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**

We find that (a + b)^{7} - a^{7} - b^{7} = 7ab(a + b)(a^{2} + ab + b^{2})^{2}. So we must find a, b such that a^{2} + ab + b^{2} is divisible by 7^{3}.

At this point I found a = 18, b = 1 by trial and error.

A more systematic argument turns on noticing that a^{2} + ab + b^{2} = (a^{3} - b^{3})/(a - b), so we are looking for a, b with a^{3} = b^{3} (mod 7^{3}). We now remember that a^{φ(m)} = 1 (mod m). But φ(7^{3}) = 2·3·49, so a^{3} = 1 (mod 343) if a = n^{98}. We find 2^{98} = 18 (343), which gives the solution 18, 1.

This approach does not give a flood of solutions. n^{98} = 0, 1, 18, or 324. So the only solutions we get are 1, 18; 18, 324; 1, 324.

Solutions are also available in Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.

© John Scholes

jscholes@kalva.demon.co.uk

19 Oct 1998