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Proposition 6.1.1. Let \( \ell \geq 3 \) be prime and let \( C \) be the affine equation \( {y}^{2} = {x}^{\ell } + t \) for some fixed \( t \in \mathbb{Z} \). This equation has a solution in \( {\mathbb{Z}}_{p} \) for all \( p \) if and only if it has one for primes \( p \) of the form \( p = {2k}\ell + 1 \) with \( 1... | Proof. Consider first the corresponding projective curve over \( {\mathbb{F}}_{p} \). Even though the equation is singular at infinity when \( \ell \geq 5 \), if \( t \neq 0 \) in \( {\mathbb{F}}_{p} \) and the characteristic of \( {\mathbb{F}}_{p} \) is different from 2 and \( \ell \) the Weil bounds apply, and since ... | No |
Corollary 6.1.2. Let \( \ell \) be a prime such that \( 3 \leq \ell \leq {31} \) . The equation \( {y}^{2} = {x}^{\ell } + t \) has a solution in \( {\mathbb{Z}}_{p} \) for all \( p \) if and only if the following conditions are satisfied:\n\n(1) For \( \ell = 3,7,{13},{17},{19} \), and 31, no condition.\n\n(2) For \( ... | Proof. Since the above proposition reduces the problem to a reasonably small finite computation, this corollary is proved by a simple computer search, and can be extended at will. | No |
Proposition 6.2.2. Let \( A \) be an \( m \times n \) integer matrix, and let \( B \) be an \( m \) - component integer column vector. There exists a matrix \( U \in {\mathrm{{GL}}}_{n}\left( \mathbb{Z}\right) \) and a matrix \( H \) in HNF such that \( {AU} = \left( {0 \mid H}\right) \) . If \( k \) is the number of z... | Proof. Recall that \( {\mathrm{{GL}}}_{n}\left( \mathbb{Z}\right) \) denotes the group of integer matrices that are invertible, i.e., of determinant \( \pm 1 \) . The first statement is proved in a manner very similar to the existence of the column echelon form (proved using Gaussian elimination). Here, we must perform... | Yes |
Proposition 6.3.1. The Diophantine equation \( f\left( {{x}_{1},\ldots ,{x}_{n}}\right) = 0 \) has a nontrivial rational solution if and only if it has a nontrivial solution in every \( {\mathbb{Q}}_{p} \) for every \( p \) such that \( p \mid {2D} \) . | Proof. The necessity of the conditions is clear. Conversely, assume that they are satisfied. By the Chevalley-Warning Theorem 2.5.2, if \( n \geq 3 \) the equation has a nontrivial solution \( {X}_{0} = {\left( {x}_{0,1},\ldots ,{x}_{0, n}\right) }^{t} \in {\mathbb{F}}_{p}^{n} \) for all \( p \) . Now if \( p \nmid {2D... | Yes |
Proposition 6.3.2. Let \( {X}_{0} = {\left( {x}_{1,0},\ldots ,{x}_{n,0}\right) }^{t} \) be a nontrivial solution to the Diophantine equation \( f\left( {{x}_{1},\ldots ,{x}_{n}}\right) = {X}^{t}{QX} = 0 \) . (1) The general solution \( X \) to the equation in rational numbers such that \( {X}^{t}Q{X}_{0} \neq 0 \) is g... | Proof. (1). Since \( f \) is homogeneous, we consider nontrivial solutions of \( f = 0 \) as elements of the projective space \( {\mathbb{P}}_{n}\left( \mathbb{Q}\right) \) . The parametric equation of a general line passing through \( {X}_{0} \) is \( X = u{X}_{0} + {vR} \) with \( \left( {u, v}\right) \in {\mathbb{P}... | Yes |
Lemma 6.3.3. Let \( Q \) be a nonsingular \( 3 \times 3 \) real symmetric matrix, and let \( {X}_{0} \) be a nonzero real vector such that \( {X}_{0}^{t}Q{X}_{0} = 0 \) . For \( X \in {\mathbb{R}}^{3},{X}^{t}{QX} = \) \( {X}^{t}Q{X}_{0} = 0 \) is equivalent to \( X = \lambda {X}_{0} \) for some \( \lambda \in \mathbb{R... | Proof. By diagonalizing \( Q \), we may assume that \( Q \) is a diagonal matrix with real nonzero diagonal entries \( a, b, c \) . Thus \( {X}^{t}{QX} = {X}^{t}Q{X}_{0} = 0 \) is equivalent to \( a{x}^{2} + b{y}^{2} + c{z}^{2} = {ax}{x}_{0} + {by}{y}_{0} + {cz}{z}_{0} = 0 \) . Since \( \left( {{x}_{0},{y}_{0},{z}_{0}}... | Yes |
Proposition 6.3.4. Let \( {X}_{0} = \left( {{x}_{0},{y}_{0},{z}_{0}}\right) \) be a nontrivial solution of the Diophantine equation \( f\left( {x, y, z}\right) = {X}^{t}{QX} = 0 \) and let \( M \) be a matrix in \( {\mathrm{{GL}}}_{3}\left( \mathbb{Q}\right) \) whose last column \( {M}_{3} \) is equal to \( {X}_{0} \) ... | Proof. By Proposition 6.3.2, the above parametrization with \( {R}^{t}Q{X}_{0} \neq 0 \) gives all solutions such that \( {X}^{t}Q{X}_{0} \neq 0 \) . Furthermore, if \( R = {s}_{1}{M}_{1} + {s}_{2}{M}_{2} \) for some \( {s}_{1} \) and \( {s}_{2} \) in \( \mathbb{Q} \) not both \( 0 \), setting \( u = \gcd \left( {{s}_{... | Yes |
Corollary 6.3.5. Let \( f\left( {x, y, z}\right) \) be a nonsingular rational quadratic form in three variables. There exist three polynomials \( {P}_{x},{P}_{y},{P}_{z} \) with integer coefficients that are homogeneous of degree 2 in two variables (i.e., integral binary quadratic forms) such that the general rational ... | Proof. Clear from the above proposition. Note that by multiplying \( d \) by a suitable rational number we may indeed assume that the polynomials \( {P}_{x} \) , \( {P}_{y} \), and \( {P}_{z} \) have integral coefficients. | Yes |
Assume that \( {ABC} \neq 0 \), let \( \left( {{x}_{0},{y}_{0},{z}_{0}}\right) \) be a particular nontrivial solution of \( A{x}^{2} + B{y}^{2} = C{z}^{2} \), and assume that \( {z}_{0} \neq 0 \) . The general solution in rational numbers to the equation is given by\n\n\[ x = d\left( {{x}_{0}\left( {A{s}^{2} - B{t}^{2}... | Proof. We apply the above proposition to the diagonal quadratic form with diagonal \( \left( {A, B, - C}\right) \), and to the particular solution \( \left( {-{x}_{0}, - {y}_{0},{z}_{0}}\right) \) (so as to obtain a parametrization with fewer minus signs). Since \( {z}_{0} \neq 0 \), we may\n\nchoose\n\[ M = \left( \be... | Yes |
Corollary 6.3.8. In the parametrization given by Corollary 6.3.6, if \( x, y \) , and \( z \) are integers then \( d = u/v \) with \( v \mid {2BC}{z}_{0}^{2} \) . | Proof. If we simply used the above proposition, we would obtain \( v \) \( {4ABC}{z}_{0}^{4} \) . However, we can do better (also in the general case) by using reduced resultants. Without entering into this theory, we simply note that if we set \( {P}_{x}\left( {S, T}\right) = {x}_{0}\left( {A{S}^{2} - B{T}^{2}}\right)... | Yes |
Lemma 6.3.9. Let \( M \in {\mathcal{M}}_{n}\left( \mathbb{Z}\right) \), let \( p \) be a prime number, and let \( d = \) \( {\dim }_{{\mathbb{F}}_{p}}\left( {\operatorname{Ker}\left( \bar{M}\right) }\right) \), where \( \bar{M} \) denotes the reduction of \( M \) modulo \( p \) . Then \( {p}^{d} \) \( \det \left( M\rig... | Proof. Let \( \overline{{U}_{1}},\ldots ,\overline{{U}_{d}} \) be an \( {\mathbb{F}}_{p} \) -basis of \( \operatorname{Ker}\left( \bar{M}\right) \), let \( \overline{{U}_{1}},\ldots ,\overline{{U}_{n}} \) be a completion to an \( {\mathbb{F}}_{p} \) -basis of \( {\mathbb{F}}_{p}^{n} \), let \( \bar{U} \in {\mathrm{{GL}... | Yes |
Lemma 6.3.10. For any \( \bar{M} \in {\mathrm{{SL}}}_{n}\left( {\mathbb{Z}/p\mathbb{Z}}\right) \) there exists a lift \( M \) such that \( M \in {\mathrm{{SL}}}_{n}\left( \mathbb{Z}\right) \) ; in other words, the natural reduction map from \( {\mathrm{{SL}}}_{n}\left( \mathbb{Z}\right) \) to \( {\mathrm{{SL}}}_{n}\lef... | Proof. The following proof is taken from [Shi]. We prove this by induction on the size \( n \) of the matrix, the result being trivial for \( n = 1 \) . Assume \( n > 1 \) and the result true for \( n - 1 \), and let \( N \) be any lift to \( {\mathcal{M}}_{n}\left( \mathbb{Z}\right) \) of the matrix \( \bar{M} \) . By... | Yes |
