On the Diophantine Equation x 2 = 4q n 4q m + 9

Transcription

1 JKAU: Sci., Vol. 1 No. 1, : (009 A.D. / 1430 A.H.) On the Diohantine Equation x = 4q n 4q m + 9 Riyadh University for Girls, Riyadh, Saudi Arabia abumuriefah@yahoo.com Abstract. In this aer, we study the title equation with q any rime and n > m 0, and we give a comlete solution when m > 0. Keywords: Diohantine equation, rimitive divisor, Lehmer air. Introduction In 1913 the Indian mathematician S. Ramanujan [1] conjectured that the equation x = n 7, had the only following solutions: n = x = This conjecture was first roved by Nagell [] in There followed during the eriod , a number of roofs based on a variety of methods (see for examle [3,4] ). Ramanujan equation has the general form x q n q D = 4 4 m +, where D is any odd integer. The urose of this aer is to solve n m x q q = (1) with x > 0, n m 0, where q is any rime, it is clear that x is an odd integer. To solve equation (1) we will use unique factorization of ideals along with linear recurrences and congruences and the BHV Theorem [5]. We start with the case n = 1 and m =

2 136 Case I Let (n, m) = (1,0), in equation (1), then we have x 5 q =. 4 Since x is odd, let x = c + 1, then we get q c c = + 1, () where c is a ositive integer. It is not known if equation () has infinitely many solutions. Case II Let d = g.c.d(m,n), q 1 = q d, n 1 = n/d, m 1 = m/d, in equation (1), then we get the same kind of equation (1) n m 1 1 x q q 1 1 = , with n 1,m 1 are corime. So we shall suose (n, m) = 1, which means that n m. Case III If m = 0, and n is an even, then equation (1) has no solutions. So we shall exclude all the above cases. Now we suose the case m > 0, and get the following: Theorem The diohantine equation n m x q q has the following two cases: = , n > m, (3) i. m =1: When q=, then it has a unique solution given by (x,n) = (3,1), otherwise it has at most two solutions

3 On the Diohantine Equation x =4q n -4q m ii. m > 1: (a) When m is odd, it has solutions only if q = 3, and these solutions are given by (x,n,m) = (93,7,3), (.3 m 1 3, (m-1), m). (b) When m is even, it has solutions only if q = and m =, and these solutions are given by (x,n) = (1,1), (3,), (5,3), (11,5), (181,13). Proof (i) Let m = 1 in (3). If q =, then we get equation x 1= n + solution (x,q) = (3,). If q, then the equation at most two solutions [6]. (ii) Let m > 1 in (3). We start by writing m n, which is clear has a unique x q n q = , has 4q 9= 4q x = Aa. (4) Where A 1 is an odd square free. Suose divides m, where m is odd and ut q 1 =q m/, we get: 1 4q 9= Aa (5) If A = 1, then a 1(mod 4), but 1 is not quadratic residuce modulo 4, therefore A 1. Now if A = 3, then q = 3, and dividing equation (4) by 3, we get the equation y =4q n 4q m +1, which have been solved by Luca [7], and the only solution in our case (q = 3) is n = 7, m = 3 and y = 31, so x = 93. Also Luca refer to the case n = (m ) as the trivial solution of this equation, and this will give us the solution x =.3 m 1 3, as desired. Hence we shall suose that q 1 5, therefore A 5 and A / 0(mod 3). We write (5) as

4 138 Suose q 1 (3 + Aa) (3 Aa) q =. (6) 1 = ππ, where π is a rime ideal, therefore the two algebraic integers aearing in the right-hand side of (6) are corime in the ring Q( A). Then This imlies that π 3+ Aa 3 π π Aa =. 3+ Aa = and 3 π Aa =. So π is a rincial ideal which imlies O(π ), hence π is a rincial ideal. and Let c+ b A z =, where c b(mod ), is a generator of π then we get q = z z z 3+ Aa =, 3 Aa z =. Since the units in the field Q( A) are ± 1, therefore Hence 3 + Aa 3 Aa ± z =, ± z =. u u = z + z =± 3. (7) From (7) we get that u =± 3 u which imlies that u has no rimitive divisors.

5 On the Diohantine Equation x =4q n -4q m Let = 3 in equation (7), then we get Or c ± = z + z = Acb. 4 ± 1 = cc ( 3 Ab) (8) If c is even, then so b is even, and in this case the right-hand side of (8) is a multile of 8, which is imossible. Thus c is an odd divisor of 1, therefore c = ± 3, ± 1. From equation (8) we now conclude that 3Ab = 5, 11, ±13, which is obviously imossible. Assume now that 5, in this case, u has no rimitive divisors. From Table and 3 in [5] and a few excetional values of z. None of the excetional Lehmer terms from that Table leads to a value of z Q( A). Thus equation (3) has no solutions when m is odd and q 5. Now let us suose that m is even, say m = k, and k is a ositive integer. From equation (4), we get 9 4 n 4 k x = q q, which imlies that x + 3 x 3 = k n k. q ( q 1). Since the two factors in the left hand side are corime we get q = and m =. Substituting in (4) we find the famous equation of Ramanujan x = n+ 7, which has only the following solutions [] This concludes the roof. (x,n) = (1,1), (3,), (5,3), (11,5), (181,13). References [1] Ramanujan, S., Collected Paers of Ramanujan, (Cambridge Univ. Press, Cambridge, (197). [] Nagell, T., "The diohantine equation x + 7 = n ", Nordisk. Mat. Tidsker, 30, 6-64, Ark. Mat., 4 (1960), [3] Mordell, L. J., Diohantine Equations, (Academic Press, London, (1969)). [4] Skolem, Th., Chowla, S. and Lewis, J., "The diohantine equation x = n+ 7 and related roblems", Proc. Amer. Math. Soc., 10 (1959),

6 140 [5] Bilu, Y., Hanrot, G. and Voutier, P.M., "Existence of rimitive divisor of Lucas and Lehmer numbers", J. Reine Angew. Math., 539 (001), [6] Maohua, L., "On the diohantine equation x + D = 4 n ", J. Number Theory, 41 (199), [7] Luca, F., "On the diohantine equation x = 4q n 4q m +1, Proc. Amer. Math. Soc., 131 (00),

7 On the Diohantine Equation x =4q n -4q m x =4q n -4q m +9 abumuriefah@yahoo.com :. x =4q n -4q m +9 n > m 0 q. m>0