25th Junior Autumn 2003 Problem 3 Find all positive integers k such that x(x+k) = y(y+1) has a solution in positive integers x, y.
k = 1 or k ≥ 4. Solution k = 1 is obvious. So suppose k > 1. We have (2x+k) 2 - (2y+1) 2 = k 2 - 1, so (2x + 2y + k + 1)(2x - 2y + k - 1) = k 2 - 1. For k odd we could take 2x - 2y + k - 1 = 2 and 2x + 2y + k + 1 = (k 2 - 1)/2, giving x = (k 2 - 4k + 3)/8, y = (k 2 - 9)/8. Note that these are positive integers for k ≥ 5 since odd squares = 1 mod 8. For k even we could take 2x - 2y + k - 1 = 1 giving x = (k 2 - 2k)/4, y = (k 2 - 4)/4, which are both positive integers for k ≥ 4. That leaves the cases k = 2 and k = 3. (Note that we are not required to find all solutions, only to show which k have solutions.). For k = 2 we have (2x+2) 2 - (2y+1) 2 = 3. But the only squares differing by 3 are 1 and 4, so there are no solutions in positive integers x, y. Similarly for k = 3 we have (2x+3) 2 - (2y+1) 2 = 8. But the only squares differing by 8 are 1 and 9 and again there are no solutions in positive integers x, y. 25th Junior Autumn 2003 © John Scholes jscholes@kalva.demon.co.uk 3 January 2004 Last corrected/updated 3 Jan 04