/compile/output.dvi On the Exponential Diophantine Equation (Mpq) x + (Mpq + 1)y = z2 Azizul Hoque Department of Mathematics, Gauhati University, Guwahati, India-781014 Abstract. In this paper, we consider the number Mpq = pq−1, where p > 0 and q > 1 are integers, and the Exponential Diophantine equation (Mpq) x +(Mpq+1)y = z2, where x , y and z are positive integers. We find the solutions to the title equation expect the case only when both p and y are odd integers. 2010 Mathematics Subject Classifications: 11D61; 11D41 Key Words and Phrases: Exponential Diophantine Equation 1. Introduction Sroysang [2] established that the Exponential Diophantine equation 31x+32y = z2 has no non-negative solution. Recently, Sroysang [3] also showed that the Exponential Diophantine equation 7x+8y = z2 has only one solution, that is (x , y, z) = (0,1,3). Sroysang [3] introduced an open problem regarding the set of all solutions (x , y, z) for the Exponential Diophantine equation px + (p+ 1)y = z2, where x , y and z are non-negative integers. In this paper, we consider the number Mpq = pq − 1, where p > 0 and q > 1 are integers, and the Exponential Diophantine equation (Mpq) x + (Mpq + 1)y = z2, where x , y and z are positive integers. We show that (Mpq, x , y, z) = (7,0,1,3) and (Mpq, x , y, z) = (3,2,2,5) are the only solutions to the above equation except the case when both p and y are odd integers. 2. Main Results In this article, we use Catalan’s conjecture [1], which states that the only solution in inte- gers a > 1, b > 1, x > 1, y > 1 to the equation ax − b y = 1 is (a, b, x , y) = (3,2,2,3). We shall now solve the exponential Diophantine equation (Mpq) x+(Mpq+1)y = z2, where x , y, z, p,q are non-negative integers and Mpq = pq − 1 with q > 1. We exclude the case when both p and y are odd positive integers. Email address: ahoque.ms@gmail.com http://www.ejpam.com 240 c© 2016 EJPAM All rights reserved. EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 9, No. 2, 2016, 240-243 ISSN 1307-5543 – www.ejpam.com A. Hoque / Eur. J. Pure Appl. Math, 9 (2016), 240-243 241 Theorem 1. The Exponential Diophantine equation (Mpq) x + (Mpq + 1)y = z2 (1) except the case when both p and y are odd positive integers, has only two solutions in non-negative integer, (Mpq, x , y, z) = (7,0,2,3) and (Mpq, x , y, z) = (3,2,2,5). Proof. We divide the problem into two cases. Case 1: Let p be an even positive integer. Then Mpq ≡ 3(mod 4). From Eq.(1) we observe that z must be odd, and thus z2 ≡ 1(mod 4) and Mpq + 1 ≡ 0(mod 4). Sub-case 1.1: Let x = 0, then Eq. (1) becomes 1+ (Mpq + 1)y = z2. This gives pq y = z2−1 and thus pq y = (z+1)(z−1). Hence there exists non-negative integers m and n such that pm = z + 1 and pn = z − 1, where m> n and m+ n= q y (2) Now we have, pn(pm−n − 1) = pm − pn = (z + 1)− (z − 1) = 2. This implies p = 2, n= 1 and m= 2. Thus Eq. (2) gives q y = 3 and hence either q = 1, y = 3 or q = 3, y = 1. Since q > 1, so that q = 3, y = 1. Now z = pn + 1 = 3 and Mpq = 7. Hence (Mpq, x , y, z) = (7,0,1,3) is the only solution to the Eq. (1) in this sub-case. Sub-case 1.2: Let