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

$$
(x-y) f(x+y)-(x+y) f(x-y)=4 x y\left(x^{2}-y^{2}\right)
$$

## Solutions
To clarify the problem a bit, we set \(x+y=a\) and \(x-y=b\). Thus, \(x^2 - y^2 = ab\), \(2x = a + b\), and \(2y = a - b\). We obtain for all real numbers \(a, b: b f(a) - a f(b) = ab(a^2 - b^2)\). By taking \(a = 0\), \(b f(0) = 0\) for all \(b\), hence \(f(0) = 0\).

Next, for \(a, b \neq 0\), we divide by \(ab: \frac{f(a)}{a} - \frac{f(b)}{b} = a^2 - b^2\), which simplifies to \(\frac{f(a)}{a} - a^2 = \frac{f(b)}{b} - b^2\). The quantity \(\frac{f(a)}{a} - a^2\) is independent of the chosen parameter! We call this \(c = \frac{f(a)}{a} - a^2\) (valid for all non-zero \(a\)). Thus, for all \(x \neq 0\), \(f(x) = x^3 + cx\), but this remains valid for \(x = 0\) since \(f(0) = 0\). It only remains to verify that these functions work!

## 6 Thursday afternoon, 25th: Thomas Budzinski

This combinatorics course revisits exercises that you will find elsewhere in other Animath polycopied materials.

## V. Last day: end-of-stage shibboleth

## 1 Statements

## 1 Beginners