Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$
(x+y)(f(x)-f(y))=(x-y) f(x+y)
$$

for all real numbers $x, y$.
(Solved by Adrien Lemercier)

With $(x, 1),(x, 2)$ and $(x+1,1)$, we obtain:

$$
\begin{aligned}
(x+1)(f(x)-f(1)) & =(x-1) f(x+1) \\
(x+2)(f(x)-f(2)) & =(x-2) f(x+2) \\
(x+2)(f(x+1)-f(1)) & =x f(x+2)
\end{aligned}
$$

Then, for any real $x$, and by cleverly eliminating the terms in $f(x+1)$ and $f(x+2)$:

$$
x(x+2)(f(x)-f(2))(x-1)=(x-2)(x+2)((x+1)(f(x)-f(1))-(x-1) f(1))
$$

Therefore:

$$
f(x)=\frac{f(2)}{2}\left(x^{2}-x\right)-\left(x^{2}-2 x\right) f(1)
$$

Thus $f$ is a polynomial of degree 2 over $\mathbb{R}$, with a zero constant term: $f(x)=a x^{2}+b x$. By reinserting this into the functional equation, we verify conversely that these functions are solutions.