Silk Road Mathematics Competition

Using the identity $x^2+xy=(-x-y{)}^2+(-x-y)y$ we obtain the following identity for the function $f$: $$(f(x){)}^2+xf(y)=(f(-x-y){)}^2-(x+y)f(y).$$ Let's write down some particular cases of the last identity: $$(1)(f(x){)}^2+xf(-x)=(f(0){)}^2\text{ (it is obtained by putting }y=-x)$$ $$(2)(f(x){)}^2+xf(0)=(f(-x){)}^2-xf(0)\text{ (it is obtained by putting }y=0).$$ (1) implies $(f(x){)}^2+xf(-x)=(f(-x){)}^2-xf(x)$ which is equivalent to $$(f(x)+f(-x))(f(x)-f(-x)+x)=0$$ Suppose that for some $a\ne 0$ we have $f(a)+f(-a)+a=0$. Then by (2) $(f(a){)}^2+2af(0)=(f(-a){)}^2=(f(a)+a{)}^2$, it follows that $$(3)2f(0)=2f(a)+a.$$ Eliminating $f(0)$ and $f(-a)$ in (1) by $a,f(a)$, we obtain $a^2=a^2/4$, hence, $a=0$. The contradiction says that $f(x)+f(-x)=0$ for all $x\ne 0$. Now from (3) we get $f(0)=0$. The identity (1) is transformed to $f(x)(f(x)-x)=0$. If we put $x=0$ in the original identity we obtain $f(f(y))=y$, which means that $f$ is injective. Hence, for $x\ne 0$ we have $f(x)\ne 0$. So, $f(x)=x$, and it obviously satisfies the given identity.

---

Alternative solution.

For $x=0$ we get $f(f(y))=y+(f(0){)}^2$, which implies that $f$ is injective and subjective. Let $f(a)=0$, then the substitution $x=y=a$ gives that $f(2a^2)=a$, which implies that $2a^2+(f(0){)}^2=0$ or $a=f(0)=0$. Particularly, $f(f(y))=y$. Substitution $y=0$ now gives $f(x^2)=(f(x){)}^2$, from which using that $f$ is injective we obtain $f(x)=-f(-x)$. Substitution $y=-x$ gives $f(x)(f(x)-x)=0$, and also using that $f$ is injective and applying $x\ne 0\Rightarrow f(x)\ne 0$, we have $f(x)=x$.