On the Exponential Diophantine Equation $$(m^2+m+1)^x+m^y=(m+1)^z $$ ( m 2 + m + 1 ) x + m y = ( m + 1 ) z
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On the Exponential Diophantine Equation (m2 + m + 1)x + my = (m + 1)z Murat Alan Abstract. Let m ≥ 1 be a positive integer. We show that the exponential Diophantine equation (m2 + m + 1)x + my = (m + 1)z has no positive integer solutions other than (x, y, z) = (1, 1, 2) when m ∈ {1, 2, 3}. Mathematics Subject Classification. Primary 11D61; Secondary 11D75. Keywords. Exponential Diophantine equations, m-adic valuation.
1. Introduction Let u, v, w be relatively prime positive integers greater than one and assume that the exponential Diophantine equation ux + v y = w z
(1.1)
in positive integers x, y, z has a solution (x0 , y0 , z0 ). Two famous conjectures related to uniqueness of this solution (x0 , y0 , z0 ) are due to Je´smanowicz and Terai with some restriction on (1.1). In 1956, Je´smanowicz conjectured that if u, v and w are any Pythagorean triples, i.e., positive integers satisfying u2 + v 2 = w2 , then the solution (x0 , y0 , z0 ) = (2, 2, 2) is the unique solution of (1.1) [5]. Another similar conjecture is proposed by Terai which states that if u, v, w, p, q, r are fixed positive integers satisfying up + v q = wr with u, v, w, p, q, r ≥ 2, then the Eq. (1.1) has unique positive integer solution (x0 , y0 , z0 ) = (p, q, r) [19,20]. Exceptional cases are listed explicitly in [24]. Although both conjectures are proved to be true in many special cases, see for example [1,3,4,6,8,10–13,18,21–23,25], they are still remain an unsolved problem yet. We refer to [9,17] for a detailed information on these two conjectures. In this note we study the exponential Diophantine equation (m2 + m + 1)x + my = (m + 1)z
(1.2)
where m > 1 is a positive integer, and we prove the following. Theorem 1.1. Let m > 1 be a positive integer. If m > 3 then the Eq. (1.2) has only the positive integer solution (x, y, z) = (1, 1, 2). For m = 2 and m = 3 0123456789().: V,-vol
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the Eq. (1.1) has exactly two solutions, namely (x, y, z) = (1, 1, 2), (2, 5, 4) and (x, y, z) = (1, 1, 2), (1, 5, 4), respectively. In the above theorem, we exclude the case m = 1 just for preserving the exponent in the expression my . In fact, it is easy to see that the equation 3x +1 = 2z has only the positive integer solution (x, z) = (1, 2) by considering it modulo 8. For the next two values of m, the Eq. (1.2) turns into the equations 7x + 2y = 3z and 13x + 3y = 4z , for which both of them have more than one solution [14,26]. So the aim of this study is to give an answer to the question whether or not the Eq. (1.2) has any positive integer solutions other than (x, y, z) = (1, 1, 2) when m > 3. The proof depends on elementary congruence considerations and some results on linear forms in two m−adic logarithms.
2. Proof of Theorem 1.1 Lemma 2.1. Let (x, y, z) be a positive integer solution of the Eq. (1.2). The following conditions hold. 1. y is odd. 2. There exists an integer t such that |x − y| = (m + 1)t. Proof. 1. By reducing Eq. (1.2) modulo m + 1 we get that 1 + (−1)y ≡ 0 (mod (m + 1)) which implies that y is odd since m
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