14th Australian 1993 Problem A3 Find all sets of three positive integers such that each integer divides the sum of the other two.
(a,a,a), (a,a,2a), (a,2a,3a) for any positive integer a. Solution Suppose the numbers are a ≤ b ≤ c. Then a + b ≤ 2c with equality iff a = b = c. Obviously (a,a,a) is a possible solution. So assume the numbers are not all equal. Then a+b is a multiple of c less than 2c, so a+b=c. Now b divides a+(a+b), so it must divide 2a. But b ≥ a, so b = a or 2a. If b = 2a, then c = 3a and we have the solution (a,2a,3a). If b = a, then c = 2a and we have the solution (a,a,2a). 14th Aus 1993 © John Scholes jscholes@kalva.demon.co.uk 19 Oct 2003 Last updated/corrected 19 Oct 03