2211 i \u2208 Finset.range k, 1 / (2 * \u2191i + 1) + (\u2191k + 1) / (2 * \u2191k + 1)\nh4 : (Finset.range (m + 1)).Nonempty\nh5 : \u2211 x \u2208 Finset.range (m + 1), 1 / \u2191(2 * (x + 1) + 1 + 1) < \u2211 x \u2208 Finset.range (m + 1), 1 / \u2191(2 * (x + 1) + 1)\nh6 :\n \u2211 k \u2208 Finset.range (m + 1), 1 / (2 * \u2191(k + 1) + 1 + 1) + 1 / (2 * \u21910 + 1 + 1) =\n \u2211 i \u2208 Finset.range k, 1 / (2 * \u2191i + 1 + 1)\nh7' : 2 * \u2191k \u2265 4\nh7 : 1 / (2 * \u2191k + 2) < 1 / 2\nh8 : (\u2191k + 1) / (2 * \u2191k + 2) < (\u2191k + 1) / (2 * \u2191k + 1)\n\u22a2 \u2211 i \u2208 Finset.range k, 1 / (2 * \u2191i + 1 + 1) + ((\u2191k + 1) / (2 * \u2191k + 2) + 1 / (2 * \u2191k + 2)) =\n \u2211 i \u2208 Finset.range k, 1 / (2 * \u2191i + 1 + 1) + (\u2191k + 2) / (2 * \u2191k + 2)", "unsolved goals\ncase succ.succ.succ\nm : \u2115\nih :\n (\u2191m.succ.succ + 1) * \u2211 i \u2208 Finset.range m.succ.succ, 1 / (2 * \u2191i + 2) <\n \u2191m.succ.succ * \u2211 i \u2208 Finset.range m.succ.succ, 1 / (2 * \u2191i + 1)\nk : \u2115 := m.succ.succ\nh1 : 1 / (2 * \u21910 + 1) = 1\nh2 : \u2200 k' \u2208 Finset.range (m + 1), 1 / (2 * \u2191(k' + 1) + 1 + 1) < 1 / (2 * \u2191(k' + 1) + 1)\nh3 :\n \u2211 k \u2208 Finset.range (m + 1), 1 / (2 * \u2191(k + 1) + 1) + (1 + (\u2191k + 1) / (2 * \u2191k + 1)) =\n \u2211 i \u2208 Finset.range k, 1 / (2 * \u2191i + 1) + (\u2191k + 1) / (2 * \u2191k + 1)\nh4 : (Finset.range (m + 1)).Nonempty\nh5 : \u2211 x \u2208 Finset.range (m + 1), 1 / \u2191(2 * (x + 1) + 1 + 1) < \u2211 x \u2208 Finset.range (m + 1), 1 / \u2191(2 * (x + 1) + 1)\nh6 :\n \u2211 k \u2208 Finset.range (m + 1), 1 / (2 * \u2191(k + 1) + 1 + 1) + 1 / (2 * \u21910 + 1 + 1) =\n \u2211 i \u2208 Finset.range k, 1 / (2 * \u2191i + 1 + 1)\nh7' : 2 * \u2191k \u2265 4\nh7 : 1 / (2 * \u2191k + 2) < 1 / 2\nh8 : (\u2191k + 1) / (2 * \u2191k + 2) < (\u2191k + 1) / (2 * \u2191k + 1)\nh9 :\n \u2211 i \u2208 Finset.range k, 1 / (2 * \u2191i + 1) + (\u2191k + 1) / (2 * \u2191k + 1) >\n \u2211 i \u2208 Finset.range k, 1 / (2 * \u2191i + 1 + 1) + (\u2191k + 2) / (2 * \u2191k + 2)\n\u22a2 (\u2191(m + 1).succ.succ + 1) * \u2211 i \u2208 Finset.range (m + 1).succ.succ, 1 / (2 * \u2191i + 2) <\n \u2191(m + 1).succ.succ * \u2211 i \u2208 Finset.range (m + 1).succ.succ, 1 / (2 * \u2191i + 1)"], "timeout_s": 600.0, "latency_s": 5.2525, "verified_at": "2026-03-26T18:16:01.338447+00:00"}} | false | true | false | 5.2525 |
compfiles_Canada1998P5 | compfiles | Copyright (c) 2023 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
Canadian Mathematical Olympiad 1998, Problem 5
Let m be a positive integer. Define the sequence {aₙ} by a₀ = 0,
a₁ = m, and aₙ₊₁ = m²aₙ - aₙ₋₁ for n ≥ 1. Prove that an ordered pair
(a,b) of nonegative integers, with a ≤ b, is a solution of the equation
(a² + b²) / (ab + 1) = m²
if an only if (a,b) = (aₙ,aₙ₊₁) for some n ≥ 0.
-/
namespace Canada1998P5
def A (m : ℕ) (hm : 0 < m) : ℕ → ℤ
| 0 => 0
| 1 => (↑m)
| n + 2 => (m : ℤ)^2 * A m hm (n + 1) - A m hm n
theorem canada1998_p5 (m : ℕ) (hm : 0 < m) (a b : ℕ) (hab : a ≤ b) :
a^2 + b^2 = m^2 * (a * b + 1) ↔
∃ n : ℕ, (a:ℤ) = A m hm n ∧ (b:ℤ) = A m hm (n + 1) := by
sorry
end Canada1998P5 | /- | /-
Copyright (c) 2023 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
Canadian Mathematical Olympiad 1998, Problem 5
Let m be a positive integer. Define the sequence {aₙ} by a₀ = 0,
a₁ = m, and aₙ₊₁ = m²aₙ - aₙ₋₁ for n ≥ 1. Prove that an ordered pair
(a,b) of nonegative integers, with a ≤ b, is a solution of the equation
(a² + b²) / (ab + 1) = m²
if an only if (a,b) = (aₙ,aₙ₊₁) for some n ≥ 0.
-/
namespace Canada1998P5
def A (m : ℕ) (hm : 0 < m) : ℕ → ℤ
| 0 => 0
| 1 => (↑m)
| n + 2 => (m : ℤ)^2 * A m hm (n + 1) - A m hm n
theorem canada1998_p5 (m : ℕ) (hm : 0 < m) (a b : ℕ) (hab : a ≤ b) :
a^2 + b^2 = m^2 * (a * b + 1) ↔
∃ n : ℕ, (a:ℤ) = A m hm n ∧ (b:ℤ) = A m hm (n + 1) := by
sorry
end Canada1998P5
| false | null | null | v4.29.0-rc6 | test | [
"competition"
] | null | {"filename": "Canada1998P5.lean"} | {"v4.21.0": {"is_valid": false, "has_sorry": true, "errors": [], "timeout_s": 600.0, "latency_s": 0.0672, "verified_at": "2026-03-26T18:15:56.404638+00:00"}} | false | true | true | 0.0672 |
compfiles_Egmo2023P1 | compfiles | Copyright (c) 2025 The Compfiles Contributors. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Ansar Azharov
-/
import Mathlib
/-!
# European Girls' Mathematical Olympiad 2023, Problem 1
There are n ≥ 3 positive real numbers a_1, a_2, . . . , a_n. For each 1 ≤ i ≤ n we let
b_i = (a_(i-1) + a_(i+1)) / a_i (here we define a_0 to be a_n and a_(n+1) to be a_1).
Assume that for all i and j in the range 1 to n, we have a_i ≤ a_j if and only if b_i ≤ b_j.
Prove that a_1 = a_2 = ... = a_n.
-/
namespace Egmo2023P1
theorem egmo2023_p1 (n : ℕ) [NeZero n] (_ : n ≥ 3) (a : Fin n → ℝ) (ha : ∀ i, a i > 0)
(b : Fin n → ℝ) (hb : ∀ i, b i = (a (i - 1) + a (i + 1)) / (a i))
(h : ∀ i j, a i ≤ a j ↔ b i ≤ b j) : ∀ i j, a i = a j := by
-- The solution was adapted from here: https://www.youtube.com/watch?v=PNFh789P31E.
