Diophantine Equations

Transcription

1 Diophantine Equations Michael E. Pohst Institut für Mathematik Technische Universität Berlin May 24, 2013

2 Mordell s equation y 2 = x 3 + κ is one of the classical diophantine equations. In his famous book Mordell already carries out investigations on determining all integer solutions x, y for given κ Z. Mordell observed that the discriminant of the cubic polynomial t 3 3xt 2y in the variable t is := 108κ. Hence, it suffices to determine all monic cubic polynomials g(t) = t 3 + at 2 + bt + c Z[T ] of discriminant.

3 The case of irreducible polynomials g(t) Any zero ρ in Q generates a cubic extension F = Q(ρ). The discriminant d(g) of the minimal polynomial g(t) of ρ coincides with the discriminant of the equation order Z[ρ]. We get = d F λ 2 for the discriminant d F of F and some λ Z. This yields a usually small list of candidates for F.

4 Example I When do a square and a cube only differ by 1? This means to solve y 2 = x 3 ± 1. In this case we have = 108. For > 0 there are only two totally real cubic fields with discriminants below 108, d F {49, 81}. In each case the quotient /d F is not an integer. For = 108 there is exactly one cubic field F such that /d F is a square, namely F = Q( 3 2) with d F =. Hence, we need to determine all integers of F of index ±1. We note that 1, α = 3 2, α 2 is an integral basis of F.

5 Index form equations For the candidates F of that list we need to test whether o F contains elements α whose minimal polynomials have a discriminant λ 2 d F. Also, the trace tr(α) must be divisible by 3. We can generate F by adjoining an element ρ o F to Q such that a Z-Basis of o F is of the form Let ω 1 = 1, ω 2 = ρ, ω 3 = (ρ 2 + Aρ + B)/N. be the minimal polynomial of ρ. m ρ (t) = t 3 Ut 2 + Vt W

6 Index form equations We can assume that the candidates for α are of the form α = xω 2 + yω 3. The corresponding discriminants satisfy ( d(α) = d(ρ) x N A 21x 2 y + 1 N 2 A 12xy ) 2 N 3 A 03y 3. with A 21 = 2U + 3A, A 12 = 3A 2 + 4Au + U 2 + V, A 03 = A(A + U) 2 + V (A + U) W. We note that d E N 2 = d(ρ). We set z = Nx and need to calculate solutions of the index form equation z 3 + A 21 z 2 y + A 12 zy 2 + A 03 y 3 = ±λn 2. This equation is a so called Thue equation. It has only finitely many solutions (z, y) Z 2.

7 Example of F (x, y) = λ over Z x x 2 y xy y 3 =

8 Example of Mordell s equation The curve y 2 = x has the solutions (x, ±y) { ( 1000, 5) ( 170, 31545) (1271, 55256) (2614, ) ( , ) }

9 Thue equations We discuss the computation of the solutions of an equation of the form F (x, y) = m in x, y Z for given m Z. Here, F (x, y) Z[x, y] is an irreducible homogeneous polynomial of degree n 3. For simplicity s sake we assume that the coefficient of the leading term x n is 1, hence z = x. If α (1),..., α (n) denote the zeros of F (x, 1) in Q then the polynomial F (x, y) splits as follows: n F (x, y) = (x α (j) y). j=1

10 Thue equations We only consider the most difficult case in which F (x, 1) is irreducible in Z[x]. For solutions (x, y) Z 2 we set β (j) = x α (j) y (1 j n). We let β (i) = min 1 j n β(j). In order to make the presentation easier we assume i = 1 in the following.

11 Thue equations Because of β (1) n m we obtain for 1 < j n: β (j) β (j) β (1) β (1) α (j) α (1) y n m α (j) α (1) y /2, if y is large enough, i.e. for y > 2 n m / α (j) α (1). This implies α(1) x y < c 1 y n with c 2 n 1 m 1 = n j=2 α(j) α (1).

12 Thue equations What can we say about about the values α (j) x/y for j > 1? We start with the elementary inequality log(x) < 2 x 1 which holds for x 1 < Then we get log α (j) x/y α (1) α (j) = log 1 α(1) x/y α (1) α (j) < 2 α (1) x/y α (1) α (j), and consequently for sufficiently large y : log x α (j) y < 2 log y.

13 Thue s Theorem Theorem that For every ε > 0 there exists a constant c > 0 such α (j) x y > c (1 j n). y 0.5(n+2)+ε

14 Thue equations The calculation of all solutions became feasible only via Baker s lower estimates for linear forms in logarithms, about 60 years after Thue s result. We let F = Q(α (1) ). By ε i (1 i r) we denote a full set of fundamental units of F. For all non associate elements µ o F of norm ±m and all roots of unity ξ TU F we then need to test, whether β = ξµε a 1 1 εar r is a solution of the Thue equation.

