IMO 1996 Problem B1 The positive integers a, b are such that 15a + 16b and 16a - 15b are both squares of positive integers. What is the least possible value that can be taken on by the smaller of these two squares?
481 2 . Solution Put 15a ± 16b = m 2 , 16a - 15b = n 2 . Then 15m 2 + 16n 2 = 481a = 13·37a. The quadratic residues mod 13 are 0, ±1, ±3, ±4, so the residues of 15m 2 are 0, ±2, ±5, ±6, and the residues of 16n 2 are 0, ±1, ±3, ±4. Hence m and n must both be divisible by 13. Similarly, the quadratic residues of 37 are 0, ±1, ±3, ±4, ±7, ±9, ±10, ±11, ±12, ±16, so the residues of 15m 2 are 0, ±2, ±5, ±6, ±8, ±13, ±14, ±15, ±17, ±18, and the residues of 16n 2 are 0, ±1, ±3, ±4, ±7, ±9, ±10, ±11, ±12, ±16. Hence m and n must both be divisible by 37. Put m = 481m', n = 481n' and we get: a = 481(15m' 2 + 16n' 2 ). We also have 481b = 16m 2 - 15n 2 and hence b = 481(16m' 2 - 15n' 2 ). The smallest possible solution would come from putting m' = n' = 1 and indeed that gives a solution. This solution is straightforward, but something of a slog - all the residues have to be calculated. A more elegant variant is to notice that m 4 + n 4 = 481(a 2 + b 2 ). Now if m and n are not divisible by 13 we have m 4 + n 4 = 0 (mod 13). Take k so that km = 1 (mod 13), then (nk) 4 = -(mk) 4 = -1 (mod 13). But that is impossible because then (nk) 12 = -1 (mod 13), but x 12 = 1 (mod 13) for all non-zero residues. Hence m and n are both divisible by 13. The same argument shows that m and n are both divisible by 37. 37th IMO 1996 © John Scholes jscholes@kalva.demon.co.uk 22 Oct 1998 Last corrected/updated 21 Aug 03