-- Note that indexing starts from 0 in the formalized statement.
obtain ⟨m, hm⟩ := Finite.exists_min a
obtain ⟨M, hM⟩ := Finite.exists_max a
have h1 : b m ≥ 2 := calc b m
_ ≥ (a m + a m) / (a m) := by
rw [hb m]
gcongr
· exact le_of_lt (ha m)
· exact hm (m - 1)
· exact hm (m + 1)
_ ≥ 2 := by
rw [← two_mul, mul_div_cancel_right₀]
positivity [ha m]
have h2 : b M ≤ 2 := calc b M
_ ≤ (a M + a M) / (a M) := by
rw [hb M]
gcongr
· exact le_of_lt (ha M)
· exact hM (M - 1)
· exact hM (M + 1)
_ ≤ 2 := by
rw [← two_mul, mul_div_cancel_right₀]
positivity [ha M]
have h3 : ∀ i, b i = 2 := by
intro i
apply eq_of_le_of_ge
· exact le_trans ((h i M).mp (hM i)) h2
· exact le_trans h1 ((h m i).mp (hm i))
intro i j
apply eq_of_le_of_ge
· rw [h i j, h3 i, h3 j]
· rw [h j i, h3 j, h3 i]
end Egmo2023P1 | /- | /-
Copyright (c) 2025 The Compfiles Contributors. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Ansar Azharov
-/
import Mathlib
/-!
# European Girls' Mathematical Olympiad 2023, Problem 1
There are n ≥ 3 positive real numbers a_1, a_2, . . . , a_n. For each 1 ≤ i ≤ n we let
b_i = (a_(i-1) + a_(i+1)) / a_i (here we define a_0 to be a_n and a_(n+1) to be a_1).
Assume that for all i and j in the range 1 to n, we have a_i ≤ a_j if and only if b_i ≤ b_j.
Prove that a_1 = a_2 = ... = a_n.
-/
namespace Egmo2023P1
theorem egmo2023_p1 (n : ℕ) [NeZero n] (_ : n ≥ 3) (a : Fin n → ℝ) (ha : ∀ i, a i > 0)
(b : Fin n → ℝ) (hb : ∀ i, b i = (a (i - 1) + a (i + 1)) / (a i))
(h : ∀ i j, a i ≤ a j ↔ b i ≤ b j) : ∀ i j, a i = a j := by
-- The solution was adapted from here: https://www.youtube.com/watch?v=PNFh789P31E.
-- Note that indexing starts from 0 in the formalized statement.
obtain ⟨m, hm⟩ := Finite.exists_min a
obtain ⟨M, hM⟩ := Finite.exists_max a
have h1 : b m ≥ 2 := calc b m
_ ≥ (a m + a m) / (a m) := by
rw [hb m]
gcongr
· exact le_of_lt (ha m)
· exact hm (m - 1)
· exact hm (m + 1)
_ ≥ 2 := by
rw [← two_mul, mul_div_cancel_right₀]
positivity [ha m]
have h2 : b M ≤ 2 := calc b M
_ ≤ (a M + a M) / (a M) := by
rw [hb M]
gcongr
· exact le_of_lt (ha M)
· exact hM (M - 1)
· exact hM (M + 1)
_ ≤ 2 := by
rw [← two_mul, mul_div_cancel_right₀]
positivity [ha M]
have h3 : ∀ i, b i = 2 := by
intro i
apply eq_of_le_of_ge
· exact le_trans ((h i M).mp (hM i)) h2
· exact le_trans h1 ((h m i).mp (hm i))
intro i j
apply eq_of_le_of_ge
· rw [h i j, h3 i, h3 j]
· rw [h j i, h3 j, h3 i]
end Egmo2023P1
| true | Copyright (c) 2025 The Compfiles Contributors. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Ansar Azharov
-/
import Mathlib
/-!
# European Girls' Mathematical Olympiad 2023, Problem 1
There are n ≥ 3 positive real numbers a_1, a_2, . . . , a_n. For each 1 ≤ i ≤ n we let
b_i = (a_(i-1) + a_(i+1)) / a_i (here we define a_0 to be a_n and a_(n+1) to be a_1).
Assume that for all i and j in the range 1 to n, we have a_i ≤ a_j if and only if b_i ≤ b_j.
Prove that a_1 = a_2 = ... = a_n.
-/
namespace Egmo2023P1
theorem egmo2023_p1 (n : ℕ) [NeZero n] (_ : n ≥ 3) (a : Fin n → ℝ) (ha : ∀ i, a i > 0)
(b : Fin n → ℝ) (hb : ∀ i, b i = (a (i - 1) + a (i + 1)) / (a i))
(h : ∀ i j, a i ≤ a j ↔ b i ≤ b j) : ∀ i j, a i = a j := by
-- The solution was adapted from here: https://www.youtube.com/watch?v=PNFh789P31E.
-- Note that indexing starts from 0 in the formalized statement.
obtain ⟨m, hm⟩ := Finite.exists_min a
obtain ⟨M, hM⟩ := Finite.exists_max a
have h1 : b m ≥ 2 := calc b m
_ ≥ (a m + a m) / (a m) := by
rw [hb m]
gcongr
· exact le_of_lt (ha m)
· exact hm (m - 1)
· exact hm (m + 1)
_ ≥ 2 := by
rw [← two_mul, mul_div_cancel_right₀]
positivity [ha m]
have h2 : b M ≤ 2 := calc b M
_ ≤ (a M + a M) / (a M) := by
rw [hb M]
gcongr
· exact le_of_lt (ha M)
· exact hM (M - 1)
· exact hM (M + 1)
_ ≤ 2 := by
rw [← two_mul, mul_div_cancel_right₀]
positivity [ha M]
have h3 : ∀ i, b i = 2 := by
intro i
apply eq_of_le_of_ge
· exact le_trans ((h i M).mp (hM i)) h2
· exact le_trans h1 ((h m i).mp (hm i))
intro i j
apply eq_of_le_of_ge
· rw [h i j, h3 i, h3 j]
· rw [h j i, h3 j, h3 i]
end Egmo2023P1 | null | v4.29.0-rc6 | test | [
"competition"
] | null | {"filename": "Egmo2023P1.lean"} | {"v4.21.0": {"is_valid": false, "has_sorry": false, "errors": ["failed to prove positivity/nonnegativity/nonzeroness", "unsolved goals\ncase intro.intro\nn : \u2115\ninst\u271d : NeZero n\nx\u271d : n \u2265 3\na : Fin n \u2192 \u211d\nha : \u2200 (i : Fin n), a i > 0\nb : Fin n \u2192 \u211d\nhb : \u2200 (i : Fin n), b i = (a (i - 1) + a (i + 1)) / a i\nh : \u2200 (i j : Fin n), a i \u2264 a j \u2194 b i \u2264 b j\nm : Fin n\nhm : \u2200 (x : Fin n), a m \u2264 a x\nM : Fin n\nhM : \u2200 (x : Fin n), a x \u2264 a M\nh1 : b m \u2265 2\n\u22a2 \u2200 (i j : Fin n), a i = a j", "unexpected token '['; expected command"], "timeout_s": 600.0, "latency_s": 0.2509, "verified_at": "2026-03-26T18:15:56.655693+00:00"}} | false | true | false | 0.2509 |
compfiles_Hungary1998P6 | compfiles | Copyright (c) 2023 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
# Hungarian Mathematical Olympiad 1998, Problem 6
Let x, y, z be integers with z > 1. Show that
(x + 1)² + (x + 2)² + ... + (x + 99)² ≠ yᶻ.
-/
namespace Hungary1998P6
lemma sum_range_square_mul_six (n : ℕ) :
(∑i ∈ Finset.range n, (i + 1)^2) * 6 = n * (n + 1) * (2 * n + 1) := by
induction n with
| zero => rfl
| succ n ih =>
rw [Finset.sum_range_succ, add_mul, ih]
ring
lemma sum_range_square (n : ℕ) :
∑i ∈ Finset.range n, (i + 1)^2 = n * (n + 1) * (2 * n + 1)/6 :=
by rw [← sum_range_square_mul_six n, Nat.mul_div_cancel]
norm_num
lemma cast_sum_square (n : ℕ) :
∑i ∈ Finset.range n, ((i:ℤ)+1)^2 =
(((∑i ∈ Finset.range n, (i+1)^2) : ℕ) : ℤ) := by norm_cast
theorem hungary1998_p6 (x y : ℤ) (z : ℕ) (hz : 1 < z) :
∑ i ∈ Finset.range 99, (x + i + 1)^2 ≠ y^z := by
-- Follows the proof in _Mathematical Olympiads 1998-1999_
-- by Titu Andreescu and Zuming Feng.
