ON THE EQUATION Xn− 1= BZn

arXiv:1412.5798v1 [math.NT] 18 Dec 2014 ON THE EQUATION X n − 1 = BZ n B. BARTOLOMÉ AND P. MIHĂILESCU Abstract. We consider the Diophantine equation X n − 1 = BZ n , where B ∈ Z is understood as a parameter. We prove that if the equation has a solution, then either the Euler totient of the radical, ϕ(rad (B)), has a common divisor with the exponent n, or the exponent is a prime and the solution stems from a solution to the diagonal case of the Nagell–Ljunggren n −1 equation: XX−1 = ne Y n , e ∈ {0, 1}. This allows us to apply recent results on this equation to the binary Thue equation in question. In particular, we can then display parametrized families for which the Thue equation has no solution. The first such family was proved by Bennett in his seminal paper on binary Thue equations [BE]. 1. Introduction Let B, n ∈ N>1 be such that (1) ϕ∗ (B) := ϕ(rad (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. More generally, for fixed B ∈ N>1 we let (2) k N (B) = {n ∈ N>1 | ∃ k > 0 such that n|ϕ∗ (B) }. 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 subfield of the cyclotomic field is not divisible by p. √ (ii) The index of irregularity of p is large, namely i(r) > p − 1, so there are i(r) odd integers k < p such that the Bernoulli number Bk ≡ 0 mod p. It is known from recent computations of Buhler and Harvey [BH] that the condition CF is satisfied by primes up to 163 · 106 . We consider the binary Thue equation (3) Xn − 1 = B · Z n, where solutions with Z ∈ {−1, 0, 1} are considered to be trivial. It is a special case of the general Pillai conjecture (Conjecture 13.17 of [BBM]). This equation is encountered as a particular case of binary Thue equations of the type (4) aX n − bY n = c, see [BGMP]. In a seminal paper [BE], Michael Bennett proves that in the case of c = ±1 there is at most one solution for fixed (a, b; n) and deduces that the Date: December 19, 2014. 1 2 B. BARTOLOMÉ AND P. MIHĂILESCU parametric family (a + 1, a; n) has the only solution (1, 1) for all n. The equation (3) inserts naturally in the family of equations (4), with a = c = ±1. Current results on (3) are restricted to values of B which are built up from small primes p ≤ 13 [G]. If expecting that the equation has no solutions, – possibly with the exception of some isolated examples – it is natural to consider the case when the exponent n is a prime. Of course, the existence of solutions (X, Z) for composite n imply the existence of some solutions with n prime, by raising X, Z to a power. In this vein, 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 (3) has a non trivial integer solution different 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: (A) n > 163 · 106 . (B) X − 1 = ±B/ne and B < nn . (C) If u 6= ±1 then and condition CF (i) holds for n and 2n r n−1 ≡ 3n ≡ 1 ≡ 1 mod n2 , mod n2 and for all r|X(X 2 − 1). If u = ±1, then Condition CF (ii) holds for n. The particular solution n = 3 and (X, Z; B) = (18, 7; 17) is reminiscent of a solution of the diagonal Nagell solution; it is commonly accepted that the existence of non trivial solutions tends to render equations more difficult to solve. Based on Theorem 1 we prove nevertheless: Theorem 2. If the equation (3) has a solution for fixed B, then either n ∈ N (B) or there is a prime p coprime to ϕ∗ (B) and a m ∈ N (B) such that n = p · m. Moreover X m , Y m are a solution of (3) for the prime exponent p and thus verify the conditions in Theorem 1. 2. Preliminary results The proof of this Theorem is based on results from [Mi]. It has been pointed out that the proof of Theorem 3 in [Mi] may require some more detailed explanation in the case of a singular system of equations in the proof of Lemma 15 of [Mi]. We provide here a comprehensive proof which confirms the indications given in [Mi]; this will be done in form of an elementary lemma, in the next section. 