Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$,

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

## Solution of exercises
Let $x=y$, then we get $f(0)=0$. Let $y=0$, then we get $f\left(x^{2}\right)=x f(x)$, so $x f(x)=-x f(-x)$, thus $f(-x)=-f(x)$ for all non-zero real $x$, and since $f(0)=0$, we deduce that the function $f$ is odd. Returning to the original equation and substituting $y$ with $-y$, the equation becomes $f\left(x^{2}-y^{2}\right)=(x+y)(f(x)-f(y))$. Thus, for all real $x, y: (x-y)(f(x)+f(y)) = (x+y)(f(x)-f(y)) \Longrightarrow x f(y)-y f(x)=-x f(y)+y f(x) \Longrightarrow 2 x f(y)=2 y f(x) \Longrightarrow x f(y)=y f(x)$. If $x, y$ are two non-zero real numbers, $\frac{f(x)}{x}=\frac{f(y)}{y}$, so there exists a constant $c$ such that $f(x)=c x$ for all non-zero real $x$, and since $f(0)=0, f(x)=c x$ for all real $x$. $f\left(x^{2}-y^{2}\right)=c\left(x^{2}-y^{2}\right)$ and $(x-y)(f(x)+f(y))=c\left(x^{2}-y^{2}\right)$, so the functions $f: x \mapsto c x$ are solutions for all real $c$.

## 8th Geometry TD (Martin & Maryam)

In this TD, we started by reviewing the concepts covered in the previous lectures through the first five exercises. We then presented the South Pole configuration available in the intermediate geometry handout on the POFM website, which was useful for exercises 6 to 11.

## Exercises