a little fiddling shows that we can replace the above equation with the equation f(x) - f(y) = f(x-y) for the analytic function f(z) = (1 - cos z) / z
the reasoning is something like this:
then the LHS - RHS difference is analytic since f(z) is analytic, and any infinite set of real solutions must have an accumulation point (since the set of real z for which |f(z)| < B is bounded for every B > 0), and i think that means the difference is identically zero for all x and y, which leads to x==y and to the solution provided in the question
details involve the difference between real analytic and analytic functions
If the bonded analytic function A(x,y)=f(x)-f(y)+f(x-y) is equal zero at the sequence of points xn, yn that are accumulated to a finite point then the function A is identically zero. But if the points xn, yn accumulated at infinity then A(x,y) may not be identically zero. For example sin(z) . So I think your nice idea does not work.
Yeah, i was worried about sin z - i think the issue is whether or not there are many solutions with all three f(x) and f(y) and f(x-y) nonzero (the known and obvious infinite sets of solutions either have f(x) = f(y) = f(x-y) = 0, as in the solution given in the problem, or they have x=0 and any y, or y=0 and any x, or any x=y, and in each of the latter three cases, one of the terms is zero)
maybe also, by taking w = x - y, we could get something from transforming the equation into f(w+y) = f(w) + f(y), to imply that unless f(w) = f(y) = f(w+y) = 0, there are some serious linearity consequences, or something