2.1. Clarification on the singular case of Theorem 3 in [Mi]. Let m ∈ Z>0 be a positive integer, K a field, V = Km as a K-vector space and let f be an homomorphism from V into V , of rank m − r, with r ≤ m. Let L = Ker (f ) ⊂ V be the kernel; so r = dimK (L). In a complex-vector space inPfinite dimension, we choose the standard Hermitian inner-product (< x, y >= k xk ȳk ). Fix an orthonormal basis EL = (e1 , e2 , · · · , er ) of L, which we complete to an orthonormal basis of V , E = (e1 , e2 , · · · , er , er+1 , · · · , em ), and assume that there exists at least one vector w1 ∈ L which is free of 0-entries over the base E. For (x, y) ∈ V 2 , the Hadamard product is defined by [x, y] = Diag(x) · y ∈ V , where Diag(x) is the diagonal matrix built from the vector x. For any subspace W ⊂ V we define the W -bouquet of L by LW = [ { [w, x] : w ∈ W, x ∈ L } ]K , ON THE EQUATION X n − 1 = BZ n 3 the K-span of all the Hadamard products of elements in W by vectors from L. Lemma 1. Let A2 = [{a1 , a2 }]K a set of two linear independent vectors which are free of 0-entries over e, such that the components of each ai (1 < i ≤ l) are pairwise distinct over E and such that a1 = (1, 1, . . . , 1) over e. Let LA2 be the resulting A2 -bouquet. Then dim(LA2 ) > dim(L). Proof. Obviously, L ⊂ LA2 (as a1 ∈ A). We would like to show that LA2 6= L. We know that the system (w1 , [w1 , a2 ], [w1 , a22 ], · · · , [w1 , am−1 ]) (the notion of 2 power of a vector here is to be understood as an ”Hadamard power”) is free (as it induces a Vandermonde matrix over E, neither w1 nor a2 have any zero among their components and all components of a2 are pairwise distinct). We know that w1 ∈ L, let us assume that [w1 , ai2 ] ∈ L for i ≤ j (we know that j ≤ r ≤ m). Then, [w1 , aj2 ] ∈ L and [w1 , aj+1 / L. However, the Hadamard product of [w1 , aj2 ] ∈ L 2 ] ∈  by a2 , that is [w1 , aj+1 2 ], belongs to LA2 . Thus, dim LA2 > dim L. 2.2. Application of Lemma 1 to the proof of the singular case in Lemma 15 of [Mi]. We apply here the lemma in the first case (that is x 6≡ s mod p, where s ∈ {−1, 0, 1}), the application to the second case being identical. Let all notations be like in Lemma 15 in [Mi]. As in [Mi], we will assume
(p−1)/2 that A = ζ −κac /a a,c=1 (where κac are the Galois exponents) is singular. Let m = (p−1)/2, K = Q(ζp ) and r = rank (A) < (p−1)/2. Without loss of generality, we assume that a regular r-submatrix of A is built with the first r rows and the first r columns. Therefore, the first r rows of A are independent, and we denote by W the sub-space of V = Km generated by the first r row vectors w1 , · · · , wr of A. For a1 = (1, 1, . . . , 1), we let a2 be the vector of V whose components are (η(σc θ))(p−1)/2 and A2 = {a1 , a2 }. Then, according to Lemma 1, there exists at c=1 least one vector ~v ∈ LA2 which is independent on the first r vectors of A. Let S be the (r + 1) × (r + 1) submatrix of A comprising the first r rows and r + 1 columns of A, to which we have added an additional row: the first r + 1 r+1 terms of ~v . Let ~λ′ be the vector solution of A~λ′ = d~′ , where d~′ = (δc,r+1 )c=1 . We ′ ′ know that ~λ 6= ~0, as S is regular and d~ is not the null vector. For 1 ≤ c ≤ r + 1, by Cramer’s rule, λc = SSc , where Sc are the determinants of some minors of S obtained by replacing the c-column by d~′ , and S = det S. Let ~λ ∈ V be a vector whose first r + 1 coordinates are those of ~λ′ and the others ~ are 0. Let (δc,r+1 )m . Then, ~λ verifies: A~λ = d.
