SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Denote ($\ast$) as the given condition. Let $a=f(0)$. Taking $x=y$ in ($\ast$), we get $$f(2x)=a+4f(x)$$ for any $x\ne 0$. Now, by replacing $x,y$ by $2x,2y$ in ($\ast$) and using the above property (here we choose $x,y$ such that $x+y,x-y,x\ne 0$), we get $$2x[4f(x+y)-4f(x-y)]=8y[a+4f(x)],$$ or $$f(x+y)-f(x-y)=\frac{ay}{x}+\frac{4y}{x}f(x).$$ Combining this with ($\ast$), we get $a=0$. It follows that $f(0)=0$ and $f(2x)=4f(x)$ for any $x$. Replacing $y=-x$ in ($\ast$), we have $$f(-2x)=4f(x)=f(2x)$$ for any $x\ne 0$. Since $f(0)=0$, we conclude that $f$ is an even function. Now, replacing $x$ by $x$ in (1), we get $$x[f(-x-y)-f(y-x)]=4yf(-x)$$ or $$x[f(y+x)-f(y-x)]=4yf(x).$$ It follows that $$xy[f(y+x)-f(y-x)]=4y^{2}f(x).$$ From ($\ast$), we get $y[f(y+x)-f(y-x)]=4xf(y)$. Combine with the above equation, we get $$x^{2}f(y)=y^{2}f(x)$$ for any real numbers $x,y$. This shows that $$f(x)=a{x}^2$$ where $a$ is a constant, which is truly a solution.