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Irish Mathematical Olympiad, 2014

Number theory Difficulty 4.8 AIME Prove it Ireland

Prove for all integers $N > 1$ that $(N^2)^{2014} - (N^5)^{106}$ is divisible by $N^3 - 1$ .

Solution

Observe that $2 \cdot 2014 - 5 \cdot 106 = 3498 = 3 \cdot 1166$ , hence
$(N^2)^{2014} - (N^5)^{106} = (N^5)^{106} ((N^3)^{1166} - 1) \\ = (N^5)^{106} (N^3 - 1) ((N^3)^{1165} + (N^3)^{1164} + \dots + 1)$
is divisible by $N^3 - 1$ .

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