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Sum of bad integers to the power of 2019

Source: China TST 2019 Test 2 Day 2 Q6

March 11, 2019
number theoryChina TST

Problem Statement

Given coprime positive integers p,q>1p,q>1, call all positive integers that cannot be written as px+qypx+qy(where x,yx,y are non-negative integers) bad, and define S(p,q)S(p,q) to be the sum of all bad numbers raised to the power of 20192019. Prove that there exists a positive integer nn, such that for any p,qp,q as described, (p−1)(q−1)(p-1)(q-1) divides nS(p,q)nS(p,q).
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