The Nagell-Ljunggren equation via Runge's method
aa r X i v : . [ m a t h . N T ] D ec THE NAGELL-LJUNGGREN EQUATION VIA RUNGE’SMETHOD
MICHAEL A. BENNETT AND AARON LEVIN
Abstract.
The Diophantine equation x n − x − = y q has four known solutions inintegers x, y, q and n with | x | , | y | , q > n >
2. Whilst we expect that thereare, in fact, no more solutions, such a result is well beyond current technology.In this paper, we prove that if ( x, y, n, q ) is a solution to this equation, then n has three or fewer prime divisors, counted with multiplicity. This improves aresult of Bugeaud and Mih˘ailescu. Introduction
The
Nagell-Ljunggren equation x n − x − y q , in integers | x | > , | y | > , n > , q ≥ x n − y q = 1 . Whilst Catalan’s conjecture was proven by Mih˘ailescu [11] in 2004, the Nagell-Ljunggren equation has, in a certain sense, outlived its more illustrious cousin, inthat it remains unknown to date whether the number of solutions to (1) in thefour variables x, y, n and q is finite. Indeed, such a conclusion is beyond currenttechnology even in the restricted case where n = q . Equation (1) does have preciselyfour known solutions:3 − − , − − , − − and ( − − − − ( N L )and there is an impressively large literature providing constraints upon any hithertounknown ones. In particular, combining results from [5], [6], [9], [11], [12] and [13],we have
Proposition 1. If ( x, y, n, q ) is a solution of equation (1) not in ( N L ) , then • q ≥ is odd, • The least prime divisor p of n satisfies p ≥ , • | x | ≥ and x has a prime divisor p ≡ q ) . Date : November 5, 2018.Supported in part by a grant from NSERC.Supported in part by NSF grant DMS-1102563.
We can in fact say rather more if we additionally assume that x is positive,improving the lower bounds on both | x | and the least prime divisor of n in this case(see [2], [3]); though similar arguments can be applied to negative values of x tosharpen Proposition 1, we will not have need of such a result here.One feature distinguishing equation (1) from (2) is that the reduction to the casewhere n is prime is without loss of generality in (2), but not in (1). That beingsaid, there is some degree of control over how composite the exponent n can be.Specifically, writing ω ( n ) for the number of distinct prime divisors of n , an argumentof Shorey [15], together with the results of [1], implies that ω ( n ) ≤ q − x both positive and negative, though this is stated in [15] only for the former). In [7],Bugeaud and Mih˘ailescu sharpened this substantially, proving that if ( x, y, n, q ) isa solution to (1), with x >
1, then necessarily ω ( n ) ≤ Ω( n ) ≤ , where Ω( n ) is the total number of prime divisors of n , counted with multiplicity.Our main result is the following improvement of this: Theorem 2. If ( x, y, n, q ) is a solution of (1) , then ω ( n ) ≤ Ω( n ) ≤ . The proof of this theorem relies upon a careful application of the classical methodof Runge to an equation of the shape f ( x ) = g ( y ), where f and g are polynomialswith integer coefficients. The result of Bugeaud and Mih˘ailescu [7] depends funda-mentally upon earlier work of Mih˘ailescu [10], which one might view, in part, asan application of Runge’s method over cyclotomic fields. The proof of the crucialinequality from [10], given in Theorem 2 of that paper, uses tools from cyclotomicfield theory and techniques developed in the proof of Catalan’s conjecture. Whatwe prove, in an essentially elementary fashion and independent of the results of[10], is that solutions to (1) necessarily satisfy ω ( n ) ≤ Ω( n ) ≤
3, with the possibleexception of the special case where ( n, q ) = ( p ,
5) and p | x −
1, for p prime. Weeliminate this remaining exceptional case by appealing to [10].Our main tools in proving Theorem 2 are the following pair of results. Theorem 3.
Suppose that q, r, s, x, y and z are integers with q prime and ≤ r < s ,such that (3) y q = x r − x − and (4) z q = x s − x − . Then | x | < (cid:26) , sqr (cid:27) q r − s − rq − . (5) Theorem 4.
