ON THE EQUATION X n 1 = BZ n. 1. Introduction

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1 ON THE EQUATION X n 1 = BZ n PREDA MIHĂILESCU Abstract. We consider the Diophantine equation X n 1 = BZ n ; here B Z is understood as a parameter. We give new conditions for existence of solutions to this equation. These reduce the possible solutions to at most one explicite pair X, Z) for given n, B). The proof uses recent results on the diagonal Nagell Ljunggren equation X p 1 X 1 = pe Y p, e {0, 1} together with a new estimate approach, specific for the equation under consideration. 1) Let B, n N >1 be such that 1. Introduction ϕ B) := ϕrad B)) and n, ϕ B)) = 1. Here rad B) is the radical of B and the condition implies that B has no prime factors t 1 mod n. In particular, none of its prime factors split completely in the n th cyclotomic field. We consider the binary Thue equation 2) X n 1 = B Z n, where solutions with Z { 1, 0, 1} are considered to be trivial. This equation is encountered as a particlar case of binary Thue equations of the type ax n by n = c, see [BGMP]. Current results on 2) are restricted to values of B which are built up from small primes p 13. It is natural to consider the case when the exponent n is a prime, since existence of solutions X, Z) for composite n imply the existence of some solutions with n prime, by raising X, Z to a power. We prove the following: Theorem 1. Let n be a prime and B > 1 an integer with φ B), n) = 1. Suppose that the equation 2) has a non trivial integer solution different from n = 3 and X, Z; B) = 18, 7; 17). Let X mod n u, 0 u < p and e = 0 if u 1 and e = 1 otherwise. Then: A. n > B. X 1 = ±B/n e and B < n n. C. If u ±1 then 2 n 3 n 1 mod n 2, and r n 1 1 mod n 2 for all r XX 2 1). If u = ±1, then n h + n, the class number of the totally real subfield of the n th cyclotomic extension. D. Furthermore, if u = 1 then B/2) n 1 ±1 mod n 2. Date: Version 3.0 November 2, The research was completed while the author is holding a research chair sponsored by the Volkswagen Stiftung. 1

2 2 PREDA MIHĂILESCU The proof of this Theorem is given below on base of results from [Mi1]. 2. Proof of Theorem 1 We relate 2) to the diagonal Nagell Ljunggren equation { X n 1 X 1 = ne Y n 0 if X 1 mod n 3), e =, 1 otherwise. ) X Note that D = n 1 n X 1, X 1 and D = n exactly when X 1 mod n. Indeed, from the expansion D = Xn 1 X 1 = X 1) + 1)n 1 = n + kx 1), X 1 for some k Z, one deduces the claim D n. If D 1, then n X 1) must hold. Conversely, inserting X 1 mod n in the previous expression shows that in this case D = n. The case X 1 mod n corresponds to e = 1; it follows that any integer solution of 2) induces one of 3), both for e = 0 and for e = 1. Indeed, if ζ C is a primitive n th root of unity, then the numbers α c = X ζ c 1 ζ c ) Z[ζ], by definition of e and α e c, n) = 1. Since for distinct c, d 0 mod n we have 1 ζ c ) e α d 1 ζ d ) e α c = ζ d ζ c, it follows that α c, α d ) 1 ζ) and in view of α c, n) = 1, it follows that the α c are coprime. But p 1 α c = Xn 1 X n 1 n e X 1) n e X 1) and thus the rational primes r are all split in the n th cyclotomic extension. For such a prime r, if X, Z; B) is a solution of 2), it follows from 1) that r, B) = 1 and thus r Z; furthermore, the equation 2) implies that r in Z n, i > 0 and thus r in X n 1 n e X 1). This holds for all primes r X n 1 n e X 1) and it follows that 3) is verified for Y = r ri X, Y Z, where the product ranges over all the primes r n 1 n e X 1). If X, Z) is a solution of 2) for the prime n and Z = C Y with Y as above, then we proved that 4) 5) X n 1 n e = Y n and X 1) X 1 = B C n /n e. Conversely, if X, Y ) is a solution of 3), then X, Y ; n e X 1)) is a solution of 2). For instance, the particular solution X, Y ; B) = 18, 7; 17) of 2) stems from 6) = 73, which is supposed to be the only non trivial solution of 3). It is proved in [Mi1] that 3) has no solutions for n except 6). For values of n > , either n h + n or 2 n 1 3 n 1 1 mod n 2 must hold, if there is a non trivial solution. This confirms the respective claims of the Theorem. In the sequel we shall show that the only possible solutions are X = B/n e ± 1. Remark 1. Note that if X, Z) verify 2), then X, Z) is a solution of Un +1 U+1 = BZ n, so the results apply also to the equation: 7) U n + 1 = BZ n.

