Find one pair of positive integers a, b such that ab(a+b) is not divisible by 7, but (a+b)7 - a7 - b7 is divisible by 77.
Solution
We find that (a + b)7 - a7 - b7 = 7ab(a + b)(a2 + ab + b2)2. So we must find a, b such that a2 + ab + b2 is divisible by 73.
At this point I found a = 18, b = 1 by trial and error.
A more systematic argument turns on noticing that a2 + ab + b2 = (a3 - b3)/(a - b), so we are looking for a, b with a3 = b3 (mod 73). We now remember that aφ(m) = 1 (mod m). But φ(73) = 2·3·49, so a3 = 1 (mod 343) if a = n98. We find 298 = 18 (343), which gives the solution 18, 1.
This approach does not give a flood of solutions. n98 = 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