# 1984 IMO Problems/Problem 2

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## Problem

Find one pair of positive integers $a,b$ such that $ab(a+b)$ is not divisible by $7$, but $(a+b)^7-a^7-b^7$ is divisible by $7^7$.

## Solution

So we want $7 \nmid ab(a+b)$ and $7^7 | (a+b)^7-a^7-b^7 = 7ab(a+b)(a^2+ab+b^2)^2$, so we want $7^3 | a^2+ab+b^2$. Now take e.g. $a=2,b=1$ and get $7|a^2+ab+b^2$. Now by some standard methods like Hensels Lemma (used to the polynomial $x^2+x+1$, so $b$ seen as constant from now) we get also some $\overline{a}$ with $7^3 | \overline{a}^2+\overline{a}b+b^2$ and $\overline{a} \equiv a \equiv 2 \mod 7$, so $7\nmid \overline{a}b(\overline{a}+b)$ and we are done. (in this case it gives $\overline{a}=325$)

This solution was posted and copyrighted by ZetaX. The original thread for this problem can be found here: [1]