Proposition 6.3.12. Assume that \( n \leq 5 \), that \( Q \) is a quadratic form with integral entries such that \( \det \left( Q\right) = \pm 1 \), and choose \( \gamma \) such that \( 4/3 < \gamma < \) \( {2}^{2/\left( {n - 1}\right) } \) . Then either we find a \( {\mathbf{b}}_{j}^{ * } \) such that \( Q\left( {\mat... | Proof. It follows from the inequality for \( \left| {\left( {\mathbf{b}}_{1}\right) }^{2}\right| \) that \( 1 \leq \left| {\left( {\mathbf{b}}_{1}\right) }^{2}\right| < 2 \), so that \( {\left( {\mathbf{b}}_{1}\right) }^{2} = \pm 1 \) since it is an integer. Since \( {\mathbf{b}}_{1}^{ * } = {\mathbf{b}}_{1} \), for \(... | Yes |
Corollary 6.3.13. Up to exchange of \( x \) and \( y \) the general integral solution to the Pythagorean equation \( {x}^{2} + {y}^{2} = {z}^{2} \) is given by \( x = d\left( {{s}^{2} - {t}^{2}}\right), y = {2dst} \) , \( z = d\left( {{s}^{2} + {t}^{2}}\right) \), where \( s, t \) are coprime integers of opposite parit... | Proof. Using Corollary 6.3.6 with the particular solution \( \left( {1,0,1}\right) \), we obtain the formulas of the corollary. If \( s \equiv t\left( {\;\operatorname{mod}\;2}\right) \), we set \( {s}_{1} = \left( {s + t}\right) /2 \) and \( {t}_{1} = \left( {s - t}\right) /2 \), so that \( s = {s}_{1} + {t}_{1}, t = ... | Yes |
Corollary 6.3.14. (1) Let \( p = \pm 2 \) . The general integral solution of \( {x}^{2} + p{y}^{2} = {z}^{2} \) with \( x \) and \( y \) coprime is given by \( x = \pm \left( {{s}^{2} - p{t}^{2}}\right), y = {2st} \), \( z = \pm \left( {{s}^{2} + p{t}^{2}}\right) \), where \( s \) and \( t \) are coprime integers with ... | Proof. Using Corollary 6.3.6 with the particular solution \( \left( {1,0,1}\right) \) we obtain the formulas \( x = d\left( {{s}^{2} - p{t}^{2}}\right), y = {2dst}, z = d\left( {{s}^{2} + p{t}^{2}}\right) \), where \( s \) and \( t \) are coprime integers and \( d \in \mathbb{Q} \) . We consider two cases.\n\nIf \( 2 \... | Yes |
Corollary 6.3.15. Let \( p \) be a positive or negative prime number with \( p \neq 2 \) . The general integral solution of \( {x}^{2} + p{y}^{2} = {z}^{2} \) with \( x \) and \( y \) coprime is given by one of the following two disjoint parametrizations.\n\n(1) \( x = \pm \left( {{s}^{2} - p{t}^{2}}\right), y = {2st},... | Proof. Using Corollary 6.3.6 with the particular solution \( \left( {1,0,1}\right) \) we again obtain the formulas \( x = d\left( {{s}^{2} - p{t}^{2}}\right), y = {2dst}, z = d\left( {{s}^{2} + p{t}^{2}}\right) \), whnere \( s \) and \( t \) are coprime integers and \( d \in \mathbb{Q} \) . We consider two cases.\n\nIf... | Yes |
If \( D > 0 \) is not a square and is congruent to 0 or 1 modulo 4 the Pell equation \( {x}^{2} - D{y}^{2} = \pm 4 \) has an infinity of solutions given in the following way. If \( \left( {{x}_{0},{y}_{0}}\right) \) is a solution with the least strictly positive \( {y}_{0} \) (and \( {x}_{0} > 0 \), say), the general s... | The equation can be written \( \mathcal{N}\left( \varepsilon \right) = \pm 1 \), with \( \varepsilon = \left( {x + y\sqrt{D}}\right) /2 \) , and since \( x \) and \( y \) are integers and \( x \equiv {Dy}\left( {\;\operatorname{mod}\;2}\right) ,\varepsilon \) is an algebraic integer of norm equal to \( \pm 1 \), hence ... | No |
Lemma 6.3.17. The set of \( k \in \mathbb{Z} \) such that \( \left( {{\varepsilon }_{1}^{k} - {\overline{{\varepsilon }_{1}}}^{k}}\right) /\sqrt{{D}_{1}} \in p\mathbb{Z} \) has the form \( {k}_{0}\mathbb{Z} \), where \( {k}_{0} \mid p - \left( \frac{{D}_{1}}{p}\right) \) . | Proof. We may clearly assume that \( p \nmid {y}_{1} \), since otherwise we can choose \( {k}_{0} = 1 \) . Assume first that \( p \neq 2 \) . If \( p \mid {D}_{1} \), expanding the right-hand side gives \( \pm {2}^{k - 1}{py} \equiv k{x}_{1}^{k - 1}{y}_{1}\left( {\;\operatorname{mod}\;p}\right) \), so that the set of s... | Yes |
Theorem 6.4.2. Let \( a, b \), and \( c \) be nonzero pth power-free integers such that \( \gcd \left( {a, b, c}\right) = 1 \), let \( \ell \) be a prime number, and denote by \( {S}_{\ell } \) the statement that the equation \( a{x}^{p} + b{y}^{p} + c{z}^{p} = 0 \) has a nontrivial solution in \( {\mathbb{Q}}_{\ell } ... | Proof. Let \( \left( {x, y, z}\right) \) be a nontrivial solution in \( {\mathbb{Q}}_{\ell } \), where we may clearly assume that \( x, y \), and \( z \) are in \( {\mathbb{Z}}_{\ell } \) and that at least one of them is an \( \ell \) -adic unit, so that in fact at least two are \( \ell \) -adic units since the coeffic... | Yes |
Lemma 6.4.3. Let \( a, b \), and \( c \) be pth power-free integers. If the equation \( a{x}^{p} + b{y}^{p} + c{z}^{p} = 0 \) has a nontrivial solution with \( x, y, z \) in \( \mathbb{Q} \), it has a nontrivial solution with \( x, y, z \) in \( \mathbb{Z} \) pairwise coprime. Furthermore, if in addition \( a, b \), an... | Proof. Easy and left to the reader; see Exercise 13. | No |
Lemma 6.4.5. Let \( K \) be a number field, and let \( \mathfrak{a},\mathfrak{b},{\mathfrak{c}}_{1} \), and \( {\mathfrak{c}}_{2} \) be integral ideals of \( K \) . Assume the following:\n\n(1) We have an ideal equality \( {\mathfrak{c}}_{1}{\mathfrak{c}}_{2} = \mathfrak{b}{\mathfrak{a}}^{p} \).\n\n(2) We have \( \gcd ... | Proof. If we set \( \mathfrak{d} = \gcd \left( {{\mathfrak{c}}_{1},{\mathfrak{c}}_{2}}\right) \) then by assumption \( \mathfrak{d} \) and \( \mathfrak{a} \) are coprime; hence \( {\mathfrak{d}}^{2} \mid \mathfrak{b} \), so that\n\n\[ \left( {{\mathfrak{c}}_{1}{\mathfrak{d}}^{-1}}\right) \left( {{\mathfrak{c}}_{2}{\mat... | Yes |
Lemma 6.4.7. Let \( c \) be an integer not equal to a pth power and such that \( {c}^{p - 1} ≢ 1\left( {\;\operatorname{mod}\;{p}^{2}}\right) \), set \( \theta = {c}^{1/p} \), and let \( K = \mathbb{Q}\left( \theta \right) \) . Let \( b \) be a nonzero integer, let \( x, y \), and \( z \) be pairwise coprime integers s... | Proof. (1). We note first that \( p \nmid y \) : indeed, if \( p \mid y \) then \( p \nmid {xz} \) by coprimality, so \( c \equiv {\left( -x/z\right) }^{p}\left( {\;\operatorname{mod}\;{p}^{2}}\right) \), and hence \( {c}^{p - 1} \equiv {\left( -x/z\right) }^{\phi \left( {p}^{2}\right) } \equiv 1 \) \( \left( {\;\opera... | Yes |
Theorem 6.4.8. Let \( b \) and \( c \) be pth power-free integers not equal to \( \pm 1 \) , set \( \theta = {c}^{1/p} \), and let \( K = \mathbb{Q}\left( \theta \right) \) . Assume that the following conditions are satisfied:\n\n(1) We have \( {c}^{p - 1} ≢ 1\left( {\;\operatorname{mod}\;{p}^{2}}\right) \) .\n\n(2) Th... | Proof. By Lemma 6.4.3 it is sufficient to prove that the equation has no solution with \( x, y, z \) in \( \mathbb{Z} \) pairwise coprime. By Lemma 6.4.7 there exist integral ideals \( {\mathfrak{a}}_{i} \) and \( {\mathfrak{b}}_{i} \) of \( K \) such that \( L{\mathbb{Z}}_{K} = {\mathfrak{b}}_{1}{\mathfrak{a}}_{1}^{p}... | Yes |
