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Let n be a positive integer. In this paper, using the results on the existence of primitive divisors of Lucas numbers and some properties of quadratic and exponential diophantine equations, we prove that if n ≡ 3 (mod 6), then the equation x2 + (3n2 + 1)y = (4n2 + 1)z has only the positive integer solutions (x, y, z) = (n, 1, 1) and (8n3 + 3n, 1, 3).

The exponential Diophantine equation x2 + (3n 2 + 1)y = (4n 2 + 1)z