Fadwa S. Abu Muriefah. ON THE DIOPHANTINE EQUATION x fc = y n

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1 DEMONSTRATIO MATHEMATICA Vol. XXXIX No Fadwa S. Abu Muriefah ON THE DIOPHANTINE EQUATION x fc = y n Abstract. In this paper we prove that the title equation where k > 0 and n > 3, may have a solution in integers (x, y, k,n) only if 5\x and p \ k, where p any odd prime dividing n, by using a recent result of Bilu, Hanrot and Voutier [3], 1. Introduction The first results regarding the diophantine equation x fe = y n, for general n is due to Lebesgue [4], who proved that when k = 0, this equation has no solution in positive integers x,y and n > 3. When k = 1, Ljunggren [5], proved that it has no solution when (x,5) = 1. In [2], Bender and Herzberg prove that the equation ax k d = p n, where (x, 5) = 1, p an odd prime, d > 1, positive integer and max (d,p) > 7, has at most one integer solution. Recently we proved in [1] with Arif, that if the tittle equation has a solution, where (x, 5) = 1, then n = p = 19(mod20). Our aim in this paper is to generalize these results. 2. Preliminaries We start by giving some important definitions DEFINITIONS. A Lucas pair is a pair (a, (i) of algebraic integers such that a + P and a(3 are non -zero co-prime rational integers and a/ 3 is not a root of unity. Given a Lucas pair (a, 3) one defines the corresponding sequences of Lucas numbers by a n - (3 n u n (a, 3) = n > 0. a (3 A prime number p is a primitive divisor of u n (a, 3) if p divides u n, but does not divides (a [3) 2 U\U2... u n -\ Mathematics Subject Classification: 11D41, 11D61. Key words and phrases: Diophantine equations; Lucas sequence; primitive divisor.

2 286 F. S. Abu Muriefah A Lucas pair (a, /3) such that u n (a, 3) has no primitive divisor will be called n defective Lucas pair. Now we reproduce the following Lemma for future use LEMMA 1 (Bilu, Hanrot and Voutier [3]). For n > 30, the n-th term of any Lucas sequences has a primitive divisor. For 3< n < 30, all values of pairs have been listed in [3] for which the n th term of Lucas sequences has no primitive divisor. THEOREM 1. The diophantine equation (1) X k = y n, k> 0 and n > 3, where (5, x) = 1, has no solution in integers. Proof. Since the result is known when k = 0 (see [4]), we shall suppose k > 0. If x is odd and y even we get x k = 2(mod8), but y n = 0(mod8). So we assume that y is odd. First let n be odd. Then there is no lose of generality in considering only n = p an odd prime. Thus factorising equation (1) in Q(i), we get (x + 5 k i){x-5 k i)=y p. Since the factors on the left hand side are co-prime and any unit of Q(i) is a p th power, we can write (2) x + 5 k i = (a + ib) p, where a,b G Z and y = a? + b 2 is odd, so a and b have the opposite parity. From (2) we get (3) x = a a p ~ 3 b 2 + ± So (5, a) = 1. Also from (2) we have 2=1 If b = ±1, then (4) becomes We have two cases:

3 On the diophantine equation x k = y n If a 2 = l(mod5), then considering equation (5) modulo 5, we get p-i 2 + V 21 4 which is a contradiction, since the right hand side does not congruent to 0 modulo If a 2 = l(mod5) then considering equation (5) modulo 5 we get p o r=o \2r + 1/ again contradiction, since the right hand side does not congruent to 0 modulo 5. So 6=±5 J, in equation (4), where 0 < j < k. Hence (a,b) = 1 and a even. Dividing (4) by S 7, we get Erl (6) = If k j > 1, considering (6) modulo 5, implies p = 5. Substituting in (6) we get (7) iô*^- 1 = a 4-2 x 5 2j x a 2 + Since (5, a) = 1, equation (7) is impossible modulo 5, unless k j l = 0 and the positive sign holds. So 1 = a 4-2 x 5 2 ( fc_1 ) x a fe+3, this equation is impossible modulo 8, since a even. Therefore b = ±5 k. Let From (2) we get a = a + &>/ 1 3 = a - b\f-i. Now ^ is a root of the equation yz 2 2(a 2 b 2 )Z + y = 0. Also 0/) 2(a 2 ft 2 )) = (a 2 + b 2, a 2 b 2 ) = 1, and y > 1, so ^ is not a root of unity. Moreover {a + (3, a/3) - (2a,a ) = 1. Therefore the sequences ut = a azp i i > 0 are a Lucas sequences. So from (8) we get u p (a,f3) = ±1.

