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

Hi Chris,

Thank you for your beautiful thought.

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.

If x>0>y then the equation turns to the sum of positive fractions which leads to the solution provides by the question.

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)

It likely works the same if x < 0 < y

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

I reformulate my question in a simpler form and Dr. Metafune constructed the nice irrational solutions!