
A1. Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying

$$
f\left(x^{2}+x y+f\left(y^{2}\right)\right)=x f(y)+y^{2}+f\left(x^{2}\right)
$$

for all real numbers $x$ and $y$.

## United Kingdom, Joe Benton


Solution 1. We claim that the only solutions are $f(x)=x$ and $f(x)=-x$; it is easy to verify that these are indeed solutions.

Now suppose $f$ satisfies the given equation. Studying $(x, 0)$ and $(-x, 0)$ leads to

$$
x f(0)+f\left(x^{2}\right)=f\left(x^{2}+f(0)\right)=-x f(0)+f\left(x^{2}\right)
$$

and so $f(0)=0$. Then, studying $(0, y)$ and $(-y, y)$ leads to

$$
y^{2}=f\left(f\left(y^{2}\right)\right)=-y f(y)+y^{2}+f\left(y^{2}\right)
$$

and therefore

$$
f\left(y^{2}\right)=y f(y)
$$

Given $(x, y)$, take $z=-(x+y)$, so that $x^{2}+x y=z^{2}+z y$. Then $f\left(x^{2}+x y+f\left(y^{2}\right)\right)=$ $f\left(z^{2}+z y+f\left(y^{2}\right)\right)$, hence from the original equation,

$$
\begin{aligned}
x f(y)+y^{2}+f\left(x^{2}\right) & =z f(y)+y^{2}+f\left(z^{2}\right) \\
x f(y)-z f(y) & =f\left(z^{2}\right)-f\left(x^{2}\right)
\end{aligned}
$$

Eliminating $z$ and using (1),

$$
2 x f(y)=f\left(z^{2}\right)-f\left(x^{2}\right)-y f(y)=f\left((x+y)^{2}\right)-f\left(x^{2}\right)-f\left(y^{2}\right)
$$

But the RHS of (2) is symmetric in $x, y$ and so $x f(y)=y f(x)$. In particular $f(x)=x f(1)$. Writing $c=f(1)$, and substituting back into the original equation, all the terms cancel except $c^{2} y^{2} \equiv y^{2}$, hence $c= \pm 1$, so $f(x)=x$ and $f(x)=-x$ are the only solutions.