-- Suppose (x + 1)² + (x + 2)² + ... + (x + 99)² = yᶻ.
intro he
-- We notice that
-- y^z = (x + 1)² + (x + 2)² + ... + (x + 99)²
-- = 99x² + 2(1 + 2 + ... + 99)x + (1² + 2² + ... + 99²)
-- = 99x² + 2[(99 ⬝ 100)/2]x + (99 ⬝ 100 ⬝ 199)/6
-- = 33(3x² + 300x + 50 ⬝ 199).
have h2 : ∑ i ∈ Finset.range 99, x^2 = 99 * x^2 := by norm_num
have h4 : ∑ i ∈ Finset.range 99, ((i:ℤ) + 1) =
∑ i ∈ Finset.range 100, (i:ℤ) := by
rw [Finset.sum_range_succ']
simp [Finset.sum_range_succ]
have h5 : ∑ i ∈ Finset.range 100, (i:ℤ) = 99 * 100 / 2 := by
rw [←Nat.cast_sum, Finset.sum_range_id]; decide
have h6 : ∑ i ∈ Finset.range 99, ((i:ℤ) + 1)^2 = (99 * 100 * 199)/6 := by
rw [cast_sum_square, sum_range_square]; decide
have hnn1 : (99:ℤ) * 100 / 2 = 99 * 50 := by decide
have hnn2 : ((99:ℤ) * 100 * 199)/6 = 33 * 50 * 199 := by decide
have h7 := calc y^z
= ∑ i ∈ Finset.range 99, ((x + i) + 1)^2 := he.symm
_ = ∑ i ∈ Finset.range 99,
(x^2 + 2 * x * ((i:ℤ) + 1) + ((i:ℤ) + 1)^2) :=
by with_reducible congr; funext; ring
_ = ∑ i ∈ Finset.range 99, (x^2 + 2 * x * ((i:ℤ) + 1)) +
∑ i ∈ Finset.range 99, ((i:ℤ) + 1)^2 := Finset.sum_add_distrib
_ = ∑ i ∈ Finset.range 99, x^2 +
∑ i ∈ Finset.range 99, (2 * x * ((i:ℤ) + 1)) +
∑ i ∈ Finset.range 99, (((i:ℤ) + 1)^2) := by rw [Finset.sum_add_distrib]
_ = 99 * x^2 + 2 * x * (99 * 100 / 2) + (99 * 100 * 199)/6
:= by rw [h2, ←Finset.mul_sum, h4, h5, h6]
_ = 3 * (11 * (3 * x^2 + 300 * x + 50 * 199)) := by rw [hnn1, hnn2]; ring
-- which implies that 3∣y.
have h8 : 3 ∣ y^z := Dvd.intro _ (Eq.symm h7)
have h9 : 3 ∣ y := Prime.dvd_of_dvd_pow Int.prime_three h8
obtain ⟨k, hk⟩ := h9
rw [hk] at h7
cases z with | zero => exact Nat.not_lt_zero 1 hz | succ z =>
cases z with | zero => exact Nat.lt_asymm hz hz | succ z =>
rw [pow_succ, pow_succ] at h7
-- Since z ≥ 2, 3²∣yᶻ, but 3² does not divide
-- 33(3x² + 300x + 50 ⬝ 199), contradiction.
omega
end Hungary1998P6 | /- | /-
Copyright (c) 2023 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
# Hungarian Mathematical Olympiad 1998, Problem 6
Let x, y, z be integers with z > 1. Show that
(x + 1)² + (x + 2)² + ... + (x + 99)² ≠ yᶻ.
-/
namespace Hungary1998P6
lemma sum_range_square_mul_six (n : ℕ) :
(∑i ∈ Finset.range n, (i + 1)^2) * 6 = n * (n + 1) * (2 * n + 1) := by
induction n with
| zero => rfl
| succ n ih =>
rw [Finset.sum_range_succ, add_mul, ih]
ring
lemma sum_range_square (n : ℕ) :
∑i ∈ Finset.range n, (i + 1)^2 = n * (n + 1) * (2 * n + 1)/6 :=
by rw [← sum_range_square_mul_six n, Nat.mul_div_cancel]
norm_num
lemma cast_sum_square (n : ℕ) :
∑i ∈ Finset.range n, ((i:ℤ)+1)^2 =
(((∑i ∈ Finset.range n, (i+1)^2) : ℕ) : ℤ) := by norm_cast
theorem hungary1998_p6 (x y : ℤ) (z : ℕ) (hz : 1 < z) :
∑ i ∈ Finset.range 99, (x + i + 1)^2 ≠ y^z := by
-- Follows the proof in _Mathematical Olympiads 1998-1999_
-- by Titu Andreescu and Zuming Feng.
-- Suppose (x + 1)² + (x + 2)² + ... + (x + 99)² = yᶻ.
intro he
-- We notice that
-- y^z = (x + 1)² + (x + 2)² + ... + (x + 99)²
-- = 99x² + 2(1 + 2 + ... + 99)x + (1² + 2² + ... + 99²)
-- = 99x² + 2[(99 ⬝ 100)/2]x + (99 ⬝ 100 ⬝ 199)/6
-- = 33(3x² + 300x + 50 ⬝ 199).
have h2 : ∑ i ∈ Finset.range 99, x^2 = 99 * x^2 := by norm_num
have h4 : ∑ i ∈ Finset.range 99, ((i:ℤ) + 1) =
∑ i ∈ Finset.range 100, (i:ℤ) := by
rw [Finset.sum_range_succ']
simp [Finset.sum_range_succ]
have h5 : ∑ i ∈ Finset.range 100, (i:ℤ) = 99 * 100 / 2 := by
rw [←Nat.cast_sum, Finset.sum_range_id]; decide
have h6 : ∑ i ∈ Finset.range 99, ((i:ℤ) + 1)^2 = (99 * 100 * 199)/6 := by
rw [cast_sum_square, sum_range_square]; decide
have hnn1 : (99:ℤ) * 100 / 2 = 99 * 50 := by decide
have hnn2 : ((99:ℤ) * 100 * 199)/6 = 33 * 50 * 199 := by decide
have h7 := calc y^z
= ∑ i ∈ Finset.range 99, ((x + i) + 1)^2 := he.symm
_ = ∑ i ∈ Finset.range 99,
(x^2 + 2 * x * ((i:ℤ) + 1) + ((i:ℤ) + 1)^2) :=
by with_reducible congr; funext; ring
_ = ∑ i ∈ Finset.range 99, (x^2 + 2 * x * ((i:ℤ) + 1)) +
∑ i ∈ Finset.range 99, ((i:ℤ) + 1)^2 := Finset.sum_add_distrib
_ = ∑ i ∈ Finset.range 99, x^2 +
∑ i ∈ Finset.range 99, (2 * x * ((i:ℤ) + 1)) +
∑ i ∈ Finset.range 99, (((i:ℤ) + 1)^2) := by rw [Finset.sum_add_distrib]
_ = 99 * x^2 + 2 * x * (99 * 100 / 2) + (99 * 100 * 199)/6
:= by rw [h2, ←Finset.mul_sum, h4, h5, h6]
_ = 3 * (11 * (3 * x^2 + 300 * x + 50 * 199)) := by rw [hnn1, hnn2]; ring
-- which implies that 3∣y.
have h8 : 3 ∣ y^z := Dvd.intro _ (Eq.symm h7)
have h9 : 3 ∣ y := Prime.dvd_of_dvd_pow Int.prime_three h8
obtain ⟨k, hk⟩ := h9
rw [hk] at h7
cases z with | zero => exact Nat.not_lt_zero 1 hz | succ z =>
cases z with | zero => exact Nat.lt_asymm hz hz | succ z =>
rw [pow_succ, pow_succ] at h7
-- Since z ≥ 2, 3²∣yᶻ, but 3² does not divide
-- 33(3x² + 300x + 50 ⬝ 199), contradiction.
omega
end Hungary1998P6
| true | Copyright (c) 2023 David Renshaw. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: David Renshaw
-/
import Mathlib
/-!