Pr+1 c=1 Let δ = c=1 λc · βc + λc · βc . Using Hadamard’s inequality, we bound |Sc | ≤
p−3
p−1 p−3 p−1 4 4 = D and |S| ≤ = D0 . Then, using the fact that the choice 1 2 2 of λc eliminates the first term in the expansion of fc , we find that |S| · |δ| ≤ P (p−1)/2p 2x(p−1)/2p · r+1 . With the same c=1 |Sc ||Rc,0 (x)|, where Rc,0 (x) = fc (x) − x arguments as in [Mi], we deduce: |Sδ| < 2(p − 1)D1 · 1 . |x|(p+1)/2p This inequality holds for all conjugates σc (δ), thus leading to: (p−1)/2 |N (Sδ)| < (2(p − 1)D1 ) · 1 |x| (p−1)(p+1) 4p . 4 B. BARTOLOMÉ AND P. MIHĂILESCU If δ 6= 0, then |N (Sδ)| ≥ 1 and thus |x| ≤ 25−p P(p−1)/2 Sc R0,c , and thus: S · |x|(p−1)/2 − c=1 |x| ≤ X c |Sc |/|S| < (p − 1)D1 < 3  p 2
p2 . If δ = 0, then 0 = Sδ = p−3 2 (p−3)/2 . These bounds are better than the ones in [Mi], and this concludes the clarification. 2.3. Link of (3) with the diagonal Nagell – Ljunggren equation. In order to use the results from [Mi], we relate (3) to the diagonal Nagell – Ljunggren equation ( 0 if X 6≡ 1 mod n, Xn − 1 e n =n Y , e= (5) X −1 1 otherwise.   n −1 , X − 1 n and δ = n exactly when X ≡ 1 mod n. First, note that δ = XX−1 Indeed, from the expansion ((X − 1) + 1)n − 1 Xn − 1 = = n + k(X − 1), X −1 X −1 with k ∈ Z, one deduces the claim δ n. If D 6= 1, then δ = n and thus n|(X − 1) must hold. Conversely, inserting X ≡ 1 mod n in the previous expression shows that in this case δ = n. We first show that any solution of (3) leads to a solution of (5). For this, let QK n n −1 −1 ) = i=1 pi . Obviously, rad ( neX(X−1) ) | rad (X n − 1). Let ζ ∈ rad ( neX(X−1) c X−ζ C be a primitive n−th root of unity. Then the numbers αc = (1−ζ c )e ∈ Z[ζ], by definition of e and (αc , n) = 1. Since for distinct c, d 6≡ 0 mod n we have (1 − ζ d )e · αd − (1 − ζ c )e · αc = ζ c − ζ d , it follows that (αc , αd ) | (1 − ζ) and in Qn−1 n −1 view of (αc , n) = 1, it follows that the αc are coprime. But c=1 αc = neX(X−1) . n −1 , all of its conjugates are coprime and thus Thus, for any rational prime pi | neX(X−1) different: their decomposition group in the n−th cyclotomic extension is trivial and they are all totally split. We remind here that such a prime pi also belongs to rad (X n −1). For such a prime pi , if (X, Z; B) is a solution of (3), it follows from (1) that (pi , B) = 1, and thus pi |Z; furthermore, (3) implies that there exists ji > 0 such n n −1 −1 that piji n ||Z n and thus piji n || neX(X−1) . This holds for all primes pi | rad ( neX(X−1) ). QK ji It follows that (5) is verified for Y = i=1 pi and Y | Z. We have proved that if (X, Z) is a solution of (3) for the prime n, then there exists C ∈ Z such that Z = C · Y with Y as above, and: (6) (7) Xn − 1 = ne (X − 1) X −1 = Yn and B · C n /ne . It follows that any integer solution of (3) induces one of (5). Conversely, if (X, Y ) is a solution of (5), then (X, Y ; ne (X − 1)) is a solution of (3). For instance, the particular solution (X, Y ; B) = (18, 7; 17) of (3) stems from 183 − 1 = 73 , 18 − 1 which is supposed to be the only non trivial solution of (5). ON THE EQUATION X n − 1 = BZ n Remark 1. Note that if (X, Z) verify (3), then (−X, Z) is a solution of BZ n , so the results apply also to the equation: 5 X n +1 X+1 = X n + 1 = BZ n . 2.4. Bounds to the solutions of Equation (5). We shall use the following Theorem from [Mi]: Theorem 3. Suppose that X, Y are integers verifying (5) 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: 
