Functional equation $(x-y)(f(x)+f(y))=(x+y)f(x-y)$

Question

I have proved that the functional equation, $(x-y)(f(x)+f(y))=(x+y)f(x-y)$, has the following results:

• $f(0)=0$,
• $f$ is an odd function.

It is clear that $f(x)=cx$ for $c \in \mathbb R$ is a solution. But I could not conclude my solution.

Answer 1

Let $y = x - 1$. This gives us that $$ f(x) + f(x - 1) = (2x - 1) f(1). \quad (\star) $$

This implies $$ f(x + 1) + f(x) = (2x + 1) f(1) $$ and adding this to Equation $(\star)$ gives us $$ 2f(x) + f(x + 1) + f(x - 1) = 4x f(1). \quad (\star \star) $$

Letting $x = x + 1$ and $y = x - 1$ in the functional equation gives us that $$ 2(f(x + 1) + f(x - 1)) = 2x f(2) \implies f(x + 1) + f(x - 1) = xf(2). $$

Substituting this into Equation $(\star \star)$ gives us $$ f(x) = \frac{1}{2} x \left(4f(1) - f(2) \right) $$ for all $x$.