IMO 1991 SL 17

Find all positive integer solutions x, y, z of the equation 3x +

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IMO 1991 SL 17

Origin: HKG

Problem

Find all positive integer solutions x, y, z of the equation 3x + 4y = 5z.

Solution

Taking the equation 3x + 4y = 5z (x, y, z > 0) modulo 3, we get that 5z \equiv1 (mod 3), hence z is even, say z = 2z1. The equation then becomes 3x = 52z1 −4y = (5z1 −2y)(5z1 + 2y). Each factor 5z1 −2y and 5z1 + 2y is a power of 3, for which the only possibility is 5z1 +2y = 3x and 5z1 −2y =

  1. Again modulo 3 these equations reduce to (−1)z1 + (−1)y = 0 and (−1)z1 −(−1)y = 1, implying that z1 is odd and y is even. Particularly, y \geq2. Reducing the equation 5z1 + 2y = 3x modulo 4 we get that 3x \equiv1, hence x is even. Now if y > 2, modulo 8 this equation yields 5 \equiv5z1 \equiv 3x \equiv1, a contradiction. Hence y = 2, z1 = 1. The only solution of the original equation is x = y = z = 2.