
Problem 3. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$
f(y f(x+y)+f(x))=4 x+2 y f(x+y)
$$

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

Origin. Netherlands (Birgit van Dalen).


Solution 1 (submitter). Setting $y=0$ yields

$$
f(f(x))=4 x
$$

from which we derive that $f$ is a bijective function. Also, we find that

$$
f(0)=f(4 \cdot 0)=f(f(f(0)))=4 f(0)
$$

hence $f(0)=0$. Now set $x=0$ and $y=1$ in the given equation and use (1) again:

$$
4=f(f(1))=2 f(1)
$$

so $f(1)=2$ and therefore also $f(2)=f(f(1))=4$. Finally substitute $y=1-x$ in the equation:

$$
f(2(1-x)+f(x))=4 x+4(1-x)=4=f(2) \quad \text { for all } x \in \mathbb{R}
$$

As $f$ is injective, from this it follows that $f(x)=2-2(1-x)=2 x$. It is easy to see that this function satisfies the original equation. Hence the only solution is the function defined by $f(x)=2 x$ for all $x \in \mathbb{R}$.
