29th IMO 1988 shortlist Problem 19 f has positive integer values and is defined on the positive integers. It satisfies f( f(m) + f(n) ) = m + n for all m, n. Find all possible values for f(1988).
1988. Solution f is obviously surjective. If f(a) = f(b), then c + a = f( f(c) + f(a) ) = f( f(c) + f(b) ) = c + b, so a = b. Hence f is injective. So the inverse f -1 exists. The functional equation gives immediately f(m + n) = f -1 (m) + f -1 (n) and f -1 (m + n) = f(m) + f(n). f(2n) = f -1 (2n-1) + f -1 (1) = f(2n-2) + f(1) + f -1 (1). Iterating, f(2n) = n f(1) + n f -1 (1). Similarly, f(2n+1) = n f(1) + (n+1) f -1 (1). So we putting f(1) = h, f -1 (1) = k, we have f(1) = h, f(2) = h + k, f(3) = h + 2k, f(4) = 2h + 2k, f(5) = 2h + 3k, ... . But h, k ≥ 1 and f is surjective, so we must have h = k = 1. Hence f(n) = n for all n. 29th IMO shortlist 1988 © John Scholes jscholes@kalva.demon.co.uk 30 Dec 2002 Last corrected/updated 30 Dec 02