n+2 n−3 2  if u 6∈ {−1, 0, 1}  4· 2 n−1 (8) E= 2 if u = 0 , (4n)   n 4 · (n − 2) otherwise. By comparing the bounds (8) with (7), it follows that |C| < 2n. In particular, the primes dividing C do not split completely in Q[ζn ] – 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, r2 6≡ 1 mod n. Thus the 2 - part of the group < r mod n > has at least four elements. 2.5. A combinatorial lemma. Lemma 2. Let p be an odd prime, k ∈ N with 1 < k < log2 (p) and P = {1, 2, . . . , p − 1}. If S = {a1 , a2 , . . . , ak } ⊂ P is a set of distinct numbers and A = ⌈p1/k ⌉, then there are k numbers bi ∈ Z, i = 1, 2, . . . , k, not all zero, with 0 ≤ |bi | ≤ A and such that k X ai bi ≡ 0 mod p. i=1 Proof. Let T = {1, 2, . . . , A} ⊂ P . Consider the functional f : T k → Z/(p · Z) given by k X ti ai mod p, with ~t = (t1 , t2 , . . . , tk ) ∈ T k . f (~t) ≡ i=1 Since |T k | > p, by the pigeon hole principle there are two vectors ~t 6= ~t′ such that f (~t) ≡ f (~t′ ) mod p. Let bi = ti − t′i ; by construction, 0 ≤ |bi | ≤ A and not all bi Pk are zero, since ~t 6= ~t′ . The choice of these vectors implies i=1 ai bi ≡ 0 mod p, as claimed.  2.6. Auxiliary facts on the Stickelberger module. The following results are deduced in [Mi] and they shall only be mentioned here without proof. Let ζ be a primitive n−th root of unity, K = Q(ζ) the n−th cyclotomic field and G = Gal(K/Q) the Galois group. The automorphisms σa ∈ G are given by ζ 7→ ζ a , a = 1, 2, . . . , n − 1; complex conjugation is denoted by  ∈ Z[G]. The Pn−1 Stickelberger module is I = ϑ · Z[G] ∩ Z[G], where ϑ = n1 c=1 c · σc−1 is the Stickelberger element. For θ ∈ I we have the relation θ + θ = ς(θ) · N, where ς(θ) ∈ Z is called the relative weight of θ. The Fueter elements are n−1 X  (k + 1)c   kc  ψk = (1 + σk − σk+1 ) · ϑ = − · σc−1 , 1 ≤ k ≤ (n − 1)/2. n n c=1 6 B. BARTOLOMÉ AND P. MIHĂILESCU Together with the norm, they generate I as a Z - module (of rank (n + 1)/2) and ς(ψk ) = 1 for all k. The Fuchsian elements are n−1 X  kc  Θk = (k − σk ) · ϑ = · σc−1 , 2 ≤ k ≤ n. n c=1 They also generate I as a Z - module. Note that Θn is the norm, and that we have the following relationship between the Fueter and the Fuchsian elements: ψ1 ψk = Θ2 and = Θk+1 − Θk , k ≥ 2 The Fermat quotient map I → Z/(n · Z), given by ϕ:θ= n−1 X c=1 nc σc−1 7→ n−1 X nc /c mod n, c=1 is a linear functional, with kernel If = {θ ∈ I : ζ θ = 1} (the Fermat module), and enjoys the properties: ζθ = ζ ϕ(θ) , θ (1 + ζ) = ζ ϕ(θ)/2 , (1 − ζ)θ = ζ ϕ(θ)/2 ·  −1 n  ς(θ)/2 n ,
where −1 is the Legendre symbol. n The last relation holds up toq a sign which depends on the embedding of ζ. For
−1 θ ϕ(θ)/2 ν. Note a fixed embedding, we let ν = n n be defined by (1 − ζ) = ζ that for θ ∈ I with ς(θ) = 2 we have (1 − ζ)2θ = ζ ϕ(θ) · n2 for any embedding. We define the following additive maps: ρ0 : θ= and F [G] Pnn−1 c=1 nc σc 7→ Q[ζ] P 7→ c∈P nc 1−ζ c , 7→ Z[ζ] 7 → (1 − ζ) · ρ0 [θ]. Pn−1 The (i)-th moment of an element θ = c=1 nc σc−1 of Z[G] is defined as: ρ: Fn [G] θ φ(i) (θ) = n−1 X n c ci mod n. c=1 Note that φ(1) is the Fermat quotient map: φ(1) = ϕ. X−ζ Let α = (1−ζ) e ∈ Z[ζ], as before, and define cx ≡ 1/(X − 1) mod n if e = 0 and cx = 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 (1 − ζ)2 [Mi]. The definition of ς(θ) implies that (9) β[θ] · β[θ] = NK/Q (α)ς(θ) = Y ς(θ) . ON THE EQUATION X n − 1 = BZ n 7 We have for any θ ∈ I, n θ cx  θ X −1 ) · 1+ 1−ζ 1−e θ cx β[θ] = (ζ α) = (ζ (1 − ζ) (10) Lemma 3. For any θ ∈ 2 · If , for any prime ideal P | D, there is a κ = κP (θ) ∈ Z/(n · Z) such that β[θ] ≡ β[θ] ≡ ζκ · Y ζ −κ ς(θ) 2 ·Y mod P ς(θ) 2 and mod P, the second relation not requiring that P be fixed under complex conjugation. Proof. Let θ0 be an element of If , and let θ = 2θ0 . Note that from (9) we have
n Y ς(θ0 )n = β[θ0 ]n · β[θ0 ]n . Thus β[θ]n = β[θ0 ]2n = Y ς(θ0 )n · β[θ0 ]/β[θ0 ] . Using (10) and the previous observations, we find: (θ0 −θ0 ) 
