problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
31.51. Prove that $\sum_{n=1}^{p-2}\left(\frac{n(n+1)}{p}\right)=-1$. | 31.51. For each natural number $n \leqslant p-1$, there exists a unique natural number $\bar{n} \leqslant p-1$ such that $n \bar{n} \equiv 1 \pmod{p}$. In this case, $\overline{p-1}=p-1$. Therefore, when $n$ runs through the numbers from 1 to $p-2$, $\bar{n}$ also runs through the numbers from 1 to $p-2$ (in a differen... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,077 |
31.52. For each natural $n \leqslant p-2$, the pair $(n, n+1)$ is of one of four types: $(R, R), (N, N), (N, R), (R, N)$, where $R$ stands for a residue and $N$ for a non-residue. Let $RR, NN, NR, RN$ be the number of all pairs of the corresponding type.
a) Prove that $RR + NN - RN - NR = 1$.
b) Let $\varepsilon = (-... | 31.52. a) It is clear that $\left(\frac{n}{p}\right)\left(\frac{n+1}{p}\right)=1$ in the cases $R R$ and $N N$, and in the cases $N R$ and $R N$ this product equals -1. Therefore, $R R + N N - R N - N R = \sum_{n=1}^{p-2}\left(\frac{n(n+1)}{p}\right)$. It remains to use the result of problem 31.51.
b) The number of re... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,078 |
31.53. Let $p=4k+1$ be a prime number.
a) Prove that there exists an integer $x$ such that $x^{2}+1$ is divisible by $p$.
b) Prove that one can choose integers $0 \leqslant r_{1}, r_{2} < <\sqrt{p}$ and $0 \leqslant s_{1}, s_{2} < \sqrt{p}$ such that the numbers $r_{1} x+s_{1}$ and $r_{2} x+s_{2}$ will give the same ... | 31.53. a) It follows directly from problem 31.36.
b) If an integer $r$ satisfies the inequalities $0 \leqslant r<\sqrt{p}$, then $r$ can take more than $\sqrt{p}$ different values (since $r$ can take the value 0). Thus, the number of different permissible pairs $(r, s)$ is greater than $\sqrt{p} \cdot \sqrt{p}=p$. The... | u^{2}+v^{2}=p | Number Theory | proof | Yes | Yes | olympiads | false | 39,079 |
31.54. Prove that any prime number $p=4k+1$ can be represented as the sum of squares of two integers, using problem 17.13. | 31.54. According to problem 31.36, one can choose a natural number $q$ such that $q^{2} \equiv-1(\bmod p)$. Consider the number $\alpha=q / p$. Let $C=\sqrt{p}$. According to problem 17.13, one can choose a natural number $x<C=\sqrt{p}$ and an integer $y$ such that $|x \alpha-y| \leqslant 1 / C$, i.e., $\left|x \frac{q... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,080 |
31.55. Let $p=4k+1$ be a prime number.
a) Prove that the equation $x^{2}+y^{2}=m p$ has a solution in natural numbers $x, y, m$.
b) Prove that if $m>1$, then a solution with a smaller $m$ can be constructed. | 31.55. a) According to problem 31.36, we can choose a natural number $x$ such that $x^{2} \equiv-1(\bmod p)$, i.e., $x^{2}+1=m p$. Thus, we have found the required solution, even with the additional condition $y=1$.
b) Let $m_{0}$ be the smallest natural number for which the equality
$$
x^{2}+y^{2}=m_{0} p
$$
holds.... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,081 |
31.56. Prove that the representation of a prime number $p=$ $=4 k+1$ as a sum of two squares of integers is unique. (We do not distinguish representations $p=x^{2}+y^{2}$ that differ only by the permutation of $x$ and $y$ or by the replacement of the signs of $x$ and $y$.
### 31.11. Sums of four squares | 31.56. Suppose that $p=x^{2}+y^{2}=a^{2}+b^{2}$. The congruence $z^{2} \equiv$ $\equiv-1(\bmod p)$ has exactly two solutions: $z \equiv \pm h(\bmod p)$. Therefore, $x \equiv \pm h y(\bmod p)$ and $a \equiv \pm h b(\bmod p)$. The signs of $x$ and $a$ can be changed, so we will assume that $x \equiv h y(\bmod p)$ and $a ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,082 |
31.57. Prove that if each of the numbers $a$ and $b$ is the sum of four squares of integers, then their product $a b$ is also the sum of four squares of integers. | 31.57. Let $a=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}$ and $b=y_{1}^{2}+y_{2}^{2}+y_{3}^{2}+y_{4}^{2}$. It is not difficult to verify that $a b=z_{1}^{2}+z_{2}^{2}+z_{3}^{2}+z_{4}^{2}$, where
$$
\begin{aligned}
& z_{1}=x_{1} y_{1}+x_{2} y_{2}+x_{3} y_{3}+x_{4} y_{4} \\
& z_{2}=x_{1} y_{2}-x_{2} y_{1}+x_{3} y_{4}-x_{4}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,083 |
31.58. Let $p$ be an odd prime number. Prove that there exist integers $x, y$ and $m$, such that $1 + x^{2} + y^{2} = m p$, where $0 < m < p$. | 31.58. The numbers $x^{2}$, where $x$ is an integer and $0 \leqslant x \leqslant \frac{p-1}{2}$, give different remainders when divided by $p$. Indeed, if $x_{1}^{2} \equiv x_{2}^{2}(\bmod p)$, then $x_{1} \pm x_{2} \equiv 0(\bmod p)$. But in the considered situation $0<x_{1}+x_{2}<p$. Similarly, the numbers $-1-y^{2}$... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,084 |
31.59. Let $p$ be an odd prime number.
a) Prove that one can choose a natural number $m<p$ and integers $x_{1}, x_{2}, x_{3}$ and $x_{4}$ such that $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=$ $=m p$.
b) Prove that the smallest such $m$ is odd.
c) Prove that the smallest such $m$ is equal to 1. | 31.59. a) It directly follows from problem 31.58: we can set \(x_{1}=1, x_{2}=x, x_{3}=y\) and \(x_{4}=0\).
b) Suppose the number \(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}\) is even. Then the number of odd numbers among \(x_{1}, x_{2}, x_{3}\) and \(x_{4}\) is even. Therefore, we can assume that the numbers \(x_{1}\) ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,085 |
31.63. a) Prove that for each exponent $d$, there are no more than $\varphi(d)$ residues modulo $p$.
b) Prove that for each exponent $d$ dividing the number $p-1$, there are exactly $\varphi(d)$ residues modulo $p$.
c) Prove that for any prime number $p$, there are exactly $\varphi(p-1)$ primitive roots. | 31.63. a) Let the remainder $x$ belong to the exponent $d$. Then the $d$ remainders $1, x, x^{2}, x^{3}, \ldots, x^{d-1}$ are distinct and all satisfy the equation $X^{d} \equiv 1(\bmod p)$. Therefore, there are no other remainders that satisfy this equation of degree $d$ (Problem 31.33). Any remainder $y$ that belongs... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,089 |
31.64. a) Given a natural number $n \geqslant 2$. Prove that a natural number $d$, for which $x^{d+1} \equiv x(\bmod n)$ for all integers $x$, exists if and only if $n=p_{1} \ldots p_{k}$, where $p_{1}, \ldots, p_{k}$ are pairwise distinct prime numbers.
b) Let $n=p_{1} \ldots p_{k}$, where $p_{1}, \ldots, p_{k}$ are p... | 31.64. a) Suppose first that $n=p^{2} q$. Let $x=p q$. Then $x \not \equiv 0(\bmod n)$, but $x^{d+1} \equiv 0(\bmod n)$ for any natural number $d$.
Now suppose that $n=p_{1} \ldots p_{k}$, where $p_{1}, \ldots, p_{k}$ are pairwise distinct prime numbers. Then if the number $x^{d+1}-x$ is divisible by $p_{1}, \ldots, p... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,090 |
31.65. Let $p-$ be an odd prime number.
a) Prove that the odd prime divisors of the number $a^{p}-1$ divide $a-1$ or are of the form $2 p x+1$.
b) Prove that the odd prime divisors of the number $a^{p}+1$ divide $a+1$ or are of the form $2 p x+1$. | 31.65. a) Let $q$ be an odd prime divisor of the number $a^{p}-1$. Then $a^{p} \equiv 1(\bmod q)$. Therefore, according to problem 31.62, the order $d$ of the number $a$ modulo $q$ is a divisor of the number $p$, i.e., $d=1$ or $d=p$. If $d=1$, then $a \equiv 1(\bmod q)$, so $q$ is a divisor of the number $a-1$. If $d=... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,091 |
31.66. Let $p$ be an odd prime number. Prove that there are infinitely many prime numbers of the form $2 p x + 1$.
保留源文本的换行和格式，直接输出翻译结果。 | 31.66. First, note that such prime numbers exist: according to problem 31.65 a) all prime divisors of the number $2^{p} - 1$ have this form. Suppose there are only a finite number of prime numbers of the form $2 p x + 1$, namely, the numbers $p_{1}, \ldots, p_{n}$. Consider the number $\left(p_{1} \ldots p_{n}\right)^{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,092 |
31.68. Prove that 3 is a primitive root modulo the prime number $p=2^{n}+1$, where $n>1$. | 31.68. For $n>1$ the numbers 3 and $p$ are coprime. According to problem 31.46 b) $\left(\frac{3}{p}\right)=-1$, therefore $3^{2^{n-1}} \equiv-1\left(\bmod 2^{n}+1\right)$. The exponent of the number 3 modulo $2^{n}+1$ is a divisor of the number $2^{n}$, and it is greater than $2^{n-1}$. Therefore, the exponent of the ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,094 |
31.69. Let $p$ be a prime number and $S=1^{n}+2^{n}+\ldots+(p-1)^{n}$. Prove that
$$
S \equiv \begin{cases}-1(\bmod p), & \text { if } n \text { is divisible by } p-1 \\ 0(\bmod p), & \text { if } n \text { is not divisible by } p-1\end{cases}
$$
### 31.13. Primitive Roots Modulo a Composite Number
Primitive roots c... | 31.69. If $n$ is divisible by $p-1$, then $a^{n} \equiv 1(\bmod p)$ for any $a$ that is coprime with $p$. Therefore, $S \equiv p-1 \equiv -1(\bmod p)$.
