Let $\alpha$ be a non-zero real number. Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$
f(f(x+y))=f(x+y)+f(x) f(y)+\alpha x y
$$

for all $x, y \in \mathbb{R}$.

(Walther Janous)

Answer. For $\alpha=-1$, the only solution is the function $f(x)=x$ for all $x$. For other $\alpha$, there is no solution.
We set $y=0$ in the functional equation and obtain with $c=f(0)$ the equation

$$
f(f(x))=f(x)(1+c) .
$$

We use this equation with $x+y$ instead of $x$ in the original equation and obtain

$$
f(x+y)(1+c)=f(f(x+y))=f(x+y)+f(x) f(y)+\alpha x y
$$

and thus

$$
f(x+y) c=f(x) f(y)+\alpha x y .
$$

We now apply this equation in two different ways to $f(x+y+z)$. We obtain

$$
c^{2} f(x+y+z)=c f(x+y) f(z)+c \alpha(x+y) z=f(x) f(y) f(z)+\alpha x y f(z)+c \alpha(x+y) z
$$

and analogously by swapping the roles of $x$ and $z$ the equation

$$
c^{2} f(x+y+z)=f(x) f(y) f(z)+\alpha z y f(x)+c \alpha(z+y) x .
$$

Thus, we have

$$
x y f(z)+c y z=z y f(x)+c y x
$$

and with $y=z=1$ we obtain

$$
x f(1)+c=f(x)+c x
$$

and thus $f(x)=a x+c$. Substituting into equation (1) yields

$$
a(a x+c)+c=(a x+c)(1+c),
$$

so $a^{2}=a(1+c)$ and $a c+c=c(1+c)$.

For $a=0$, the function is constant, which is clearly not a solution to the original equation, so $a=1+c$ and $a c=c$.

This implies $c=0$ and $a=1$. Therefore, $f(x)=x$ for all $x$. This works precisely for $\alpha=-1$.