Suppose that integers x, y, z and u and odd primes p and q satisfy py q = x p − x − ,p z q = x p − x − HE NAGELL-LJUNGGREN EQUATION VIA RUNGE’S METHOD 3 and p u q = x p − x − . Then | x | < p (cid:18) max (cid:26) , pq (cid:27)(cid:19) q p + p − p − / ( q − . (6) 2. Preliminaries
Before proceeding with the proof of Theorem 2, we require a pair of lemmata ofa combinatorial nature. Given solutions to (1), these will enable us to deduce theexistence of simultaneous solutions to a number of related equations. The first is aresult of Shorey [15, Lemma 7]. For a positive integer n , let Q n = φ ( G ( n )), where G ( n ) is the square-free part of n and φ is Euler’s function. Lemma 5.
Let ( x, y, n, q ) be a solution of (1) with n odd. If the divisor D of n satisfies ( D, n/D ) = (
D, Q n/D ) = 1 , then there exist integers y and y with y y = y and y q = ( x D ) n/D − x D − ,y q = x D − x − . We use this to prove the following
Lemma 6.
Let ( x, y, n, q ) be a solution of (1) with n odd and write n = p α · · · p α k k , where α i ∈ N and the p i are primes with p < p < · · · < p k . Then, for each ≤ s ≤ k and ≤ t ≤ α s − , writing M s,t = p ts p α s +1 s +1 · · · p α k k , there exists an integer y s,t and δ s ∈ { , } such that we have (7) x p s M s,t − x M s,t − p δ s s y qs,t . If δ s = 1 , then q | α s .Proof. Observe that, by assumption,(8) k Y s =1 α s − Y t =0 x p s M s,t − x M s,t − y q . Let us suppose first that k = 1. Since, given integers | z | > m ≥
1, we havethat gcd (cid:18) z m − z − , z − (cid:19) divides m, it follows, writing X = x M ,t = x p t for t ∈ { , , . . . , α − } , thatgcd (cid:18) X p − X − , X − (cid:19) ∈ { , p } . MICHAEL A. BENNETT AND AARON LEVIN
Moreover, if X ≡ p ), say X − p a with a ∈ Z , then X p − + X p − + · · · + 1 = (1 + p a ) p − + (1 + p a ) p − + · · · + 1 ≡ p + p − X i =1 ip a (mod p ) ≡ p + a ( p − p / ≡ p (mod p ) , and, in particular, repeatedly applying Fermat’s Little Theorem, we haveord p x p t +11 − x p t − ! = (cid:26) x ≡ p )0 otherwise,for each 0 ≤ t ≤ α −
1. There thus exist integers y t , again for 0 ≤ t ≤ α −
1, suchthat x p t +11 − x p t − p δ y qt , where δ = (cid:26) x ≡ p )0 otherwise.Since, from (8), we have α δ = ord p α − Y t =0 x p t +11 − x p t − q ord p y ≡ q ) , the desired result follows.Suppose now that the stated conclusion is true for solutions to (1) with ω ( n ) = k −
1, where k ≥
2. Assuming that ( x, y, n, q ) is a solution of (1) with n odd and ω ( n ) = k , say, n = p α · · · p α k k , we may thus apply Lemma 5 with D = n/p α to deduce the existence of integers Z and W such that(9) X − x − Z q and X p α − X − W q , where X = x n/p α . As previously, we find that q ord p W = ord p α − Y t =0 X p t +11 − X p t − (cid:26) α if X ≡ p )0 otherwise,whereby we have (7) for s = 1. Applying our inductive hypothesis to the firstequation in (9) (where we have ω ( n/p α ) = k − (cid:3) Proof of Theorem 3
The idea of the proof is to apply Runge’s method to an appropriate curve. Theclassical Runge’s method is used to bound integer points on superelliptic curves ofthe form y q = f ( x ), where f ∈ Z [ x ] is monic and q | deg f . In order to construct asituation where this version of Runge’s method applies, we will consider a productof powers of x r − x − and x s − x − to obtain a polynomial f of degree divisible by q andthen examine integer solutions to w q = f ( x ). HE NAGELL-LJUNGGREN EQUATION VIA RUNGE’S METHOD 5
