NEW BOUNDS AND CONDITIONS FOR THE EQUATION OF NAGELL LJUNGGREN

Transcription

1 Ma ora capisco il mio non capire, Che una risposta non ci sarà, Che la risposta sull avvenire È in una voce che chiederà: Shomèr ma mi-llailah, shomèr ma mi-lell, Shomèr ma mi-llailah, ma mi-lell 1 To A. Granville NEW BOUNDS AND CONDITIONS FOR THE EQUATION OF NAGELL LJUNGGREN PREDA MIHĂILESCU Abstract. The general case of the Nagell-Ljunggren equation is x p 1 x 1 = pe y q, with x, y Z, e {0, 1}, and p q odd primes. In this paper we derive a new bound q > f(p for which there are no solutions. For small q the problem is harder and we achieve a conditional result: q h p for q < g(p and some additional condition on (p, q must hold. Both functions f, g are quadratic and they leave a small gap for p 257, on which we have no general result. The picture is completed by some explicit, unconditional upper bounds for the absolute values x, y. Contents 1. Introduction 2 2. Classical general results Notations and general facts Cyclotomic Properties of the Equation Binomial Power Series 5 3. Local Methods 6 4. Global methods Proof of Theorem 2 10 References 12 1 Shomèr ma mi-llailah, by Francesco Guccini. 1

2 2 PREDA MIHĂILESCU 1. Introduction The general case of the Nagell-Ljunggren equation is x p 1 (1 x 1 = pe y q with x, y Z e {0, 1}, and p q, p, q 3. and p q, odd primes. There are no non-trivial solutions known for distinct odd exponents and it is conjectured that (1 has in fact no solutions. However, it is not even proved that (1 has only finitely many solution. Currently the most effective methods used for the analysis of (1 are tools of Diophantine approximation and linear forms in logarithms; in their recent paper [BHM], Bugeaud, Hanrot and Mignotte prove with these methods that (1 has no non trivial solutions for p 29 by using methods which go back to [BiH1], [BiH2]. Furthermore, Bugeaud and Mignotte proved in [BM] that (1 has no solutions with q 1 mod p; this will allow us to assume subsequently that q 1 mod p for any presumable solution. We refer to this paper and Ribenboim s book [Ri] for further references on the history of research on the equation under consideration. One easily verifies that, if the pair (x, y is a solution for a given prime p, then e = 1 when x 1 mod p and e = 0 otherwise. Theorem 1. Suppose that x, y are coprime integers and p, q are distinct odd primes for which (1 holds. Then q < (p 1 2. Furthermore, if q h p, the relative class number of the p th cyclotomic extension, and (1 has a non trivial solution, then either (p, q S with (2 S = { (p, q Z 2 : 31 p 257; 2 log 2 (p 1 > log 2 (q > (p 1 or q < (p 12 and: (3 x p q 1 1 mod q 2 or ( 1 ζ 1 q (4 Tr (1 + ζ q 0 mod q and x 1 mod q. (1 ζ 1 }, The bound q < (p 1 2 improves the one found in [BHM] for all p such that p 1 mod 8. The constant being good, it serves eliminating small values of p 1 mod 8 even in cases when the sharper asymptotic bound of [BHM] holds. Finally, we establish the following unconditional upper bound for the absolute value of x in a non trivial solution of (1: Theorem 2. Suppose that x, y are coprime integers and p, q are odd primes with q < (p 1 2, for which (1 holds, and x, y are a non trivial solution. Then (5 min{ x, y } < q 10 p2. 2. Classical general results Let p, q be distinct odd primes. The study of the equation (1 is intimately related to properties of the p th cyclotomic field. We develop in this section the notation used subsequently and present some classical results on this equation and on the group rings acting on possible solutions thereof. Most of these results have been either known for a long time or are exposed in depth in some of our recent papers. We give here a succinct overview and refer the reader to [Ri], [BHM], [Mi1], [Mi2] for more details.

