38th Swedish 1998 Problem 1 Find all positive integers a, b, c, such that (8a-5b) 2 + (3b-2c) 2 + (3c-7a) 2 = 2.
(a,b,c) = (3,5,7), (12,19,28), Solution We must have two of |8a-5b|, |3b-2c|, |3c-7a| equal to 1 and the other 0. So there are 6 cases: (1) 8a- 5b=1, 3b-2c=1, 3c-7a=0, solving (a,b,c) = (12,19,28) . Hence the "negative" equations ... = -1, ... = -1, ... = 0 give a= -12, b= - 19, c= -28, which is invalid; (2) 8a-5b=1, 3b-2c=-1, 3c-7a=0, solving a = -3, b = -5, c = -7, invalid , so negative gives (a, b, c) = (3,5,7) ; (3) 8a-5b=1, 3b-2c=0, 3c-7a=1, solving a = 6 1/2, b = 10 1/5, c = 15 3/10, invalid (not integral), so negative equation also has invalid solutions; (4) 8a-5b = 1, 3b-2c=0, 3c-7a= -1, solving a = -1/2, b = -1, c= -3/2, invalid and negative also; (5) 8a-5b = 0, 3b-2c=1, 3c-7a=1, solving a=12 1/2, b=20, c=29 1/2, invalid and negative also; (6) 8a-5b=0, 3b-2c=1, 3c-7a= -1, solving a = -2 1/2, b=4, c= 5 1/2, invalid and negative also. 38th Swedish 1998 © John Scholes jscholes@kalva.demon.co.uk 16 January 2004 Last corrected/updated 16 Jan 04