Let q, r, s, x, y, z ∈ Z be a solution to (3) and (4), as in the statement of thetheorem. We may assume throughout that | x | >
2. Note that the main result ofBennett [1] implies that the Diophantine equation x t X q − ( x − Y q = 1has, if t = 1, precisely the solution X = Y = 1 in nonzero integers X and Y , and,for a fixed positive integer 2 ≤ t ≤ q −
1, at most a single nonzero solution
X, Y .Choosing integers 0 ≤ r , s < q such that r ≡ r (mod q ) and s ≡ s (mod q ) andrewriting (3) and (4) as x r (cid:16) x ( r − r ) /q (cid:17) q − ( x − y q = 1 and x s (cid:16) x ( s − s ) /q (cid:17) q − ( x − z q = 1 , we may thus conclude that 1 , r and s are pairwise incongruent modulo q (whereby,since Proposition 1 implies that 3 ∤ rs , q ≥ r q )and s q ), we may define an integer a with 0 < a < q , via a ≡ − ( r − − ( s −
1) (mod q ) . Since r s (mod q ), it follows that a = q −
1, whence 1 ≤ a ≤ q −
2. Let us nowdefine N = a ( r −
1) + s − , so that N ≡ − ( r − − ( s − r −
1) + ( s − ≡ q ) , whereby N/q is an integer.We consider the equation w q = (cid:18) x r − x − (cid:19) a (cid:18) x s − x − (cid:19) , (10)where now Runge’s condition is satisfied. This equation has the integer solution( w, x ) = ( y a z, x ). Consider the Laurent series expansion (cid:20)(cid:18) x r − x − (cid:19) a (cid:18) x s − x − (cid:19)(cid:21) q = ( x r − a/q ( x s − /q ( x − ( − a − /q = x N/q (cid:18) − x r (cid:19) a/q (cid:18) − x s (cid:19) /q (cid:18) − x (cid:19) ( − a − /q = x N/q ∞ X i =0 ( − i (cid:18) a/qi (cid:19) x − ri ! ∞ X j =0 ( − j (cid:18) /qj (cid:19) x − sj ∞ X k =0 ( − k (cid:18) − ( a + 1) /qk (cid:19) x − k ! = x N/q ∞ X n =0 a n x − n , where a n = X ri + sj + k = ni,j,k ≥ (cid:18) a/qi (cid:19)(cid:18) /qj (cid:19)(cid:18) − ( a + 1) /qk (cid:19) ( − i + j + k . MICHAEL A. BENNETT AND AARON LEVIN
Note that the series converges if | x | >
1. It follows that, for any solution ( w, x ) of(10) with | x | > w = ζx N/q ∞ X n =0 a n x − n , for some q th root of unity ζ . In our case, w and x are both real, so we must have(11) w = ± x N/q ∞ X n =0 a n x − n . Before we proceed further, we need to understand the coefficients a n somewhatbetter. Lemma 7.
Let n be a nonnegative integer. Then we have ord q a n = − n − ord q ( n !) , (12) q n + [ nq − ] a n ∈ Z , (13) and (14) | a n | ≤ ([ n/r ] + 1)([ n/s ] + 1) . Proof. If l is a nonnegative integer and m is an integer, coprime to a given prime q , then we have ord q (cid:18) m/ql (cid:19) = − l − ord q ( l !) . Let n ≥ i, j and k are nonnegative integers for which ri + sj + k = n . We thus haveord q (cid:18)(cid:18) a/qi (cid:19)(cid:18) /qj (cid:19)(cid:18) − ( a + 1) /qk (cid:19)(cid:19) = − i − j − k − ord q ( i ! j ! k !) . It follows, if i and j are nonnegative integers, not both zero, that − ord q (cid:0) − ( a +1) /qn (cid:1) + ord q (cid:16)(cid:0) a/qi (cid:1)(cid:0) /qj (cid:1)(cid:0) − ( a +1) /qk (cid:1)(cid:17) = ( r − i + ( s − j + ord q (cid:16) ( ri + sj + k )! i ! j ! k ! (cid:17) ≥ r − , since ( ri + sj + k )! i ! j ! k ! is an integer. We thus haveord q a n = ord q (cid:18)(cid:18) − ( a + 1) /qn (cid:19) ( − n (cid:19) = − n − ord q ( n !) , whereby q n +ord q ( n !) a n ∈ Z . Statement (13) follows upon observing thatord q ( n !) = ∞ X k =1 (cid:20) nq k (cid:21) ≤ (cid:20) nq − (cid:21) . To prove inequality (14), recall that 1 ≤ a ≤ q − | c | ≤ q implies | (cid:0) c/qi (cid:1) | ≤ i ≥ (cid:3) HE NAGELL-LJUNGGREN EQUATION VIA RUNGE’S METHOD 7