3 NAGELL LJUNGGREN Notations and general facts. Let K = Q(ζ = Q[X]/(Φ p (X be the p th cyclotomic extension, with Φ p (X = Xp 1 X 1. We let P = {1, 2,..., p 1} and σ c G = Gal (Q(ζ/Q be the automorphism of Q(ζ with ζ ζ c, for c P. Complex conjugation is denoted by j G, so α = σ p 1 (α = jα. We write λ = (1 ζ for an algebraic integer generating the unique ramified prime ideal above p in K. If θ = c P n cσ c Z[G], we denote its weight by w(θ = c P n c and its absolute value θ = c n c ; if θ Z 0 [G], we shall say that θ is positive and for such elements w(θ = θ. In general θ = c n cσ c Z[G] can always be written as difference of two positive elements: θ = θ + θ = n c σ c ( n c σ c, c;n c 0 c;n c<0 so θ = w(θ + + w(θ. The Stickelberger element is ϑ = 1 p c P cσ 1 c and the Stickelberger module is I = Z[G] (ϑz[g]. It has the property of annihilating the class group and since (6 θ + jθ = ς(θ N Q(ζ/Q, for some integer multiple ς(θ Z, annihilation is trivial for real classes. In particular ( A A θ = N(A ς(θ for any θ I. We shall write (7 I + = I Z 0 [G], I n = { θ I + : ς(θ = n }, ψ n = (σ n+1 1 σ n ϑ I 1. Then the set {ψ n : n = 1, 2,..., (p 1/2} forms a Z - base for I. Further useful properties (e.g. [Mi1] are (8 ζ ψn = ζ ϕ(n+1 with ϕ(x xp 1 1 p (1 ζ θ = ζ ϕ(θ p ς(θ/2 for θ I 2n, n > 0. mod p, In the second relation, the Fermat quotient map ϕ was extended from the base ψ n to the whole Stickelberger module, by additivity. Finally, we shall need the following computational ( Lemma 1. Let τ ±n = Tr ζ n Q(ζ/Q for n = 0, 1,...,p 1. Then (9 (10 τ n p 1 2 τ n p 1 2 = + p = 1±ζ { p if n > 0 and n 0 mod 2 0 otherwise, and { p if n = 0 and n otherwise. Proof. The value τ 0 = (p 1/2 can be computed directly from the minimal polynomial of the unit 1/(1 + ζ. Then τ 1 = Tr (1 1/(1 + ζ = (p 1/2. For n = 2 we write ζ2 1+ζ = ζ ζ = ζ ζ and thus τ 2 = τ 0 p. For n > 0 we have the induction: ζ n ζ = ζn+2 ζ n + ζ n 1 + ζ = ζ n (ζ 1 + ζn 1 + ζ,

4 4 PREDA MIHĂILESCU and thus τ n+2 = τ n, since Tr(ζ n+1 = Tr(ζ n whenever n < p 1. The second identity is similar Cyclotomic Properties of the Equation. If x, y, p, q, e verify (1, we shall write α = x ζ (1 ζ and α e c = σ c (α. In order to avoid the case distinction on e, we shall be often able to reduce our analysis to α = (x ζ = α (1 ζ e. With this, (1 becomes N(α = y q and there is an ideal A = (α, y Z[ζ] with (11 A q = (α. This follows from the fact that for distinct a, b P, the ideals σ a (A, σ b (A are coprime [Mi1]: (12 (σ a (α, σ b (α = 1 for a, b P, a b. The relation (11 suggests that the group ring R = Z[G]/ ( q,n Q(ζ/Q acts upon A and it will be an important object to consider. Unless otherwise specified, we shall identify elements θ = c ν cσ c R with their lifts θ = c n cσ c Z 0 [G] defined by 0 n c < q, n c ν c mod q and w ( θ is minimal under these conditions (the last restriction allows to use the relation in R generated by setting the norm to 0. Thus the notation θ will be mostly skipped and it will be understood that θ R acts upon numbers and ideals via the given lift. The following crucial remark guides all the various approaches which we shall take in this paper. Lemma 2. Notations being like above and assuming that N(α = y q, there is an F q [G] module P R such that α θ (Q(ζ q for all θ P. There is a module N R such that (1 jθ P, θ N and such that N A, the annihilator of A in the relative class group of Q(ζ. Furthermore (13 α (1 jθ (Z[ζ] (1 jq for all θ N. Proof. The