Find all functions $f$ such that $f(xf(y)+y)+f(xy+x)=f(x+y)+2xy$.

Question

I know that $f(x)=x$ and f(x)=-2x are solutions to this functional equation and I doubt there are any other solutions but I'm not entirely sure. Also I don't really know how to prove that if $f$ is a solution to this equation, then $f$ must be a linear function.

Answer 1

Suppose $f(a) = 1$ for some $a$ (i will show the existence of $a$ in the end), note that $a \neq 0$ since plugging $x = 0$ in the original equation gives $f(0) = 0$.

Plug $y = a$ in the original equation to obtain:

$$f(x+a)+f(xa+x)=f(x+a)+2xa $$ equivalently: $$f(x(a + 1))=2xa, \text{ or just: } f(x) = x\frac{2a}{a + 1}$$ Here $a\neq -1$ because otherwise we arrive at $f(0) = 2xa \neq 0$.

Setting $x = a$ gives: $$ 1 = f(a) = \frac{2a^2}{a + 1}$$ It follows that $a = 1$ or $a = -\frac{1}{2}$, which gives $f(x) = x$ or $f(x) = -2x$, exactly as you wanted.

Proof, that there exists $a$ such that $f(a) = 1$:

Put $x = 1$ to obtain: $f(f(y)+y)=2y$.

I conclude that $a = f(\frac{1}{2}) + \frac{1}{2}$ perfectly fits.