Now suppose that $n$ is not divisible by $p-1$. Let $x$ be a primitive root modulo $p$. Then according to problem 31.62, $x^{n} \not \equiv 1(\bmod p)$. The number $x$... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,095 |
31.70. Prove that $(1+k m)^{m^{\alpha-1}} \equiv 1\left(\bmod m^{\alpha}\right)$ for any $m$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 31.70. Note that if $r \geqslant 1$, then $\left(1+k m^{r}\right)^{m} \equiv 1\left(\bmod m^{r+1}\right)$. Indeed, $\left(1+k m^{r}\right)^{m}=1+C_{m}^{1} k m^{r}+C_{m}^{2} k^{2} m^{2 r}+\ldots$ In this sum, the term $C_{m}^{1} k m^{r}=k m^{r+1}$ is divisible by $m^{r+1}$. Subsequent terms are divisible by $m^{2 r}$, s... | Combinatorics | MCQ | Yes | Yes | olympiads | false | 39,096 | |
31.71. Let $p-$ be a prime number. Prove that a primitive root modulo $p^{\alpha}$ is also a primitive root modulo $p$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 31.71. If $x$ is a primitive root modulo $p^{\alpha}$, then there does not exist a natural number $s < p^{\alpha-1}(p-1)$ such that $x^{s} \equiv 1\left(\bmod p^{\alpha}\right)$. Suppose that $x^{t} \equiv 1(\bmod p)$ for some natural number $t < p-1$. Then according to problem $31.70$, $x^{t p^{\alpha-1}} \equiv 1\lef... | Combinatorics | MCQ | Yes | Yes | olympiads | false | 39,097 | |
31.72. Let $x$ be a primitive root modulo a prime $p$. Suppose that $x^{p^{\alpha-2}(p-1)} \not \equiv 1\left(\bmod p^{\alpha}\right)$, where $\alpha \geqslant 2$.
Prove that then $x$ is a primitive root modulo $p^{\alpha}$. | 31.72. Let $k$ be the smallest natural number for which $x^{k} \equiv 1\left(\bmod p^{\alpha}\right)$. Then $p^{\alpha-1}(p-1)$ is divisible by $k$ (Problem 31.12). It is clear that $x^{k} \equiv 1(\bmod p)$, so $k$ is divisible by $p-1$. Therefore, $k=p^{\beta}(p-1)$, where $0 \leqslant \beta \leqslant \alpha-1$. Supp... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,098 |
31.73. Let $x$ be a primitive root modulo an odd prime $p$. Prove that at least one of the numbers $x$ and $x+p$ is a primitive root modulo $p^{2}$. | 31.73. Suppose that the numbers $x$ and $x+p$ are not primitive roots modulo $p^{2}$. Both these numbers are primitive roots modulo $p$, so according to problem $31.72$, $x^{p-1} \equiv 1\left(\bmod p^{2}\right)$ and $(x+p)^{p-1} \equiv 1\left(\bmod p^{2}\right)$. Therefore, the number $(x+p)^{p-1}-x^{p-1}=(p-1) x^{p-2... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,099 |
31.74. Prove that if $x$ is a primitive root modulo $p^{2}$, where $p$ is an odd prime number, then $x$ is a primitive root modulo $p^{\alpha}$ for any $\alpha \geqslant 2$. | 31.74. The numbers $x$ and $p$ are coprime, so according to Fermat's Little Theorem, $x^{p-1} \equiv 1 \pmod{p}$, i.e., $x^{p-1} = 1 + p t$. The smallest natural number $k$ for which $x^{k} \equiv 1 \pmod{p^2}$ is $p(p-1)$, so $t$ is not divisible by $p$.
Raise the equation $x^{p-1} = 1 + p t$ to the power $n = p^{\al... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,100 |
31.75. Let $p$ be an odd prime number. Prove that for any natural $\alpha$ there exists a primitive root modulo $2 p^{\alpha}$. | 31.75. It is clear that $\varphi\left(2 p^{\alpha}\right)=\varphi\left(p^{\alpha}\right)$. Let $x$ be a primitive root modulo $p^{\alpha}$. By replacing $x$ with $x+p^{\alpha}$ if necessary, we can assume that $x$ is odd. It suffices to prove that $x^{h} \equiv 1\left(\bmod p^{\alpha}\right)$ if and only if $x^{h} \equ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,101 |
31.76. Prove that a primitive root modulo $2^{n}$ exists if and only if $n \leqslant 2$. | 31.76. The numbers 1 and 3 are primitive roots modulo 2 and 4. It remains to prove that if $n \geqslant 3$, then a primitive root modulo $2^{n}$ does not exist. Since $\varphi\left(2^{n}\right)=2^{n-1}$, it is sufficient to prove that $x^{2^{n-2}} \equiv 1\left(\bmod 2^{n}\right)$ for any odd $x$ when $n \geqslant 3$. ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,102 |
32.1. Let $f$ and $g$ be polynomials and $\gamma$ be a closed non-self-intersecting curve in the complex plane. Prove that if
$$
|f(z)-g(z)|<|f(z)|+|g(z)|
$$
for all $z \in \gamma$, then inside the curve $\gamma$ there is the same number of roots of the polynomials $f$ and $g$, counting multiplicities (Rouche's theor... | 32.1. Consider vector fields $v(z)=f(z)$ and $w(z)=g(z)$ on the complex plane. From condition (1), it follows that the vectors $v$ and $w$ are not oppositely directed at any point on the curve $\gamma$.
The *index* of a curve $\gamma$ relative to a vector field $v$ is the number of rotations of the vector $v(z)$ as th... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,107 |
32.2. Let $f(z)=z^{n}+a_{1} z^{n-1}+\ldots+a_{n}$, where $a_{i}$ are complex numbers. Prove that then inside the circle $|z|=1+\max _{i}\left|a_{i}\right|$ there are exactly $n$ roots of the polynomial $f$ (counting their multiplicity).
## 32.1. Separation of Roots
Here we will discuss various statements that allow u... | 32.2. Let $a=\max _{i}\left|a_{i}\right|$. The polynomial $g(z)=z^{n}$ has a root 0 of multiplicity $n$ inside the considered circle. Therefore, it is sufficient to check that if $|z|=1+a$, then $|f(z)-g(z)|<|f(z)|+|g(z)|$. We will even prove that $|f(z)-g(z)|<|g(z)|$, i.e.,
$$
\left|a_{1} z^{n-1}+\ldots+a_{n}\right|<... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,108 |
32.3. Let $N(x)$ be the number of sign changes in the sequence $f(x), f'(x), \ldots, f^{(n)}(x)$, where $f$ is a polynomial of degree $n$. Prove that the number of roots of the polynomial $f$ (counting multiplicities) between $a$ and $b$, where $f(a) \neq 0$, $f(b) \neq 0$ and $a < b$, does not exceed $N(a) - N(b)$, an... | 32.3. Let the point $x$ move along the segment $[a, b]$ from $a$ to $b$. The number $N(x)$ changes only when $x$ passes through a root of the polynomial $f^{(m)}$ for some $m \leqslant n$.
Consider first the case when the point $x$ passes through an $r$-fold root $x_{0}$ of the polynomial $f(x)$. In the neighborhood o... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,109 |
32.4. a) Prove that the number of positive roots of the polynomial $f(x)=a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n}$, where $a_{n} \neq 0$, does not exceed the number of sign changes in the sequence $a_{0}, a_{1}, \ldots, a_{n}$ (Descartes' rule).
b) Prove that the number of negative roots of the polynomial $f(x)=a_{0} x^... | 32.4. a) Since $f^{(r)}(0)=r!a_{n-r}$, $N(0)$ coincides with the number of sign changes in the sequence of coefficients of the polynomial $f$. It is also clear that $N(+\infty)=0$.
b) It is sufficient to apply Descartes' rule to the polynomial $f(-x)=b_{0} x^{n}+b_{1} x^{n-1}+\ldots+b_{n}$, where $b_{k}=(-1)^{n-k} a_{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,110 |
32.5. Prove that if in the polynomial $f(x)=a_{0} x^{n}+$ $+a_{1} x^{n-1}+\ldots+a_{n}$, where $a_{n} \neq 0$, there are $2 m$ consecutive terms missing (i.e., the coefficients of these terms are zero), then this polynomial has at least $2 m$ imaginary (non-real) roots, and if there are $2 m+1$ consecutive terms missin... | 32.5. Let's estimate the number of positive and negative roots of the given polynomial according to the rule formulated in problem 32.4. Suppose that between two terms \(a_{n-k} x^{k}\) and \(a_{n-k+2 m+1} x^{k-2 m-1}\), there are \(2 m\) intermediate terms missing. If they were not missing, they would provide \(2 m+1\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,111 |
32.6. Let $w(x)$ be the number of sign changes in the sequence $f(x), f_{1}(x), \ldots, f_{n}(x)$. Prove that the number of roots of the polynomial $f$ (not counting multiplicities), lying between $a$ and $b$, where $f(a) \neq 0, f(b) \neq 0$ and $a<b$, is exactly $w(a)-w(b)$ (Sturm). | 32.6. Let us first consider the case when the polynomial $f$ has no multiple roots (i.e., the polynomials $f$ and $f'$ have no common roots). In this case, $f_n$ is some non-zero constant.