existence of P is a simple computation: if θ 1, θ 2 P, then one easily verifies that aθ 1 + bθ 2 P for any a, b Z/(q Z. Thus P is a F q - module; this does however not imply that P is not trivial! For the second claim, we let A = { θ R : A θ = (ρ B, ρ Z(ζ, B Z[ζ + ζ] } R, which is the annihilator of A in the relative class group C of Q(ζ. Then for θ A we have A qθ = (α θ = (ρ q B q ; thus, there is a unit η such that ( eθ α (1 jθ 1 ζ = α (1 jθ = η (ρ/ρ q. 1 ζ Since η η = 1, it follows from Dedekind s unit theorem that it must be a root of unity; the only such roots in Q(ζ are the 2p th ones and they are all q th powers. It follows that (α /α θ = ±(ν/ν q for all θ A, where ν Z[ζ] differs from ρ by some root of unity. This proves the assertion of the lemma. The fact that N is a module is a simple verification, but we have proved more, namely that it is not trivial.

5 NAGELL LJUNGGREN Binomial Power Series. The common idea of our various methods starts with the equation α θ = ρ q for θ P and consists in comparing the value ρ to some power series estimation of α θ/q. This leads us to closer look at the power series of interest. For r R, the series of Abel (or generalized binomial series is f(z = ( r z k k ; k 0 it converges uniformly for z < 1 and the sum verifies f(z = (1 + z r. We want to investigate series expansions of algebraic numbers µ Z[ζ] which verify µ q = (1 ζ/x θ, where θ = c n cσ c P. Thus, we consider the formal power series for (1 ζ/x θ and start with some definitions. Definition 1. Let R = Q(ζ[[T]] be the ring of formal powers series over the p th cyclotomic field. The elementary q th root series is defined as: f(t = ( 1/q ( ζ T k k = a k T k R, k 0 k 0 ( 1/q where a k = ( ζ k k. Let 0 < n < q and σ G; then σ acts upon ζ and not upon the formal variable T. Furthermore, we define: f[nσ](t = σ (f(t n = f nσ (T = ( n/q (ζ k σ T k = a k (nσ T k R, k 0 k 0 ( n/q and a k (nσ = ( ζ k kσ. The map f[x] is then extended by additivity to a map f : R R, such that (f[θ](z q = (1 z ζ θ for any value z C for which f[θ] converges. For θ = c n cσ c, we define f[θ](t = n 0 a n (θ T n = c P f[n c σ c ](T. An interesting property of the series f[θ] is that its coefficients are galois covariant and they naturally do not depend upon the evaluation value z; they also have interesting arithmetic properties, Both will be of importance for subsequent estimations. We gather the main results on f[θ] in one proposition; a proof can be found in [Mi1]. Proposition 1. For k 0, let E(k = k + v q (k!. Then E(k is strictly monotonous, verifies E(k < k q q 1 and the arithmetic behavior of the coefficients a k(θ is given by (14 b k (θ := q E(k a k (θ Z[ζ] and E(k = v q (a k (θ. and the integer multiples are bounded by: (15 ( θ /q b k (θ < q E(k m

6 6 PREDA MIHĂILESCU Let θ = c P n cσ c R have absolute value θ = H. If z C, z < 1, and S m [θ](z is the m th partial sum of the series f[θ](z and R m [θ](z is the corresponding remainder, then ( R m [θ](z = f[θ](z S m (Θ; z ( H/q m + 1 z m+1 (1 z m+1+h/q. Finally, if (θ mod (N, q P, then α θ = β qθ = A (f[θ](x q, for some β(θ Q(ζ and A Z. In particular, if ξ is a primitive q th root of unity, there is a map κ : R Z/(q Z, which we shall denote by Galois exponent map, and such that f[θ](x A 1/q = ξ κ(θ β(θ. One easily verifies the action of complex conjugation: κ(θ + κ(jθ = 0. The signature of θ is the vector (κ(σθ Z/(q Z σ G and its most useful characteristic is the order of