This article is from the Puzzles FAQ, by Chris Cole firstname.lastname@example.org and Matthew Daly email@example.com with numerous contributions by others.
Suppose f is non-decreasing with
f(x+y) = f(x) + f(y) + C for all real x, y.
Prove: there is a constant A such that f(x) = Ax - C for all x.
(Note: continuity of f is not assumed in advance.)
By induction f(mx) = m(f(x)+C)-C. Let x=1/n, m=n and find that
f(1/n) = (1/n)(f(1)+C)-C. Now let x=1/n and find that f(m/n) =
(m/n)(f(1)+C)-C. f(-x+x) = f(-x) + f(x) + C ==> f(-x) = -2C - f(x)
(since f(0) = -C) ==> f(-m/n) = -(m/n)(f(1)+C)-C. Since f is
monotonic ==> f(x) = x*(f(1)+C)-C for all real x (Squeeze Theorem).