Lemma 6.4.9. Keep the above notation and assume that \( p \nmid f = \left\lbrack {{\mathbb{Z}}_{K} : \mathbb{Z}\left\lbrack \theta \right\rbrack }\right\rbrack \) . Then if \( \alpha \in {\mathbb{Z}}_{K} \) we have \( {\alpha }^{p} \equiv \mathcal{N}\left( \alpha \right) \left( {{\;\operatorname{mod}\;p}{\mathbb{Z}}_{K... | Proof. By definition of \( f \) we have \( {f\alpha } \in \mathbb{Z}\left\lbrack \theta \right\rbrack \), so that \( {f\alpha } = \mathop{\sum }\limits_{{0 \leq j < p}}{a}_{j}{\theta }^{j} \) with \( {a}_{j} \in \mathbb{Z} \) . Thus \( {\left( f\alpha \right) }^{p} \equiv \mathop{\sum }\limits_{{0 \leq j < p}}{a}_{j}^{... | Yes |
Theorem 6.4.10. Let \( b \) and \( c \) be pth power-free integers, set \( \theta = {c}^{1/p} \), let \( K = \mathbb{Q}\left( \theta \right), f = \left\lbrack {{\mathbb{Z}}_{K} : \mathbb{Z}\left\lbrack \theta \right\rbrack }\right\rbrack \), denote by \( e = e\left( K\right) \) the exponent of the class group of \( K \... | Proof. First note that it follows from the Dedekind criterion (see Theorem 6.1.4 of \( \left\lbrack \mathrm{{Coh0}}\right\rbrack \) ) that the condition on \( c \) in (1) is indeed equivalent to \( p \nmid f \) . Thus, by elementary algebraic number theory the prime ideal decomposition of \( p \) in the extension \( K/... | Yes |
Corollary 6.4.11. The equations \( {x}^{p} + b{y}^{p} + c{z}^{p} = 0 \) have a nontrivial solution in every completion of \( \mathbb{Q} \) but no nontrivial solutions in \( \mathbb{Q} \) : | Proof. Since we have at our disposal a CAS that can say immediately whether the conditions of the theorems are satisfied, a simple computer program proves the corollary, which can of course be extended at will. | No |
Corollary 6.4.12 (Selmer). The equation \( 3{x}^{3} + 4{y}^{3} + 5{z}^{3} = 0 \) has a nontrivial solution in every completion of \( \mathbb{Q} \) but no nontrivial solutions in \( \mathbb{Q} \) . | Proof. Multiplying the equation by 2 and setting \( \left( {X, Y, Z}\right) = \left( {{2y}, x, z}\right) \) , it is clear that its solubility in any field of characteristic 0 is equivalent to that of the equation \( {X}^{3} + 6{Y}^{3} + {10}{Z}^{3} = 0 \), and it is easy to check that the conditions of Theorem 6.4.10 a... | Yes |
Proposition 6.4.13. Assume that \( p \) is odd. If \( a{x}^{p} + b{y}^{p} + c{z}^{p} = 0 \) with \( x \neq 0 \) then\n\n\[{Y}^{2} = {X}^{p} + {a}^{2}{\left( bc\right) }^{p - 1}/4,\;\text{ with }\n\]\n\[X = - {bcyz}/{x}^{2}\;\text{ and }\;Y = {\left( -bc\right) }^{\left( {p - 1}\right) /2}\left( {b{y}^{p} - c{z}^{p}}\ri... | Proof. The proof is of course a simple verification: using the given formulas for \( X \) and \( Y \), since \( p \) is odd we have\n\n\[4{x}^{2p}\left( {{Y}^{2} - {X}^{p}}\right) = {\left( bc\right) }^{p - 1}{\left( b{y}^{p} - c{z}^{p}\right) }^{2} + 4{\left( bc\right) }^{p}{y}^{p}{z}^{p}\]\n\n\[= {\left( bc\right) }^... | Yes |
Assume that neither \( b/a, c/a \), nor \( c/b \) is the cube of \( a \) rational number. If the elliptic curve \( E \) with affine equation \( {Y}^{2} = {X}^{3} + {\left( 4abc\right) }^{2} \) has zero rank then the equation \( a{x}^{3} + b{y}^{3} + c{z}^{3} \) has no nontrivial rational solutions. | This is essentially a restatement of Corollary 8.1.15, and also immediately follows from the above proposition after rescaling. Note that Proposition 8.4.3 tells us that the elliptic curve \( {Y}^{2} = {X}^{3} - {432}{\left( abc\right) }^{2} \) is 3 -isogenous with the elliptic curve \( {Y}^{2} = {X}^{3} + {\left( 4abc... | No |
Proposition 6.4.15. Let \( \mathcal{C} \) be the cubic curve with projective equation \( a{x}^{3} + b{y}^{3} + c{z}^{3} = 0 \), let \( E \) be the elliptic curve with projective equation \( {Y}^{2}Z = {X}^{3} + {\left( 4abc\right) }^{2}{Z}^{3} \), and define \[ \phi \left( {x, y, z}\right) = \left( {-{4abcxyz}, - {4abc... | Proof. (1). As for Proposition 6.4.13, of which up to rescaling this is a special case, the proof is a simple verification: setting \( \phi \left( {x, y, z}\right) = \left( {X, Y, Z}\right) \), then if \( a{x}^{3} + b{y}^{3} + c{z}^{3} = 0 \) we check that \( {Y}^{2}Z - {\left( 4abc\right) }^{2}{Z}^{3} = {X}^{3} \). Fu... | Yes |
Proposition 6.4.16. The equation \( {x}^{2} - {3xy} + 3{y}^{2} = {z}^{3} \) with \( x \) and \( y \) coprime integers has the three disjoint parametrizations\n\n\[ \left( {x, y, z}\right) = \left( {{s}^{3} + 3{s}^{2}t - {6s}{t}^{2} + {t}^{3},{3st}\left( {s - t}\right) ,{s}^{2} - {st} + {t}^{2}}\right) ,\]\n\n\[ \left( ... | Proof. Let \( \rho = \left( {-1 + \sqrt{-3}}\right) /2 \) be a primitive cube root of unity. In the principal ideal domain \( \mathbb{Z}\left\lbrack \rho \right\rbrack \) our equation factors as \( \left( {x - \left( {1 - \rho }\right) y}\right) (x - (1 - \) \( \left. {\left. {\rho }^{2}\right) y}\right) = {z}^{3} \) .... | Yes |
Theorem 6.4.17. (1) Let \( p \) be a prime number such that \( p \equiv 2\left( {\;\operatorname{mod}\;3}\right) \). The equation \( {x}^{3} + {y}^{3} + c{z}^{3} = 0 \) has no solutions in nonzero integers \( x, y \), and \( z \) when \( c = 1,3, p \), or \( {p}^{2} \) with \( p \equiv 2 \) or 5 modulo 9, except for \(... | Proof. We may clearly assume that \( x, y \), and \( z \) are pairwise coprime integers. We prove all the results simultaneously, and consider two cases.\n\nCase 1: \( 3 \mid {cz} \). Then \( 3 \mid x + y \); hence \( 3 \mid \left( {{x}^{2} - {xy} + {y}^{2}}\right) = \left( {{\left( x + y\right) }^{2} - {3xy}}\right) \... | Yes |
An integer \( n \geq 1 \) is a sum of two rational cubes if and only if it is a sum of two nonnegative rational cubes. | We may evidently assume that \( n \) is not a cube, so let \( n = {x}_{0}^{3} + {y}_{0}^{3} \) be a decomposition of \( n \) as a sum of two cubes. If \( {x}_{0} \) and \( {y}_{0} \) are nonnegative there is nothing to prove, so we may assume without loss of generality that \( {y}_{0} > 0 \) and \( {x}_{0} < 0 \) . It ... | Yes |
Proposition 6.4.19. There are infinitely many integers that are not the sum of two cubes of rational numbers. | Proof. Indeed, Theorem 6.4.17 tells us for instance that odd primes congruent to 2 or 5 modulo 9 cannot be the sum of two cubes of rational numbers, and by Dirichlet's theorem on primes in arithmetic progression (Theorem 10.5.30) there are infinitely many such primes. | Yes |
An integer \( n \) is a sum of two cubes of (positive or negative) integers if and only if there exists a positive divisor \( d \) of \( n \) such that \( \left( {{4n}/d - {d}^{2}}\right) /3 \) is the square of an integer. | Left to the reader (Exercise 18). | No |
Theorem 6.4.23. Assume the Birch and Swinnerton-Dyer conjecture (Conjecture 8.1.7), and let \( p \) be a prime number. Then \( p \) is the sum of two cubes of rational numbers if and only if one of the following conditions is satisfied:\n\n(1) \( p = 2 \) .\n\n(2) \( p \equiv 4,7 \), or 8 modulo 9 .\n\n(3) \( p \equiv ... | The proof of this theorem uses the classical theory of modular forms, complex multiplication, and central values of \( L \) -series. For example, it implies that the only \( p \leq {100} \) such that \( p \equiv 1\left( {\;\operatorname{mod}\;9}\right) \) that are sums of two cubes of rational numbers are \( p = {19} \... | No |