Let us now define P ( x ) = q N/q + [ Nq ( q − ] x N/q N/q X n =0 a n x − n . Note that, by Lemma 7, P ( x ) ∈ Z [ x ] and, since ord q a n +1 < ord q a n ≤ n , q N/q + [ Nq ( q − ] w ∓ P ( x ) = 0 . Let m = N/q + 1. Then from our definitions, Lemma 7 and (11), (cid:12)(cid:12)(cid:12) q N/q + [ Nq ( q − ] w ∓ P ( x ) (cid:12)(cid:12)(cid:12) = q N/q + [ Nq ( q − ] (cid:12)(cid:12) x N/q P ∞ n = m a n x − n (cid:12)(cid:12) ≤ q N/ ( q − | x | N/q P ∞ n = m ([ n/r ] + 1)([ n/s ] + 1) | x | − n . Since we have that ([( n + 1) /r ] + 1)([( n + 1) /s ] + 1)([ n/r ] + 1)([ n/s ] + 1) ≤ , for n ≥ m ≥ s > r ≥
3, where the inequality is sharp (corresponding to( r, s, n ) = ( r, r, r − (cid:12)(cid:12)(cid:12) q N/q + [ Nq ( q − ] w ∓ P ( x ) (cid:12)(cid:12)(cid:12) is bounded above by q N/ ( q − ([ m/r ] + 1)([ m/s ] + 1) | x | − P ∞ n =0 n | x | − n = q N/ ( q − ([ m/r ] + 1)([ m/s ] + 1) | x |− and hence | x | < q N/ ( q − ([ m/r ] + 1)([ m/s ] + 1) . From N = a ( r −
1) + s − ≤ ( q − r −
1) + s − , we find that mr < sqr and ms < , whereby the result follows. 4. Proof of Theorem 4
Let x, y, z, u, p and q be as in the statement of Theorem 4. We choose a and b to be the smallest nonnegative integers satisfying a ≡ p + 1 (mod q ) and b ≡ − p − q ) . Then we have, writing N = a ( p −
1) + b ( p −
1) + ( p − N ≡ a + 2 b + 3 ≡ q )and hence(15) (cid:18) x p − x − (cid:19) a x p − x − ! b x p − x − ! = w q , MICHAEL A. BENNETT AND AARON LEVIN for some integer w , where the left hand side is a polynomial in x with degreedivisible by q . We have (cid:18) x p − x − (cid:19) a x p − x − ! b x p − x − ! q = ( x p − a/q ( x p − b/q ( x p − /q ( x − ( − a − b − /q = x N/q (cid:18) − x p (cid:19) a/q (cid:18) − x p (cid:19) b/q (cid:18) − x p (cid:19) /q (cid:18) − x (cid:19) ( − a − b − /q = x N/q ∞ X n =0 b n x − n , where P ∞ n =0 b n x − n is the product of the four series ∞ X i =0 ( − i (cid:18) a/qi (cid:19) x − pi , ∞ X j =0 ( − j (cid:18) b/qj (cid:19) x − p j , ∞ X k =0 ( − k (cid:18) /qk (cid:19) x − p k and ∞ X l =0 ( − l (cid:18) − ( a + b + 1) /ql (cid:19) x − l . We may thus write b n = X pi + p j + p k + l = ni,j,k,l ≥ (cid:18) a/qi (cid:19)(cid:18) b/qj (cid:19)(cid:18) /qk (cid:19) (cid:18) − ( a + b + 1) /ql (cid:19) ( − i + j + k + l . Note here that we have 0 ≤ a, b ≤ q −
1, so that a + b + 1 ≤ q −
1. It follows that (cid:12)(cid:12)(cid:12)(cid:12)(cid:18) − ( a + b + 1) /ql (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) < l + 1and hence(16) | b n | < (cid:18)(cid:20) np (cid:21) + 1 (cid:19) (cid:18)(cid:20) np (cid:21) + 1 (cid:19) (cid:18)(cid:20) np (cid:21) + 1 (cid:19) ( n + 1) . The series P ∞ n =0 b n x − n thus converges if | x | > w, x ) of (15) with | x | >
1, we can write w = ± x N/q ∞ X n =0 b n x − n . Arguing as in the proof of Lemma 7, we find thatord q (cid:18)(cid:18) a/qi (cid:19)(cid:18) b/qj (cid:19)(cid:18) /qk (cid:19) (cid:18) − ( a + b + 1) /ql (cid:19)(cid:19) ≥ − i − j − k − l − ord q ( i ! j ! k ! l !) . We assume now that q ∤ ( a + b + 1) (or equivalently, p = q ). It will be clear fromour argument that one obtains even stronger bounds when q | ( a + b + 1) (note thatin this case, ( x p − a ( x p − b ( x p −