3 ON THE EQUATION X n 1 = BZ n 3 The points A. and C. of Theorem 1 are proved in [Mi1], in connection with 3), so we may assume in particular that n > Let X, Z) be an integer solution of 2) verifying 1) for a prime n > and s = ±1 be the sign of X. The point D. is known in the literature see e.g. [BGMP] but we prove it briefly here. If X 1 mod n, then BZ n = X n 1 2 mod n 2 and thus B/2) n 1 1/Z) nn 1) 1 mod n 2, which implies the claim. We shall use the following Theorem from [Mi1]: Theorem 2. Suppose that X, Y are integers verifying 3) with n 17 being a prime. Let u = X mod n). Then there is a D R + such that X < D. The values of D in the various cases of the equation are the following: 4 ) n+2 n if u { 1, 0, 1} 8) D = 4n) n 1 2 if u = 0, 4 n 2) n otherwise. By comparing the bounds 8) with the second equation in 4) it follows that C < 2n. In particular, the primes dividing C do not split completely in Q[ζ n ] since a prime splitting in this field has the form r = 2kn + 1 > 2n. Remark 2. Note that C < 2n implies a fortiori that for all primes r C, r 2 1 mod n. Thus the 2 - part of the group < r mod n > has at least for elements. The rest of this paper gives the proof that C = ±1. If C = ±1, then X 1 = ±B/n e, as stated in point B. of the Theorem and X is a solution of 3). The bounds on X in 8) imply B < n n, the second claim of B. We begin with a combinatorial Lemma 1. Let p be an odd prime, k N with 1 < k < log 2 p) and P = {1, 2,..., p 1}. If S = {a 1, a 2,..., a k } P is a set of distinct numbers and A = p 1/k, then there are k numbers b i Z, i = 1, 2,..., k, not all zero, with 0 b i A and such that k a i b i 0 mod p. i=1 Proof. Let T = {1, 2,..., A} P. Consider the functional f : Ti k Z/p Z) given by k f t) t i a i mod p, with t = t 1, t 2,...,t k ) T k. i=1 Since T k > p, by the pigeon hole principle there are two vectors t t such that f t) f t ) mod p. Let b i = t i t i ; by construction, 0 b i A and not all b i are zero, since t t. The choice of these vectors implies k i=1 a ib i 0 mod p, as claimed. Assume that X, Z; B) is an integer solution of 2) for the exponent n and X is given by 4). It follows that 3) holds for Y = Z/C. We shall use a series of results on Stickelberger elements and their relations to solutions of 2). All these results are deduced in [Mi1] and they shall only be mentioned here without proof.