vanishing which is an even number. z(θ = {σ G : κ(σθ = 0, } 3. Local Methods The base for a local approach for the study of the Nagell-Ljunggren equation will be a Theorem which we adapt by restriction from [Mi3]: Theorem 3. If the equation (1 has a solution (x, y; p, q and the exponents are such that max{p, p(p 20 } > q and q h p, with h p the relative class number of the p th cyclotomic extension, then (17 x f 0 mod q 2 with f { 1, 0, 1}. Under the condition q h p one notes that the solutions to (1 give raise to the existence of some units in Z[ζ] with particular properties. The relations (17 allow to expand these units q - adically and it turns out that the norm of those expansions cannot be 1, a contradiction which shows that in the given range for p, q there are no solutions with q h p. The next Lemma defines the units above mentioned. Lemma 3. Let x, y; p, q be a solution of (1 with odd prime exponents and assume the q h p. Then there is a ν Z[ζ] such that x ζ ( ν q x ζ = ± = µ q (18 µ = ±ν/ν Q(ζ. ν Furthermore, if φ = x ζ (ζ ζ q (µ 1q, then φ Z[ζ]. Proof. The existence of µ has been shown in the proof of Lemma 1 above. In order to show that φ is a unit, one first verifies, according to a well established pattern [BH], [Mi1], that φ Z Q(ζ = Z[ζ]. Next, we consider ( q x ζ Φ = (µ q 1 q = 1, ζ ζ equality which follows from (18. It is easily shown that φ Φ, so φ is in fact a unit. We now prove the main theorem of this section:

7 NAGELL LJUNGGREN 7 Theorem 4. If the equation (1 has non-trivial solutions for odd prime exponents p, q with q < (p 12 and q h p, then either x pq 1 1 mod q 2 or (4 holds. Proof. We prove the statement by contraposition. Since it is known [BHM] that there are no solutions for p, q < 29, we may assume that p 29 and then (p 12 < p 2 20 and, assuming that q h p, the congruence (17 must hold. This allows for q - adic expansions of ν/ν in (18 and, a fortiori, for the unit φ. Let us first consider the case when f = 0 so x 0 mod q 2 and set r Z such that r q 1 mod p. Then µ = ζ 2r (1 (ζ ζx/q + O ( (x/q 2, and ( ζ 2r q ( 1 φ = ( ζ(1 ζx 1 ζ ζ ζ ζ ζ 2r 1 x + O(x2 /q. After dividing out some units, we find Let χ = ζ2r 1 ζ ζ 2r 1 φ = 1 ζ2r 1 ζ ζ 2r 1 x + O(qx Z[ζ]. = ζr 1 ζ r 1 ζ r ζ r so 1 = N(φ = 1 x Tr(χ + O(xq and thus Tr(χ 0 mod q or equivalently Tr(σ q (χ 0 mod q. Note that if r = 1 and thus q 1 mod p, χ vanishes and we will have to use higher order terms. If q 1 mod p, for the estimate of the last trace we let 0 < n < p be such that n q 2 mod p: Tr (ζ q 1 ζ q 1 ζ ζ ( = Tr ζ q 2 ζ2(q 1 1 ζ 2 = Tr 1 { (n + 1 if n is odd = p (n + 1 otherwise. ( ζ n n ζ 2i We first assume q < p and thus n = q 2 is odd, so Tr(χ 1 mod q. If q > p, let 0 n = q 2 mp < p, with m 0. If n is odd, then Tr(χ = mp (q 1 mp+1 mod q; by definition of m, we have 1 mp+1 < q and thus Tr(χ 0 mod q. Finally, if n is even and Tr(χ = p (n + 1 = (m + 1p + 1 q 0 mod q, then (m + 1p + 1 = q, so q 1 mod p, a case which we still have to consider. In this case r = 1 and after some algebraic manipulations which we omit here, one finds the simple expression ( 1/q φ = 1 x 2 + O ( (x/q 3. 2 Since the remainder is q - adically strict smaller than (x/q 2 and p 1 mod q, this can not be a unit for any value 0 x 0 mod q 2, and this completes the discussion of the case f = 0. In the remaining cases we have (19 fζ x ζ x ζ = x ζ 1 fζ : x ζ 1 fζ = Since x f mod q 2, the expansion for µ lays at hand. i=0 ( ( x f x f 1 fζ + f : 1 fζ + f.

