1984 IMO Problems/Problem 2

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 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!

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1984 IMO Problems/Problem 2

Contents

Problem

Solution 1

Solution 2

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