Proposition 6.4.25. Every integer (and in fact every rational number) is the sum of three cubes of rational numbers. For instance, we have \( n = {x}^{3} + \) \( {y}^{3} + {z}^{3} \), with | \[ x = m - 1, y = \frac{3\left( {{m}^{2} + m}\right) }{{m}^{2} + m + 1}, z = \frac{-{m}^{3} + {3m} + 1}{{m}^{2} + m + 1}, \] where we have set \( m = n/9 \) . Proof. Just check. Of course this does not explain how to obtain such identities or why they exist. | No |
Proposition 6.4.27. Every integer is a sum of five cubes of integers, where we can in fact assume that at least two are equal. In other words, every integer has the form \( 2{x}^{3} + {y}^{3} + {z}^{3} + {t}^{3} \) . | Proof. The identity \( {6x} = {\left( x - 1\right) }^{3} + {\left( -x\right) }^{3} + {\left( -x\right) }^{3} + {\left( x + 1\right) }^{3} \) shows that every multiple of 6 is a sum of four cubes of which two are equal. If \( n \) is an integer, \( n - {n}^{3} \) is divisible by 6 hence is a sum of four cubes, so \( n =... | Yes |
Proposition 6.4.29. Up to permutation of the variables the equation \( {w}^{3} + {x}^{3} + {y}^{3} + {z}^{3} = 0 \) in \( \mathbb{Q} \) has the trivial parametrization \( x = - w, z = - y \), and the parametrization\n\n\[ \nw = - d\left( {\left( {s - {3t}}\right) \left( {{s}^{2} + 3{t}^{2}}\right) + 1}\right) ,\]\n\n\[... | Proof. If we set \( W = \left( {w + x}\right) /2, X = \left( {x - w}\right) /2, Y = \left( {y + z}\right) /2 \), and \( Z = \left( {z - y}\right) /2 \) the equation is equivalent to \( W\left( {{W}^{2} + 3{X}^{2}}\right) = - Y\left( {{Y}^{2} + 3{Z}^{2}}\right) \) . Excluding the trivial parametrization we have \( W \ne... | Yes |
Corollary 6.4.31. For \( d = 1,2,7,9,{17},{19},{20},{26},{28},{37},{43},{63},{65} \), and 91 the equation \( {x}^{3} + d{y}^{3} = 1 \) has a (necessarily unique) integral solution with \( y \neq 0 \), given respectively by \( \left( {x, y}\right) = \left( {0,1}\right) ,\left( {-1,1}\right) ,\left( {2, - 1}\right) ,\lef... | Proof. Clear by direct check, the uniqueness coming from Skolem's theorem. | No |
Corollary 6.4.32. The only integral solutions to the equation \( {y}^{2} = {x}^{3} + 1 \) are \( \left( {x, y}\right) = \left( {-1,0}\right) ,\left( {0, \pm 1}\right) \), and \( \left( {2, \pm 3}\right) \). | Proof. We rewrite the equation as \( \left( {y - 1}\right) \left( {y + 1}\right) = {x}^{3} \). If \( y \) is even, \( y - 1 \) and \( y + 1 \) are coprime, hence are both cubes, and since the only two cubes that differ by 2 are -1 and 1 we deduce that \( \left( {x, y}\right) = \left( {-1,0}\right) \). Otherwise \( y \)... | Yes |
Lemma 6.4.35. Let \( K = \mathbb{Q}\left( \sqrt{-3}\right) \) and \( \alpha \in {\mathbb{Z}}_{K} \) . Then \( \alpha \bar{\alpha } \) is a square in \( \mathbb{Z} \) if and only if there exist \( n \in \mathbb{Z} \) and \( \beta \in {\mathbb{Z}}_{K} \) such that \( \alpha = n{\beta }^{2} \) . | Proof. Since \( {\mathbb{Z}}_{K} \) is a principal ideal domain, simply decompose \( \alpha \) into a product of a root of unity and a product of powers of prime elements of \( {\mathbb{Z}}_{K} \) . The details are left to the reader (Exercise 22). | No |
The only rational solution to the equation \( {y}^{2} = {x}^{3} - 1 \) is \( \left( {x, y}\right) = \left( {1,0}\right) \), and the only rational solutions to the equation \( {y}^{2} = {x}^{3} + 1 \) are \( \left( {x, y}\right) = \left( {-1,0}\right) ,\left( {0, \pm 1}\right) \), and \( \left( {2, \pm 3}\right) \). | Proof. Write \( x = m/n \) with \( \gcd \left( {m, n}\right) = 1 \) . Multiplying the equation \( {y}^{2} = {x}^{3} + \varepsilon \) by \( {n}^{4} \) we see that \( n\left( {{m}^{3} + \varepsilon {n}^{3}}\right) \) is a square, and if we set \( c = m + {\varepsilon n} \) this means that \( {nc}\left( {{m}^{2} - {\varep... | Yes |
Corollary 6.5.2. Assume that \( c \) is squarefree. The equation \( {x}^{4} + {y}^{4} = c{z}^{2} \) is everywhere locally soluble if and only if \( c > 0 \) and the odd prime divisors of \( c \) are congruent to 1 modulo 8. | Proof. Indeed, by the proposition the equation is locally soluble in \( {\mathbb{Q}}_{p} \) for an odd prime divisor \( p \) of \( c \) if and only if -1 is a fourth power in \( {\mathbb{F}}_{p}^{ * } \), hence if \( p \equiv 1\left( {\;\operatorname{mod}\;8}\right) \), and it is locally soluble in \( {\mathbb{Q}}_{2} ... | Yes |
Proposition 6.5.3 (Fermat). Let \( \varepsilon = \pm 1 \) . The Diophantine equation \( {x}^{4} + \) \( \varepsilon {y}^{4} = {z}^{2} \) has no solutions with \( {xyz} \neq 0 \) . | Proof. We may clearly assume that \( x, y \), and \( z \) are pairwise coprime. Assume first that \( z \) is even. This can happen only if \( \varepsilon = - 1 \), since otherwise we would get a contradiction modulo 8 . Writing the equation as \( {y}^{4} + {z}^{2} = {x}^{4} \) with \( y \) odd, by Corollary 6.3.13 we o... | Yes |
Proposition 6.5.4. The Diophantine equation \( {x}^{4} + 2{y}^{4} = {z}^{2} \) has no solution with \( y \neq 0 \) . | Proof. Let \( \left( {x, y, z}\right) \) be integers such that \( {x}^{4} + 2{y}^{4} = {z}^{2} \), where we may assume that \( x, y \), and \( z \) are pairwise coprime. By Corollary 6.3.14 there exist coprime integers \( s \) and \( t \) with \( s \) odd such that \( {x}^{2} = \pm \left( {{s}^{2} - 2{t}^{2}}\right) \)... | Yes |
Proposition 6.5.5. Let \( E \) be the elliptic curve with projective equation \( {Y}^{2}Z = {X}^{3} + {ab}{c}^{2}X{Z}^{2} \), and define \( \phi \left( {x, y, z}\right) = \left( {-{bcx}{y}^{2}, b{c}^{2}{yz},{x}^{3}}\right) \). (1) The map \( \phi \) sends the set of nonzero rational solutions of \( a{x}^{4} + b{y}^{4} ... | Proof. Setting \( X = - {bcx}{y}^{2}, Y = b{c}^{2}{yz} \), and \( Z = {x}^{3} \), it is clear that \( P = \) \( \phi \left( \left( {x, y, z}\right) \right) = \left( {X, Y, Z}\right) \in E\left( \mathbb{Q}\right) \), and \( P \) is unchanged as a point in projective space if \( \left( {x, y, z}\right) \) is replaced by ... | Yes |
Proposition 6.5.6. Let \( c \) be a nonzero integer. The equation \( {x}^{4} - {y}^{4} = c{z}^{2} \) has a solution with \( {xyz} \neq 0 \) if and only if \( \left| c\right| \) is a congruent number. More precisely, if \( {x}^{4} - {y}^{4} = c{z}^{2} \) with \( {xyz} \neq 0 \) then \( {Y}^{2} = X\left( {{X}^{2} - {c}^{... | Proof. These formulas can of course be checked directly, but are not inverse of each other. More precisely the formula for \( \left( {X, Y}\right) \) comes directly from Proposition 6.5.5, and the formula for \( \left( {x, y, z}\right) \) is the inverse of the formula giving twice the point \( \left( {X, Y}\right) \) p... | Yes |
Corollary 6.5.7. Assume that \( c \in {\mathbb{Z}}_{ \geq 3} \) is squarefree and that the equation \( {x}^{4} + {y}^{4} = c{z}^{2} \) is everywhere locally soluble, in other words that the odd prime divisors of \( c \) are congruent to 1 modulo 8, and let \( {E}_{c} \) be the elliptic curve with affine equation \( {Y}... | Proof. It follows from Proposition 6.5.5 that the torsion points of \( {E}_{c} \) cannot come from nonzero solutions to our equation and that \( T \) is the only nontrivial torsion point, proving (1). Note that for \( c = 2 \) the torsion subgroup of \( {E}_{c} \) has order 4, generated by \( P = \left( {2,4}\right) \)... | No |
