Let N be the set of positive integers and R+ the positive reals. f : N → R+ satisfies f(n) → 0 as n → ∞. Show that there are only finitely many solutions to f(a) + f(b) + f(c) = 1.
Solution
We show first that there are only finitely many solutions to f(a) + f(b) = k (> 0). Take a to be < b. Since f(n) → 0 we can find N such that f(n) < k/2 for n > N. So a must be ≤ N. Now for each such a0, we can find M such that f(n) < k - f(a0) for n ≥ M, so there are only finitely many b such that f(a0) + f(b) = k. Hence there are only finitely many solutions to f(a) + f(b) = k.
Now consider f(a) + f(b) + f(c) = 1. We can find N such that f(n) < 1/3 for n > N. So at least one of a, b, c must be ≤ N. But for each such a we then have only finitely many solutions to f(b) + f(c) = 1 - f(a). Hence there are only finitely many solutions in total.
© John Scholes
jscholes@kalva.demon.co.uk
25 Feb 2002