On the Diophantine Equation ax2 + (3a+1)m = (4a+1)n

Let a be a positive integer, and let p be an odd prime such that p does not divide a(3a +1) and 4a + 1 is a power of p. In this paper, by the deep result of Bilu, Hanrot and Voutier, i.e. the existence of primitive prime factors of Lucas and Lehmer sequences, by the computation of Jacobi's symbol and by elementary arguments, we prove that: if a not equal 1, 2, then the Diophantine equation of the title has at most two positive integer solutions (x, m, n). Moreover, the diophantine equations x(2) + 4(m) = 5(n) and 2x(2) + 7(m) = 9(n) have precisely three positive integer solutions (x, m, n).