Corollary 6.5.8. (1) For any \( a \in \mathbb{Z} \smallsetminus \{ 0\} \) the equation \( {x}^{4} + a{y}^{4} = {z}^{2} \) has solutions with \( {xy} \neq 0 \) if and only if the elliptic curve \( {Y}^{2} = {X}^{3} + {aX} \) has nonzero rank, and in that case it has infinitely many inequivalent solutions. | Proof. (1). The solutions with \( {xy} = 0 \) correspond to the points \( \mathcal{O} \) and \( T \) on the elliptic curve \( E \) . Thus by Proposition 6.5.5 our equation has solutions with \( {xy} \neq 0 \) if and only if there exists a nontorsion point \( G \) such that \( \alpha \left( G\right) = 1 \) modulo square... | Yes |
Proposition 6.5.9. Let \( E \) be the elliptic curve with projective equation \( {Y}^{2}Z = {X}^{3} - a{b}^{2}{c}^{3}{Z}^{3}. \) | Proof. (1). Setting \( X = - {bcxy}, Y = b{c}^{2}z \), and \( Z = {x}^{3} \), it is clear that \( P = \left( {X, Y, Z}\right) \in E\left( \mathbb{Q}\right) \), and \( P \) is unchanged as a point in projective space if \( \left( {x, y, z}\right) \) is replaced by a twisted projectively equivalent solution. Conversely, ... | Yes |
Proposition 6.6.1. Assume that \( c \) is 4th power-free.\n\n(1) We have \( {\mathcal{C}}_{c}\left( {\mathbb{Q}}_{2}\right) \neq \varnothing \) if and only if \( c \equiv 1 \) or 2 modulo 16 . | Proof. If \( \left( {x : y : z}\right) \in {\mathcal{C}}_{c}\left( {\mathbb{Q}}_{p}\right) \), we may clearly assume that \( x, y \), and \( z \) are \( p \) -adic integers and that at least one is a \( p \) -adic unit. If \( p \nmid c \), reduction modulo \( p \) gives a projective curve \( \overline{{\mathcal{C}}_{c}... | Yes |
Lemma 6.6.2. Let \( p \nmid {2c} \) be a prime number. Then \( {\mathcal{C}}_{c}\left( {\mathbb{Q}}_{p}\right) \neq \varnothing \) if and only if \( \overline{{\mathcal{C}}_{c}}\left( {\mathbb{F}}_{p}\right) \neq \varnothing \) . In particular, if \( p ≢ 1\left( {\;\operatorname{mod}\;8}\right) \) then \[ {\mathcal{C}}... | Proof. One direction is clear. Conversely, assume that \( \overline{{\mathcal{C}}_{c}}\left( {\mathbb{F}}_{p}\right) \neq \varnothing \), and let \( \left( {{x}_{0} : {y}_{0} : {z}_{0}}\right) \) with \( {x}_{0},{y}_{0} \), and \( {z}_{0} \) not all divisible by \( p \) such that \( {x}_{0}^{4} + {y}_{0}^{4} \equiv c{z... | Yes |
Corollary 6.6.3. The curve \( {\mathcal{C}}_{c} \) is everywhere locally soluble (i.e., has points in every completion of \( \mathbb{Q} \) ) if and only if \( c > 0 \) and the following conditions are satisfied:\n\n(1) \( c \equiv 1 \) or 2 modulo 16 .\n\n(2) \( p \mid c, p \neq 2 \) implies \( p \equiv 1\left( {\;\ope... | Proof. Clear. | No |
Corollary 6.6.4. For all primes \( p \) such that \( p \equiv 1\left( {\;\operatorname{mod}\;{1160}}\right) \) the curve \( {\mathcal{C}}_{{p}^{2}} \) is everywhere locally soluble, but is not globally soluble, so is a counterexample to the Hasse principle. | Proof. It is clear that the above conditions are satisfied modulo 16, 5, and 29, and also modulo 13 since 7, 8, and 11 are quadratic nonresidues modulo 13. On the other hand, by Fermat's Proposition 6.5.3, the equation \( {x}^{4} + {y}^{4} = {Z}^{2} \) does not have any nontrivial solutions, so this is in particular th... | Yes |
Proposition 6.6.8. Let \( c \geq 3 \) be 4th power-free. A necessary condition for the global solubility in \( \mathbb{Q} \) of \( {x}^{4} + {y}^{4} = c{z}^{4} \) is the existence of a normalized divisor \( \gamma \) of \( c \) in \( \mathbb{Z}\left\lbrack \zeta \right\rbrack \) of the form \( \gamma = A + {B\zeta } + ... | Proof. (1) and (2). Factoring our equation in \( {\mathbb{Z}}_{K} \) gives\n\n\[ {x}^{4} + {y}^{4} = \left( {x + {\zeta y}}\right) \left( {x + {\zeta }^{3}y}\right) \left( {x + {\zeta }^{5}y}\right) \left( {x + {\zeta }^{7}y}\right) = c{z}^{4}. \]\n\nAssume that \( \pi \) is a prime element such that \( \pi \mid \gcd \... | No |
Lemma 6.6.9. Let \( c \equiv 1\left( {\;\operatorname{mod}\;{16}}\right) \) (respectively \( c \equiv 2\left( {\;\operatorname{mod}\;{16}}\right) \) ). An element \( \gamma \) is normalized and satisfies conditions (1) to (6) of Proposition 6.6.8 if and only if \( {\sigma }_{5}\left( \gamma \right) \) (respectively \( ... | Proof. Since \( {\sigma }_{5}\left( {A + {B\zeta } + C{\zeta }^{2} + D{\zeta }^{3}}\right) = A - {B\zeta } + C{\zeta }^{2} - D{\zeta }^{3} \) and \( \zeta {\sigma }_{7}(A + \) \( \left. {{B\zeta } + C{\zeta }^{2} + D{\zeta }^{3}}\right) = B + {A\zeta } - D{\zeta }^{2} - C{\zeta }^{3} \), the lemma is clear. | Yes |
Corollary 6.6.11. Assume that the equation \( {x}^{4} + {y}^{4} = c{z}^{4} \) is everywhere locally soluble, in other words that c satisfies the conditions of Corollary 6.6.3. Let \( \mathcal{F} \) be a set of representatives of suitable divisors \( \gamma \) of \( c \) modulo multiplication by powers of \( {\varepsilo... | Proof. Indeed, it is clear that multiplication of \( \gamma \) by a power of \( {\varepsilon }^{4} \) does not change the conditions in (6). Furthermore, it is immediate to check that multiplication by \( {\varepsilon }^{4} \) does not change \( {AC} - {BD}{\;\operatorname{mod}\;8} \) or \( C\left( {A + C}\right) - D(B... | Yes |
Proposition 6.6.12. Let \( c \geq 3 \) be a 4th power-free integer. If either \( {E}_{c}\left( \mathbb{Q}\right) \) or \( {F}_{c}\left( \mathbb{Q}\right) \) has rank 0 then \( {\mathcal{C}}_{c}\left( \mathbb{Q}\right) = \varnothing \) . | Proof. Note first that the trivial 2-torsion point \( \left( {X, Y}\right) = \left( {0,0}\right) \) on \( {E}_{c} \) corresponds to \( x = 0 \) on \( {\mathcal{C}}_{c} \), hence to \( c = {y}^{4} \), which is absurd by assumption. We now use Proposition 8.1.14, which we will prove in Chapter 8. It tells us that there c... | Yes |
Theorem 6.6.13 (Dem'yanenko). If \( c \geq 3 \) is a 4th power-free integer and if the rank of \( {E}_{c}\left( \mathbb{Q}\right) \) is less than or equal to 1 then \( {\mathcal{C}}_{c}\left( \mathbb{Q}\right) = \varnothing \) . | The existence of the maps \( \phi \) and \( \psi \), together with the map \( \bar{\phi } \) from \( {\mathcal{C}}_{c} \) to \( {E}_{c} \) defined by \( \bar{\phi }\left( \left( {x, y}\right) \right) = \left( {-{y}^{2}, y{x}^{2}}\right) \), means that the Jacobian of \( {\mathcal{C}}_{c} \) is isogenous to \( {E}_{c} \... | Yes |
The integer \( c = {5906} \) is the smallest integer that is the sum of two fourth powers of rational numbers, and not the sum of two fourth powers of integers. | Indeed, for all smaller values of \( c \) except \( c = {4481} \) we see that either the equation \( {x}^{4} + {y}^{4} = c \) has no rational solutions, or it has an integral solution. It is an immediate verification that 5906 is not a sum of two fourth powers of integers, and it is the sum of the two fourth powers of ... | No |