(θ0 −θ0 ) X −1 · 1+ β[θ]n = Y ς(θ0 )n · ζ cx · (1 − ζ)1−e 1−ζ (11) β[θ]n = Y ς(θ0 )n · ζ (2cx +1)ϕ(θ0 ) · (1 + D/(1 − ζ))(θ0 −θ0 ) (θ0 −θ0 )  D = Y ς(θ0 )n · 1 + . 1−ζ Thus for any prime ideal P | D there is a κ = κP (θ) ∈ Z/(n · Z) such that β[θ] β[θ] ≡ ζ κ · Y ς(θ0 ) mod P and ≡ ζ −κ · Y ς(θ0 ) mod P. The second relation follows from (9) and does not require that P be fixed under complex conjugation. Note that the two relations (10) and (11) lead to a binomial series expansion for β[θ].  We will also use the Voronoi identities – see Lemma 1.0 in [Jha] –, which we remind here for convenience: Lemma 4. Let m be an even integer such that 2 ≤ m ≤ n − 1. Let a be an integer, coprime to n. Then am n−1 X j=1  aj m−1 (am+1 − a)Bm j ≡ n m mod n, where Bm is the m-th Bernoulli number. In particular, for m = n − 1, we get n−1 X j=1  aj n−2 an − a j ≡ n n mod n, which is the Fermat quotient map of the a-th Fuchsian element, ϕ(Θa ). Lemma 5. Let ψk be the k-th Fueter element. Then, there exists no constant A such that, for all k > 1, φ(−1) (ψk ) = A · φ(1) (ψk ). 8 B. BARTOLOMÉ AND P. MIHĂILESCU Proof. By linearity, we can prove it for the Fuchsian elements and use the definition of the Fueter elements as a function of the Fuchsian elements. Let ϕ be the Fermat quotient map. For any integer 1 < k < n − 1, we have: (n − k)n − (n − k) ≡ −k n − n + k
n ≡ −n k n−k + 1 Dividing both terms by n we find: mod n2 mod n2 . ϕ(Θn−k ) = n − (1 + ϕ(Θk )) . (12) With this result in mind, assume that such a constant A exists and remember that φ(1) = ϕ. Using Lemma 4 with m = 2, we find that: k3 − k B2 mod n 2k 2 However, B2 = 1/6, and thus the equation becomes:   1 1 k− mod n, for all 1 < k < n − 1. ϕ(Θk ) ≡ 12A k φ(−1) (Θk ) = A · ϕ(Θk ) ≡ This equation is also true for   1 1 n−k− ϕ(Θn−k ) ≡ 12A n−k mod n, for all 1 < k < n − 1. Equation (12) shows that in this case, A would depend on k and n.  The following two lemmas yield computational details for binomial series developments that we shall use. θ/n  Pn−1 D . Then, Lemma 6. Let θ = c=1 nc σc ∈ Z[G] and f [θ] = 1 + 1−ζ f [θ] = 1 + N X ak [θ] k=1 where ak [θ] = k!nk ρk0 [θ] Dk + O(DN +1 ), +O  n (1 − ζ)k  . Proof. We prove this result by induction on the number of components of θ. First, note that   1 nkc nc /n = · (1 + O(n)). · k k! nk Thus, if θ = nc σc , then: f [θ] = 1 + n−1 X k=1 where N X ak [θ] 1 nkc Dk =1+ Dk + O(DN +1 ), · k · (1 + O(n)) · k k! n (1 − ζ) k!nk k=1  n . ak [θ] = +O (1 − ζ)k Assume that θ = θ1 + θ2 with θi ∈ I, i = 1, 2 for which the Lemma holds. Then,  θ1 /n  θ2 /n D D f [θ] = 1 + 1−ζ · 1 + 1−ζ PN = 1 + k=1 αk [θ]Dk + O(DN +1 ), ρk0 [θ]  ON THE EQUATION X n − 1 = BZ n where = Pk = 1 k!nk αk [θ] = 9 aj [θ1 ] ak−j [θ2 ] k−l · (1 + j=1 nj j!(1−ζ)l · nk−j (k−j)!(1−ζ)   k n 1 (ρ [θ ] + ρ [θ ]) + O 0 1 0 2 k k k!nk k!n (1−ζ)  · ρk0 [θ] + O n k!nk (1−ζ)k O(n)) . This proves the claim by complete induction.  As a consequence, we may deduce that matrices built from the first coefficients occurring in some binary series developments are regular. Pn−1 Lemma 7. Let θ = c=1 nc σc ∈ Z[G] such that φ(−1) (θ) 6≡ 0 mod n, let f [θ] = θ/n  D and 0 < N < n be a fixed integer. Then, 1 + 1−ζ f [θ] = 1 + N X k=1 and the matrix . bk [θ] Dk + O(DN +1 ), k!nk (1 − ζ)k with bk [θ] ∈ Z[ζ], AN = (bk [θ])N c,k=1 ∈ GL(K, N ) Proof. Let λ = 1 − ζ; we show that the determinant of AN is not zero modulo λ. Using Lemma 6, we know that we have a development of symbolic power series f [θ] = 1 + N X ak [θ] k=1 where k!nk Dk + O(DN +1 ),  n . ak [θ] = +O (1 − ζ)k By definition, (1−ζ)k ·ak [σc θ] ∈ Z[ζ] for all σc ∈ G. Let bk [θ] = (1−ζ)k ·ak [θ] ∈ Z[ζ]. Then, according to Lemma 6,    n bk [σc θ] = (1 − ζ)k · ρk0 [σc θ] + O (1−ζ) k = ρk [σc θ] + O(n) k P n−1 1−ζ = + O(n) l=1 nl · 1−ζ lc k P n−1 nl mod λ ≡ l=1 lc  (−1) k φ [θ] mod λ. ≡ c  k N φ(−1) [θ] mod λ. We obtained a Vandermonde deterThus, det AN ≡ c ρk0 [θ]  k,c=1 minant:  N (N −1)/2 1≤i,j≤N Y 1 1 det AN ≡ φ(−1) [θ] · − i j mod λ. i6=j The assumption that φ(−1) [θ] 6≡ 0 mod n, together with 1/i 6≡ 1/j mod n for Q1≤i,j≤N 1 1  0 < i 6= j < n, imply finally that i6=j − ≡ 6 0 mod n, which confirms i j our claim.  