1984 IMO Problems/Problem 2


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 1

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]

Solution 2

Lemma: $a^{7} \equiv a (mod 7)$.

Proof: Recall that if $7 \nmid x$, then $x^{3} \equiv 1, -1 (mod 7)$.

Therefore, $x^{6} \equiv 1 (mod 7)$.

$\implies$ $a^{7} \equiv a (mod 7)$ $\forall$ $a \not\equiv 0 (mod 7)$.

However, if $7|x$, then $a^{7} \equiv a (mod 7)$.

So now, we need to find one pair of integers (a, b) such that $(a+b)^{7} - a^{7} - b^{7} \equiv 0 (mod 7)$. This means $(a+b)^{7} \equiv a^{7} + b^{7} (mod 7)$. $\implies (a+b) \equiv a+b (mod 7)$. But this is true for all pairs of integers (a, b). So any random pair of integers would work.

Footnote: Even a pair of integers (a, b) which satisfies $7|ab(a+b)$ would work. So the condition given is irrelevant. Try it!

See Also

1984 IMO (Problems) • Resources
Preceded by
Problem 1
1 2 3 4 5 6 Followed by
Problem 3
All IMO Problems and Solutions