8 8 PREDA MIHĂILESCU If f = 1 we have (20 ζ r µ = 1 x ζ q 1 + ζ + 1 ( 2 x + 1 (1 ζ2 + q (ζ q (1 + ζ 2 + ( (x O. q Consequently ( ζ r+1 1 φ = ε 1 (x + 1 (ζ + 1(ζ r + O(q(x + 1, 1 for some unit ε Z[ζ]. Like previously, by letting χ = ζ r+1 1 (ζ+1(ζ r 1 we find that Tr(χ = 0 which is (4 (compare with Tr(σ q (χ. This explains the case when χ 0. Since χ = 0 implies r = 1 and q 1 mod p it follows that q > p in this case; one computes under this assumption and modulo cyclotomic units, that φ = 1 (x+12 2q χ + O ( (x with χ = ζ (1+ζ ; since Tr(χ = p it follows that Tr(χ 0 mod q if p ±1 mod q and in particular p > q. This is a contradiction showing that φ cannot be a unit in this case. 1 In the case when f = 1, we distinguish also according to the value of e. If e = 0, then the development of φ derived from (19 easily leads to 1 = N(φ p q 1 1 mod q 2. If e = 0 this is also all the information we can gain. Local developments of N(φ then only yield higher terms in the q - adic expansion of p q 1 1. If e = 1 we can however use p - adic expansion. Let in this case x = 1 +up k with (u, p = 1 and k 1. From (19 we have now: φ = x ζ 1 ζ = (1 + ζ r q ( ( q (1 + ζ r 1 + (x 1(1 + ζr 1 q(1 ζ(1 + ζ r + O((x 12 ( 1 + ( ζ x 1 1 ζ ζr ζ r + O((x 1 2 Like previously, we find N(φ = 1 + (x 1Tr(χ + O((x 1 2, where χ = 1 1+ζ r and Tr(χ = p mod p. This is a contradiction indicating that φ cannot in fact be a unit in Z[ζ], which completes the proof of the theorem. 4. Global methods Proposition 2. Suppose that q h p and (1 has a non trivial solution and let p 1 l = ; then p, q are related by the following inequality: 2log 2 (q (21 q < (p 12 2l + (p 1. 1 The condition (4 is a less usual condition in p, q, yet quite easily computable, either directly, or by tackling formulae as prepared in Lemma 1; a simple closed formula does apparently not exist. It is also not clear and is probably a harder question to estimate the proportion of prime pairs for which the condition is fulfilled. It is certainly not an impossible condition, but it is not more than a necessary one for existence of solutions to the Diophantine equation.