10 B. BARTOLOMÉ AND P. MIHĂILESCU 3. Proof of Theorem 1 It is proved in [Mi] that (5) has no solutions when (CF) holds; as a consequence of the computations in [BH], it has no solution for n ≤ 163 · 106 , except for (8). For values of n > 163 · 106 , either (CF) does not hold or 2n−1 ≡ 3n−1 ≡ 1 mod n2 , if there is a non trivial solution. This confirms respectively claims (A) and (C) from Theorem 1. In the sequel we shall show that the only possible solutions are X = ±B/ne + 1. We may assume in particular that n > 163 · 106 . We have already proved that X −1 = B·C n /ne . If C = ±1, then X −1 = ±B/ne , as stated in point (B) of the Theorem and X is a solution of (5). The bounds on |X| in (8) imply |B| < nn , the second claim of (B). Consequently, Theorem (3) will follow if we prove that C = ±1; we do this in this chapter. Assume that there is a prime p|C with pi ||C and will show that p = 1. Let P ⊂ Z[ζ] be a prime ideal lying above p and let d(p) ⊂ G be its decomposition group. The proof uses local approximations based on expansions in binomial series. We derive the relations which lead to these expansions. Let D = B · C n /ne = X − 1, with C defined by (7). Note that (7) implies that either (n, D) = 1, or n2 |B and (n, C) = 1, the last relation following from the bounds C n ≤ E < 4(n − 2)n , hence |C| < n. Then 1/(1 − ζ) is congruent to an algebraic integer modulo D · Z[ζ]. We claim that there are at least two elements, σ1 , σ2 ∈ d(p) such that σ1 6=  · σ2 . If d(p) is not a pure 2 - group, it contains a non trivial subgroup of odd order, which fixes the claim. If d(p) is a 2 - group and thus  ∈ d(p), it has, by Remark 2, at least 4 elements, which implies our claim in this case, too. Assume that there exist l ≥ 2 elements {σ1 . . . σl } ∈ d(p) such that σi 6=  · σj for i 6= j ∈ {1 . . . l}. Let σi (ζ) = ζ ci , ci ∈ Z/(n·Z). It follows from Lemma 2 that there are h1 , . . . hl ∈ Z with Pl Pl |hi | ≤ n1/l and i=1 hi ci ≡ 0 mod n. Let µ = i=1 hi σi ∈ Z[d(p)] ⊂ Z[G]. By construction, ζ µ = 1. Since K/Q is abelian and all the primes P|(p) have the same decomposition group d(p), µ has the following stronger property: let P|(p), S ⊂ G be a set of representatives of G/d(p) and γ ∈ Z[ζ] be such that γ ≡ ζ cσ mod σ(P) for all σ ∈ S. Then γ µ ≡ 1 mod pZ[ζ], as follows directly from ζ µ ≡ 1 mod σ(P). We next build θ0 ∈ If such that ς(θ0 ) = 2 and φ(−1) (θ0 ) 6≡ 0 mod n. Let u, v ∈ {2, · · · , n − 1}, u 6= v and let ψu , ψv be the associated Fueter elements. By Lemma 5, the system:  w · ϕ(ψu ) + z · ϕ(ψv ) = 0 w · φ(−1) (ψu ) + z · φ(−1) (ψv ) 6= 0 has a non-trivial solution in Fn × Fn , which we denote by (w0 , z0 ). Then, by setting θ0 = σw0 · ψu + σz0 · ψv , be obtain a Stickelberger element θ0 which verifies our three conditions, namely: ϕ(θ0 ) = 0, ς(θ0 ) = 2 and φ(−1) (θ0 ) 6≡ 0 mod n. Let l X (−)̺i hi σi (()̺i θ0 ), Θ=2· i=1 where ̺i = 0 if hi ≥ 0 and ̺i = 1 if hi < 0. Pl By construction, Θ ∈ 2 · If and φ(−1) (Θ) 6= 0. Let h = 2 · i=1 |hi | = ς(Θ)/2. From subsection 2.6, we know that there exists a Jacobi integer β[Θ] ∈ Z[ζ] such ON THE EQUATION X n − 1 = BZ n  that β[Θ]n = (ζ cx (1 − ζ)1−e )Θ · 1 + κ X−1 1−ζ h in both cases we have β[Θ] ≡ µ(ζ ) · Y Θ 11 (see (10)). It follows from (11) that mod P and thus, by Lemma 3, β[Θ] ≡ Y h mod pZ[ζ]. (13) Pn−1 Let Θ = 2 c=1 nc σc ; for any prime P|(p), the binomial series of the n−th root of the right hand side in (11) converges in the P - adic valuation and its sum is equal to β[Θ] up to a possible n−th root of unity ζ c . Here we make use of the choice of Θ: comparing (13) with the product above, it follows that ζ c = 1 for all primes P | (p). For any N > 0, we have pinN |DN and thus k !  −1  n−1 Y NX n /n D c mod pinN . β[Θ] ≡ Y h c 1 − ζ k c=1 k=0 We develop the product in a series, obtaining an expansion which converges uniformly at primes above p and is Galois covariant: ! N −1 X bk [σd Θ] h k 1+ β[σd Θ] = Y + O(pinN ). ·D (1 − ζ)k nk k! k=1 for any N > 0 and bk [Θ] ∈ Z[ζ]. Let I ⊂ {1, 2, · · · , n − 1} of cardinal N > 1 and J ⊂ Z[G] of the Galois automorphisms of K indexed by I: J = {σc }c∈I . Consider P the linear combination ∆ = σ∈J λσ · β[σ · Θ] where λσ ∈ Q[ζ] verify the linear system: X λσ · bk [σ · Θ] = 0, for k = 1, . . . , N, k 6= ⌈N/2⌉ and σ∈J (14) X σ∈J λσ · b⌈N/2⌉ [σ · Θ] = 1. Applying Lemma 6 we observe that this system is regular for any N < n. There exists therefore a unique solution P which is not null. +∞ We recall that a power series k=0 ak X k ∈ C[[X]] is dominated by the series P+∞ k k=0 bk X ∈ R[[X]] with non-negative coefficients, if for all k ≥ 0, we have |ak | ≤ bk . The dominance relation is preserved by addition and multiplication of power series. Following the proof of Proposition 8.2.1 in [Bilu], one shows that if r ∈ R>0 and χ ∈ C, with |χ| ≤ 1, then the binomial series (1 + χT )r is dominated by (1 − T )−r . From this, we obtain that (1 + T )Θ/n is dominated by (1 − T )−w(Θ)/n , where w(Θ) is the absolute weight √ of Θ. Applying this to our selected Θ, whose absolute weight is bounded by 4n n, we find after some computations that |bk [σ · Θ]| < (2n)2k . N Let A = det (bk [σc · Θ])c∈I,k=1 6= 0 be the determinant of the matrix of the
N system (14), which is non vanishing, as noticed above. Let d~ = δc,⌈N/2⌉ c=1 where δi,j is Kronecker’s symbol. The solution to our system is λσ = Aσ /A, where N Aσ ∈ Z[ζ] are the determinants of some minors of (bk [σc · Θ])c∈I,k=1 obtained by ~ Hadamard’s inequality implies that replacing the respective column by d. |Aσ | ≤ |A| ≤ 2 (2n)2N (N −1) · (N − 1)(N −1)/2 ≤ (2n)2N · N N/2 N N/2 ·n N (N −1) and 12 Let δ = A · ∆ ∈ Z[ζ], B. BARTOLOMÉ AND P. MIHĂILESCU δ= X σ∈J Aσ · β[σ · Θ] ∈ Z[ζ]. We claim that δ 6= 0. Indeed, by choice of the λ’s, we have δ = A + pinN z for some z ∈ Z[ζ], and thus A = −pinN z. The upper bound for |A| implies a fortiori that vp (A) ≤ ⌈N/2 · log2 N + (N − 1) · log2 n⌉ and the last identity would require that 3/4 log2 n−2 log2 n−n7/4 ≥ 0, vp (A) ≥ nN . The two inequalities for vp (A) imply 11 4 n which is impossible n > 4. Therefore δ 6= 0. 2 Given the bounds on Aσ , we obtain |δ| ≤ N Y h (2n)2N · N N/2 and using the fact 1/l n that h < 2ln and Y < n (Theorem 1.(B)), we find n−1  (l+1)/l 2 . |NK/Q (δ)| < n2ln · (2n)2N · N N/2+1 We can set N = ⌈n3/4 ⌉ and then, the above bound becomes  3/2 n−1 (l+1)/l +2n3/2 + 38 n3/4 + 34 ) |NK/Q (δ)| < 22n · n(2ln (15) . The initial homogenous conditions in (14) imply δ ≡ 0 mod pin⌈N/2⌉ , therefore |NK/Q (δ)| ≥ pin(n−1)N/2 . Combining this inequality with (15) and n ≥ 163 · 106 , one finds for l = 2 (which is possible for all p) that log p < 1.02. This shows that n p = 1 or 2. However, if p = 2, one could choose l = 3 < 27 < log log 2 and then find log p < 0.39, which would be a contradiction. Thus, C = ±1, which completes the proof of Theorem 1. 