Problem 1357

Let $P(x, y)$ denote the assertion given in the statement of the problem. If there exists $a$ such that $f(a) = 0$, $P(a, 0)$ results in $a = 0$. Now
$$P(x, x-y) \rightarrow f(2x-y)f(x^2-xy+y^2) = (2x-y)(x^2-xy+y^2)$$
By dividing this equation by $P(x, y)$ (where $x + y \neq 0$), it is obtained that
$$\frac{f(2x - y)}{f(x + y)} = \frac{2x - y}{x + y} \qquad (1)$$
In equation (1) put $y = 1-x$. This will result in $f(3x-1) = (3x-1)f(1)$ and because $3x-1$ is surjective over real numbers, we conclude that $f(x) = xf(1)$. By putting this equality in the original equality, it is deduced that $f(1) = \pm 1$ so functions $f(x) = x$ and $f(x) = -x$ are the only answers. ■