Proposition 6.7.3. Assume \( H\left( {p, t}\right) \), and define \( {A}_{p}\left( t\right) \) to be the (possibly empty) set of nonnegative integers a such that\n\n\[ \mathop{\sum }\limits_{{k = 0}}^{{\left( {p - 1}\right) /2}}\left( \begin{matrix} p \\ {2k} \end{matrix}\right) {a}^{2k}{t}^{\left( {p - 1}\right) /2 - ... | Proof. Let \( \left( {x, y}\right) \) be a solution to the equation \( {y}^{2} = {x}^{p} + t \) . In the quadratic field \( K = \mathbb{Q}\left( \sqrt{t}\right) \) we can write \( \left( {y - \sqrt{t}}\right) \left( {y + \sqrt{t}}\right) = {x}^{p} \) . I claim that the ideals generated by the two factors on the left ar... | Yes |
Corollary 6.7.4. Let \( p \geq 3 \) be a prime, let \( x \) and \( y \) be integers, and assume that \( t = {y}^{2} - {x}^{p} \) satisfies \( H\left( t\right) \) (so that in particular \( x \) and \( y \) are coprime and \( x \) is odd). Assume in addition that \( t + x \) is not a square, and furthermore when \( p = 3... | Proof. Clear since \( x = {a}^{2} - t \), except in the given special case. | No |
Proposition 6.7.5. Assume \( H\left( {3, t}\right) \). (1) When \( t \equiv 2 \) or 3 modulo 4 then if \( t \) is not of the form \( t = - \left( {3{a}^{2} \pm 1}\right) \) the equation \( {y}^{2} = {x}^{3} + t \) has no integral solutions. If \( t = - \left( {3{a}^{2} + \varepsilon }\right) \) with \( \varepsilon = \p... | Proof. This case is especially simple since the equations defining the sets \( {A}_{3}\left( t\right) \) are linear in \( t \). We obtain \( t = - \left( {3{a}^{2} \pm 1}\right), x = {a}^{2} - t \), and \( y = \pm \left( {{a}^{3} + {3at}}\right) \), giving the solutions of the proposition, to which we must add the solu... | Yes |
Proposition 6.7.7. Assume \( H\left( {5, t}\right) \) . The only values of \( t \) for which the equation \( {y}^{2} = {x}^{5} + t \) has a solution are \( t = - 1 \) (with only solution \( \left( {x, y}\right) = \left( {1,0}\right) \) ), \( t = - {19} \) (with only solutions \( \left( {x, y}\right) = \left( {{55}, \pm... | Proof. The equation defining the set \( {A}_{5}\left( t\right) \) of Proposition 6.7.3 is \( {t}^{2} + \) \( {10}{a}^{2}t + 5{a}^{4} \pm 1 = 0 \) . This has a rational solution in \( t \) if and only if the discriminant is a square, hence if and only if \( {20}{a}^{4} \mp 1 = {b}^{2} \) for some integer \( b \) . Looki... | Yes |
Corollary 6.7.8. Let \( x \) and \( y \) be integers such that the pair \( \left( {x, y}\right) \) is not equal to \( \left( {1,0}\right) ,\left( {{55}, \pm {22434}}\right) \), or \( \left( {{377}, \pm {2758646}}\right) \) . Assume that \( t = {y}^{2} - {x}^{5} \) satisfies \( H\left( t\right) \) (so that in particular... | Proof. Clear. | No |
Theorem 6.7.10. Let \( {u}_{n} = {u}_{n}\left( {\alpha ,\beta }\right) \) be a Lucas sequence as above. Then\n\n(1) If \( n > {30},{u}_{n} \) always has a primitive divisor.\n\n(2) If \( 5 < n < {30} \) is prime and \( {u}_{n}\left( {\alpha ,\beta }\right) \) has no primitive divisors, then either \( n = 7 \) and \( \l... | This theorem solves a century-old problem, and its proof involves both very delicate estimates on linear forms in logarithms and new algorithms for solving Thue equations. It is thus a beautiful mixture of difficult mathematics with an extensive rigorous computer computation (as the epigraph of the paper remarkably ill... | No |
Corollary 6.7.11. Let \( p \geq 7 \) be prime, and assume \( H\left( {p, t}\right) \) . The only value of \( t \) for which the equation \( {y}^{2} = {x}^{p} + t \) has a solution is \( t = - 1 \) with only solution \( \left( {x, y}\right) = \left( {1,0}\right) \). | Proof. Indeed, the equation defining the set \( {A}_{p}\left( t\right) \) is \( \left( {{\alpha }^{p} - {\beta }^{p}}\right) /\left( {\alpha - \beta }\right) = \pm 1 \) with \( \alpha = a + \sqrt{t} \) and \( \beta = a - \sqrt{t} \) ; hence with the notation of the above definition \( {u}_{p}\left( {\alpha ,\beta }\rig... | No |
Proposition 6.7.12 (V.-A. Lebesgue). For \( p \geq 3 \) prime the only integral solution to the equation \( {y}^{2} = {x}^{p} - 1 \) is \( \left( {x, y}\right) = \left( {1,0}\right) \) . | Proof. First note that since the class number of \( \mathbb{Q}\left( \sqrt{-1}\right) \) is equal to 1 the condition \( H\left( {p, - 1}\right) \) is satisfied for all \( p \) . Furthermore, \( t = - 1 \) does not occur in the special cases, so we must show that 0 is the only element of \( {A}_{p}\left( t\right) \) . T... | Yes |
Theorem 6.7.13. Assume \( H\left( t\right) \), in other words that \( t \) is a squarefree negative integer such that \( t ≢ 1\left( {\;\operatorname{mod}\;8}\right) \) . For \( n \geq 3 \) and \( - {100} \leq t \leq - 1 \) the Diophantine equations \( {y}^{2} = {x}^{n} + t \) do not have any integral solutions except ... | Proof. For \( t = - 1 \), we have the trivial solutions \( \left( {x, y}\right) = \left( {1,0}\right) \) if \( n \) is odd, \( \left( {x, y}\right) = \left( {\pm 1,0}\right) \) if \( n \) is even. Thanks to Proposition 6.7.1 we know that for \( - {100} < t < - 1 \) the only ones for which there are solutions with \( n ... | No |
Corollary 6.7.15 (Nagell). Let \( p \geq 3 \) be a prime. The only integer solutions to the equation \( {y}^{2} + y + 1 = 3{x}^{p} \) are \( \left( {x, y}\right) = \left( {1,1}\right) \) and \( \left( {x, y}\right) = \left( {1, - 2}\right) \). | Proof. This equation is equivalent to \( {\left( 2y + 1\right) }^{2} + 3 = {12}{x}^{p} \) ; it follows that \( 3 \mid \left( {{2y} + 1}\right) \), so setting \( {2y} + 1 = {3z} \) with \( z \) odd we obtain \( 3{z}^{2} + 1 = 4{x}^{p} \), and the result follows from the proposition. | No |
Proposition 6.8.1. (1) \( {L}_{-n} = {\left( -1\right) }^{n}{L}_{n},{F}_{-n} = {\left( -1\right) }^{n - 1}{F}_{n},{L}_{n}^{2} - 5{F}_{n}^{2} = \) \( 4{\left( -1\right) }^{n},2{L}_{m + n} = 5{F}_{m}{F}_{n} + {L}_{m}{L}_{n},2{F}_{m + n} = {F}_{m}{L}_{n} + {F}_{n}{L}_{m},{L}_{2m} = \) \( {L}_{m}^{2} + 2{\left( -1\right) }... | Proof. The formulas of (1) are proved by direct computation from the definitions in terms of \( \alpha \) and \( \beta \), which can be summarized by the equality \( \left( {{L}_{n} + {F}_{n}\sqrt{5}}\right) /2 = {\alpha }^{n} \) . | Yes |
Theorem 6.8.2. (1) For \( n \geq 0 \) we have \( {L}_{n} = {x}^{2} \) with \( x \in \mathbb{Z} \) if and only if \( n = 1 \) or \( n = 3 \) . | Proof. Since \( {L}_{2m} = {L}_{m}^{2} + 2{\left( -1\right) }^{m - 1},{L}_{2m} = {x}^{2} \) implies that \( \left| {{x}^{2} - {L}_{m}^{2}}\right| = 2 \) , which is impossible. Thus we may assume that \( n \) is odd. Clearly \( {L}_{1} = 1 \) and \( {L}_{3} = 4 \) are squares, so we may assume that \( n > 3 \) . We can ... | Yes |
Theorem 6.8.3. (1) We have \( {F}_{n} = {x}^{2} \) with \( x \in \mathbb{Z} \) if and only if \( n = 0, \pm 1 \) , 2, or 12. | Proof. The proof of this theorem is similar and left to the reader (see Exercise 43). | No |
Corollary 6.8.4. Consider the Diophantine equation\n\n\[ \n{y}^{2} = 5{x}^{4} + a \n\]\n\n(1) For \( a = 1 \), the only integral solutions are \( \left( {x, y}\right) = \left( {0, \pm 1}\right) \) and \( \left( {\pm 2, \pm 9}\right) \).\n\n(2) For \( a = - 1 \), the only integral solutions are \( \left( {x, y}\right) =... | Proof. If we write the equations as \( {y}^{2} - 5{x}^{4} = \varepsilon {b}^{2} \) with \( \varepsilon = \pm 1 \) and \( b = 1 \) or \( b = 2 \), we see that we can apply the solution to the Pell equation \( {y}^{2} - 5{X}^{2} = \pm 1 \) or \( \pm 4 \). The fundamental unit of the order \( \mathbb{Z}\left\lbrack \sqrt{... | Yes |