x ≥ 1. Since (Mpq + 1)y ≡ 0(mod 4) and z2 ≡ 1(mod 4), the Eq. (1) gives (Mpq) x ≡ 1(mod 4). Again since Mpq ≡ 3(mod 4), x must be even. Let x = 2k for some integer k ≥ 1. Then Eq. (1) implies (Mpq) 2k + pq y = z2. This gives pq y = z2 − (Mpq k)2 and thus pq y = (z + Mpq k)(z − Mpq k). Hence there exists non- negative integers i and j such that pi = z +Mpq k and p j = z −Mpq k, where i > j and i + j = q y. (3) Now we have p j(pi− j − 1) = pi − p j = 2(Mpq) k. Since p is even, let p = 2t for some positive integer t. Then we have 2 j−1 t j(pi− j − 1) = (Mpq) k. (4) If t > 1 then t | (Mpq) k and hence p | 2(Mpq) k. Since gcd(p, Mpq) = 1, we have p | 2, a contradiction. Hence t = 1 and p = 2. A. Hoque / Eur. J. Pure Appl. Math, 9 (2016), 240-243 242 Now Eq. (4) gives, j = 1 and it becomes, pi−1 − 1= (Mpq) k (5) By using Catalan’s Conjecture, the equation pi−1 − (Mpq) k = 1 has only one solution (p, Mpq, i − 1, k) = (3,2,2,3) only when i > 2 and k > 1. But since p = 2, Eq. (5) has no solution only when i > 2 and k > 1. It is now remaining to examine only when either i ≥ 2 or k ≥ 1. But we have i > 1, q > 1, k ≥ 1 and Eq. (3) gives i + 1= q y . Thus we get i = 2, q = 3, y = 1 or k = 1. Now if i = 2, q = 3 and y = 1, then Eq.(5) gives, p− 1= (Mpq) k⇒ 1= (Mpq) k⇒ k = 0 This contradicts to k ≥ 1. Hence Eq. (1) has no solution in this case. Again, if k = 1, then Eq. (5) gives pi−1 − 1=Mpq ⇒ pi−1 − 1= pq − 1 ⇒ i − 1= q ⇒ q y − 2= q ⇒ q(y − 1) = 2 ⇒ q = 2, y = 2. Thus we have Mpq = 3, x = 2k = 2, y = 2 and z = p j + (Mpq) k = 5. Therefore (Mpq, x , y, z) = (3,2,2,5) is the only solution to Eq. (1) in this sub-case. Case 2: Let p be an odd positive integer. Then Mpq ≡ 0(mod 4). From Eq. (1) we observe that z must be odd, and thus z2 ≡ 1(mod 4). Sub-case 2.1: Let y = 0. Then Eq. (1) becomes (Mpq) x + 1= z2. This implies (Mpq) x = (z+1)(z−1) and thus there exists non-negative integers a, b such that (Mpq) a = z+1 and (Mpq) b = z − 1, where a > b and x = a+ b. Now (Mpq) b(M a−b pq − 1) = (Mpq) a − (Mpq) b = 2. This gives 2≡ 0(mod 4), an absurdity. Thus there is no solution to Eq. (1) in this sub-case. Sub-case 2.2: Let y ≥ 1 even integer and let y = 2k. Then Eq. (1) becomes (Mpq) x + (Mpq + 1)2k = z2. This equation implies (Mpq) x = z2 − (pkq)2 = (z + pkq)(z − pkq). Thus there are non-negative integers c, d such that (Mpq) c = z + pkq and (Mpq) d = z − pkq, where c > d and c + d = x . Now, (Mpq) d(M c−d pq − 1) = (Mpq) c − (Mpq) d = 2pkq = 2(Mpq + 1)k. This implies 0≡ 2(mod 4). This is an absurdity. Hence Eq. (1) has no solution in this case. REFERENCES 243 ACKNOWLEDGEMENTS The Author acknowledges UGC for JRF Fellowship (No.GU/ UGC/VI(3)/JRF/2012 /2985). References [1] P. Mihailescu. Primary cyclotomic units and a proof of Catalan’s conjecture, Journal für die reine und angewandte Mathematik, 27, 167-195. 2004. [2] B. Sroysang. On the Diophantine equation 31x + 32y = z2, International Journal of Pure and Applied Mathematics, 81, 609-612. 2012. [3] B. Sroysang. On the Diophantine equation 7x +8y = z2, International Journal of Pure and Applied Mathematics, 84, 111-114. 2013.