IMO 1988 SL 19

Origin: MEX

Problem

Let f(n) be a function defined on the set of all positive integers and having its values in the same set. Suppose that f(f(m)+f(n)) = m+n for all positive integers n, m. Find all possible values for f(1988).

Solution

We will show that f(n) = n for every n (thus also f(1988) = 1988). Let f(1) = r and f(2) = s. We obtain respectively the following equalities: f(2r) = f(r + r) = 2; f(2s) = f(s + s) = 4; f(4) = f(2 + 2) = 4r; f(8) = f(4+4) = 4s; f(5r) = f(4r+r) = 5; f(r+s) = 3; f(8) = f(5+3) = 6r+s. Then 4s = 6r + s, which means that s = 2r. Now we prove by induction that f(nr) = n and f(n) = nr for every n \geq4. First we have that f(5) = f(2 + 3) = 3r + s = 5r, so that the statement is true for n = 4 and n = 5. Suppose that it holds for n −1 and n. Then f(n + 1) = f(n −1 + 2) = (n −1)r + 2r = (n + 1)r, and f((n + 1)r) = f((n −1)r + 2r) = (n −1) + 2 = n + 1. This completes the induction. Since 4r \geq4, we have that f(4r) = 4r2, and also f(4r) = 4. Then r = 1, and consequently f(n) = n for every natural number n. Second solution. f(f(1)+n+m) = f(f(1)+f(f(n)+f(m))) = 1+f(n)+ f(m), so f(n) + f(m) is a function of n + m. Hence f(n + 1) + f(1) = f(n)+f(2) and f(n+1)−f(n) = f(2)−f(1), implying that f(n) = An+B for some constants A, B. It is easy to check that A = 1, B = 0 is the only possibility.