4 4 PREDA MIHĂILESCU 2.1. Auxiliary facts on the Stickelberger module. Let ζ be a primitive n th root of unity, K = Qζ) the n th cyclotomic field and G = Gal K/Q) the Galois group. The automorphisms σ a G are given by ζ ζ a, a = 1, 2,...,p 1; complex conjugation is denoted by j Z[G]. The Stickelberger module is I = ϑ Z[G] Z[G], where ϑ = 1 n 1 n c σ 1 c is the Stickelberger element. For Θ I we have the relation Θ + jθ = ςθ) N, where ςθ) Z is the relative weight of Θ. The Fueter elements are ψ n = 1 + σ n σ n+1 )ϑ, 1 n < p 1, They generate I as a Z - module of rank p + 1)/2) and ςψ n ) = 1 for all n. The Fermat quotient map I Z/n Z) is given by and enjoys the properties: 9) p 1 ϕ : Θ = n c σc 1 ζ Θ = ζ ϕθ), 1 + ζ) Θ = ζ ϕθ)/2 ) Θ = ζ ϕθ)/2 p 1 n c /c mod p ) ςθ)/2 1 n). n The last relation holds up to a sign which depends on the embedding of ζ. For a fixed embedding, we let ν = 1 ) n n be defined by ) Θ = ζ ϕθ)/2 ν. Note that for Θ I with ςθ) = 2 we have ) 2Θ = ζ ϕθ) n 2 for any embedding. Let α = X ζ 1 ζ) Z[ζ], as before, and define c e x 1/X 1) mod n if e = 0 and c x = 0 if e = 1. For any θ I, there is a Jacobi integer β[θ] Z[ζ] such that β[θ] n = ζ cx α) θ, normed by β[θ] 1 mod ) 2 [Mi1]. The definition of ςθ) implies that 10) β[θ] β[θ] = N K/Q α) ςθ) = Y ςθ) The proof of C = ±1. We assume there is a prime r C with r in C. Let R Z[ζ] be a prime ideal lying above r, let dr) G be its decomposition group and d be its size. The proof uses local approximations based on expansions in binomial series. We next derive the relations which lead to these expansions. Let D = B C n with C defined by 4) and note that p, D) = 1 in this case, so 1/) is congruent to an algebraic integer modulo D Z[ζ]. We have for any θ I, β[θ] n = ζ cx α) θ = ζ cx ) 1 e ) θ 1 + X 1 ) θ 11) Let θ 0 I be such that ϕθ 0 ) = 0 and ςθ 0 ) = 2, so by the relation 9) and the remark after it, ) 2θ0 = n 2 ζ ϕθ0) = n 2. Such an element can be easily constructed: indeed, let a = ϕψ 2 ), b = ϕψ 3 ), with ψ i being the Fueter elements above. If a = 0 or b = 0, then one takes the double of the respective element. Otherwise, set θ 0 = σ b ψ 2 + σ p a ψ 3 : it follows from the definition of the Fermat quotient that ϕθ 0 ) = 0. Finally, we let θ = 2θ 0 and note that from 10) we have Y 2n = β[θ 0 ] n β[θ 0 ] n here we use again the fact that θ is

5 ON THE EQUATION X n 1 = BZ n 5 the double of a group ring element. Thus β[θ] n = Y 2n β[θ 0 ]/β[θ 0 ]. Using 11) and the previous observations, we find: β[θ] n = Y 2n ζ cx ) 1 e) θ 0 jθ X 1 ) θ0 jθ 0 = ζ 2cx+1)ϕθ0) 1 + D/)) θ0 jθ0 = 1 + D/)) θ0 jθ0. Thus 12) β[θ] n = Y 2n 1 + X 1 ) θ0 jθ 0 = Y 2n 1 + D ) θ0 jθ 0, and for any prime ideal R D there is a κ = κ R θ) Z/n Z) such that 13) β[θ] ζ κ Y 2 mod R and β[θ] ζ κ Y 2 mod R. The second relation follows from 10) and does not require that R be fixed under complex conjugation. Note that the two relations 12) and 13) lead to a binomial series expansion for β[θ]. The typical obstruction in this case is the presence of the unknown exponent κ; it can be overcome here by using the decomposition group. We claim that there are σ 1, σ 2 dr) such that σ 1 j σ 2. If dr) is not a pure 2 - group, it contains a non trivial subgroup of odd order, which fixes the claim. If dr) is a 2 - group and thus j dr), it has, by Remark 2, at least 4 elements; this implies our claim in this case too. Let σ i ζ) = ζ ci, c i Z/p Z). It follows from Lemma 1 that there are a, b Z with a, b p 1/2 and ac 1 + bc 2 0 mod p; after eventually changing the signs of a and b, we may assume that a > 0. Let µ = aσ 1 + bσ 2 Z[dr)] Z[G]. By construction, µζ) = 1. Since K/Q is abelian and all the primes R r) have the decomposition group dr), µ has the following stronger property: let R r), S G be a set of representatives of G/dr) and γ Z[ζ] be such that γ ζ cσ mod σr) for all σ S. Then µγ) 1 mod rz[ζ], as follows directly from µζ) 1 mod σr). With θ like above, we now let 14) Θ = Θ 0 = Θ/2 I. { aσ 1 θ + bσ 2 θ if b 0 aσ 1 θ bσ 2 jθ) otherwise. Let h = 2a + b ) = ςθ 0 ) < 4n 1/2. It follows from 13) that in both cases we have β[θ] µζ κ ) Y h mod R and thus 15) β[θ] Y h mod rz[ζ]. Let Θ 0 = p 1 n cσ c ; for any prime R r), the binomial series of the n th root of the right hand side in 12) converges in the R - adic valuation and its sum is equal to β[θ] up to a possible n th root of unity ζ c. For any N > 0, we have R inn D N