First, let us check that if we pass through one of the roots of the polynomials $f_1, \ldots, f_{n-1}$, the number of sign changes... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,112 |
32.7. Prove that the roots of the derivative of a polynomial $P$ belong to the convex hull of the roots of the polynomial $P$ itself (Gauss-Lucas theorem). | 32.7. Let $P(z)=\left(z-z_{1}\right) \ldots\left(z-z_{n}\right)$. It is easy to verify that
$$
\frac{P^{\prime}(z)}{P(z)}=\frac{1}{z-z_{1}}+\ldots+\frac{1}{z-z_{n}}
$$
Suppose that $P^{\prime}(w)=0, P(w) \neq 0$ and $w$ does not belong to the convex hull of the points $z_{1}, \ldots, z_{n}$. Then a line can be drawn ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 39,113 |
32.9. Prove that if the polynomial $q r$ is divisible by an irreducible polynomial $p$, then one of the polynomials $q$ and $r$ is divisible by $p$. | 32.9. Let the polynomial $q$ not be divisible by $p$. Then $\gcd(p, q)=1$, i.e., there exist polynomials $a$ and $b$ such that $a p + b q = 1$. Multiplying both sides of this equation by $r$, we get $a p r + b q r = r$. The polynomials $p r$ and $q r$ are divisible by $p$, so $r$ is divisible by $p$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,115 |
32.10. Let $k$ be a field. Prove that a polynomial $f(x)$ with coefficients from $k$ has a factorization into irreducible factors, and this factorization is unique.
For the ring of integers, the irreducibility of polynomials is defined in the same way as in the case of a field, i.e., a polynomial $f(x)$ with integer c... | 32.10. The existence of the factorization is easily proved by induction on $n=\operatorname{deg} f$. First of all, note that for an irreducible polynomial $f$, the required factorization consists of the polynomial $f$ itself. For $n=1$, the polynomial $f$ is irreducible. Suppose the factorization exists for any polynom... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,116 |
32.11. Prove that $\operatorname{cont}(f g)=\operatorname{cont}(f) \operatorname{cont}(g)$ (Gauss's lemma). | 32.11. It is sufficient to consider the case when $\operatorname{cont}(f)=\operatorname{cont}(g)=1$. Indeed, the coefficients of the polynomials $f$ and $g$ can be divided by $\operatorname{cont}(f)$ and $\operatorname{cont}(g)$ respectively.
Let $f(x)=\sum a_{i} x^{i}, g(x)=\sum b_{i} x^{i}, f g(x)=\sum c_{i} x^{i}$.... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,117 |
32.13. Let $f(x)=a_{0}+a_{1} x+\ldots+a_{n} x^{n}$ be a polynomial with integer coefficients, and let $p$ be a prime number such that the coefficient $a_{n}$ is not divisible by $p$, the coefficients $a_{0}, \ldots, a_{n-1}$ are divisible by $p$, but the coefficient $a_{0}$ is not divisible by $p^{2}$. Prove that then ... | 32.13. Suppose that
$$
f=g h=\left(\sum b_{k} x^{k}\right)\left(\sum c_{l} x^{l}\right)
$$
where $g$ and $h$ are polynomials of positive degree with integer coefficients. The number $b_{0} c_{0}=a_{0}$ is divisible by $p$, so one of the numbers $b_{0}$ and $c_{0}$ is divisible by $p$. Let, for definiteness, $b_{0}$ b... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,119 |
32.14. Prove that if $p$ is a prime number, then the polynomial $f(x)=x^{p-1}+x^{p-2}+\ldots+x+1$ is irreducible.
## 32.3. Symmetric Polynomials
A polynomial $f\left(x_{1}, \ldots, x_{n}\right)$ is called symmetric if for any permutation $\sigma$ the equality
$$
f\left(x_{\sigma_{(1)}}, \ldots, x_{\sigma_{(n)}}\righ... | 32.14. For the polynomial
$$
f(x+1)=\frac{(x+1)^{p}-1}{(x+1)-1}=x^{p-1}+C_{p}^{1} x^{p-2}+\ldots+C_{p}^{p-1}
$$
Eisenstein's criterion can be applied, since all numbers $C_{p}^{1} x^{p-2}, \ldots, C_{p}^{p-1}$ are divisible by $p$ (Problem 14.30). | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,120 |
32.15. Prove that if $a+b+c+d=2$ and $1 / a+1 / b+1 / c+$ $+1 / d=2$, then
$$
\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}+\frac{1}{1-d}=2
$$
$$
\% * *
$$ | 32.15. Let $\sigma_{k}$ be the $k$-th elementary symmetric function of $a, b, c, d$. Given that $\sigma_{1}=2$ and $\sigma_{3}=2 \sigma_{4}$. Therefore,
$$
\begin{aligned}
\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}+\frac{1}{1-d}=\frac{4-3 \sigma_{1}+2 \sigma_{2}-\sigma_{3}}{1-\sigma_{1}+\sigma_{2}-\sigma_{3}+\sigma_{4}... | 2 | Algebra | proof | Yes | Yes | olympiads | false | 39,121 |
32.16. Prove that
$$
\sum_{r=0}^{n}(-1)^{r} \sigma_{r} p_{n-r}=0
$$ | 32.16. The generating functions $\sigma(t)$ and $p(t)$ are related by the equation $\sigma(-t) p(t)=1$. By equating the coefficients of $t^{n}, n \geqslant 1$, on both sides, we obtain the required result. | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,122 |
32.17. Prove that
$$
n p_{n}=\sum_{r=1}^{n} s_{r} p_{n-r}
$$ | 32.17. The generating function $s(t)$ is expressed through $p(t)$ as follows:
$$
s(t)=\frac{d}{d t} \ln p(t)=\frac{p^{\prime}(t)}{p(t)}, \quad \text { i.e. } \quad s(t) p(t)=p^{\prime}(t)
$$
By equating the coefficients of $t^{n-1}$, we obtain the required result. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 39,123 |
32.18. Let $s_{k}=x_{1}^{k}+\ldots+x_{n}^{k}$. Prove that
$$
n \sigma_{n}=s_{1} \sigma_{n-1}-s_{2} \sigma_{n-2}+\ldots+(-1)^{n-1} s_{n} \sigma_{0}
$$
(Newton's formulas). | 32.18. First solution. The required equality can be rewritten as
$$
s_{0} \sigma_{n}-s_{1} \sigma_{n-1}+s_{2} \sigma_{n-2}+\ldots+(-1)^{n} s_{n} \sigma_{0}=0
$$
The product $s_{n-k} \sigma_{k}$ consists of terms of the form $x_{i}^{n-k} x_{j_{1}} \ldots x_{j_{k}}$. If $i$ coincides with one of the numbers $j_{1}, \ld... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,124 |
32.19. The sum of three integers $x, y, z$ is zero. Prove that $2\left(x^{4}+y^{4}+z^{4}\right)$ is a square of an integer. | 32.19. Let $\sigma_{1}=x+y+z, \sigma_{2}=x y+y z+z x, \sigma_{3}=x y z$ and $s_{k}=x^{k}+$ $+y^{k}+z^{k}$. We will write Newton's formulas for $n=1,2$ and 4, taking into account that $\sigma_{1}=s_{1}=0$. As a result, we get $2 \sigma_{2}=-s_{2}$ and $s_{4}+s_{2} \sigma_{2}=0$. Therefore, $2 s_{4}=-s_{2}\left(2 \sigma_... | 2s_{4}=s_{2}^2 | Algebra | proof | Yes | Yes | olympiads | false | 39,125 |
32.20. The integers $x_{1}, \ldots, x_{5}$ are such that $x_{1}+\ldots+x_{5}$ and $x_{1}^{2}+\ldots+x_{5}^{2}$ are divisible by an odd number $n$. Prove that $x_{1}^{5}+\ldots+x_{5}^{5}-5 x_{1} \ldots x_{5}$ is also divisible by $n$.