# Hungarian Mathematical Olympiad 1998, Problem 6
Let x, y, z be integers with z > 1. Show that
(x + 1)² + (x + 2)² + ... + (x + 99)² ≠ yᶻ.
-/
namespace Hungary1998P6
lemma sum_range_square_mul_six (n : ℕ) :
(∑i ∈ Finset.range n, (i + 1)^2) * 6 = n * (n + 1) * (2 * n + 1) := by
induction n with
| zero => rfl
| succ n ih =>
rw [Finset.sum_range_succ, add_mul, ih]
ring
lemma sum_range_square (n : ℕ) :
∑i ∈ Finset.range n, (i + 1)^2 = n * (n + 1) * (2 * n + 1)/6 :=
by rw [← sum_range_square_mul_six n, Nat.mul_div_cancel]
norm_num
lemma cast_sum_square (n : ℕ) :
∑i ∈ Finset.range n, ((i:ℤ)+1)^2 =
(((∑i ∈ Finset.range n, (i+1)^2) : ℕ) : ℤ) := by norm_cast
theorem hungary1998_p6 (x y : ℤ) (z : ℕ) (hz : 1 < z) :
∑ i ∈ Finset.range 99, (x + i + 1)^2 ≠ y^z := by
-- Follows the proof in _Mathematical Olympiads 1998-1999_
-- by Titu Andreescu and Zuming Feng.
-- Suppose (x + 1)² + (x + 2)² + ... + (x + 99)² = yᶻ.
intro he
-- We notice that
-- y^z = (x + 1)² + (x + 2)² + ... + (x + 99)²
-- = 99x² + 2(1 + 2 + ... + 99)x + (1² + 2² + ... + 99²)
-- = 99x² + 2[(99 ⬝ 100)/2]x + (99 ⬝ 100 ⬝ 199)/6
-- = 33(3x² + 300x + 50 ⬝ 199).
have h2 : ∑ i ∈ Finset.range 99, x^2 = 99 * x^2 := by norm_num
have h4 : ∑ i ∈ Finset.range 99, ((i:ℤ) + 1) =
∑ i ∈ Finset.range 100, (i:ℤ) := by
rw [Finset.sum_range_succ']
simp [Finset.sum_range_succ]
have h5 : ∑ i ∈ Finset.range 100, (i:ℤ) = 99 * 100 / 2 := by
rw [←Nat.cast_sum, Finset.sum_range_id]; decide
have h6 : ∑ i ∈ Finset.range 99, ((i:ℤ) + 1)^2 = (99 * 100 * 199)/6 := by
rw [cast_sum_square, sum_range_square]; decide
have hnn1 : (99:ℤ) * 100 / 2 = 99 * 50 := by decide
have hnn2 : ((99:ℤ) * 100 * 199)/6 = 33 * 50 * 199 := by decide
have h7 := calc y^z
= ∑ i ∈ Finset.range 99, ((x + i) + 1)^2 := he.symm
_ = ∑ i ∈ Finset.range 99,
(x^2 + 2 * x * ((i:ℤ) + 1) + ((i:ℤ) + 1)^2) :=
by with_reducible congr; funext; ring
_ = ∑ i ∈ Finset.range 99, (x^2 + 2 * x * ((i:ℤ) + 1)) +
∑ i ∈ Finset.range 99, ((i:ℤ) + 1)^2 := Finset.sum_add_distrib
_ = ∑ i ∈ Finset.range 99, x^2 +
∑ i ∈ Finset.range 99, (2 * x * ((i:ℤ) + 1)) +
∑ i ∈ Finset.range 99, (((i:ℤ) + 1)^2) := by rw [Finset.sum_add_distrib]
_ = 99 * x^2 + 2 * x * (99 * 100 / 2) + (99 * 100 * 199)/6
:= by rw [h2, ←Finset.mul_sum, h4, h5, h6]
_ = 3 * (11 * (3 * x^2 + 300 * x + 50 * 199)) := by rw [hnn1, hnn2]; ring
-- which implies that 3∣y.
have h8 : 3 ∣ y^z := Dvd.intro _ (Eq.symm h7)
have h9 : 3 ∣ y := Prime.dvd_of_dvd_pow Int.prime_three h8
obtain ⟨k, hk⟩ := h9
rw [hk] at h7
cases z with | zero => exact Nat.not_lt_zero 1 hz | succ z =>
cases z with | zero => exact Nat.lt_asymm hz hz | succ z =>
rw [pow_succ, pow_succ] at h7
-- Since z ≥ 2, 3²∣yᶻ, but 3² does not divide
-- 33(3x² + 300x + 50 ⬝ 199), contradiction.
omega
end Hungary1998P6 | null | v4.29.0-rc6 | test | [
"competition"
] | null | {"filename": "Hungary1998P6.lean"} | {"v4.21.0": {"is_valid": false, "has_sorry": false, "errors": ["unknown tactic", "unsolved goals\ncase intro.succ.succ\nx y : \u2124\nh2 : \u2211 i \u2208 Finset.range 99, x ^ 2 = 99 * x ^ 2\nh4 : \u2211 i \u2208 Finset.range 99, (\u2191i + 1) = \u2211 i \u2208 Finset.range 100, \u2191i\nh5 : \u2211 i \u2208 Finset.range 100, \u2191i = 99 * 100 / 2\nh6 : \u2211 i \u2208 Finset.range 99, (\u2191i + 1) ^ 2 = 99 * 100 * 199 / 6\nhnn1 : 99 * 100 / 2 = 99 * 50\nhnn2 : 99 * 100 * 199 / 6 = 33 * 50 * 199\nk : \u2124\nhk : y = 3 * k\nz : \u2115\nhz : 1 < z + 1 + 1\nhe : \u2211 i \u2208 Finset.range 99, (x + \u2191i + 1) ^ 2 = y ^ (z + 1 + 1)\nh8 : 3 \u2223 y ^ (z + 1 + 1)\nh7 : (3 * k) ^ z * (3 * k) * (3 * k) = 3 * (11 * (3 * x ^ 2 + 300 * x + 50 * 199))\n\u22a2 False"], "timeout_s": 600.0, "latency_s": 1.7075, "verified_at": "2026-03-26T18:15:58.363303+00:00"}} | false | true | false | 1.7075 |
compfiles_Imo1959P1 | compfiles | Copyright (c) 2020 Kevin Lacker. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Kevin Lacker
-/
import Mathlib
/-!
# International Mathematical Olympiad 1959, Problem 1.
Prove that the fraction `(21n+4)/(14n+3)` is irreducible for every
natural number `n`.
-/
namespace Imo1959P1
lemma calculation
(n k : ℕ)
(h1 : k ∣ 21 * n + 4)
(h2 : k ∣ 14 * n + 3) :
k ∣ 1 := by
have h3 : k ∣ 2 * (21 * n + 4) := h1.mul_left 2
have h4 : k ∣ 3 * (14 * n + 3) := h2.mul_left 3
have h5 : 3 * (14 * n + 3) = 2 * (21 * n + 4) + 1 := by ring
exact (Nat.dvd_add_right h3).mp (h5 ▸ h4)
/-
Since Lean doesn't have a concept of "irreducible fractions" per se,
we just formalize this as saying the numerator and denominator are
relatively prime.
-/
theorem imo1959_p1 : ∀ n : ℕ, Nat.Coprime (21 * n + 4) (14 * n + 3) :=
fun n => Nat.coprime_of_dvd' <| λ k _ h1 h2 => calculation n k h1 h2
end Imo1959P1 | /- | /-
Copyright (c) 2020 Kevin Lacker. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Kevin Lacker
-/
import Mathlib
/-!
# International Mathematical Olympiad 1959, Problem 1.
Prove that the fraction `(21n+4)/(14n+3)` is irreducible for every
natural number `n`.
