On equation. Boris Bartolomé. January 25 th, Göttingen Universität & Institut de Mathémathiques de Bordeaux

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1 Göttingen Universität & Institut de Mathémathiques de Bordeaux January 25 th, 2016 January 25 th, / 19

2 Overview January 25 th, / 19

3 Binary Thue: prologue This is a joint work with Preda Mih ilescu. We call binomial Thue equation: A.X n B.Y n = C, where n 3 and A, B and C are non-zero integers. Thue proved in 1909 that, for a xed n, this equation has at most a nite number of solutions in integers (x, y). Currently, even the best numerical bounds on the solutions are too large for numerical resolution. This equation has been totally solved in some particular cases, always with C = ±1. We study equation: which we call Binary Thue., (1) January 25 th, / 19

4 Binary Thue: prologue Previous results: Michael Bennett [Be] proves that when C = ±1 in equation A.X n B.Y n = C, there is at most one solution for xed (A, B; n), Bassó + Bérczes + Györy + Pintér [BBGP] prove that Binary Thue has no solutions for B < 235. Buhler and Harvey recently proved [BH] that condition CF is veried for primes up to We prove that there are no solutions for families of density one of numbers B and n, and we show the form of the potential solution in the other cases. January 25 th, / 19

5 Binary Thue: prologue Let B Z, n N >1, dene ϕ (B) := ϕ(rad (B)), where rad (B) is the radical of B, and assume that: (n, ϕ (B)) = 1. More generally, for a xed B Z we let N (B) = {n N >1 k > 0 such that n ϕ (B) k }. If p is an odd prime, we shall denote by CF the combined condition requiring that I The Vandiver Conjecture holds for p, so the class number h p + of the maximal real subeld of the cyclotomic eld Q[ζ p] is not divisible by p. II The index of irregularity of p is small, namely i r (p) < p 1, so there are i r (p) odd integers k < p such that the Bernoulli number B k 0 mod p. January 25 th, / 19

6 Binary Thue Theorem (Bartolomé, Mih ilescu, 2015) Let n be a prime and B > 1 an integer with (ϕ (B), n) = 1. Suppose that BT has a non trivial integer solution dierent from n = 3 and (X, Z; B) = (18, 7; 17). Let X u mod n, 0 u < n and e = 1 if u = 1 and e = 0 otherwise. Then: 1. n > X 1 = ±B/n e and B < n n. 3. If u {0, 1, n 1}, then condition CF (II) fails for n and 2 n 1 3 n 1 1 mod n 2, and r n 1 1 mod n 2 for all r X (X 2 1). If u {0, 1, n 1}, then Condition CF (I) fails for n. January 25 th, / 19

7 Binary Thue Theorem (Bartolomé, Mih ilescu, 2015) If Equation X n 1 = B.Z n has a solution for a xed B verifying (n, ϕ (B)) = 1, then either n N (B) or there is a prime p coprime to ϕ (B) and an m N (B) such that n = p m. Moreover X m, Z m is a solution of X p 1 = B.Z p for the prime exponent p and thus veries the conditions of the previous theorem. January 25 th, / 19

8 From diagonal to binary Thue Any solution of diagonal leads to a solution of binary Thue. We remind that diagonal is: X n 1 X 1 = ne Y n, e = 0 if X 1 mod n, 1 otherwise. Let (X, Y ) be a solution of diagonal. Then (X, Y ; n e (X 1)) is a solution of binary Thue. For instance, the particular solution (X, Y ; B) = (18, 7; 17) of binary Thue stems from = 73, which is supposed to be the only non trivial solution of diagonal. (2) January 25 th, / 19

9 From binary Thue to diagonal Any solution of X n 1 = B Z n leads to a solution of diagonal. X We prove that any prime p i Z, in the radical of n 1 n e (X 1), is coprime with all of its conjugates in Z[ζ], and they are thus totally split. Such p i are also in the radical of X n 1. Therefore, if (X, Z) is a solution of binary Thue, then there exists C Z such that Z = C Y and X n 1 n e = Y n and (3) (X 1) X 1 = B C n /n e. (4) January 25 th, / 19

10 Bounds to the solutions of diagonal Theorem (Theorem 3 from [Mi]) Suppose that X, Y are integers solutions of diagonal, with n 17 being a prime. Let u = (X mod n). Then there is an E R + such that X < E. The values of E in the various cases of the equation are the following: 4 ( ) n+2 n 3 2 if u { 1, 0, 1} 2 E = (4n) n 1 2 if u = 0, 4 (n 2) n otherwise. January 25 th, / 19

11 of the main Theorem 1 Claims (1) and (3) come directly from [Mi], by applying the link between binary Thue and diagonal. 2 We have already proved that X 1 = B C n /n e. If C = ±1, then X 1 = ±B/n e, as stated in point (2) of the Theorem and X is a solution of diagonal. The bounds on X in Theorem 3 from [Mi] imply B < n n, the second claim of (2). 3 The rest of the presentation consists is proving that C = ±1. It is quite technical. January 25 th, / 19

