2nd Swedish 1962 Problem 3 Find all pairs (m, n) of integers such that n 2 - 3mn + m - n = 0.
(m,n) = (0,0) or (0,1) Solution (3m+1)(3n-1) = 3n 2 -1, so if p divides 3n-1, then it also divides 3n 2 -1, and hence also (3n 2 -1) - n(3n-1) = n-1, and hence also (3n-1) - 3(n-1) = 2. So 3n-1 = 0, ±1 or ±2. But n must be an integer, so 0, 1, -2 do not work. -1 gives n = 0 and hence m = 0, which is a solution. 2 gives n = 1 and hence m = 0, which is a solution. 2nd Swedish 1962 © John Scholes jscholes@kalva.demon.co.uk 30 December 2003 Last corrected/updated 30 Dec 03