Lemma 6.8.8. The only integral solutions to the Diophantine equation \( {y}^{2} = 8{x}^{4} + 1 \) are \( \left( {x, y}\right) = \left( {0, \pm 1}\right) \) and \( \left( {\pm 1, \pm 3}\right) \) . | Proof. We may assume \( y > 0 \) . Since \( y \) is odd we can write \( y = {2s} + 1 \) , so that \( 2{x}^{4} = s\left( {s + 1}\right) \) . If \( s \) is even there exist coprime integers \( u \) and \( v \) such that \( s = 2{u}^{4} \) and \( s + 1 = {v}^{4} \), so that \( 1 + 2{u}^{4} = {v}^{4} \) . By Proposition 6.... | Yes |
Proposition 6.8.9. The only integral solutions to the Diophantine equation \( x\left( {x + 1}\right) \left( {{2x} + 1}\right) = 6{y}^{2} \) with \( y \neq 0 \) and \( x \) even are \( \left( {x, y}\right) = \left( {{24}, \pm {70}}\right) \) . | Proof. Clearly \( x \) is nonnegative. Since \( x \) is even and \( x, x + 1 \), and \( {2x} + 1 \) are pairwise coprime, the equation implies that the odd numbers \( x + 1 \) and \( {2x} + 1 \) are either squares or triples of squares. It follows that \( x + 1 ≢ 2 \) \( \left( {\;\operatorname{mod}\;3}\right) \) and \... | Yes |
Proposition 6.8.10. (1) \( {M}_{-n} = {M}_{n},{G}_{-n} = - {G}_{n},{M}_{n}^{2} - 3{G}_{n}^{2} = 1 \) ,\n\n\[ \n{M}_{m + n} = 3{G}_{m}{G}_{n} + {M}_{m}{M}_{n},{G}_{m + n} = {G}_{m}{M}_{n} + {G}_{n}{M}_{m},{M}_{2m} = 2{M}_{m}^{2} - 1, \n\]\n\( {G}_{2m} = 2{G}_{m}{M}_{m} \)\n\n(2) \( \gcd \left( {{M}_{n},{G}_{n}}\right) =... | Proof. Essentially identical to the proof of Proposition 6.8.1, this time using \( {M}_{n} + {G}_{n}\sqrt{3} = {\alpha }^{n} \) . | No |
Lemma 6.8.11. Assume that \( n \) is even. Then \( {M}_{n} \) is odd, \( 5 \nmid {M}_{n},\left( \frac{5}{{M}_{n}}\right) = 1 \) if and only if \( 3 \mid n \) and \( \left( \frac{-2}{{M}_{n}}\right) = 1 \) if and only if \( 4 \mid n \) . | Proof. Write \( n = {2m} \) . Since \( {M}_{2m} = 2{M}_{m}^{2} - 1 \), it is clear that \( {M}_{n} \) is odd, and since \( {M}_{m}^{2} \equiv 0 \) or \( \pm 1 \) modulo \( 5,{M}_{n} \equiv - 1,1 \), or 2 modulo 5 and in particular \( 5 \nmid {M}_{n} \) . Since \( {M}_{n} \) satisfies a linear recurrence with integral c... | Yes |
Proposition 6.8.12 (Ma). If \( n \geq 0 \), then \( {M}_{n} \) has the form \( 4{x}^{2} + 3 \) if and only if \( n = 2 \), so \( {M}_{n} = 7 \) . | Proof. Assume that \( {M}_{n} = 4{x}^{2} + 3 \), so that \( {M}_{n} \equiv 3 \) or 7 modulo 8 . Since the period of \( {M}_{n} \) modulo 8 is \( \left( {1,2,7,2}\right) \), this implies that \( n \equiv 2\left( {\;\operatorname{mod}\;4}\right) \), or equivalently, that \( n \equiv \pm 2\left( {\;\operatorname{mod}\;8}\... | Yes |
Theorem 6.8.13. The only integral solutions with \( y \neq 0 \) to the Diophantine equation \( x\left( {x + 1}\right) \left( {{2x} + 1}\right) = 6{y}^{2} \) are \( \left( {x, y}\right) = \left( {1, \pm 1}\right) \) and \( \left( {x, y}\right) = \left( {{24}, \pm {70}}\right) \) . | Proof. The case that \( x \) is even has been proved in Proposition 6.8.9. So assume that \( x \) is odd. As in the even case, since \( x, x + 1 \), and \( {2x} + 1 \) are pairwise coprime, \( x \) is either a square or three times a square, so \( x ≢ 2\left( {\;\operatorname{mod}\;3}\right) \) . Since \( x + 1 \) is e... | Yes |
Corollary 6.9.2. The general solution of \( {x}^{2} + {y}^{2} = {z}^{2} \) with \( \gcd \left( {x, y}\right) = 1 \) and \( x \) odd is\n\n\[ x = {s}^{2} - {t}^{2},\;y = {2st},\;z = \pm \left( {{s}^{2} + {t}^{2}}\right) ,\]\n\nand the general solution of \( {x}^{2} - {y}^{2} = {z}^{2} \) with \( \gcd \left( {x, y}\right... | Proof. Immediate from the above proposition and the fact that two out of three sign changes can be included in the exchange of \( s \) and \( t \) or in the exchange of \( s \) with \( - s \) . | No |
Proposition 6.9.3. The following three conditions are equivalent:\n\n(1) There exist three \( p \) -adic units \( \alpha ,\beta \), and \( \gamma \) such that \( {\alpha }^{p} + {\beta }^{p} = {\gamma }^{p} \) (in other words, FLT I is soluble p-adically).\n\n(2) There exist three integers \( a, b, c \) in \( \mathbb{Z... | Proof. From the binomial theorem it is clear that if \( u \equiv 1\left( {{\;\operatorname{mod}\;p}{\mathbb{Z}}_{p}}\right) \) then \( {u}^{p} \equiv 1\left( {{\;\operatorname{mod}\;{p}^{2}}{\mathbb{Z}}_{p}}\right) \) . Thus if \( u \equiv v\left( {{\;\operatorname{mod}\;p}{\mathbb{Z}}_{p}}\right) \) and \( u \) and \(... | Yes |
Corollary 6.9.4. FLT I cannot be proved by congruence conditions (i.e., p-adically) if and only if condition (3) of the proposition is satisfied for some a such that \( 1 \leq a \leq \left( {p - 1}\right) /2 \) . | Proof. Indeed, condition (3) is invariant when we change \( a \) modulo \( p \), and also under the change \( a \mapsto p - 1 - a \), so the result is clear. | No |
Corollary 6.9.5. If for all \( a \in \mathbb{Z} \) such that \( 1 \leq a \leq \left( {p - 1}\right) /2 \) we have \( {\left( a + 1\right) }^{p} - {a}^{p} - 1 ≢ 0\left( {\;\operatorname{mod}\;{p}^{2}}\right) \), then the first case of FLT is true for \( p \) . | Proof. Indeed, if \( {a}^{p} + {b}^{p} = {c}^{p} \) with \( p \nmid {abc} \) then condition (2) of the proposition is satisfied; hence by (3), as above there exists \( a \) such that \( 1 \leq \) \( a \leq \left( {p - 1}\right) /2 \) with \( {\left( a + 1\right) }^{p} - {a}^{p} - 1 \equiv 0\left( {\;\operatorname{mod}\... | Yes |
Corollary 6.9.7. Let \( p > 2 \) be an odd prime, and assume that \( q = {2p} + 1 \) is also a prime. Then FLT I is valid; in other words, if \( {x}^{p} + {y}^{p} + {z}^{p} = 0 \) then \( p \mid {xyz} \) . | Proof. Since for \( k = 2 \) we have \( \left( {{k}^{k} - 1}\right) R\left( {{X}^{k} - 1,{\left( X + 1\right) }^{k} - 1}\right) = - {3}^{2} \) , the condition of the proposition is \( q \neq 3 \), which is always true. | No |
Corollary 6.9.10 (Wieferich). If FLT I for a prime exponent \( p \geq 3 \) has a nonzero solution then \( {2}^{p - 1} \equiv 1\left( {\;\operatorname{mod}\;{p}^{2}}\right) \) . | Proof. Indeed, if \( {x}^{p} + {y}^{p} = {z}^{p} \) with \( p \nmid {xyz} \), then exactly one of \( x, y \) , and \( z \) is even. We may thus assume that \( 2 \mid y \), so the result follows from the theorem. | No |
Proposition 6.9.11. For every prime number \( \ell \) there exist nonzero elements \( \alpha ,\beta \), and \( \gamma \) in \( {\mathbb{Z}}_{\ell } \) such that \( {\alpha \beta \gamma } \equiv 0\left( {\;\operatorname{mod}\;\ell }\right) \) and \( {\alpha }^{p} + {\beta }^{p} = {\gamma }^{p} \) . | Proof. Set \( F\left( X\right) = {X}^{p} + {\ell }^{p} - 1 \) . Assume first that \( \ell \neq p \) . Then \( F\left( X\right) \equiv \) \( \left( {X - 1}\right) \left( {{X}^{p - 1} + \cdots + 1}\right) \left( {\;\operatorname{mod}\;\ell }\right) \) . Since \( \ell \neq p \), it follows that 1 is a simple root of \( F\... | Yes |
