Solve on $\mathbb{R}$ :

$$
f\left(x^{2}-y^{2}\right)=x f(x)-y f(y)
$$
Let $f$ be a potential solution. We start by proving that $f$ is odd.

$f(0)=f\left(x^{2}-x^{2}\right)=x f(x)-x f(x)=0$ and for non-zero $y$, $x f(x)-y f(y)=f\left(x^{2}-y^{2}\right)=$ $x f(x)+y f(-y)$, hence $f(y)=-f(-y)$ for all real $y$, so $f$ is odd. Therefore, we only consider positive variables.

By taking $y=0, x f(x)=f\left(x^{2}\right)$, thus for all real $x, y$:

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

or by setting $a=\sqrt{x}$ and $b=\sqrt{y}$:

$$
f(a-b)+f(b)=f(a)
$$

With $a=2 t$ and $b=t$ and the oddness, we get for all real $t$, $2 f(t)=f(2 t)$.

With $x=t+1$ and $y=t$ in the initial equation,

$$
\begin{gathered}
f(2 t+1)=(t+1) f(t+1)-t f(t)=(t+1) f(t)+(t+1) f(1)-t f(t)=f(t)+(t+1) f(1) \\
f(2 t+1)=2 f(t)+f(1)
\end{gathered}
$$

thus by combining the lines above, for all real $t, f(t)=t f(1)$.

We verify that linear functions are indeed solutions.

## 2 Tuesday, 22 afternoon: Alexander Semenov

## Exercises done in class