Problem 4. Determine the functions $f: \mathbb{Q} \rightarrow \mathbb{Q}$ with the property that $f(x+3 f(y))=f(x)+f(y)+2 y$, for all $x, y \in \mathbb{Q}$.
Solution. The functions $f_{k}: \mathbb{Q} \rightarrow \mathbb{Q}, k=1,2$, defined by $f_{1}(x)=x$ and $f_{2}(x)=-2 x / 3$ satisfy the requirement. We will show that these are the only solutions.

2 points

For $x=y-3 f(y)$ we get $f(y-3 f(y))=-2 y, y \in \mathbb{Q}$.

1 point

By substituting $y$ with $y-3 f(y)$ in the given relation, we have $f(x-6 y)=f(x)-2 y+ 2(y-3 f(y))$, from which $f(x-6 y)=f(x)-6 f(y)$, for any $x, y \in \mathbb{Q}$.

1 point

For $x=y=0$ we get $f(0)=0$. For $x=6 y$ we get $f(6 y)=$ $6 f(y)$, for any $y \in \mathbb{Q}$.

1 point

Then $f(x-6 y)=f(x)-f(6 y)$, from which, for $u=6 y$ and $v=x-6 y$, it follows that $f(u+v)=f(u)+f(v)$, for any $u, v \in \mathbb{Q}$.

1 point

We deduce that $f(x)=x f(1), x \in \mathbb{Q}$. The relation from the statement imposes $3 f^{2}(1)- 2 f(1)-1=0$, hence $f(1)=1$ or $f(1)=-2 / 3$, which concludes the solution.