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at least 3n + 3 different solutions

Source: Bundeswettbewerb Mathematik 1988, stage 2, problem 4

September 6, 2003
number theoryDiophantine equationequationalgebraIMO Shortlist

Problem Statement

Provided the equation xyz=pn(x+y+z)xyz = p^n(x + y + z) where p≥3p \geq 3 is a prime and n∈Nn \in \mathbb{N}. Prove that the equation has at least 3n+33n + 3 different solutions (x,y,z)(x,y,z) with natural numbers x,y,zx,y,z and x<y<zx < y < z. Prove the same for p>3p > 3 being an odd integer.
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