Russian 1997 Problem 15 Find all solutions in positive integers to a + b = gcd(a,b) 2 , b + c = gcd(b,c) 2 , c + a = gcd(c,a) 2 .
(a,b,c) = (2,2,2) Solution Put d = gcd(a,b,c). Then a = Ad, b = Bd, c = Cd. Put t = gcd(A,B), r = gcd(B,C), s = gcd(A,C). If r, s have a common factor, then it divides all of A, B, C, contradicting the definition of d. So r, s, t are coprime in pairs. Hence A = Rst, B = rSt, C = rsT for some R, S, T which are relatively prime in pairs. Thus a = Rstd, b = rStd, c = rsTd. Now a + b = gcd(a,b) 2 implies A + B = t 2 d. Similarly, B + C = r2d, C + A = s2d. Adding, 2(A + B + C) = d(r 2 + s 2 + t 2 ). Now if any odd prime divides d, then it must divide A + B + C and A + B, B + C and C + A and hence each of A, B, C, which is impossible. Similarly, if 4 divides d, then 2 divides A, B, C, which is impossible. Hence d = 1 or 2. We can also write a + b = gcd(a,b) 2 etc as Rs + rS = td, St + sT = rd, Tr + Rt = sd. Suppose (without loss of generality) that r is the smallest of r, s, t. Then 2r ≥ rd = St + sT ≥ s + t ≥ 2r. So we must have equality throughout and hence d = 2, S = T = 1 and r = s = t. But r, s, t are coprime, so r = s = t = 1. Hence from Rs + rS = td, we have R = 1 also. Hence a = b = c = 2. Thanks to Suat Namli Russian 1997 © John Scholes jscholes@kalva.demon.co.uk 7 February 2004 Last corrected/updatd 7 Feb 04