-/
namespace Imo1959P1
lemma calculation
(n k : ℕ)
(h1 : k ∣ 21 * n + 4)
(h2 : k ∣ 14 * n + 3) :
k ∣ 1 := by
have h3 : k ∣ 2 * (21 * n + 4) := h1.mul_left 2
have h4 : k ∣ 3 * (14 * n + 3) := h2.mul_left 3
have h5 : 3 * (14 * n + 3) = 2 * (21 * n + 4) + 1 := by ring
exact (Nat.dvd_add_right h3).mp (h5 ▸ h4)
/-
Since Lean doesn't have a concept of "irreducible fractions" per se,
we just formalize this as saying the numerator and denominator are
relatively prime.
-/
theorem imo1959_p1 : ∀ n : ℕ, Nat.Coprime (21 * n + 4) (14 * n + 3) :=
fun n => Nat.coprime_of_dvd' <| λ k _ h1 h2 => calculation n k h1 h2
end Imo1959P1
| true | Copyright (c) 2020 Kevin Lacker. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Kevin Lacker
-/
import Mathlib
/-!
# International Mathematical Olympiad 1959, Problem 1.
Prove that the fraction `(21n+4)/(14n+3)` is irreducible for every
natural number `n`.
-/
namespace Imo1959P1
lemma calculation
(n k : ℕ)
(h1 : k ∣ 21 * n + 4)
(h2 : k ∣ 14 * n + 3) :
k ∣ 1 := by
have h3 : k ∣ 2 * (21 * n + 4) := h1.mul_left 2
have h4 : k ∣ 3 * (14 * n + 3) := h2.mul_left 3
have h5 : 3 * (14 * n + 3) = 2 * (21 * n + 4) + 1 := by ring
exact (Nat.dvd_add_right h3).mp (h5 ▸ h4)
/-
Since Lean doesn't have a concept of "irreducible fractions" per se,
we just formalize this as saying the numerator and denominator are
relatively prime.
-/
theorem imo1959_p1 : ∀ n : ℕ, Nat.Coprime (21 * n + 4) (14 * n + 3) :=
fun n => Nat.coprime_of_dvd' <| λ k _ h1 h2 => calculation n k h1 h2
end Imo1959P1 | null | v4.29.0-rc6 | test | [
"competition",
"imo"
] | null | {"filename": "Imo1959P1.lean"} | {"v4.21.0": {"is_valid": true, "has_sorry": false, "errors": [], "timeout_s": 600.0, "latency_s": 0.0842, "verified_at": "2026-03-26T18:15:57.549955+00:00"}} | true | true | false | 0.0842 |
compfiles_Imo1959P2 | compfiles | Copyright (c) 2024 Yury Kudryashov. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Yury Kudryashov
-/
import Mathlib
/-!
# International Mathematical Olympiad 1959, Problem 2
For what real values of x is
√(x+√(2x-1)) + √(x-√(2x-1)) = A,
given:
(a) A = √2
(b) A = 1
(c) A = 2,
where only non-negative real numbers are admitted for square roots?
-/
open Set Real
namespace Imo1959P2
def IsGood (x A : ℝ) : Prop :=
sqrt (x + sqrt (2 * x - 1)) + sqrt (x - sqrt (2 * x - 1)) = A ∧ 0 ≤ 2 * x - 1 ∧
0 ≤ x + sqrt (2 * x - 1) ∧ 0 ≤ x - sqrt (2 * x - 1)
variable {x A : ℝ}
/-
We follow second solution from the
[Art of Problem Solving](https://artofproblemsolving.com/wiki/index.php/1959_IMO_Problems/Problem_2)
website.
Namely, we rewrite the equation as $\sqrt{2x-1}+1+|\sqrt{2x-1}-1|=A\sqrt{2}$,
then consider the cases $\sqrt{2x-1}\le 1$ and $1 < \sqrt{2x-1}$ separately.
-/
theorem isGood_iff : IsGood x A ↔
sqrt (2 * x - 1) + 1 + |sqrt (2 * x - 1) - 1| = A * sqrt 2 ∧ 1 / 2 ≤ x := by
cases le_or_gt (1 / 2) x with
| inl hx =>
have hx' : 0 ≤ 2 * x - 1 := by linarith
have h₁ : x + sqrt (2 * x - 1) = (sqrt (2 * x - 1) + 1) ^ 2 / 2 := by
rw [add_sq, sq_sqrt hx']; field_simp; ring
have h₂ : x - sqrt (2 * x - 1) = (sqrt (2 * x - 1) - 1) ^ 2 / 2 := by
rw [sub_sq, sq_sqrt hx']; field_simp; ring
simp only [IsGood, *, div_nonneg (sq_nonneg _) (zero_le_two (α := ℝ)), sqrt_div (sq_nonneg _),
and_true]
rw [sqrt_sq, sqrt_sq_eq_abs] <;> [skip; positivity]
field_simp
| inr hx =>
have : 2 * x - 1 < 0 := by linarith
simp only [IsGood, this.not_ge, hx.not_ge]; simp
theorem IsGood.one_half_le (h : IsGood x A) : 1 / 2 ≤ x := (isGood_iff.1 h).2
theorem sqrt_two_mul_sub_one_le_one : sqrt (2 * x - 1) ≤ 1 ↔ x ≤ 1 := by
simp [sqrt_le_iff, ← two_mul]
theorem isGood_iff_eq_sqrt_two (hx : x ∈ Icc (1 / 2) 1) : IsGood x A ↔ A = sqrt 2 := by
have : sqrt (2 * x - 1) ≤ 1 := sqrt_two_mul_sub_one_le_one.2 hx.2
simp only [isGood_iff, hx.1, abs_sub_comm _ (1 : ℝ), abs_of_nonneg (sub_nonneg.2 this), and_true]
suffices 2 = A * sqrt 2 ↔ A = sqrt 2 by convert this using 2; ring
rw [← div_eq_iff, div_sqrt, eq_comm]
positivity
theorem isGood_iff_eq_sqrt (hx : 1 < x) : IsGood x A ↔ A = sqrt (4 * x - 2) := by
have h₁ : 1 < sqrt (2 * x - 1) := by simpa only [← not_le, sqrt_two_mul_sub_one_le_one] using hx
have h₂ : 1 / 2 ≤ x := by linarith
simp only [isGood_iff, h₂, abs_of_pos (sub_pos.2 h₁), add_add_sub_cancel, and_true]
rw [← mul_two, ← div_eq_iff (by positivity), mul_div_assoc, div_sqrt, ← sqrt_mul' _ zero_le_two,
eq_comm]
ring_nf
theorem IsGood.sqrt_two_lt_of_one_lt (h : IsGood x A) (hx : 1 < x) : sqrt 2 < A := by
rw [(isGood_iff_eq_sqrt hx).1 h]
refine sqrt_lt_sqrt zero_le_two ?_
linarith
theorem IsGood.eq_sqrt_two_iff_le_one (h : IsGood x A) : A = sqrt 2 ↔ x ≤ 1 :=
⟨fun hA ↦ not_lt.1 fun hx ↦ (h.sqrt_two_lt_of_one_lt hx).ne' hA, fun hx ↦
(isGood_iff_eq_sqrt_two ⟨h.one_half_le, hx⟩).1 h⟩
theorem IsGood.sqrt_two_lt_iff_one_lt (h : IsGood x A) : sqrt 2 < A ↔ 1 < x :=
⟨fun hA ↦ not_le.1 fun hx ↦ hA.ne' <| h.eq_sqrt_two_iff_le_one.2 hx, h.sqrt_two_lt_of_one_lt⟩
theorem IsGood.sqrt_two_le (h : IsGood x A) : sqrt 2 ≤ A :=
(le_or_gt x 1).elim (fun hx ↦ (h.eq_sqrt_two_iff_le_one.2 hx).ge) fun hx ↦
(h.sqrt_two_lt_of_one_lt hx).le
theorem isGood_iff_of_sqrt_two_lt (hA : sqrt 2 < A) : IsGood x A ↔ x = (A / 2) ^ 2 + 1 / 2 := by
have : 0 < A := lt_trans (by simp) hA
constructor
· intro h
have hx : 1 < x := by rwa [h.sqrt_two_lt_iff_one_lt] at hA
rw [isGood_iff_eq_sqrt hx, eq_comm, sqrt_eq_iff_eq_sq] at h <;> linarith
· rintro rfl
rw [isGood_iff_eq_sqrt]
· conv_lhs => rw [← sqrt_sq this.le]
ring_nf
· rw [sqrt_lt' this] at hA
linarith
abbrev solution_set_sqrt2 : Set ℝ := Icc (1 / 2) 1
theorem imo1959_p2a : IsGood x (sqrt 2) ↔ x ∈ solution_set_sqrt2 := by
refine ⟨fun h ↦ ?_, fun h ↦ (isGood_iff_eq_sqrt_two h).2 rfl⟩
exact ⟨h.one_half_le, not_lt.1 fun h₁ ↦ (h.sqrt_two_lt_of_one_lt h₁).false⟩
abbrev solution_set_one : Set ℝ := ∅
theorem imo1959_p2b : IsGood x 1 ↔ x ∈ solution_set_one := by
have not_isGood_one : ¬IsGood x 1 := fun h ↦
h.sqrt_two_le.not_gt <| (lt_sqrt zero_le_one).2 (by norm_num)
exact iff_of_false not_isGood_one fun a ↦ a
abbrev solution_set_two : Set ℝ := { 3 / 2 }
theorem imo1959_p2c : IsGood x 2 ↔ x ∈ solution_set_two := by
rw [mem_singleton_iff]
exact
(isGood_iff_of_sqrt_two_lt <| (sqrt_lt' two_pos).2 (by norm_num)).trans <| by norm_num
end Imo1959P2 | /- | /-
Copyright (c) 2024 Yury Kudryashov. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Yury Kudryashov
-/
import Mathlib
/-!