$$
\% * \%
$$ | 32.20. Let's write down Newton's formulas $\sigma_{1}=s_{1}$ and $2 \sigma_{2}=s_{1} \sigma_{1}-s_{2}$. By the condition, the numbers $\sigma_{1}=s_{1}$ and $s_{2}$ are divisible by $n$. Therefore, $2 \sigma_{2}$ is also divisible by $n$. Since the number $n$ is odd, $\sigma_{2}$ is divisible by $n$. Now let's write do... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,126 |
32.21. Let $x_{1}, x_{2}, x_{3}$ be the roots of the polynomial $x^{3}+p x+q$. Compute $s_{n}=x_{1}^{n}+x_{2}^{n}+x_{3}^{n}$ for $n=1,2, \ldots, 10$. | 32.21. The equality $x_{i}^{n+3}+p x_{i}^{n+1}+q x_{i}^{n}=0$ shows that the recurrence relation $s_{n+3}+p s_{n+1}+q s_{n}=0$ holds. It is also clear that $s_{0}=3$ and $s_{1}=0$. Moreover, $s_{-1}=\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}=\frac{x_{2} x_{3}+x_{1} x_{3}+x_{1} x_{2}}{x_{1} x_{2} x_{3}}=$ $=-\frac{... | s_{10}=-2p^{5}+15p^{2}q^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,127 |
32.22. Let $x_{1}=b+c+d, x_{2}=-(a+b+c), x_{3}=a-d$, $y_{1}=a+c+d, y_{2}=-(a+b+d)$ and $y_{3}=b-c$. Let, further, $t^{3}+p_{1} t+q_{1}$ and $t^{3}+p_{2} t+q_{2}$ be polynomials with roots $x_{1}, x_{2}, x_{3}$ and $y_{1}, y_{2}, y_{3}$ respectively. Prove that $p_{1}=p_{2}$ if and only if $a d=b c$. | 32.22. It is clear that $p_{1}=x_{1} x_{2}+\left(x_{1}+x_{2}\right) x_{3}=x_{1} x_{2}-x_{3}^{2}=-(b+c+d) \times$ $\times(a+b+c)-(d-a)^{2}$ and $p_{2}=-(a+c+d)(a+b+d)-(b-c)^{2}$. Therefore, $p_{1}-p_{2}=3(a d-b c)$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,128 |
32.23. Let
$$
\begin{aligned}
f_{2 n}=(b+c & +d)^{2 n}+(a+b+c)^{2 n}+ \\
& +(a-d)^{2 n}-(a+c+d)^{2 n}-(a+b+d)^{2 n}-(b-c)^{2 n}
\end{aligned}
$$
where $a d=b c$. Prove that $f_{2}=f_{4}=0$ and $64 f_{6} f_{10}=45 f_{8}^{2}$ (Ramanujan's identities).
$$
\therefore \quad * \quad *
$$ | 32.23. Let us define the numbers $x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3}$ and the polynomials $t^{3}+p_{1} t+q_{1}$ and $t^{3}+p_{2} t+q_{2}$ as in the condition of problem 32.22. According to this problem, $p_{1}=p_{2}$, since $a d=b c$. Let $p_{1}=p_{2}=p$.
Let $s_{n}=x_{1}^{n}+x_{2}^{n}+x_{3}^{n}$ and $s_{n}^{\pr... | 64f_{6}f_{10}=45f_{8}^{2} | Algebra | proof | Yes | Yes | olympiads | false | 39,129 |
32.24. a) Let $f\left(x_{1}, \ldots, x_{n}\right)$ be a symmetric polynomial. Prove that there exists a polynomial $g\left(y_{1}, \ldots, y_{n}\right)$ such that $f\left(x_{1}, \ldots, x_{n}\right)=g\left(\sigma_{1}, \ldots, \sigma_{n}\right)$. Moreover, the polynomial $g$ is unique (the fundamental theorem of symmetri... | 32.24. a) A polynomial $f\left(x_{1}, \ldots, x_{n}\right)=\sum a_{k_{1}, \ldots, k_{n}} x_{1}^{k_{1}} \ldots x_{n}^{k_{n}}$ is called a homogeneous polynomial of degree $m$ if $k_{1}+\ldots+k_{n}=m$ for all its monomials. It is sufficient to consider the case where $f$ is a homogeneous polynomial. We will say that the... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,130 |
32.25. Prove that any skew-symmetric polynomial $f\left(x_{1}, \ldots, x_{n}\right)$ can be represented in the form
$$
\Delta\left(x_{1}, \ldots, x_{n}\right) g\left(x_{1}, \ldots, x_{n}\right)
$$
where $g$ is a symmetric polynomial.
Let $\lambda=\left(\lambda_{1}, \ldots, \lambda_{n}\right)$ be a partition, i.e., a... | 32.25. It is sufficient to check that $f$ is divisible by $\Delta$. Indeed, if $f / \Delta$ is a polynomial, then this polynomial is symmetric for obvious reasons. Let's show, for example, that $f$ is divisible by $x_{1}-x_{2}$. Make the substitution $x_{1}=u+v, x_{2}=v-u$. As a result, we get
$$
f\left(x_{1}, x_{2}, ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,131 |
32.26. Prove that the inequality $M_{\lambda}(x) \geqslant M_{\mu}(x)$ holds for all $x=\left(x_{1}, \ldots, x_{n}\right)$ with positive $x_{1}, \ldots, x_{n}$ if and only if $|\lambda|=|\mu|$ and $\lambda \geqslant \mu$. In this case, equality is achieved only when $\lambda=\mu$ and $x_{1}=\ldots=x_{n}$ (Muirhead).
#... | 32.26. Suppose first that the required inequality holds for all $x>0$. Let $x_{1}=\ldots=x_{k}=a$ and $x_{k+1}=\ldots=x_{n}=1$. Then
$$
1 \leqslant \lim _{a \rightarrow \infty} M_{\lambda}(x) / M_{\mu}(x)=\lim _{a \rightarrow \infty}\left(a^{\lambda_{1}+\ldots+\lambda_{k}} / a^{\mu_{1}+\ldots+\mu_{k}}\right)
$$
There... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 39,132 |
32.27. Prove that $\cos n \varphi$ can be expressed polynomially in terms of $\cos \varphi$, i.e., there exists a polynomial $T_{n}(x)$ such that $T_{n}(x)=\cos n \varphi$ when $x=\cos \varphi$.
The polynomials $T_{n}(x)$ from problem 32.27 are called Chebyshev polynomials. | 32.27. Formula
$$
\cos (n+1) \varphi+\cos (n-1) \varphi=2 \cos \varphi \cos n \varphi
$$
shows that
$$
T_{n+1}(x)=2 x T_{n}(x)-T_{n-1}(x) .
$$
Polynomials $T_{n}(x)$, defined by this recurrence relation and initial conditions $T_{0}(x)=1$ and $T_{1}(x)=x$, have the required property. | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,133 |
32.28. Calculate the Chebyshev polynomials $T_{n}(x)$ for $n \leqslant 5$.
Compute the Chebyshev polynomials $T_{n}(x)$ for $n \leqslant 5$. | 32.28. Answer: $T_{1}(x)=x, T_{2}(x)=2 x^{2}-1, T_{3}(x)=4 x^{3}-3 x$, $T_{4}(x)=8 x^{4}-8 x^{2}+1, T_{5}(x)=16 x^{5}-20 x^{3}+5 x$. | T_{1}(x)=x,T_{2}(x)=2x^{2}-1,T_{3}(x)=4x^{3}-3x,T_{4}(x)=8x^{4}-8x^{2}+1,T_{5}(x)=16x^{5}-20x^{3}+5x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,134 |
32.29. Prove that $\left|T_{n}(x)\right| \leqslant 1$ for $x \leqslant 1$. | 32.29. This follows from the fact that $T_{n}(x)=\cos n \varphi$ when $x=\cos \varphi$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 39,135 |
32.30. Prove that $T_{n}(x)=2^{n-1} x^{n}+a_{1} x^{n-1}+\ldots+a_{n}$, where $a_{1}, \ldots, a_{n}$ are integers. | 32.30. This follows from the recurrence relation $T_{n+1}(x)=$ $=2 x T_{n}(x)-T_{n-1}(x)$, which is proved in the solution to problem 32.27. | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,136 |
32.31. Let $P_{n}(x)=x^{n}+\ldots$ be a polynomial of degree $n$ with leading coefficient 1, and suppose that $\left|P_{n}(x)\right| \leqslant \frac{1}{2^{n-1}}$ for $|x| \leqslant 1$. Then $P_{n}(x)=\frac{1}{2^{n-1}} T_{n}(x)$. (In other words, the polynomial $\frac{1}{2^{n-1}} T_{n}(x)$ is the least deviating from ze... | 32.31. We will use only one property of the polynomial $T_{n}(x)=2^{n-1} x^{n}+\ldots$, namely that $T_{n}(\cos (k \pi / n))=\cos k \pi=(-1)^{k}$ for $k=0,1, \ldots, n$. Consider the polynomial $Q(x)=\frac{1}{2^{n-1}} T_{n}(x)-P_{n}(x)$. Its degree does not exceed $n-1$, since the leading terms of the polynomials $\fra... | P_{n}(x)=\frac{1}{2^{n-1}}T_{n}(x) | Algebra | proof | Yes | Yes | olympiads | false | 39,137 |
32.32. Prove that the Chebyshev polynomials $T_{n}(x)$ and $T_{m}(x)$ have the following property: $T_{n}\left(T_{m}(x)\right)=T_{m}\left(T_{n}(x)\right)$. | 32.32. Let $x=\cos \varphi$. Then $T_{n}(x)=\cos (n \varphi)=y$ and $T_{m}(y)=$ $\cos m(n \varphi)$, so $T_{m}\left(T_{n}(x)\right)=\cos m n \varphi$. Similarly, $T_{n}\left(T_{m}(x)\right)=$ $\cos m n \varphi$. Therefore, the equality $T_{n}\left(T_{m}(x)\right)=T_{m}\left(T_{n}(x)\right)$ holds for $|x|<1$, which mea... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,138 |
32.33. a) Let $n=2 k+1$. Prove that the number $\cos \left(\frac{2 l \pi}{n}\right)$ for any integer $l$ is a root of the polynomial $T_{k+1}(x)-T_{k}(x)$
b) Let $n=2 k$. Prove that the number $\cos \left(\frac{2 l \pi}{n}\right)$ for any integer $l$ is a root of the polynomial $T_{k+1}(x)-T_{k-1}(x)$. | 32.33. a) Let $n=2 k+1$ and $\varphi=2 l \pi / n$. Then $T_{k+1}(\cos \varphi)-$ $-T_{k}(\cos \varphi)=\cos (k+1) \varphi-\cos k \varphi \cdot$ In this case, $(k+1) \varphi+k \varphi=(2 k+1) \varphi=$ $=2 l \pi$. Therefore, $\cos (k+1) \varphi=\cos k \varphi$.