1) is a perfect q th power). Then, assuming q ∤ ( a + b + 1), after a little work, we conclude thatord q b n = ord q (cid:18) − ( a + b + 1) /qn (cid:19) = − n − ord q ( n !) , HE NAGELL-LJUNGGREN EQUATION VIA RUNGE’S METHOD 9 whence b n = 0 for each n and q n +ord q ( n !) b n ∈ Z . We thus have q n + [ nq − ] b n ∈ Z , for each n ≥
0. Define P ( x ) = q N/q + [ Nq ( q − ] x N/q N/q X n =0 b n x − n . We have that P ( x ) ∈ Z [ x ] and, since ord q b n +1 < ord q b n ≤ n , q N/q + [ Nq ( q − ] w ∓ P ( x ) = 0 . Setting as before m = N/q + 1, we thus have that | q N/q + [ Nq ( q − ] w ∓ P ( x ) | = q N/q + [ Nq ( q − ] (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x N/q ∞ X n = m b n x − n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) is a positive integer. On the other hand, via (16), the right hand side here isbounded above by q N/ ( q − | x | N/q ∞ X n = m ([ n/p ] + 1)( (cid:2) n/p (cid:3) + 1)( (cid:2) n/p (cid:3) + 1)( n + 1) | x | − n which is in turn at most q N/ ( q − ([ m/p ] + 1)([ m/p ] + 1)([ m/p ] + 1)( m + 1) | x | − ∞ X n =0 (7 / n | x | − n . Here, we have used that([( n + 1) /p ] + 1)( (cid:2) ( n + 1) /p (cid:3) + 1)( (cid:2) ( n + 1) /p (cid:3) + 1)( n + 2)([ n/p ] + 1)([ n/p ] + 1)([ n/p ] + 1)( n + 1) ≤ < / , with equality in the first of these inequalities corresponding to ( n, p ) = (26 , ≤ q N/ ( q − ([ m/p ] + 1)([ m/p ] + 1)([ m/p ] + 1)( m + 1) 1 | x | − / , whereby | x | < / q N/ ( q − ([ m/p ] + 1)([ m/p ] + 1)([ m/p ] + 1)( m + 1) . If we now use the fact that N ≤ ( q − p −
1) + ( q − p −
1) + p − q − p + p −
2) + p − , we find that m < p and m < ( p + p − q ≥ p, p /q − q < p. We conclude easily that | x | <
72 + 493 p (cid:18) max (cid:26) , pq (cid:27)(cid:19) q p + p − p − / ( q − , and the result follows. Proof of Theorem 2
We now prove Theorem 2. Let ( x, y, n, q ) be a solution of (1) with n odd andΩ( n ) ≥
4. As in Lemma 6, we write n = p α · · · p α k k where α i ∈ N and the p i are primes with p < p < · · · < p k . Note that k = ω ( n ), P ki =1 α i = Ω( n ), and that, by assumption, equation (8) holds.Define δ s as in Lemma 6, for 1 ≤ s ≤ k . We begin by supposing that δ s = 0 foreach value of s . Let us set p = (cid:26) p if α ≥ ,p if α = 1 , and define the positive integer m via n = mp p . From Lemma 6, we may findintegers y and y for which x n − x m − y y ) q and x mp − x m − y q . Applying Theorem 3 with r = p , s = p p , and using the fact that | x | ≥ q + 1, wethus have 2 m q m < (cid:26) , p q (cid:27) q p − p ( p − q − . Since q ≥ p ≥ p ≥ n ) ≥ m ≥ p ), it follows that2 p q p ≤ m q m < p q p + p ( p − q − and so 2 p q p ( p − q − / ( q − < p, an immediate contradiction.Next, let us suppose that δ s = 1 for at least one value of s . Choosing s to bethe smallest such index, we thus have that q | α s and, writing X = x β where β = Y j>s p α j j , and applying Lemma 6, X p αs − is − X p αs − i − s − p s y qi , for i = 0 , , . . . , α s − y i . Setting Z = X p αs − s , we may thus write Z p js − Z − p js z qj for j = 1 , , , where the z j are integers. Theorem 4 thus implies that(17) | Z | < p s (cid:18) max (cid:26) , p s q (cid:27)(cid:19) q p s + p s − p s − / ( q − . Since | Z | ≥ | x | p αs − s , HE NAGELL-LJUNGGREN EQUATION VIA RUNGE’S METHOD 11 and Proposition 1 enables us to conclude that | x | ≥ q + 1, we thus have2 p αs − s q p αs − s < p s (cid:18) max (cid:26) , p s q (cid:27)(cid:19) q p s + p s − p s − / ( q − . From p s ≥ q | α s , we reach an easy contradiction, at leastprovided either q ≥