15 Thue equations We set A = max( a 1,..., a r ) and obtain a system of equations a 1 log ε (j) a r log ε (j) β (j) r = log (2 j r 1 + r 2 ) in the variables a 1,..., a r. µ (j) We denote the corresponding matrix of coefficients by Γ and let Γ 1 = (u kj ) with row norm c 2. This gives the following upper bound for A: β (j) A c 2 max log c 3 log y. µ (j) In view of the previous slide we can choose c 3 = 2c 2 for sufficiently large values of y.

16 Example I We recall that r = 1 and calculate a fundamental unit ε 1 = 1 α + α 2 = The absolute values of the differences of disfferent roots all are The corresponding Thue equation is x 3 + 2y 3 = ±1. Any solution β = x + αy is of the form ±ε a 1 1 and we have A = a 1. We get matrices Γ = (log ε 1 ) = (1.3474) and Γ 1 = (1/ log ε 1 ) = (0.7422). Therefore the first three constants are c 1 = 0.84, c 2 = o.742, c 3 =

17 Siegel s identity For 1 < k < l we divide Siegel s identity (α (1) α (k) )β (l) + (α (k) α (l) )β (1) + (α (l) α (1) )β (k) = 0 by the last summand on the left hand side: (α (1) α (k) )β (l) (α (1) α (l) )β (k) 1 = (α(l) α (k) )β (1) (α (1) α (l) )β (k). From this we obtain an upper bound for the linear form in logarithms Λ := log (α (1) α (k) )µ (l) (α (1) α (l) )µ (k) + a ε (l) 1 1 log a r log ε (k) 1 ε (l) r ε (k) r.

18 Siegel s identity Using equations and estimates from above we get Λ < 2 (α (1) α (k) )β (l) (α (1) α (l) )β 1 (k) = 2 (α (l) α (k) )β (1) (α (1) α (l) )β (k) c 4 exp( c 5 A) with α (k) α (l) c 4 = 4c 1 (α (1) α (k) )(α (1) α (l), c 5 = n. c 3 On the other hand, Baker s method produces a lower bound of the form exp( c 6 log A). From these two bounds we derive an upper bound M for A. The latter is quite large, e.g. >

19 Baker s method Let β be a non zero algebraic number with minimal polynomial f (x) = b 0 x m + b 1 x m b m Z[x], i.e. gcd{b 0,..., b m } = 1. In Q[x] we have f (x) = b 0 m j=1 (x β(j) ) with β = β (1). Then h(β) := 1 m log b 0 m max(1, β (j) ) j=1 is called the absolute logarithmic height of β.

20 Theorem Baker/Wüstholz Let α 1,..., α k be non zero algebraic numbers and a 1,..., a k Z. For 1 i k we choose a branch of the complex logarithm such that Λ := a 1 log(α 1 ) a k log(α k )) 0. For 1 i k we set A i = max { h(α i ), } 1 [Q(α i : Q], log(α i ) [Q(α i : Q] Theorem For D := [Q(α 1,..., α k ] : Q] and A := max{a 1,..., a k, e} we have log Λ 18(k + 1)!k k+1 (32D) k+2 A 1 A k log(a)..

21 Example I We calculate constants c 4 = 1.54, c 5 = The height h(ε) becomes and from Baker s Theorem we obtain c 6 := The inequality c 6 log(a) < log(c 4 ) c 5 A is violated for A >

22 Thue equations, Bilu and Hanrot From the system of equations a 1 log ε (j) a r log ε (j) r = log β (j) µ (j) (2 j r 1 + r 2 ) we get a k = r j=1 u kj log x α(j) y µ (j) = (log y ) r j=1 u kj + r j=1 u kj ( log + log α (i) α (j) α(j) x y ). µ (j) α(i) α (j)

23 Thue equations, Bilu and Hanrot We set r δ k := u kj, ν k := j=1 j=1 r u kj log α (i) α (j), c 6 := 2c 1 µ (j) r u kj α (i) α (j), j=1 and the equation for a k yields the inequality δ k log y a k + ν k < c 6 exp( c 5 A). From two such inequalities, say for conjugates k, l, we eliminate log y.

24 A lemma of Davenport The result is an inequality of the form a k ϑ + a l δ < c 7 exp( c 5 A) with ϑ = δ l /δ k, δ = (δ k ν l δ l ν k )/δ k, c 7 = c 6 (δ k + δ l )/δ k. Then the huge bound for A can be drastically reduced. Lemma Let M, B, q be positive integers satisfying 1 q MB, qϑ < 2(MB) 1, qδ > 3/B. Then the previous inequality has no solutions a k, a l with log(mb 2 c 7 ) c 5 A M. Appropriate starting values are the upper bound M for A and B = q can be detected as the denominator of a convergent of the continuous fraction expansion of ϑ.

25 Example of an index form equation The Thue equation x x 2 y xy y 3 = has no solutions.