Silk Road Mathematics Competition competition problems

Using the identity $x^2 + x y = (-x - y)^2 + (-x - y) y$ we obtain the following identity for the function $f$:
$$(f(x))^2 + x f(y) = (f(-x - y))^2 - (x + y) f(y).$$
Let's write down some particular cases of the last identity:
$$(1) \quad (f(x))^2 + x f(-x) = (f(0))^2 \text{ (it is obtained by putting } y = -x)$$
$$(2) \quad (f(x))^2 + x f(0) = (f(-x))^2 - x f(0) \text{ (it is obtained by putting } y = 0).$$
(1) implies $(f(x))^2 + x f(-x) = (f(-x))^2 - x f(x)$ which is equivalent to
$$(f(x) + f(-x))(f(x) - f(-x) + x) = 0$$
Suppose that for some $a \neq 0$ we have $f(a) + f(-a) + a = 0$. Then by (2) $(f(a))^2 + 2 a f(0) = (f(-a))^2 = (f(a) + a)^2$, it follows that
$$(3) \qquad 2 f(0) = 2 f(a) + a.$$
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 $x \neq 0$. 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 $x \neq 0$ we have $f(x) \neq 0$. So, $f(x) = x$, and it obviously satisfies the given identity.

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 $f(2 a^2) = a$, which implies that $2 a^2 + (f(0))^2 = 0$ 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 $x \neq 0 \Rightarrow f(x) \neq 0$, we have $f(x) = x$.