9 NAGELL LJUNGGREN 9 Proof. If q h p, then N = (1 jr in Lemma 2 and thus α (1 jθ = ν(θ q for any θ Z 0 [G]. Let J = { c P n c σ c : n c {0, 1}} Z 0 [G] and J J be a complete set of representatives of J/, where θ θ θ θ (1 + jz[g]; we set M = J = 2 (p 1/2 1. For θ J we let φ(θ = y ν(θ Z[ζ]. By the pigeon hole principle, we can find distinct θ 1, θ 2 J such that z(θ 1 θ 2 = z(θ 1 +jθ 2 2l = 2 log M/ log q p 1 log 2 (q 1. For these values, let δ = φ(θ 1 φ(θ 2 Z[ζ]: if δ = 0, then α θ1+jθ2 = α θ1+jθ2 and (12 implies that θ 1 + jθ 2 (1 + jz[g]. This contradicts the choice of J and it follows that δ 0. We shall now estimate the norm of the algebraic integer δ. Let σ G and δ = δ/y = ν(θ 1 ν(θ 2. If κ(σθ 1 = κ(σθ 2 then the developments of ν(σθ i, i = 1, 2 correspond up to the first order: σδ is small. We thus gain the following upper bounds: { 1 if κ(σθ1 κ(σθ 2 (22 σ(δ = σ (ν(θ 1 ν(θ 2 2 2p q x if κ(σθ 1 = κ(σθ 2. The first case is an obvious consequence of α/α = 1. For θ = c n cσ c we define η(θ = c n cζ c. With this we have in the second case, and since κ(σ(θ 1 θ 2 = 0: and δ = ξ κ(σ(θ1 (f[(1 jσθ 1 ](1/x f[(1 jσθ 2 ](1/x σ(δ η(σθ 1 η(σθ 2 η(σθ 1 η(σθ 2 = qx 2(p 1 q x + R 2 (1/x. The remainder is bounded by ( θi /q 1 (p 1 ((p 1 + q R 2 (1/x 2 < 2 x2 q 2 x 2 < p 1 2 q x x < p 1 1 q x q, while the estimate of the first order term is 2(p 1 η(σθ1 η(σθ2 η(σθ1 η(σθ2 qx q x. By assembling the bounds for the first order term and the remainder, we find (22. Since 0 δ = yδ, we can now apply the estimate (22 to the norm ( 2l 2p 1 N(yδ (2y p 1 < y p 1 (2p/ x 2l < y p 1 (y/ x 2l. q x The bound (21 follows from the estimate 2l > p 1 log 2 (q 1 and the simple lower ( bound x > 2q 1/(p 1. 2q+1 yq Theorem 5. The equation (1 has no solutions for q (p 1 2. Proof. We assume that q < M/2 = (p 1(p2 1 8 = p 1 2 ((p+1/2 2. This bound can be proved with the method of proof of Theorem 1 in [Mi1], using the lower bound x > p p 1 which holds whenever q > (p 1 2 as we now prove. Since y q = N(α and (σ a (α, σ b (α = 1, it follows that each prime l y splits completely in Z[ζ] and thus l 1 mod p; in particular y 2p + 1. Furthermore, from (1 and under the assumption that q > (p 1 2, we have x 1 2 yq/(p 1 y p 1 /2, which implies the claim. Let I 2 = {θ I + : w(θ = (p 1}, as before; the elements in I 2 are obviously incongruent modulo q and we claim that I 2 M. We mentioned that if υ n =

10 10 PREDA MIHĂILESCU σ n+1 (σ n + 1 Z[G], then ψ n = υ n ϑ I build a basis for I mod (N, when n = 1, 2,...,(p 1/2. Furthermore w(ψ n = (p 1/2. We certainly have I 2 S 2 = {ψ n +ψ m : 1 n, m (p 1/2} and S 2 σs 2 = for all 1 σ G, as one easily verifies. Thus I 2 G S 2 and an easy combinatorial computation shows that (p 1 S 2 = M, which confirms our claim. Since q < M/2, by the pigeon hole principle there must thus exist θ 1, θ 2 I 2 with θ 1 θ 2, such that κ(θ 1 = κ(θ 2. With these we define φ = β(θ 1 β(θ 2 = β(θ 2 (β(θ 1 /β(θ 2 1; by consequence of (12 and θ 1 θ 2, we have 0 φ Z[ζ]. We estimate the norm of φ. For this note first that β(θ i β(θ i = y 2, i = 1, 2 and thus β(θ 1 /β(θ 2 = 1. Since κ(θ 1 = κ(θ 2, an estimate similar to (22 yields: ( ( 1 N(φ = y p 1 N β(θ1 2 5(p 1 β(θ 2 1 y p 1 2 p 1, 4q x x and finally < 2 (p 1/2, which is inconsistent with x > y p 1 /2 and y y (p 1/2 2p+1. The assumption q > (p 1 2 thus leads to a contradiction, which completes the proof. If q h p, the Proposition 2 yields an upper bound for values of q such that (1 has a solution. By assuming some additional conditions on (p, q, we find in Theorem 4 a lower bound for such values q. Using also the unconditional upper bound from Theorem 5, the following corollary delimits the set in which the two conditions overlap. For this we let S be defined by (2; then: Corollary 1. Let p, q be primes such that q h p, q > (p 12. Then the equation (1 has no integer