# International Mathematical Olympiad 1959, Problem 2
For what real values of x is
√(x+√(2x-1)) + √(x-√(2x-1)) = A,
given:
(a) A = √2
(b) A = 1
(c) A = 2,
where only non-negative real numbers are admitted for square roots?
-/
open Set Real
namespace Imo1959P2
def IsGood (x A : ℝ) : Prop :=
sqrt (x + sqrt (2 * x - 1)) + sqrt (x - sqrt (2 * x - 1)) = A ∧ 0 ≤ 2 * x - 1 ∧
0 ≤ x + sqrt (2 * x - 1) ∧ 0 ≤ x - sqrt (2 * x - 1)
variable {x A : ℝ}
/-
We follow second solution from the
[Art of Problem Solving](https://artofproblemsolving.com/wiki/index.php/1959_IMO_Problems/Problem_2)
website.
Namely, we rewrite the equation as $\sqrt{2x-1}+1+|\sqrt{2x-1}-1|=A\sqrt{2}$,
then consider the cases $\sqrt{2x-1}\le 1$ and $1 < \sqrt{2x-1}$ separately.
-/
theorem isGood_iff : IsGood x A ↔
sqrt (2 * x - 1) + 1 + |sqrt (2 * x - 1) - 1| = A * sqrt 2 ∧ 1 / 2 ≤ x := by
cases le_or_gt (1 / 2) x with
| inl hx =>
have hx' : 0 ≤ 2 * x - 1 := by linarith
have h₁ : x + sqrt (2 * x - 1) = (sqrt (2 * x - 1) + 1) ^ 2 / 2 := by
rw [add_sq, sq_sqrt hx']; field_simp; ring
have h₂ : x - sqrt (2 * x - 1) = (sqrt (2 * x - 1) - 1) ^ 2 / 2 := by
rw [sub_sq, sq_sqrt hx']; field_simp; ring
simp only [IsGood, *, div_nonneg (sq_nonneg _) (zero_le_two (α := ℝ)), sqrt_div (sq_nonneg _),
and_true]
rw [sqrt_sq, sqrt_sq_eq_abs] <;> [skip; positivity]
field_simp
| inr hx =>
have : 2 * x - 1 < 0 := by linarith
simp only [IsGood, this.not_ge, hx.not_ge]; simp
theorem IsGood.one_half_le (h : IsGood x A) : 1 / 2 ≤ x := (isGood_iff.1 h).2
theorem sqrt_two_mul_sub_one_le_one : sqrt (2 * x - 1) ≤ 1 ↔ x ≤ 1 := by
simp [sqrt_le_iff, ← two_mul]
theorem isGood_iff_eq_sqrt_two (hx : x ∈ Icc (1 / 2) 1) : IsGood x A ↔ A = sqrt 2 := by
have : sqrt (2 * x - 1) ≤ 1 := sqrt_two_mul_sub_one_le_one.2 hx.2
simp only [isGood_iff, hx.1, abs_sub_comm _ (1 : ℝ), abs_of_nonneg (sub_nonneg.2 this), and_true]
suffices 2 = A * sqrt 2 ↔ A = sqrt 2 by convert this using 2; ring
rw [← div_eq_iff, div_sqrt, eq_comm]
positivity
theorem isGood_iff_eq_sqrt (hx : 1 < x) : IsGood x A ↔ A = sqrt (4 * x - 2) := by
have h₁ : 1 < sqrt (2 * x - 1) := by simpa only [← not_le, sqrt_two_mul_sub_one_le_one] using hx
have h₂ : 1 / 2 ≤ x := by linarith
simp only [isGood_iff, h₂, abs_of_pos (sub_pos.2 h₁), add_add_sub_cancel, and_true]
rw [← mul_two, ← div_eq_iff (by positivity), mul_div_assoc, div_sqrt, ← sqrt_mul' _ zero_le_two,
eq_comm]
ring_nf
theorem IsGood.sqrt_two_lt_of_one_lt (h : IsGood x A) (hx : 1 < x) : sqrt 2 < A := by
rw [(isGood_iff_eq_sqrt hx).1 h]
refine sqrt_lt_sqrt zero_le_two ?_
linarith
theorem IsGood.eq_sqrt_two_iff_le_one (h : IsGood x A) : A = sqrt 2 ↔ x ≤ 1 :=
⟨fun hA ↦ not_lt.1 fun hx ↦ (h.sqrt_two_lt_of_one_lt hx).ne' hA, fun hx ↦
(isGood_iff_eq_sqrt_two ⟨h.one_half_le, hx⟩).1 h⟩
theorem IsGood.sqrt_two_lt_iff_one_lt (h : IsGood x A) : sqrt 2 < A ↔ 1 < x :=
⟨fun hA ↦ not_le.1 fun hx ↦ hA.ne' <| h.eq_sqrt_two_iff_le_one.2 hx, h.sqrt_two_lt_of_one_lt⟩
theorem IsGood.sqrt_two_le (h : IsGood x A) : sqrt 2 ≤ A :=
(le_or_gt x 1).elim (fun hx ↦ (h.eq_sqrt_two_iff_le_one.2 hx).ge) fun hx ↦
(h.sqrt_two_lt_of_one_lt hx).le
theorem isGood_iff_of_sqrt_two_lt (hA : sqrt 2 < A) : IsGood x A ↔ x = (A / 2) ^ 2 + 1 / 2 := by
have : 0 < A := lt_trans (by simp) hA
constructor
· intro h
have hx : 1 < x := by rwa [h.sqrt_two_lt_iff_one_lt] at hA
rw [isGood_iff_eq_sqrt hx, eq_comm, sqrt_eq_iff_eq_sq] at h <;> linarith
· rintro rfl
rw [isGood_iff_eq_sqrt]
· conv_lhs => rw [← sqrt_sq this.le]
ring_nf
· rw [sqrt_lt' this] at hA
linarith
abbrev solution_set_sqrt2 : Set ℝ := Icc (1 / 2) 1
theorem imo1959_p2a : IsGood x (sqrt 2) ↔ x ∈ solution_set_sqrt2 := by
refine ⟨fun h ↦ ?_, fun h ↦ (isGood_iff_eq_sqrt_two h).2 rfl⟩
exact ⟨h.one_half_le, not_lt.1 fun h₁ ↦ (h.sqrt_two_lt_of_one_lt h₁).false⟩
abbrev solution_set_one : Set ℝ := ∅
theorem imo1959_p2b : IsGood x 1 ↔ x ∈ solution_set_one := by
have not_isGood_one : ¬IsGood x 1 := fun h ↦
h.sqrt_two_le.not_gt <| (lt_sqrt zero_le_one).2 (by norm_num)
exact iff_of_false not_isGood_one fun a ↦ a
abbrev solution_set_two : Set ℝ := { 3 / 2 }
theorem imo1959_p2c : IsGood x 2 ↔ x ∈ solution_set_two := by
rw [mem_singleton_iff]
exact
(isGood_iff_of_sqrt_two_lt <| (sqrt_lt' two_pos).2 (by norm_num)).trans <| by norm_num
end Imo1959P2
| true | Copyright (c) 2024 Yury Kudryashov. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Yury Kudryashov
-/
import Mathlib
/-!