b) Let $n=2 k$ and $\varphi=2 l \pi / n$. Then $T_{k+1}(\c... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,139 |
32.34. a) Calculate the polynomials $T_{k+1}(x)-T_{k}(x)$ for $k \leqslant 4$.
b) Prove that the numbers $\cos \frac{2 \pi}{5}$ and $\cos \frac{4 \pi}{5}$ are roots of the polynomial $4 x^{2}+2 x-1$.
c) Prove that the numbers $\cos \frac{2 \pi}{7}, \cos \frac{4 \pi}{7}$, and $\cos \frac{6 \pi}{7}$ are roots of the po... | 32.34. a) Using the result of problem 32.28, we get $T_{2}-T_{1}=2 x^{2}-x-1, T_{3}-T_{2}=4 x^{3}-2 x^{2}-3 x+1, T_{4}-T_{3}=8 x^{4}-4 x^{3}-$ $-8 x^{2}+3 x+1, T_{5}-T_{4}=16 x^{5}-8 x^{4}-20 x^{3}+8 x^{2}+5 x-1$.
b) According to problem 32.33, the numbers $\cos 0=1, \cos \frac{2 \pi}{5}$, and $\cos \frac{4 \pi}{5}$ a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,140 |
32.35. Using polynomials $P_{n}$, prove the following statement: if both numbers $\alpha$ and $\cos (\alpha \pi)$ are rational, then the number $2 \cos (\alpha \pi)$ is an integer, i.e., $\cos (\alpha \pi)=0, \pm 1 / 2$ or $\pm 1$.
See also problem 29.55.
## 32.5. Algebraic and Transcendental Numbers
A complex numbe... | 32.35. Let $\alpha=m / n$ be an irreducible fraction. Set $x_{0}=$ $=2 \cos t$, where $t=\alpha \pi$. Then $P_{n}\left(x_{0}\right)=2 \cos (n t)=2 \cos (n \alpha \pi)=2 \cos (m \pi)=$ $= \pm 2$. Therefore, $x_{0}$ is a root of the polynomial $P_{n}(x) \mp 2=x^{n}+b_{1} x^{n-1}+\ldots$ $\ldots+b_{n}$ with integer coeffi... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,141 |
32.36. Let $x_{0}$ be a root of the polynomial
$$
a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n-1} x+a_{n}=0
$$
with integer coefficients $a_{0}, a_{1}, \ldots, a_{n}$. Prove that the number $a_{0} x_{0}$ is an integer algebraic. | 32.36. The number $y_{0}=a_{0} x_{0}$ is a root of the polynomial
$$
y^{n}+a_{1} y^{n-1}+a_{2} a_{0} y^{n-2}+\ldots+a_{n-1} a_{0}^{n-2} y+a_{n} a_{0}^{n-1}=0
$$
For $a_{0}=0$ this is obvious, and for $a_{0} \neq 0$ we need to set $y=a_{0} x$; after canceling by $a_{0}^{n-1}$ we get the original polynomial. | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,142 |
32.39. a) Let $\alpha$ and $\beta$ be algebraic numbers, and $\varphi(x, y)$ be an arbitrary polynomial with rational coefficients. Prove that then $\varphi(\alpha, \beta)$ is an algebraic number.
b) Let $\alpha$ and $\beta$ be algebraic integers, and $\varphi(x, y)$ be an arbitrary polynomial with integer coefficient... | 32.39. a) Let $\left\{\alpha_{1}, \ldots, \alpha_{n}\right\}$ and $\left\{\beta_{1}, \ldots, \beta_{m}\right\}$ be sets of numbers conjugate to $\alpha$ and $\beta$ respectively. Consider the polynomial
$$
F(t)=\prod_{i=1}^{n} \prod_{j=1}^{m}\left(t-\varphi\left(\alpha_{i}, \beta_{j}\right)\right)
$$
The coefficients... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,144 |
32.40. Let $\alpha$ and $\beta$ be algebraic numbers related by the equation $\varphi(\alpha, \beta)=0$, where $\varphi$ is a polynomial with rational coefficients. Prove that then for any number $\alpha_{i}$ conjugate to $\alpha$, there exists a number $\beta_{j}$ conjugate to $\beta$ such that $\varphi\left(\alpha_{i... | 32.40. Let $\left\{\alpha_{1}, \ldots, \alpha_{n}\right\}$ and $\left\{\beta_{1}, \ldots, \beta_{m}\right\}$ be sets of numbers conjugate to $\alpha$ and $\beta$ respectively. Consider the polynomial
$$
f(x)=\prod_{j=1}^{m} \varphi\left(x, \beta_{j}\right)
$$
The coefficients of this polynomial are rational and $f(\a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,145 |
32.41. Let $\alpha$ be a root of the polynomial
$$
f(x)=x^{n}+\beta_{n-1} x^{n-1}+\ldots+\beta_{0}
$$
where $\beta_{0}, \ldots, \beta_{n-1}$ are algebraic integers. Prove that then $\alpha$ is an algebraic integer.
An algebraic number $\alpha$ is called totally real if all its conjugates are real. In other words, al... | 32.41. Consider the polynomial
$$
F(x)=\prod_{i, \ldots, l}\left(x^{n}+\beta_{n-1, i} x^{n-1}+\ldots+\beta_{0, l}\right)
$$
where $\left\{\beta_{n-1, i}\right\}, \ldots,\left\{\beta_{0, l}\right\}$ are all the numbers conjugate to $\beta_{n-1}, \ldots, \beta_{0}$ respectively. It is easy to verify that the coefficien... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,146 |
32.42. Prove that the number $\alpha=2 \cos (k \pi / n)$ is totally real.
## 32.6. Adjoining a Root of a Polynomial
Let $k$ be a subfield of the field of complex numbers, and $f(x)$ be a polynomial with coefficients in $k$. If a root $\alpha$ of the polynomial $f(x)$ does not belong to $k$, then one can consider the ... | 32.42. It is clear that $\alpha=\varepsilon+\varepsilon^{-1}$, where $\varepsilon=\exp (k \pi / n)$. Let the number $\alpha_{1}$ be conjugate to $\alpha$. From problem 32.40, it follows that $\alpha_{1}=\varepsilon_{1}+\varepsilon_{1}^{-1}$, where the number $\varepsilon_{1}$ is conjugate to $\varepsilon$. The number $... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,147 |
32.43. Let $f(x)$ be an irreducible polynomial of degree $n$ over a field $k$, and let $\alpha$ be one of its roots. Prove that the field $k(\alpha)$ consists of numbers of the form $c_{n-1} \alpha^{n-1} + c_{n-2} \alpha^{n-2} + \ldots + c_{1} \alpha + c_{0}$, where $c_{0}, \ldots, c_{n-1}$ are numbers from the field $... | 32.43. It is clear that all numbers of the given form must belong to $k(\alpha)$. Therefore, it is sufficient to prove that numbers of the given form form a field. For this, in turn, it is sufficient to prove that the product of numbers of this form has the same form and the inverse element of a number of this form als... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,148 |
32.44. Let $f(x)$ be an irreducible polynomial of degree $n$ over a field $k$, and let $\alpha$ be one of its roots. Prove that if $\sum_{m=0}^{n-1} c_{m} \alpha^{m}=\sum_{m=0}^{n-1} d_{m} \alpha^{m}$, where $c_{0}, \ldots, c_{n-1}$ and $d_{0}, \ldots, d_{n-1}$ are numbers from the field $k$, then $c_{m}=d_{m}$ for $m=... | 32.44. Let $b_{m}=c_{m}-d_{m}$. We need to prove that if $\sum_{m=0}^{n-1} b_{m} \alpha^{m}=$ $=0$, where $b_{0}, \ldots, b_{n-1}$ are numbers from the field $k$, then $b_{0}=b_{1}=\ldots=b_{n-1}=0$. The polynomial $g(x)=b_{n-1} x^{n-1}+\ldots+b_{0}$ has a common root $\alpha$ with the irreducible polynomial $f(x)$, so... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,149 |
33.1. Prove that the fraction
$$
\frac{0.1234567891011 \ldots 4748495051}{0.51504948 \ldots 4321}
$$
begins with the digits 0.239. | 33.1. Let $a=0.1234 \ldots 5051$ and $b=0.5150 \ldots 321$. It is required to prove that $0.239 b \leqslant a$, $0.24 \cdot 0.515=0.1236>a$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,150 |
33.3. Calculate with an accuracy of 0.00001 the product
$$
\left(1-\frac{1}{10}\right)\left(1-\frac{1}{10^{2}}\right)\left(1-\frac{1}{10^{3}}\right) \ldots\left(1-\frac{1}{10^{99}}\right)
$$ | 33.3. Answer: 0.89001.