12 that C = ±1: notation Assume that (X, Z, B) veries Binary Thue. Let α = X ζ (1 ζ) Z[ζ], K = Q(ζ). According to (3), N e K/Q (α) = Y n (where Z = C.Y ). Let σ c G = Gal(K/Q) : ζ ζ c, I be the Stickelberger ideal of Z[G]. The absolute weight of c n cσ c Z[G] is c n c. The Fermat quotient map I Z/(n.Z) given by n 1 n 1 ϕ : θ = n c σ c cn c c=1 c=1 mod n has kernel I f = {θ I : ζ θ = 1}. Let I + f I f the subset of positive elements of I f (that is, elements c n cσ c I f such that for all c, n c 0) Let c n c = ς(θ). n 1, where ς(θ) is the relative weight 2 of θ. Finally, let p be a prime dividing C (then, by (4), p also divides X 1). January 25 th, / 19

13 Galois exponent Suppose we know that (1 + x) is β q for some β Q(ζ p ). Then, β = ξq κ f (x) where f (x) is the sum of the binomial series. κ is unknown. When we act on (1 + x) with an element θ F q [G], the sum of the binomial series f (θx) = (1 + x) θ/q diers from β θ by a q-th root of unity depending on θ. We denote the exponent of this q-th root of unity, the Galois exponent, κ(θ), so we have β θ κ(θ)f (θx) = ξq January 25 th, / 19

14 Lemma For any θ 2.I + f, for any prime ideal P p, there is a κ = κ P (θ) Z/(n Z) such that β[θ] ζ κ Y ς(θ) 2 mod P. (5) We prove that the decomposition group of p contains at least three distinct elements. This allows us to build µ F p [G] small enough (of small absolute weight h/2), such that κ = 0. Using this µ, we then build Θ 2.I + f with also κ = 0 and some additional nice properties. With this, (5) leads to a Galois-covariant P-adic binomial series expansion for β[θ]: ) N 1 β[σθ] = Y (1 h b k [σθ] + (1 ζ) k n k.(x 1)k + O(p inn ), k! k=1 with b k [σθ] Z[ζ]. January 25 th, / 19

15 that C = ±1: linear system We then consider the following linear combination: = σ λ σ β[σ Θ] where λ σ K verify the linear system: λ σ b k [σ Θ] = 0, for k = 0,..., N 1, k N/2 σ σ λ σ b N/2 [σ Θ] = (1 ζ) N/2 n N/2 N/2!. We prove that this system is regular for N < n 1 and admits thus a non-null solution. If we let A = det (b k [σ c Θ]) N 1 k=0;c I 0, then we observe that δ = A. Z[ζ]. We set N = n 3/4 and prove that then, δ 0, but δ 0 mod p in N/2. We then bound N(δ) by above (using Hadamard's inequality) and by below (using the congruence for δ), and nd that ( p n(n 1)N/2 < N(δ) < n 11 2 n3/ n3/ ) n 1 and January 25 th, / 19

16 This double inequality, combined with (1) of the main theorem, leads to log p < 1.64, which means that the only possibilities for p would be p = 2, 3 or 5. Finally, we rule out the case p 5 as follows: in this case, p ±1 mod n and the decomposition group D(p) contains the automorphism σ p. We choose thus µ = 1 + pjσp 1 and proceeding as previously, we build another Θ with the required properties, leading to the double inequality p n(n 1)N/2 N K/Q (δ) < ( n 4(p+1)+3N2 /2 N N/2+1 ) n 1. Letting N = 48, we obtain the inequality 2 n n /24 < 64n 73 n 6 73 log(n)/ log(2), which is false for for n > 695, and a fortiori for n > We obtain a contradiction in this case, too. January 25 th, / 19

17 Solutions to the related Diagonal - equation are upper bounded by relatively small bounds. However, the method we used here is insucient to raise a nal contradiction. We proved that binary Thue equations have no solutions for for families of density one of numbers B and n. We show the form of the potential solution of binary Thue in the other cases. January 25 th, / 19

18 Publications A.Bazso, A.Bérczes, K.Györy and A.Pintér (2010). On the resolution of equations Ax n By n = C in integers x, y and n 3, II. Publicationes Mathematicae Debrecen, vol. 76, pp M. A. Bennet (2001). Rational Approximation To Algebraic Numbers Of Small Height: The Diophantine Equation ax n by n = 1. J. Reine Angew. Math., vol. 535, pp J. P. Buhler and D. Harvey (2011). Irregular primes to 163 million. Mathematics of computation, vol. 80, n 276, pp P. Mih ilescu (2008). Class Number Conditions for the Diagonal Case of the Equation of and. Diophantine Approximation, Springer Verlag, Development in Mathematics, vol. 16, pp January 25 th, / 19

19 Thank you for your attention. January 25 th, / 19