SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Let denote by $P(x,y)$ the equation $$f(xf(y)-y)+f(xy-x)+f(x+y)=2xy.$$ $P(0,y)$ gives us $f(-y)+f(y)=0,\forall y$. Thus $f$ is an odd function.

$P(-1,y)$ follows $$f(-f(y)-y)+f(-y+1)+f(-1+y)=-2y.$$ From this, since $f$ is odd, we have $f(-f(y)-y)=-2y$ and thus $f(f(y)+y)=2y$. So $f$ is surjective.

Since $f$ is surjective, there is real number $a$ such that $f(a)=-1$.

$P(x,a)$ gives us $$f(-x-a)+f(ax-x)+f(x+a)=2ax.$$ From this, again since $f$ is odd we have $f(ax-x)=2ax$ for all $x\in \mathbb{R}$.\ ()$<br/><br/>If$a=1$thenfromaboveequation,wehave$f(0)=2 x, \forall x \in \mathbb{R}$,whichisacontradiction.Thus,$a \neq 1$.<br/><br/>Replace$x=\frac{t}{a-1}$in$()$,wehave$\left(c^{2}+c-2\right) x y=0, \forall x, y \in \mathbb{R}$whichimpliesthat$c \in\{1,-2\}$.<br/><br/>So$f(x)=x$or$f(x)=-2 x$forall$x \in \mathbb{R}$.

It is easy to check that these two functions satisfy the condition.