Let $a=\left(1-\frac{1}{10}\right)\left(1-\frac{1}{10^{2}}\right) \ldots\left(1-\frac{1}{10^{5}}\right)$ and $b=\left(1-\frac{1}{10^{6}}\right) \ldots$ $\ldots\left(1-\frac{1}{10^{99}}\right)$. Direct calculations show that
$$
0.89001+\frac{1}{10^{6}}1-x-y$. Therefore, $b>1-\frac{1}{10^{6}}-\fr... | 0.89001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,152 |
33.4. Prove that $3.14<\pi<3.142$ and $9.86<\pi^{2}<9.87$.
## 33.2. Arithmetic Operations. Polynomials | 33.4. We will use the identity
$$
4 \operatorname{arctg} \frac{1}{5}-\operatorname{arctg} \frac{1}{239}=\frac{\pi}{4}
$$
(Problem 11.11) and the inequalities for arctangent proved in Problem 29.46. From these inequalities, it follows that
$$
\begin{aligned}
& \frac{\pi}{4}>4\left(\frac{1}{5}-\frac{1}{3 \cdot 5^{3}}\... | 3.1405<\pi<3.14169.862<\pi^{2}<9.8698 | Inequalities | proof | Yes | Yes | olympiads | false | 39,153 |
33.5. Given a number $a$. Prove that $a^{n}$ can be computed using no more than $2 \log _{2} n$ multiplications. | 33.5. Let the binary representation of the number $n$ be $a_{0}+a_{1} \cdot 2+\ldots$ $\ldots+a_{m} \cdot 2^{m}$, where $a_{m} \neq 0$. Then $m \leqslant \log _{2} n<m+1$, since $2^{m} \leqslant n<2^{m+1}$. The numbers $a, a^{2}, a^{4}, \ldots, a^{2^{m}}$ can be computed with $m$ multiplications. To compute $a^{n}$, on... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,154 |
33.6. To compute the value of the polynomial $P(x)=a_{n} x^{n}+$ $+a_{n-1} x^{n-1}+\ldots+a_{0}$ at $x=x_{0}$, one can compute $a_{n} x_{0}^{n}$, $a_{n-1} x_{0}^{n-1}, \ldots, a_{1} x_{0}$, and then add all the obtained numbers and $a_{0}$. This requires $2 n-1$ multiplications (computing $x_{0}^{k}$ for $k=2,3, \ldots... | 33.6. Let $b_{1}=a_{n} x_{0}+a_{n-1}, b_{2}=b_{1} x_{0}+a_{n-2}, \ldots, b_{n}=b_{n-1} x_{0}+a_{0}$. Then $b_{n}=P\left(x_{0}\right)$, since
$$
P(x)=\left(\ldots\left(\left(a_{n} x_{0}+a_{n-1}\right) x_{0}+a_{n-2}\right) \ldots\right) x_{0}+a_{0}
$$ | b_{n}=P(x_{0}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,155 |
33.8. Given $n$ pairwise distinct numbers. It is required to find the largest of them, comparing a pair of numbers at each step.
a) Prove that this can be done in $n-1$ steps.
b) Prove that this cannot be done in fewer than $n-1$ steps.
Many different sorting algorithms are known. Here we will discuss some of them.
... | 33.8. a) When comparing two numbers, we will choose the largest one. The smaller number will be crossed out from the list; it will not participate in further comparisons. Clearly, after $n-1$ comparisons, only one number will remain in the list - the largest one.
b) Each comparison eliminates only one candidate for th... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 39,157 |
33.9. Prove that for insertion sort, the number of comparisons is bounded between $n-1$ and $\frac{n(n-1)}{2}$.
| To estimate the number of comparisons sufficient for solving various sorting tasks, the result of the following problem is often used. | 33.9. The smallest number of comparisons will occur when the numbers $a_{1}, a_{2}, \ldots, a_{n}$ are already sorted in ascending order. In this case, there will be no permutations at all. The numbers $a_{1}$ and $a_{2}, a_{2}$ and $a_{3}, \ldots, a_{n-1}$ and $a_{n}$ will be compared. The largest number of comparison... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 39,158 |
33.10. There is a pile of $n \geqslant 2$ stones. On the first step, it is divided roughly in half, i.e., if $n$ is even, it is divided into two piles of $n / 2$ stones each, and if $n$ is odd, it is divided into two piles of $\frac{n-1}{2}$ and $\frac{n+1}{2}$ stones. On the second step, the same process is repeated f... | 33.10. By induction on $m$, it is easy to prove that for a pile of $2^{m-1}+1$ stones, the process stops after $m$ steps, since after the first step, the larger pile contains $2^{m-2}+1$ stones. It is also clear that for a pile of $2^{m}$ stones, the process also stops after $m$ steps.
Thus, for a pile of $n$ stones, ... | -1<\log_{2}n\leqslant | Combinatorics | proof | Yes | Yes | olympiads | false | 39,159 |
33.11. Given a sequence of $n$ numbers. Prove that for its merge sort, no more than $m n-$ $-2^{m}+1$ comparisons of pairs of numbers are required, where the number $m$ is determined by the inequalities $m-1<\log _{2} n \leqslant m$. | 33.11. According to problem 33.10, the number $m$, which is determined by the given inequalities, is the number of steps required to complete the division of sequences (the steps where all sequences obtained from the previous step are roughly halved).
To merge two sorted sequences consisting of $p$ and $q$ numbers, no... | n-2^{}+1 | Combinatorics | proof | Yes | Yes | olympiads | false | 39,160 |
33.12. In a sequence of $2 n$ numbers, it is required to simultaneously find the largest and the smallest number.
a) Prove that this can be done by comparing $3 n-2$ pairs of numbers.
b) Prove that in the general case, it is impossible to get by with fewer than $3 n-2$ comparisons of pairs of numbers. | 33.12. a) Let's divide the given numbers arbitrarily into $n$ pairs. By comparing the numbers in each pair, we select $n$ largest numbers and $n$ smallest numbers from the pairs. Clearly, the largest number must be found among the $n$ largest selected numbers. It can be found by making $n-1$ comparisons (Problem 33.8).... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 39,161 |
34.1. Find all functions $f(x)$ for which $2 f(1-x)+$ $+1=x f(x)$. | 34.1. Substituting $1-x$ for $x$, we get $2 f(x)+1=(1-x) f(1-x)$. The original equation shows that $f(1-x)=\frac{x f(x)-1}{2}$. Substituting this expression into the new relation, we get $2 f(x)+1=(1-x) \frac{x f(x)-1}{2}$, which means $f(x)=\frac{x-3}{x^{2}-x+4}$. Direct verification shows that this function satisfies... | f(x)=\frac{x-3}{x^{2}-x+4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,163 |
34.2. Find all functions $f(x)$ that are defined for $x \neq 1$ and satisfy the relation
$$
f(x) + f\left(\frac{1}{1-x}\right) = x
$$ | 34.2. Let $\varphi_{1}(x)=\frac{1}{1-x}$. Then $\varphi_{2}(x)=1-1 / x$ and $\varphi_{3}(x)=x$. Therefore, we obtain the system of equations
$$
\left\{\begin{array}{l}
f(x)+f\left(\frac{1}{1-x}\right)=x \\
f\left(\frac{1}{1-x}\right)+f\left(1-\frac{1}{x}\right)=\frac{1}{1-x} \\
f\left(1-\frac{1}{x}\right)+f(x)=1-\frac... | f(x)=\frac{1}{2}(x+1-\frac{1}{x}-\frac{1}{1-x}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,164 |
34.3. Find all functions $f(x)$ that are defined for all $x \neq 0, \pm 1$ and satisfy the relation
$$
x f(x) + 2 f\left(\frac{x-1}{x+1}\right) = 1
$$
## 34.2. Functional Equations for Arbitrary Functions | 34.3. For $\varphi_{1}(x)=\frac{x-1}{x+1}$, we sequentially find: $\varphi_{2}(x)=-1 / x$, $\varphi_{3}(x)=\frac{x+1}{1-x}$, and $\varphi_{4}(x)=x$. Therefore, we obtain the system of equations
$$
\left\{\begin{array}{l}
x f(x)+2 f\left(\frac{x-1}{x+1}\right)=1 \\
\frac{x-1}{x+1} f\left(\frac{x-1}{x+1}\right)+2 f\left... | f(x)=\frac{4x^{2}-x+1}{5(x-1)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,165 |
34.4. a) Suppose that each rational number $x$ is associated with a real number $f(x)$ such that $f(x+y)=f(x)+f(y)$ and $f(x y)=f(x) f(y)$. Prove that either $f(x)=x$ for all $x$, or $f(x)=0$ for all $x$.
b) Solve the same problem in the case where the number $f(x)$ is associated not only with rational numbers but with... | 34.4. a) Suppose that $f(x) \neq 0$ for at least one number $x$. Then from the equality $f(x \cdot 1) = f(x) f(1)$ it follows that $f(1) = 1$. Further, $f(0) = f(0 + 0) = f(0) + f(0)$, therefore $f(0) = 0$.