solutions, with the exception at most of the pairs (p, q S. Proof. Let l be like in (21 and suppose that p, q are such that: (p 1 2 2l + (p 1, so (p 1 ( 1 2l 1 < 1. Then 1 log 2(q p 1 < 1 p 1, and log 2(2q > p 1 ; (p 1 2 q < since q < (p 1 2 we may eliminate q and find 32 log 2 (p 1 p 1. One verifies that this holds iff p 257, and [BHM] have eliminated the cases p < 31. The claim follows from the previous proposition Proof of Theorem 2. Let θ = c n cσ c I have relative weight ς(θ = 2m. Then, as shown above, there is a Galois exponent κ(θ Z/(q Z together with a β(θ Z[ζ] such that the following expansion holds: ( x β(θ = ξ κ(θ p 1 m/q f[θ](1/x. p e Since the series f[θ] is independent of the value of x, this suggests the following bounding strategy: build large linear combinations of β(θ in which the corresponding expansion vanishes up to a high power of 1/x. If the combination is a non vanishing algebraic integer, this yields a good upper bound for x. The latter condition appears to be the harder one and we can only prove it under some defensive premises, which ponder the quality of the resulting upper bounds. We assume first that p < q, and shall choose ς(θ = 6; we know already that q < (p 1 2. Since I 6 ( (p 1/2+5 6 > 3(p 1(q 1 for p 53, we can choose a set J 0 I 6 with J 0 = 3(q 1, ς(θ = 6 and θ Gθ for any pair θ, θ J 0. We

11 NAGELL LJUNGGREN 11 then let J = G J 0 = {σθ : σ G, θ J 0 } and set δ = θ J λ(θβ(θ, for some unknowns λ(θ Q(ζ, ξ +, for all θ J. We then require that δ vanishes up to the sixth order, together with all but one of its conjugates. We obtain the following linear system: (23 λ(θ (a k (θξ κ(σaθ/b + a k (θξ κ(σaθ/b = C δ ((a, b, k, (0, 0, 0, θ J where a P, b Q, 0 k < 6, the symbol on the right hand of the equation is the Kronecker - delta in three variables and C is a constant, whose value we shall still fix. Let A = ( ( a θ,(a,b,k = a k (θξ κ(σaθ/b + a k (θξ κ(σaθ/b θ,(a,b,k be the matrix of the above linear system and let us first assume that it is a regular matrix with determinant A = det(a Z[ζ, ξ] +, since a k (θ Z[ζ]. In view of (15, the coefficients b k (θ are bounded in this case by ( k + 2 b k (θ < q E(k. 2 For values k 6 < p < q, we have k = E(k and the absolute value of A can thus be bounded using Hadamard s estimates for determinants. For this, split A in 6 matrices (A k 5 k=0 of dimension (M/2 (3M, such that A k q k (k+2 2, where M = (p 1(q 1. We then have: M/2 A (3M M/4 (q 15 (7!2 < M 6M. 8 If we set C = A in (23, then Cramer s rule shows immediately that the system has algebraic integer solutions and furthermore, using the same estimates of Hadamard, we also have λ(θ < M 6M. Consequently, ψδ MA x 3(p 1/q R 5 < 56 x 3( 2+(p 1/q MA for 1 ψ Gal (Q(ζ, ξ/q. Finally, by assuming δ 0, we may use the natural bound δ < MAx 3 and thus find the double inequality: ( 1 N(δ < 56 MAx 3(p/q 2 M. Under the chosen premises, q > p > 50 and we find the bounds x < M 2M < (p 1 6(p 13 < q 3pq and y < M 2M(p 1/(q 1 = M 2(p 12 < (p 1 6(p 12 < q 3p2. If δ = 0, then the inhomogeneous condition implies 0 = δ = A + R 5 and thus R 5 /A = 1 Z. But the estimate 0 R 5 < 56Ax 3 then leads to a much better estimate than the previous. We still have to consider the case when the matrix A is not regular. In that case, let k = rank (A < M/2. We can drop M/2 k equations and unknowns, in such a way that the resulting system is regular. Then the same steps like in the previous case will lead to bounds on x, y which improve the respective ones deduced for det(a 0. In the case q < p, we can apply the same idea, the main difference consisting in the fact that the size of A may grow, since we shall need to annulate the expansion of δ to a higher degree of 1/x. Leaving to the reader some details which are similar to the case discussed above, here is a sketch of the proof for this case.