7, or q ∈ { , } and α s = q . We may thus suppose that( q, α s ) = (3 ,
3) or (5 ,
5) (whereby, from the fact that ω ( n ) ≤ q −
2, we have k = 1and k ≤
3, respectively). We therefore have either n = p for p prime (so thatΩ( n ) < q, α s ) = (5 , s < k , then | Z | = | x | βp s > | x | p s ≥ p s , contradicting (17). We may thus assume that s = k . We will show that necessarily k = 1. Suppose otherwise. Then setting X = x p k and applying Lemma 6, we thushave X p k − X − p k y ,X p k − X − p k y and X p k − p k − X − p k y , for integers y , y and y . We choose integers a and b , with 0 ≤ a, b ≤
4, such that a ( p k −
1) + b ( p k −
1) + p k − p k − ≡ a + 2 b + 2 ≡ . This is always possible as det (cid:18) p k − p k −
11 2 (cid:19) = − ( p k − and p k (cid:18) X p k − X − (cid:19) a X p k − X − ! b X p k − p k − X − ! / = X N/ ∞ X n =0 c n X − n , where now N = a ( p k −
1) + b ( p k −
1) + p k − p k − . Here, we have c n = X (cid:18) a/ i (cid:19)(cid:18) b/ j (cid:19)(cid:18) / l (cid:19) (cid:18) − ( a + b + 1) / m (cid:19) ( − i + j + l + m , where the sum is over nonnegative integers i, j, l and m with p k i + p k j + p k − p k l + m = n. We note that a + b + 1 ≡ p k ≡ p k − ≡ p k − ≥
5, and thesecond, from using again Theorem 1.1 of [1]).As previously, ord c n = − n − ord ( n !) , n + [ n ] c n ∈ Z \ { } , for each n ≥
0, and(18) | c n | < (cid:18)(cid:20) np k (cid:21) + 1 (cid:19) (cid:18)(cid:20) np k (cid:21) + 1 (cid:19) (cid:18)(cid:20) np k − p k (cid:21) + 1 (cid:19) ( n + 1) . We have that w = p ( a +2 b +2) / k y a y b y = ± X N/ ∞ X n =0 c n x − n and hence writing P ( X ) = 5 N/ N/ x N/ N/ X n =0 c n X − n , that 5 N/ N/ w ∓ P ( X ) is a nonzero integer. Writing m = N/ ≤ N/ | X | N/ ∞ X n = m ([ n/p k ] + 1)( (cid:2) n/p k (cid:3) + 1)( (cid:2) n/p k − p k (cid:3) + 1)( n + 1) | X | − n . Writing T n = ([ n/p k ] + 1)( (cid:2) n/p k (cid:3) + 1)( (cid:2) n/p k − p k (cid:3) + 1)( n + 1), we have that T n +1 /T n ≤ / < , provided n ≥ ≤ p k − < p k (with equality in the first inequality if n =244 , p k − = 5 and p k = 7). We thus have that1 ≤ N/ ( m/p k + 1)( m/p k + 1)( m/p k − p k + 1)( m + 1) | X | − ∞ X n =0 n | X | − n , whence | X | < N/ ( m/p k + 1)( m/p k + 1)( m/p k − p k + 1)( m + 1) . Since N ≤ p k + 4 p k + p k − p k − ≤ p k − , we have | X | < p k (cid:18) p k + 1 (cid:19) (cid:18) p k + 1 (cid:19) · · (cid:18) p k + 1 (cid:19) . On the other hand, | X | = | x | p k ≥ p k , a contradiction, since p k ≥
7. Our conclusion is thus that(19) n = p for p prime, q = 5 and x ≡ p ) . To finish the proof of Theorem 2, it remains to treat case (19); we are currentlyunable to do so without appeal to the results of [10]. If we have a solution to x p − x − y in integers x and y and prime p with | x | > x ≡ p ), then x p k − x p k − − p y k for integers y k , 1 ≤ k ≤
5. Appealing to Theorem 2 of Mih˘ailescu [10] and the factthat | x | > , it follows that 10 p < | x | p < p , HE NAGELL-LJUNGGREN EQUATION VIA RUNGE’S METHOD 13 a contradiction which completes the proof of Theorem 2.
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