Let a and b be positive integers. When a2 + b2 is divided by a + b, the quotient is q and the remainder is r. Find all pairs a, b such that q2 + r = 1977.
Solution
a2 + b2 >= (a + b)2/2, so q ≥ (a + b)/2. Hence r < 2q. The largest square less than 1977 is 1936 = 442. 1977 = 442 + 41. The next largest gives 1977 = 432 + 128. But 128 > 2.43. So we must have q = 44, r = 41. Hence a2 + b2 = 44(a + b) + 41, so (a - 22)2 + (b - 22)2 = 1009. By trial, we find that the only squares with sum 1009 are 282 and 152. This gives two solutions 50, 37 or 50, 7.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
12 Oct 1998