24th Australian 2003 Problem A1 Find all solutions p, q, r in prime numbers to p(q - r) = q + r.
(p, q, r) = (5, 3, 2). Solution Note that we must have q > r since q + r is positive. p cannot be 2, because then q = 3r is composite. So p must be an odd prime. We have (p - 1)q = (p + 1)r, so r must divide p - 1 and q must divide p + 1. If p - 1 = r, then r = 2, p = 3. But that gives q = 4, which is not prime. So p - 1 >= 2r. If p + 1 = q, then p must be 2 and q must be 3, but we have already shown that p cannot be 2. So p + 1 >= 2q. If p - 1 = 2r and p + 1 = 2q, then p = q + r. Hence one of q, r must be even. But q > r, so r = 2. Hence p = 5 and q = 3. It is easy to check that is a solution. If p + 1 = 2q, p - 1 > 2r, then q + r  1. So p(q - r) > p > q + r. Contradiction. But if p + 1 > 2q, then p > 2q - 1. But p is odd, so p > 2q. But q - r >= 1, so p(q - r) > 2q > q + r. Contradiction. 24th Aus 2003 © John Scholes jscholes@kalva.demon.co.uk 15 June 2003 Last updated/corrected 15 Jun 03