6 6 PREDA MIHĂILESCU and thus β[θ] p 1 N 1 ζ c Y h p 1 N 1 k=0 k=0 nc /n k nc /n ) ) ) k D k c ) ) ) k D c mod R inn. Here we make use of the choice of Θ: comparing 15) with the product above, it follows that ζ c = 1 for all primes R r). Thus the congruence above holds modulo r inn. We transform the produc into a sum and obtain a uniform expansion in powers of r which is Galois covariant: ) N 1 σ d β[θ]) = Y 1 h + σ d γ k [Θ]) D k + Or inn ), k=1 for any N > 0 and γ k [Θ] K. It can be shown by using the results and methods in [Mi1] that for Ek) = k n+1 n 1, the numbers Γ k = n Ek) γ k Z[ζ] and Γ k < n 2k ; the previous equation becomes ) N 1 σ d β[θ]) = Y 1 h σ d Γ k [Θ]) 16) + D k + Or inn ). n Ek) k=1 We let J G be a set of A = 4h logn) < 16 n logn) < n 3/5 automorphisms and consider the linear combination = σ J λσ)β[σ Θ] where λσ) Z[ζ] are subject to the linear system 17) λσ)γ k σ Θ) = 0, for k = 0, 1,..., A 2 and 18) σ J λσ)γ A 1 σ Θ) 0 mod r. σ J Results from [Mi1] based on estimates for solutions of linear systems, imply that the system has a solution with λσ) Λ = n 2A 1) A ) A. Since β[σ Θ] = Y h < n nh and given the bound h < 4n 1/2, we find that σ J λσ) n nh < AΛn nh < n 2nh, and thus 19) N K/Q ) < n 2nn 1)h. The homogeneous conditions 17) imply 0 mod r A 1)n and it follows from 18) that 0. Hence N K/Q ) 0 mod r A 1)nn 1) > n 4nn 1)h, in contradiction with 19). This shows that C = ±1, which completes the proof of Theorem 1. Acknowledgments: I am grateful to Professor Kálmán Győry for having pointed my attention to this problem and for an emulating exchange on this topic. Thanks go to the anonymous referee for helpful suggestions.

7 ON THE EQUATION X n 1 = BZ n 7 References [BGMP] [BHM] [Mi1] M. A. Bennett, K. Győry, M. Mignotte and Á. Pintér: Binomial Thue equations a nd polynomial powers, Compositio Math ), pp Y. Bugeaud, G. Hanrot and M. Mignotte: Sur l équation diophantienne xn 1 x 1 = yq, III, Proc. London. Math. Soc ), pp P. Mihăilescu: Class Number Conditions for the Diagonal Case of the Equation of Nagell Ljunggren, To appear in the Wolfgang Schmidt - Jubilaeum Volume, Springer Verlag, New York. P. Mihăilescu) Mathematisches Institut der Universität Göttingen address, P. Mihăilescu: preda@uni-math.gwdg.de