From the equality $f(x + y) = f(x) + f(y)$ it follows that if $n$ is a natural number, then $f(nx) = n f(x)$. He... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,166 |
34.5. Find all functions $f(x)$ that are defined for all $x$ and satisfy the relation
$$
x f(y)+y f(x)=(x+y) f(x) f(y)
$$
for all $x, y$. | 34.5. Let $x=y$. Then we get $2 x f(x)=2 x(f(x))^{2}$. If $x \neq 0$, then $f(x)=0$ or 1.
Suppose that $f(a)=0$ for some $a \neq 0$. Setting $x=a$, we get $a f(y)=0$ for all $y$, i.e., $f=0$.
Suppose that $f(a)=1$ for some $a \neq 0$. Setting $x=a$, we get $a f(y)+y=(a+y) f(y)$, i.e., $y=y f(y)$. Thus, $f(y)=1$ for a... | f(x)=0forallx,orf(x)=1forallx\neq0f(0)= | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,167 |
34.6. Find the function $f(x)$, which is defined for all $x$, is non-zero at some point, and for all $x, y$ satisfies the equation $f(x) f(y)=f(x-y)$. | 34.6. Let's take a point $x_{0}$ for which $f\left(x_{0}\right) \neq 0$, and set $y=0$. Then $f\left(x_{0}\right) f(0)=f\left(x_{0}\right)$, so $f(0)=1$. Setting $x=y$, we get $(f(x))^{2}=f(0)=1$. Therefore, $f(x)= \pm 1$. Finally, setting $y=x / 2$, we obtain $f(x) f(x / 2)=f(x / 2)$, and since $f(x / 2)= \pm 1 \neq 0... | f(x)=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,168 |
34.7. Prove that there does not exist a function $f(x)$ defined for all $x$ that satisfies the relation $f(f(x)) = x^2 - 2$.
## 34.3. Functional Equations for Continuous Functions | 34.7. Consider the functions $g(x)=x^{2}-2$ and $h(x)=g(g(x))=x^{4}-$ $-4 x^{2}+2$. The roots of the equation $g(x)=x$ are -1 and 2. Both these numbers are also roots of the equation $h(x)=x$. To find the other roots of this equation, we divide $x^{4}-4 x^{2}-x+2$ by $x^{2}-x-2$. The result is the quadratic polynomial ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,169 |
34.8. A continuous function $f(x)$ is defined for all $x$ and satisfies the relation
$$
f(x+y)=f(x)+f(y)
$$
Prove that $f(x)=C x$, where $C=f(1)$. | 34.8. It is clear that $f(2 x)=f(x+x)=f(x)+f(x)=2 f(x), f(3 x)=$ $=f(2 x+x)=f(2 x)+f(x)=2 f(x)+f(x)=3 f(x)$. Similarly, it can be proved that $f(n x)=n f(x)$ for any natural number $n$. Further, $f(x)=f(x+0)=f(x)+f(0)$, so $f(0)=0$, and therefore $f(x)+f(-x)=$ $=f(x-x)=f(0)=0$. Thus, the equality $f(n x)=n f(x)$ holds ... | f(x)=xf(1) | Algebra | proof | Yes | Yes | olympiads | false | 39,170 |
34.9. Find all continuous functions that are defined for all $x$ and satisfy the relation $f(x)=$ $=a^{x} f(x / 2)$, where $a$ is a fixed positive number. | 34.9. It is clear that $f(x)=a^{x} f(x / 2)=a^{x} a^{x / 2} f(x / 4)=a^{x} a^{x / 2} a^{x / 4} f(x / 8)=$ $=\ldots=a^{x\left(1+1 / 2+1 / 4+\ldots+1 / 2^{k}\right)} f\left(x / 2^{k+1}\right)$. If $k \rightarrow \infty$, then $f\left(x / 2^{k+1}\right) \rightarrow f(0)$, since the function $f$ is continuous. Moreover, $1... | f(x)=C^{2x} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,171 |
34.10. A continuous function $f(x)$ is defined for all $x$, and $f\left(x_{0}\right) \neq 0$ for some $x_{0}$. Prove that if the relation
$$
f(x+y)=f(x) f(y)
$$
holds, then $f(x)=a^{x}$ for some $a>0$. | 34.10. The equality $f(x) f\left(x_{0}-x\right)=f\left(x_{0}\right)$ shows that $f(x) \neq 0$ for all $x$. But then $f(x)>0$ for all $x$, since $f(x)=f(x / 2) f(x / 2)$.
By induction on $n$, it is easy to prove that
$$
f(n x)=(f(x))^{n}
$$
for all natural $n$. It is also clear that $f(0)=1$, since $f(x)=f(x+0)=f(x) ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,172 |
34.11. A continuous function $f(x)$ is defined for all $x>0$ and satisfies the relation
$$
f(x y)=f(x)+f(y)
$$
a) Prove that if $f(a)=1$ for some $a$, then $f(x)=\log _{a} x$.
b) Prove that if $f\left(x_{0}\right) \neq 0$ for some $x_{0}$, then $f(a)=1$ for some $a$. | 34.11. a) Let's make the substitution $u=\log _{a} x$, i.e., $x=a^{u}$, and consider the function $g(u)=f(x)=f\left(a^{u}\right)$. The function $g$ is also continuous. It satisfies the relation $g(u+v)=g(u)+g(v)$. Indeed, $g(u+v)=f\left(a^{u+v}\right)=f\left(a^{u} a^{v}\right)=f\left(a^{u}\right)+f\left(a^{v}\right)=g(... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,173 |
34.12. Find all continuous solutions of the functional equation
$$
f(x+y)=f(x)+f(y)+f(x) f(y)
$$ | 34.12. It is clear that
$$
f(2 x)=2 f(x)+(f(x))^{2}=(1+f(x))^{2}-1
$$
We will prove by induction on $n$ that for any natural $n$ the equality holds
$$
f(n x)=(1+f(x))^{n}-1
$$
Indeed, from this equality it follows that
$$
\begin{aligned}
& f((n+1) x)=f(n x+x)=f(n x)+f(x)+f(x) f(n x)= \\
& \quad=(1+f(x))^{n}-1+f(x)... | f(x)\equiv-1,orf(x)\equiv0,orf(x)=k^x-1,wherek>0,k\neq1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,174 |
34.13. Find all continuous solutions of the functional equation
$$
f(x+y) f(x-y)=(f(x))^{2}
$$
(Lobachevsky).
## 34.4. Functional equations for differentiable functions | 34.13. Let $f(0)=a$ and $f(1)=b$. By setting $x=y=t / 2$ in the original functional equation, we get $f(t / 2)=\sqrt{a f(t)}$, so $f(1 / 2)=\sqrt{a b}=a(a / b)^{1 / 2}=a c^{1 / 2}$, where $c=b / a, f(1 / 4)=\sqrt{a f(1 / 2)}=a c^{1 / 4}$, and generally, $f\left(1 / 2^{n}\right)=a c^{1 / 2^{n}}$ (if $a=0$, we still get ... | f(x)=^x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,175 |
34.14. Find all differentiable functions $f$ for which $f(x) f^{\prime}(x)=0$ for all $x$.
## 34.5. Functional equations for polynomials | 34.14. An equivalent condition is that $\left(f^{2}(x)\right)^{\prime}=0$, hence the function $f^{2}(x)$ is constant. From the continuity of the function $f$, it follows that the function $f(x)$ is also constant. | f(x)= | Calculus | math-word-problem | Yes | Yes | olympiads | false | 39,176 |
34.15. Find all polynomials $P(x)$ for which the identity $x P(x-1)=(x-26) P(x)$ holds. | 34.15. For $x=0$ we get $0=-26 P(0)$, i.e. $P(0)=0$. For $x=1$ we get $P(0)=-25 P(1)$, i.e. $P(1)=0$. Next, we set $x=2,3, \ldots, 25$ and sequentially obtain $2 P(1)=-24 P(2), \ldots, 25 P(24)=-P(25)$. Therefore, $P(0)=P(1)=\ldots=P(25)=0$. This means that $P(x)=x(x-1)(x-2) \ldots(x-25) Q(x)$, where $Q(x)$ is some pol... | P(x)=\cdotx(x-1)(x-2)\ldots(x-25) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,177 |
34.16. The polynomial $P(x, y)$ has the following property: $P(x, y)=P(x+1, y+1)$ for all $x$ and $y$. Prove that $P(x, y)=\sum_{k=0}^{n} a_{k}(x-y)^{k}$.
According to problem 28.77, for a polynomial $f$ of degree $n+1$, the following equality holds:
$$
f(x)=f(y)+(x-y) f^{\prime}(y)+\ldots+(x-y)^{n+1} \frac{f^{(n+1)}... | 34.16. Let $t=y-x$. Then $P(x, t+x)=P(x+1, t+x+1)$ for all $x$ and $t$. Consider the polynomial $Q(t, x)=P(x, t+x)$. It has the following property: $Q(t, x)=Q(t, x+1)$ for all $x$. Therefore, $Q(t, x)$ does not depend on $x$. Indeed, fix $t=t_{0}$ and consider the polynomial $g(x)=Q\left(t_{0}, x\right)$. Then $g(x+1)-... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,178 |
34.19. Prove that the functional equation
$$
\frac{f(x)-g(y)}{x-y}=\varphi\left(\frac{x+y}{2}\right)
$$
can be reduced to the functional equation (2) | 34.19. The point is that if equality (3) holds for all real $x, y, x \neq y$, then
$$
\varphi\left(\frac{x+y}{2}\right)=\frac{\varphi(x)+\varphi(y)}{2}
$$
To prove this, replace $x$ with $x+y$ and $y$ with $x-y$ in (3). As a result, we get
$$
\frac{f(x+y)-g(x-y)}{2 y}=\varphi(x)
$$
for all real $x, y, y \neq 0$. Se... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,181 |
34.20. a) Find the polynomial solutions of the functional equation $f(\alpha x+\beta)=f(x)$ for $\alpha= \pm 1$.
b) Prove that if the solution of the functional equation $f(\alpha x+\beta)=f(x)$ is a polynomial of degree $n$, then $\alpha^{n}=1$. (In particular, if $\alpha \neq \pm 1$, then $n \geqslant 3$ ). | 34.20. а) If $\alpha=1$ and $f(x)=a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n}$, where $a_{0} \neq 0$, then we obtain the identity
$$
f(x)=a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n}=a_{0}(x+\beta)^{n}+a_{1}(x+\beta)^{n-1}+\ldots+a_{n}
$$
Such an identity is possible only if $a_{1}=a_{1}+a_{0} n \beta$, i.e., $\beta=0$.
