MathDB
Number of solutions of a diophantine equation

Source: Czech and Slovak Olympiad 1987, National Round, Problem 2

April 10, 2020
number theoryDiophantine equationprime numbersnumber of solutionsnational olympiad

Problem Statement

Given a prime p>3p>3 and an odd integer n>0n>0, show that the equation xyz=pn(x+y+z)xyz=p^n(x+y+z) has at least 3(n+1)3(n+1) different solutions up to symmetry. (That is, if (x′,y′,z′)(x',y',z') is a solution and (x′′,y′′,z′′)(x'',y'',z'') is a permutation of the previous, they are considered to be the same solution.)
Back to ProblemsView on AoPS