Lemma 6.9.12. Let \( p \) be a prime number, let \( \zeta = {\zeta }_{p} \) be a primitive pth root of unity, let \( K = \mathbb{Q}\left( \zeta \right) \), and let \( \pi = 1 - \zeta \) generate the prime ideal \( \mathfrak{p} \) of \( {\mathbb{Z}}_{K} \) such that \( p{\mathbb{Z}}_{K} = {\mathfrak{p}}^{p - 1} \) . Let... | Proof. Assume first that \( {\alpha }_{0}^{p} \equiv \beta \left( {\;\operatorname{mod}\;{\mathfrak{p}}^{p + 1}}\right) \) . Since the absolute ramification index \( e = e\left( {\mathfrak{p}/p}\right) \) is equal to \( p - 1 \), Lemma 4.1.41 with \( r = 2 \) tells us that there exists a \( \mathfrak{p} \) -adic unit \... | Yes |
Corollary 6.9.13. Let \( p \) be a regular prime, and let \( \varepsilon \) be a unit of \( K \) such that the congruence \( \varepsilon \equiv {\alpha }^{p}\left( {\;\operatorname{mod}\;{\pi }^{p}}\right) \) has a solution in \( {\mathbb{Z}}_{K} \) . Then \( \varepsilon = {u}^{p} \) for some \( u \in {\mathbb{Z}}_{K} ... | Proof. Assume the contrary, and consider the extension \( L = K\left( {\varepsilon }^{1/p}\right) \) . Since \( \varepsilon \) is not a \( p \) th power and \( \zeta \in K \) it follows that \( L/K \) is a cyclic extension of degree \( p \) (the simplest case of a Kummer extension). The relative ideal discriminant of t... | Yes |
Lemma 6.10.1. Let \( S\left( X\right) = \mathop{\sum }\limits_{{k \geq 0}}{s}_{k}{X}^{k} \) be a power series with integral coefficients such that \( {s}_{0} = 1 \) and not identically equal to 1, let \( d \geq 2 \) be an integer, and write \( S{\left( X\right) }^{1/d} = \mathop{\sum }\limits_{{k \geq 0}}{a}_{k}{X}^{k}... | Proof. (1). Set \( g\left( X\right) = \mathop{\sum }\limits_{{k \geq 1}}{s}_{k}{X}^{k} \) and write \( g{\left( X\right) }^{j} = \mathop{\sum }\limits_{{k \geq j}}{g}_{j, k}{X}^{k} \) . We have\n\n\[ S{\left( X\right) }^{1/d} = {\left( 1 + g\left( X\right) \right) }^{1/d} = 1 + \mathop{\sum }\limits_{{j \geq 1}}\left( ... | Yes |
Lemma 6.10.2. Assume that for some integer \( d \geq 1 \) and real numbers \( a \) , \( b \), and \( r \) we have the inequality \( {\left( a - r\right) }^{d} < b < {\left( a + r\right) }^{d} \) . Then we have \( \left| {\operatorname{sign}\left( r\right) {b}^{1/d} - a}\right| < \left| r\right| \) . | Proof. If \( d \) is odd, then \( a - r < {b}^{1/d} < a + r \) (where from now on, \( {b}^{1/d} \) denotes the unique \( d \) th root of \( b \) when \( d \) is odd), so that \( r > 0 \) and \( \left| {{b}^{1/d} - a}\right| < r \) . If \( d \) is even then \( b > 0 \) and \( \left| {a - r}\right| < {b}^{1/d} < \left| {... | Yes |
Proposition 6.10.3. Let \( f\left( X\right) = \mathop{\sum }\limits_{{0 \leq i \leq n}}{f}_{i}{X}^{i} \in \mathbb{Z}\left\lbrack X\right\rbrack \) be a monic polynomial of degree \( n \), let \( r \geq 2 \) be an integer, set \( d = \gcd \left( {r, n}\right), m = n/d \), and let \( h\left( X\right) \) be the polynomial... | Proof. By definition of \( h\left( x\right) \) we formally have \( f{\left( X\right) }^{1/d} = h\left( X\right) + O\left( {1/X}\right) \) ; hence for \( a = \pm 1 \) we obtain \( {g}_{a}\left( X\right) = \left( {{da}/{D}_{m}}\right) {X}^{n - m} + O\left( {X}^{n - m - 1}\right) \), so that the degree of \( {g}_{a}\left(... | Yes |
Corollary 6.10.4. (1) The only integer solutions to \( {y}^{2} = {x}^{4} + {x}^{3} + {x}^{2} + x + 1 \) are \( \left( {x, y}\right) = \left( {-1, \pm 1}\right) ,\left( {0, \pm 1}\right) \), and \( \left( {3, \pm {11}}\right) \) . | Proof. For (1), we easily find that \( L = - 1 \) and \( U = 3 \), and the second condition is satisfied with \( {k}_{0} = 3 \) . Thus we need only look at \( - 1 \leq x \leq 3 \) to prove the corollary, which is immediate. | No |
Theorem 6.11.2 (Mihăilescu). Let \( p \) and \( q \) be distinct primes, and let \( x \) and \( y \) be nonzero integers such that \( {x}^{p} - {y}^{q} = 1 \) . Then \( p = 2, q = 3, x = \pm 3 \) , and \( y = 2 \) . | Note that we have already proved this theorem for \( q = 2 \) (see Proposition 6.7.12). We will prove it for \( p = 2 \) below (see Theorem 6.11.8).\n\nSince we can write \( {y}^{q} = \left( {x - 1}\right) \left( {\left( {{x}^{p} - 1}\right) /\left( {x - 1}\right) }\right) \), we can expect as usual that each factor on... | No |
Lemma 6.11.3. Let \( p \) be prime, let \( x \in \mathbb{Z} \) be such that \( x \neq 1 \), and set \( {r}_{p}\left( x\right) = \left( {{x}^{p} - 1}\right) /\left( {x - 1}\right) \n\n(1) If \( p \) divides one of the numbers \( \left( {x - 1}\right) \) and \( {r}_{p}\left( x\right) \) it divides both.\n\n(2) If \( d = ... | Proof. Expanding \( {r}_{p}\left( x\right) = \left( {{\left( x - 1 + 1\right) }^{p} - 1}\right) /\left( {x - 1}\right) \) by the binomial theorem we can write\n\n\[ \n{r}_{p}\left( x\right) = {\left( x - 1\right) }^{p - 1} + p + \left( {x - 1}\right) \mathop{\sum }\limits_{{k = 1}}^{{p - 2}}\left( \begin{matrix} p \\ k... | Yes |
Corollary 6.11.4. Let \( \left( {x, y, p, q}\right) \) be such that \( {x}^{p} - {y}^{q} = 1 \) . Then \( \gcd \left( {{r}_{p}\left( x\right), x - }\right. \) 1) \( = p \) if \( p \mid y \) and \( \gcd \left( {{r}_{p}\left( x\right), x - 1}\right) = 1 \) otherwise. | Proof. Since \( {y}^{q} = \left( {x - 1}\right) {r}_{p}\left( x\right) \) it follows that \( p \mid y \) if and only if \( p \) divides either \( x - 1 \) or \( {r}_{p}\left( x\right) \), hence by the above lemma, if and only if \( \gcd \left( {{r}_{p}\left( x\right), x - }\right. \) 1) \( = p \) . | Yes |
Corollary 6.11.6. If \( x \) and \( y \) are nonzero integers and \( p \) and \( q \) are odd primes such that \( {x}^{p} - {y}^{q} = 1 \), there exist nonzero integers \( a \) and \( b \) and positive integers \( u \) and \( v \) with \( q \nmid u \) and \( p \nmid v \) such that\n\n\[ x = {qbu}, x - 1 = {p}^{q - 1}{a... | Proof. Since \( p \mid y \), by the above corollary we have \( \gcd \left( {{r}_{p}\left( x\right), x - 1}\right) = \) \( p \), so by Lemma 6.11.3 (3) we have \( {r}_{p}\left( x\right) \equiv p\left( {\;\operatorname{mod}\;{p}^{2}}\right) \), and in particular \( {v}_{p}\left( {{r}_{p}\left( x\right) }\right) = 1 \) . ... | Yes |
Theorem 6.11.8 (Ko Chao). If \( q \) is prime there are no nonzero solutions to the equation \( {x}^{2} - {y}^{q} = 1 \) apart from \( \left( {x, y}\right) = \left( {\pm 3,2}\right) \) for \( q = 3 \) . | Proof. We may clearly assume \( q \neq 2 \) . Furthermore, we have proved in Corollary 6.4.32 that there are no solutions for \( q = 3 \) apart from the given ones. We may thus assume that \( q \geq 5 \) . By Nagell’s result, we know that \( x \) is odd, and we may of course assume \( x > 0 \) . Choose \( \varepsilon =... | Yes |
Lemma 6.11.9. Set \( w\left( j\right) = j + {v}_{q}\left( {j!}\right) \). Then \( {q}^{w\left( j\right) }\left( \begin{matrix} p/q \\ j \end{matrix}\right) \) is an integer not divisible by \( q \), and \( w\left( j\right) \) is a strictly increasing function of \( j \). | Proof. By Lemma 4.2.8 (2) (a) and (c), we know that \( \left( \begin{matrix} p/q \\ j \end{matrix}\right) \) is an \( \ell \) -adic integer for \( \ell \neq q \), and that its \( q \) -adic valuation is equal to \( - w\left( j\right) \), proving the first assertion. Since \( w\left( {j + 1}\right) - w\left( j\right) = ... | Yes |
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