摘要

Let a, b, c, m be positive integers such that a + b = c(2), 2 inverted iota c, m > 1 and m equivalent to +/- 1 (mod c). We prove that if a 4 or 5 (mod 8), ((a + 1)/c) = -1 and m > 6c(2) log c, where ((a + 1)/c) is the Jacobi symbol, then the equation (am(2) + 1)(x) + (bm(2) - 1)(y) = (cm)(z) only has the positive integer solution (x, y, z) = (1,1, 2).

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