4. Consequences for the general case of the binary Thue equation (3) In this section we derive the Theorem 2. For this we assume that (3) has a solution with (ϕ∗ (B), n) = 1, since our results only hold in this case, a fact which is reflected also in the formulation of the Theorem 2. Consider the case when n = p · q is the product of two distinct primes. If (n, B) = 1, then Theorem 1 holds for both p and q with the value e = 0. If X, Z is a solution, then Theorem 1 – (B) implies that X p = ±B + 1 and X q = ±B + 1. Consequently either X p + X q = 2 or X p − X q = 2. This is impossible for |X| > 2 and a simple case distinction implies that there are no solutions. As a consequence, Corollary 1. Consider Equation (3) for fixed B and suppose that n is an integer which has two distinct prime divisors p > q > 2 with (p, B) = (q, B) = 1. Then (3) has no solutions for which (1) holds. If all divisors of n are among the primes dividing B, we are led to the following equation: p(X q − 1) = q(X p − 1), which has no solutions in the integers other than 1. Indeed, assume X to be a solution of the equation (X 6= 1), and q = p + t, t ≥ 0. The real function f (t) = p(X p+t − 1) − (p + t)(X p − 1) is strictly monotonous and f (0) = 0. Therefore, the equation p(X q − 1) = q(X p − 1) has no solutions. There is only the case left in which n is built from two primes, one dividing B and one not. In this case, one obtains that equation p(X q − 1) = X p − 1 which can also be shown not to have non trivial solutions, using the above remark, this time with f (t) = p(X p+t − 1) − (X p − 1). Hence: ON THE EQUATION X n − 1 = BZ n 13 Corollary 2. The equation (3) has no solutions for exponents n which are divisible by more than one prime and for B such that (1) holds. We are left to consider the case of prime powers n = pc with c > 1. If p ∤ B, c−1 we obtain X n/p − 1 = B/pe , so in particular B/pe + 1 ≥ 2p is a pc−1 −th power. Since in this case, (3) has in particular a solution for the exponent p, the Theorem 1 implies that B < pp ; combining this with the previous upper bound implies that there are no solutions for c > 2. For c = 2, we deduce that |X| < p and, after applying the Theorem 1 again and letting ξ = ζ 1/p be a primitive p2 −th root of unity, we obtain the equation 2 Y p2 Xp − 1 = NQ[ξ]/Q (α) = e p p (X − 1) α= X −ξ . (1 − ξ)e Like usual, the conjugates of the ideal (α) are pairwise coprime. We let A = (Y, α), an ideal with N (A) = (Y ); moreover, if L|A is a prime ideal and N (L) = (ℓ), then the rational prime ℓ is totally split in Q[ξ], the factors being the primes (ℓ, σc (α)). Being totally split, it follows in particular that ℓ ≡ 1 mod p2 so Y ≥ ℓ > 2p2 , in contradiction with Y < X < p + 1. This shows that there are no solutions for n = p2 . Corollary 3. If the Equation (3) in which n = pc is a prime power has non trivial solutions for which (1) holds, then c = 1. The primes dividing the exponent n used in the above corollaries are by definition coprime to ϕ∗ (B). As a consequence, if n is an exponent for which (3) has a solution and m|n is the largest factor of n with m ∈ N (B) – as defined in (2) – then the corollaries imply that there is at most one prime dividing n/m and the exponent of this prime in the prime decomposition of n must be one. This is the first statement of Theorem 2, which thus follows from these corollaries and Theorem (3). Acknowledgments: The first author is grateful to the Universities of Bordeaux and Göttingen for providing a stimulating environment during the development of this work. Both authors thank Mike Bennett and Kalman Győry for suggesting this interesting problem for an algebraic investigation. References [BE] [BGMP] [BBM] [Bilu] [BHM] [BH] [G] [Jha] M. A. Bennet: Rational Approximation To Algebraic Numbers Of Small Height: The Diophantine Equation |axn − by n | = 1, J. Reine Angew. 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