# International Mathematical Olympiad 1959, Problem 2
For what real values of x is
√(x+√(2x-1)) + √(x-√(2x-1)) = A,
given:
(a) A = √2
(b) A = 1
(c) A = 2,
where only non-negative real numbers are admitted for square roots?
-/
open Set Real
namespace Imo1959P2
def IsGood (x A : ℝ) : Prop :=
sqrt (x + sqrt (2 * x - 1)) + sqrt (x - sqrt (2 * x - 1)) = A ∧ 0 ≤ 2 * x - 1 ∧
0 ≤ x + sqrt (2 * x - 1) ∧ 0 ≤ x - sqrt (2 * x - 1)
variable {x A : ℝ}
/-
We follow second solution from the
[Art of Problem Solving](https://artofproblemsolving.com/wiki/index.php/1959_IMO_Problems/Problem_2)
website.
Namely, we rewrite the equation as $\sqrt{2x-1}+1+|\sqrt{2x-1}-1|=A\sqrt{2}$,
then consider the cases $\sqrt{2x-1}\le 1$ and $1 < \sqrt{2x-1}$ separately.
-/
theorem isGood_iff : IsGood x A ↔
sqrt (2 * x - 1) + 1 + |sqrt (2 * x - 1) - 1| = A * sqrt 2 ∧ 1 / 2 ≤ x := by
cases le_or_gt (1 / 2) x with
| inl hx =>
have hx' : 0 ≤ 2 * x - 1 := by linarith
have h₁ : x + sqrt (2 * x - 1) = (sqrt (2 * x - 1) + 1) ^ 2 / 2 := by
rw [add_sq, sq_sqrt hx']; field_simp; ring
have h₂ : x - sqrt (2 * x - 1) = (sqrt (2 * x - 1) - 1) ^ 2 / 2 := by
rw [sub_sq, sq_sqrt hx']; field_simp; ring
simp only [IsGood, *, div_nonneg (sq_nonneg _) (zero_le_two (α := ℝ)), sqrt_div (sq_nonneg _),
and_true]
rw [sqrt_sq, sqrt_sq_eq_abs] <;> [skip; positivity]
field_simp
| inr hx =>
have : 2 * x - 1 < 0 := by linarith
simp only [IsGood, this.not_ge, hx.not_ge]; simp
theorem IsGood.one_half_le (h : IsGood x A) : 1 / 2 ≤ x := (isGood_iff.1 h).2
theorem sqrt_two_mul_sub_one_le_one : sqrt (2 * x - 1) ≤ 1 ↔ x ≤ 1 := by
simp [sqrt_le_iff, ← two_mul]
theorem isGood_iff_eq_sqrt_two (hx : x ∈ Icc (1 / 2) 1) : IsGood x A ↔ A = sqrt 2 := by
have : sqrt (2 * x - 1) ≤ 1 := sqrt_two_mul_sub_one_le_one.2 hx.2
simp only [isGood_iff, hx.1, abs_sub_comm _ (1 : ℝ), abs_of_nonneg (sub_nonneg.2 this), and_true]
suffices 2 = A * sqrt 2 ↔ A = sqrt 2 by convert this using 2; ring
rw [← div_eq_iff, div_sqrt, eq_comm]
positivity
theorem isGood_iff_eq_sqrt (hx : 1 < x) : IsGood x A ↔ A = sqrt (4 * x - 2) := by
have h₁ : 1 < sqrt (2 * x - 1) := by simpa only [← not_le, sqrt_two_mul_sub_one_le_one] using hx
have h₂ : 1 / 2 ≤ x := by linarith
simp only [isGood_iff, h₂, abs_of_pos (sub_pos.2 h₁), add_add_sub_cancel, and_true]
rw [← mul_two, ← div_eq_iff (by positivity), mul_div_assoc, div_sqrt, ← sqrt_mul' _ zero_le_two,
eq_comm]
ring_nf
theorem IsGood.sqrt_two_lt_of_one_lt (h : IsGood x A) (hx : 1 < x) : sqrt 2 < A := by
rw [(isGood_iff_eq_sqrt hx).1 h]
refine sqrt_lt_sqrt zero_le_two ?_
linarith
theorem IsGood.eq_sqrt_two_iff_le_one (h : IsGood x A) : A = sqrt 2 ↔ x ≤ 1 :=
⟨fun hA ↦ not_lt.1 fun hx ↦ (h.sqrt_two_lt_of_one_lt hx).ne' hA, fun hx ↦
(isGood_iff_eq_sqrt_two ⟨h.one_half_le, hx⟩).1 h⟩
theorem IsGood.sqrt_two_lt_iff_one_lt (h : IsGood x A) : sqrt 2 < A ↔ 1 < x :=
⟨fun hA ↦ not_le.1 fun hx ↦ hA.ne' <| h.eq_sqrt_two_iff_le_one.2 hx, h.sqrt_two_lt_of_one_lt⟩
theorem IsGood.sqrt_two_le (h : IsGood x A) : sqrt 2 ≤ A :=
(le_or_gt x 1).elim (fun hx ↦ (h.eq_sqrt_two_iff_le_one.2 hx).ge) fun hx ↦
(h.sqrt_two_lt_of_one_lt hx).le
theorem isGood_iff_of_sqrt_two_lt (hA : sqrt 2 < A) : IsGood x A ↔ x = (A / 2) ^ 2 + 1 / 2 := by
have : 0 < A := lt_trans (by simp) hA
constructor
· intro h
have hx : 1 < x := by rwa [h.sqrt_two_lt_iff_one_lt] at hA
rw [isGood_iff_eq_sqrt hx, eq_comm, sqrt_eq_iff_eq_sq] at h <;> linarith
· rintro rfl
rw [isGood_iff_eq_sqrt]
· conv_lhs => rw [← sqrt_sq this.le]
ring_nf
· rw [sqrt_lt' this] at hA
linarith
abbrev solution_set_sqrt2 : Set ℝ := Icc (1 / 2) 1
theorem imo1959_p2a : IsGood x (sqrt 2) ↔ x ∈ solution_set_sqrt2 := by
refine ⟨fun h ↦ ?_, fun h ↦ (isGood_iff_eq_sqrt_two h).2 rfl⟩
exact ⟨h.one_half_le, not_lt.1 fun h₁ ↦ (h.sqrt_two_lt_of_one_lt h₁).false⟩
abbrev solution_set_one : Set ℝ := ∅
theorem imo1959_p2b : IsGood x 1 ↔ x ∈ solution_set_one := by
have not_isGood_one : ¬IsGood x 1 := fun h ↦
h.sqrt_two_le.not_gt <| (lt_sqrt zero_le_one).2 (by norm_num)
exact iff_of_false not_isGood_one fun a ↦ a
abbrev solution_set_two : Set ℝ := { 3 / 2 }
theorem imo1959_p2c : IsGood x 2 ↔ x ∈ solution_set_two := by
rw [mem_singleton_iff]
exact
(isGood_iff_of_sqrt_two_lt <| (sqrt_lt' two_pos).2 (by norm_num)).trans <| by norm_num
end Imo1959P2 | null | v4.29.0-rc6 | test | [
"competition",
"imo"
] | null | {"filename": "Imo1959P2.lean"} | {"v4.21.0": {"is_valid": false, "has_sorry": false, "errors": ["unknown identifier 'sqrt'", "unknown identifier 'sqrt'", "unknown identifier 'sqrt'", "unknown identifier 'sqrt'", "function expected at\n sqrt\nterm has type\n ?m.449", "function expected at\n sqrt\nterm has type\n ?m.449", "function expected at\n sqrt\nterm has type\n ?m.449", "function expected at\n sqrt\nterm has type\n x\u271d", "function expected at\n sqrt\nterm has type\n x\u271d", "unknown identifier 'sq_sqrt'", "function expected at\n sqrt\nterm has type\n x\u271d", "function expected at\n sqrt\nterm has type\n x\u271d", "unknown identifier 'sq_sqrt'", "unknown identifier 'sqrt_div'", "unknown identifier 'sqrt_sq'", "don't know how to synthesize placeholder\ncontext:\nx A : \u211d\nh : IsGood x A\n\u22a2 ?m.27224 one_half_le h", "don't know how to synthesize placeholder\ncontext:\nx A : \u211d\nh : IsGood x A\n\u22a2 Sort ?u.27207", "function expected at\n sqrt\nterm has type\n ?m.27237", "unknown identifier 'sqrt_le_iff'", "unsolved goals\nx : \u211d\nx\u271d : Sort u_1\nsqrt : x\u271d\n\u22a2 sorry () \u2264 1 \u2194 x \u2264 1", "function expected at\n Icc\nterm has type\n ?m.31210", "function expected at\n sqrt\nterm