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

$$
f(x+y)=2 f\left(\frac{x}{2}\right)+3 f\left(\frac{y}{3}\right)
$$

for all rational numbers $x$ and $y$.
Let $f$ be a function that is a solution to the problem. Then, by substituting $x=y=0$, we find $f(0)=$ $5 f(0)$, hence $f(0)=0$. By setting $x=0$, we have for all rational $y$,

$$
f(y)=3 f\left(\frac{y}{3}\right)
$$

Similarly, by setting $y=0$, we have for all rational $x$,

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

The equation can then be rewritten, for all rational $x$ and $y$,

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

We recognize the Cauchy equation, so $f$ is linear, of the form $f(x)=a x$. Conversely, linear functions are indeed solutions.