If $\a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,182 |
34.21. Let the polynomial $f$ of degree $n \geqslant 3$ satisfy the relation $f(\alpha x+\beta)=f(x)$, where $\alpha \neq \pm 1$ and $\alpha^{n}=1$. Prove that then
$$
f(x)=a_{0}\left(x+\frac{\beta}{\alpha-1}\right)^{n}+c
$$ | 34.21. It is sufficient to consider polynomials of the form $f(x)=x^{n}+$ $+a_{1} x^{n-1}+\ldots+a_{n}$. We need to prove that $a_{j}=C_{n}^{j} \frac{\beta^{j}}{(\alpha-1)^{j}}$ for $j=1, \ldots, n-1$. We will prove this by induction on $j$.
Comparing the coefficients of $x^{n-j}$ for the polynomials $f(x)$ and $f(\al... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,183 |
35.1. Prove that the continued fraction $[0 ; 1,1,1, \ldots]$ is equal to the ratio of the smaller segment to the larger in the golden section, and the continued fraction $[1 ; 1,1,1, \ldots]$ is equal to the ratio of the larger segment to the smaller. | 35.1. Let $x=[0 ; 1,1,1, \ldots]$. Then $x=\frac{1}{1+x}$, i.e., $x(1+x)=1$. It is also clear that $x>0$. Solving the quadratic equation and discarding the negative root, we get the required result.
Let $y=[1 ; 1,1,1, \ldots]$. Then $y-1=1 / y$, i.e., $y(y-1)=1$. It is also clear that $y>0$. Solving the quadratic equa... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,184 |
35.2. Prove that continued fractions are related to the Euclidean algorithm in the following way. Let $m<n$ be two natural numbers. Apply the Euclidean algorithm to them: $n=a_{0} m+r_{1}, m=a_{1} r_{1}+r_{2}, r_{1}=a_{2} r_{2}+r_{3}, \ldots, r_{s-1}=a_{s} r_{s}$. Let $x_{k}=\left[0 ; a_{k}, a_{k+1}, \ldots, a_{s}\righ... | 35.2. Consider the sequence of numbers $y_{0}=\frac{m}{n}, y_{1}=\frac{r_{1}}{m}$, $y_{2}=\frac{r_{2}}{r_{1}}, \ldots, y_{s}=\frac{r_{s}}{r_{s-1}}$. Each of these numbers is between 0 and 1. It is easy to verify that $y_{k}=\frac{1}{a_{k}+y_{k+1}}$ for $k \leqslant s$ (we assume $y_{s+1}=0$). Indeed, this equality is e... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,185 |
35.3. Let $p_{0}=a_{0}$ and $q_{0}=1$.
a) Prove that for $k \geqslant 2$ the relations $p_{k}=a_{k} p_{k-1}+p_{k-2}$ and $q_{k}=a_{k} q_{k-1}+q_{k-2}$ hold.
b) Prove that $p_{k-1} q_{k}-q_{k-1} p_{k}=(-1)^{k}$.
c) Prove that $p_{k-2} q_{k}-q_{k-2} p_{k}=(-1)^{k-1} a_{k}$. | 35.3. a) We apply induction on $k$. For $k=2$, the required relations are easily verified. If $a_{0}, a_{1}, \ldots$ are considered as independent variables, the following obvious equality holds:
$$
\left[a_{0} ; a_{1}, \ldots, a_{k+1}\right]=\left[a_{0} ; a_{1}, \ldots, a_{k}+\frac{1}{a_{k+1}}\right]
$$
Therefore, a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,186 |
35.6. Prove that
$$
\frac{p_{n}}{q_{n}}=a_{0}+\frac{1}{q_{0} q_{1}}-\frac{1}{q_{1} q_{2}}+\frac{1}{q_{2} q_{3}}-\ldots+\frac{(-1)^{n-1}}{q_{n-1} q_{n}}
$$
Consider the sequence of polynomials $K_{0}=1, K_{1}\left(x_{1}\right)=x_{1}$,
$$
K_{n}\left(x_{1}, \ldots, x_{n}\right)=x_{n} K_{n-1}\left(x_{1}, \ldots, x_{n-1}... | 35.6. The equality from problem 35.3 b) can be rewritten as $\frac{p_{k}}{q_{k}}=\frac{p_{k-1}}{q_{k-1}}+\frac{(-1)^{k-1}}{q_{k-1} q_{k}}$. Then we write $\frac{p_{k-1}}{q_{k-1}}=\frac{p_{k-2}}{q_{k-2}}+\frac{(-1)^{k-2}}{q_{k-2} q_{k-1}}$ and so on until $\frac{p_{1}}{q_{1}}=\frac{p_{0}}{q_{0}}+\frac{1}{q_{0} q_{1}}=a_... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,189 |
35.7. Prove that $K_{n}\left(x_{1}, \ldots, x_{n}\right)$ represents the sum of all monomials that are obtained from $x_{1} x_{2} \ldots x_{n}$ by erasing several non-overlapping pairs of adjacent variables $x_{i} x_{i+1}$. It is allowed to erase nothing or to erase everything; in the latter case, we consider that the ... | 35.7. We apply induction on $n$. For $n=1$, the statement is obvious. The induction step is made as follows. The monomials that result from erasing several pairs $x_{i} x_{i+1}$ can be divided into two groups: those in which the variable $x_{n}$ is not erased, and those in which the variable $x_{n}$ is erased. Since $x... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 39,190 |
35.10. Prove that the best approximations of a number $\alpha$ are the convergents of the continued fraction expansion of $\alpha$ and only they.
## 35.3. Continued Fractions and Pell's Equation
Here we will discuss another approach to constructing solutions of Pell's equation, based on the expansion of the number $\... | 35.10. Suppose first that $x / y$ is the best approximation of the number $\alpha=\left[a_{0} ; a_{1}, a_{2}, \ldots\right]$, which does not coincide with any convergent $p_{n} / q_{n}$. We will show that then $p_{0} / q_{0}\left|\alpha-a_{0}\right|$, which cannot be. If $x / y > p_{1} / q_{1} \geqslant \alpha$, then
... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,193 |
35.11. Let $d$ be a square-free natural number. Then all solutions of the equation $x^{2}-d y^{2}= \pm 1$ in natural numbers are found among the convergents of the continued fraction expansion of $\sqrt{d}$.
Now we need to determine which specific convergents correspond to the solutions of the equation $x^{2}-d y^{2}=... | 35.11. According to problem 35.10, it is sufficient to check that if $x^{2}-d y^{2}= \pm 1$, where $x$ and $y$ are natural numbers, then $x / y$ is the best approximation of the number $\sqrt{d}$ (the case $y=1$ is easily handled separately). First, note that
$$
\left|\frac{x}{y}-\sqrt{d}\right|=\frac{1}{y}\left|\frac... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,194 |
35.12. Prove that the continued fraction of a reduced quadratic irrationality $\alpha$ is purely periodic, i.e., it has the form $\left[a_{0} ; a_{1}, \ldots, a_{k}, a_{0}, a_{1}, \ldots, a_{k}, \ldots\right]$. | 35.12. The number $\alpha$ is a root of the quadratic equation $a x^{2} + b x + c = 0$, where $a, b, c$ are integers, so $\alpha = \frac{-b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{P \pm \sqrt{D}}{Q}$. By changing the sign of the number $Q = 2a$ if necessary, we can assume that $\alpha = \frac{P + \sqrt{D}}{Q}$.
Let $\a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,195 |
35.13. Let $\sqrt{d}=\left[a_{0} ; a_{1}, \ldots, a_{k-1}, 2 a_{0}, a_{1}, \ldots\right]$, i.e., the number $k$ divides the period. Prove that then the convergent $p_{k-1} / q_{k-1}$ gives a solution to the equation $x^{2}-d y^{2}=(-1)^{k}$. | 35.13. Consider the number
$$
\alpha=[\sqrt{d}]+\sqrt{d}=\left[2 a_{0} ; a_{1}, \ldots, a_{k-1}, \alpha\right]=\frac{\hat{p}_{k-1} \alpha+\hat{p}_{k-2}}{\hat{q}_{k-1} \alpha+\hat{q}_{k-2}}
$$
where $\frac{\hat{p}_{k-1}}{\hat{q}_{k-1}}=\frac{p_{k-1}}{q_{k-1}}+a_{0}$. This number satisfies the quadratic equation
$$
\h... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,196 |
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