has type\n ?m.31497", "function expected at\n sqrt\nterm has type\n x\u271d", "unsolved goals\nx A : \u211d\nx\u271d\u00b9 : Sort u_1\nIcc : x\u271d\u00b9\nx\u271d : Sort u_2\nsqrt : x\u271d\nhx : x \u2208 sorry\n\u22a2 IsGood x A \u2194 A = sorry", "function expected at\n sqrt\nterm has type\n ?m.33921", "function expected at\n sqrt\nterm has type\n x\u271d", "failed to synthesize\n AddGroup \u2115\n\nAdditional diagnostic information may be available using the `set_option diagnostics true` command.", "failed to synthesize\n AddGroup \u2115\n\nAdditional diagnostic information may be available using the `set_option diagnostics true` command.", "simp made no progress", "function expected at\n sqrt\nterm has type\n ?m.38109", "unsolved goals\nx A : \u211d\nx\u271d : Sort u_1\nsqrt : x\u271d\nh : IsGood x A\nhx : 1 < x\n\u22a2 sorry < A", "function expected at\n sqrt\nterm has type\n ?m.38605", "unknown identifier 'isGood_iff_eq_sqrt_two'", "function expected at\n sqrt\nterm has type\n ?m.39281", "function expected at\n sqrt\nterm has type\n ?m.39960", "function expected at\n sqrt\nterm has type\n ?m.40540", "unsolved goals\nx A : \u211d\nx\u271d : Sort u_1\nsqrt : x\u271d\nhA : sorry < A\n\u22a2 0 < sorry ()", "unsolved goals\nx A : \u211d\nx\u271d : Sort u_1\nsqrt : x\u271d\nhA : sorry < A\nthis : 0 < A\nh : IsGood x A\n\u22a2 Sort ?u.44022\n\nx A : \u211d\nx\u271d : Sort u_1\nsqrt : x\u271d\nhA : sorry < A\nthis : 0 < A\nh : IsGood x A\n\u22a2 ?m.44030", "unknown identifier 'sqrt_eq_iff_eq_sq'", "unknown identifier 'sqrt_sq'", "tactic 'rewrite' failed, equality or iff proof expected\n ?m.44184\nA : \u211d\nx\u271d : Sort u_1\nsqrt : x\u271d\nhA : sorry < A\nthis : 0 < A\n| A", "unknown identifier 'sqrt_lt''", "unsolved goals\nA : \u211d\nx\u271d : Sort u_1\nsqrt : x\u271d\nhA : sorry < A\nthis : 0 < A\n\u22a2 Sort ?u.44149\n\nA : \u211d\nx\u271d : Sort u_1\nsqrt : x\u271d\nhA : sorry < A\nthis : 0 < A\n\u22a2 ?m.44152", "unknown identifier 'Icc'", "function expected at\n sqrt\nterm has type\n ?m.44210", "unknown identifier 'isGood_iff_eq_sqrt_two'", "unknown identifier 'lt_sqrt'", "unsolved goals\nx : \u211d\n\u22a2 IsGood x 1 \u2194 x \u2208 solution_set_one", "unknown identifier 'mem_singleton_iff'", "unsolved goals\nx : \u211d\n\u22a2 IsGood x 2 \u2194 x \u2208 solution_set_two"], "timeout_s": 600.0, "latency_s": 0.8435, "verified_at": "2026-03-26T18:15:58.393786+00:00"}} | false | true | false | 0.8435 |
End of preview. Expand in Data Studio
ai4math-lean
21 Lean 4 formal mathematics datasets with machine-verified labels.
Every problem has been verified against lean-server v4.21.0 on UVA's HPC cluster.
Results are embedded in each row as both structured verification JSON and flattened convenience columns.
Quick Start
from datasets import load_dataset
# Load a single dataset
ds = load_dataset("charliemeyer2000/ai4math-lean", "deepseek_prover")
# Load a large dataset with streaming
ds = load_dataset("charliemeyer2000/ai4math-lean", "goedel_pset", streaming=True)
# Filter to valid proofs
valid = ds.filter(lambda x: x["v4210_is_valid"] == True)
# Filter to RL targets (compiles but no proof)
rl = ds.filter(lambda x: x["v4210_compiles"] and not x["has_proof"])
Totals
| Metric | Count |
|---|---|
| Total problems | 3,394,310 |
| With proofs (SFT) | 1,273,472 |
| Valid on v4.21.0 | 352,065 |
Datasets
| Dataset | Problems | Proofs | Valid v4.21.0 |
|---|---|---|---|
| compfiles | 297 | 259 | 67 |
| deepseek_prover | 27,503 | 27,503 | 27,126 |
| deepseek_proverbench | 325 | 0 | 0 |
| formal_conjectures | 618 | 13 | 13 |
| formalmath | 5,560 | 5,556 | 2 |
| goedel_minif2f | 244 | 0 | 0 |
| goedel_mobench | 360 | 0 | 0 |
| goedel_pset | 1,732,594 | 263,589 | 9,755 |
| goedel_test | 8 | 0 | 0 |
| hf_lean_workbook | 29,750 | 29,750 | 29,208 |
| hf_minif2f_lean4 | 231 | 0 | 0 |
| hf_minif2f_v2 | 488 | 0 | 0 |
| hf_tonic_minif2f | 488 | 0 | 0 |
| lean_proofs | 98 | 97 | 15 |
| lean_workbook_full | 140,214 | 0 | 0 |
| matholympiadbench | 360 | 0 | 0 |
| nemotron_proofs | 1,349,950 | 915,078 | 285,867 |
| numinamath_lean | 104,155 | 31,615 | 0 |
| proofnet | 371 | 0 | 0 |
| putnam2025 | 24 | 12 | 12 |
| putnam_bench | 672 | 0 | 0 |
Schema
Each row contains:
| Field | Type | Description |
|---|---|---|
id |
string | Unique problem identifier |
source |
string | Dataset source name |
formal_statement |
string | Lean 4 theorem statement (no proof body) |
header |
string | Import/setup code |
lean4_code |
string | Full runnable Lean 4 code |
has_proof |
bool | True if contains a real proof (not sorry) |
proof_body |
string? | Proof tactic block (if has_proof) |
natural_language |
string? | NL problem statement (if available) |
lean_version |
string? | Original Lean toolchain version |
split |
string? | train/test/val split |
tags |
list[string] | Tags (topic, difficulty, etc.) |
category |
string? | Math category |
verification |
string (JSON) | Full verification results per Lean version |
v4210_is_valid |
bool? | True=valid, False=errors/sorry, None=timeout |
v4210_compiles |
bool | Whether the code compiled (is_valid is not None) |
v4210_has_sorry |
bool? | Whether sorry was detected |
v4210_latency_s |
float? | Verification wall time in seconds |
Verification Details
The verification column contains a JSON string mapping Lean versions to results:
{
"v4.21.0": {
"is_valid": false,
"has_sorry": true,
"errors": [],
"timeout_s": 600.0,
"latency_s": 34.59,
"verified_at": "2026-03-26T21:57:47Z"
}
}
License
Apache 2.0
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