4 288 F. S. Abu Muriefah We conclude that our Lucas pair (a, is p defective. Using the Lemma we get p < 30. In this case Ut has a primitive divisor for all primes values of p > 13. Moreover, for p = 3,5,7,13, there are precisely 12 Lucas sequences for which Up does not have a primitive divisor, one can easily see that none of these 12 sequences has the property that the roots of characteristic equation are in Q(i) (see Table 1 and 3 in [3]). When p = 11, there is no Lucas number which has no primitive divisors. It follows that the equation (1) has no solution for all p. Now if n even then from the above it is sufficient to consider n = 4. Hence y 2 + x = 5 2k and y 2 - x = 1. Adding these two equations we get 2 y 2 = 5 2k + 1, which is impossible modulo 5. Now we consider the case 5 ar. From proof of Theorem 1 it sufficient to consider n = p an odd prime. THEOREM 2. The diophantine equation x k may have a solution only if 5 x and = y p, p an odd prime Proof. Here 5 y, so let x = 5 U X, y = 5 V Y, where u > 0,v > 0 and (5,X) = (5,y) = 1. Then (1) becomes p\k. (9) (5 U X) k = 5 pv Y p. We have three cases: 1. 2u = min(2u,pv,2k). Then canceling 5 2u in (9) we get g2(fc-u) _ gpu-2uyp If k u = 0 and pv 2u = 0, we get the famous equation of Lebesgue which has no solution [4]. If k u > 0 and pv 2u = 0 then the equation has no solution from Theorem 1. Finally if only k u = 0 we get (10) X = 5^-2 k Y p. If p\k then we can write (10) as X 2 which has no solution. 2. 2k = min(2u, pv, 2k). Then canceling 5 2fc in (9) we get + 1 = (5 V ~TY) P (5 U X) = 5Pu~ 2k Y p.

5 On the diophantine equation x k = y n 289 Considering this equation modulo 5, we get either pv 2k = 0 then (h u X) = Y p which has no solution [4], or pv 2u > 0 and then we get the same equation (10). 3. pv = mm(2u,pv,2k). This case when treated like the previous two cases, does not give us any solutions. This completes the proof. T References [1] F. S. Abu Muriefah and S. A. Arif, The diophantine equation x 2 + q 2k = y p, Arab. J. Sci. Eng. Section A. 26 (2001), [2] E. A. Bender and N. P. Herzberg. Some diophantine equations related to the quadratic form ax 2 + by 2, in Studies in Algebra and Number Theory, Adv. Math. Stud., Academic Press, New York, (1979), [3] Y. Bilu, G. Hanrot and P. M. Voutier, Existence of primitive divisors of Lucas and Lehmer numbers, J. Reine Angew. Math. 539 (2001), [4] V. A. Lebesgue, Sur l'impossible'en nombres entiers de l'équation x m = y 2 +1, Nouv. Ann. des. Math. 1 (1850), [5] W. Ljunggren, On the diophantine equation x 2 +p 2 = y n, Kong. Norsk. Vid. Selskab. Forh. Trondhein. 16 (1943), P.O. Box 60561, RIYADH 11555, SAUDI ARABIA abumuriefah@yahoo.com Received December 5, 2004; revised version May 5, 2005.

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