12 12 PREDA MIHĂILESCU Let m = p/q and M = (p 1(q 1 like before. Then y < x m and taking J 0 I 2n, we need 2mnM equations and a brief computation shows that it will suffice to take, like previously, J 0 I 6, i.e. n = 3. Then J 0 will be chosen as a set of N = 6mM elements, leading to equally many equations. One should notice that since p > q, the same weight of annihilators will provide for equations reaching up to higher values of k. We thus consider the same linear system (23, yet the range of k is now 0 k < 2nm = 6m. We assume first that (23 is regular in this case and derive the Hadamard estimates for the system determinant: ( ((6m! 2 mm A B 0 = M 6m q 3m(6m m In the case δ 0, we find the bounds y < q 10p2 ; x < q 10m p2. This bounds are improved if δ = 0 or the matrix A is singular, thus completing the proof of the theorem. Acknowledgments: I thank Victor Vuletescu for careful reading, discussions and suggestions and Yann Bugeaud, who directed my attention to this interesting equation his informative remarks. References [Be] M. Bennett: Rational approximation to algebraic numbers of small height, The equation ax m by n = 1, Journal für die reine und angewandte Mathematik, 535 (2001, pp [BiH1] Y. Bilu: Solving Thue equations of high degree., J. of Number Theory 60, (1996, pp [BiH2] Y. Bilu and G. Hanrot: Solving superelliptic Diophantine equations by Baker s method, Compositio Math. 112 (1998, pp [Bl] Ch. Blatter: Analysis III, Heidelberger Taschenbücher 153 (1974 [BH] Y. Bugeaud and G. Hanrot: Un nouveau critère pour l équation de Catalan, Matematika, 47 (2000, pp [BHM] Y. Bugeaud, G. Hanrot and M. Mignotte: Sur l équation diophantienne xn 1 x 1 = yq, III, Proc. London. Math. Soc. 84 (2002, pp [BM] Y. Bugeaud and M. Mignotte: L équation de Nagell Ljunggren xn 1 = y x 1 q, L Enseignement Mathémathique 48 (2002, pp [Ca] J. W. S. Cassels: On the equation a x b y = 1, II, Proc. Cambridge Society bf vol 56 (1960, pp Corr. Id. 57 (1961, p [Mi1] P. Mihăilescu: On the class groups of cyclotomic extensions in presence of a solution to Catalan s equation, to appear in J. of Number Theory. [Mi2] P. Mihăilescu: Primary Cyclotomic Units and a Proof of Catalan s Conjecture, Journal für die reine und angewandte Mathematik, 572 (2004, pp [Mi3] P. Mihăilescu: Cyclotomic Investigation of the Fermat - Catalan Equation, submitted [Ri] P. Ribenboim: Catalan s conjecture, Academic Press, (1994. [Wa] L. Washington: Introduction to Cyclotomic Fields, Second Edition, Springer (1996, Graduate Texts in Mathematics 83. (P. Mihăilescu Mathematisches Institut der Universität Göttingen address, P. Mihăilescu: preda@uni-math.gwdg.de