Romanian Mathematical Olympiad - District Round

We claim that the solutions are $f_1 = 1_{\mathbb{R}}$ and $f_2 = -1_{\mathbb{R}}$.

For $x = 0$ we obtain $f(f(y)) = f(f(0)) + y$, for each $y \in \mathbb{R}$, so $f$ is an one-to-one function.

For $y = 0$ we have $f(xf(x) + f(0)) = f(f(x^2))$, $\forall x \in \mathbb{R}$, which, due to $f$ being one-to-one, leads to $xf(x) + f(0) = f(x^2)$, $\forall x \in \mathbb{R}$.

Putting $x = 1$ in the last relation gives $f(0) = 0$, so the last relation can be written as $xf(x) = f(x^2)$, $\forall x \in \mathbb{R}$, while the first relation becomes $f(f(x)) = x$, $\forall x \in \mathbb{R}$.

Putting $x \to f(x)$ in the above relation, we have $f(f(x)^2) = f(f(x))f(x) = xf(x) = f(x^2)$, which, due to $f$ being one-to-one, implies that $(f(x))^2 = x^2$, so for each $x \in \mathbb{R}$ we have $f(x) = x$ or $f(x) = -x$.

Supposing that exists a pair $(x_0, y_0) \in \mathbb{R}^* \times \mathbb{R}^*$ such that $f(x_0) = x_0$ and $f(y_0) = -y_0$. Putting $x \to x_0$ and $y \to y_0$ in the given functional equation, we obtain $f(x^2 - y) = x^2 + y$, so $x^2 - y = x^2 + y$ or $x^2 - y = -x^2 - y$, implying that $y = 0$ or $x = 0$, which is a contradiction. Therefore, the claim is proved.