3. Find all functions $f: \mathbb{Z}^{+} \cup\{0\} \rightarrow \mathbb{R}$ such that

$$
f(n+m)+f(n-m)=f(3 n), \quad \forall n, m \in \mathbb{Z}^{+} \cup\{0\}, n \geq m
$$
Solution. If we set $m=0$ in the equation, we get $2 f(n)=f(3 n)$, for every $n \in \mathbb{Z}^{+}$. For $n=m=0$ we get $f(0)=0$. Furthermore, if we set $m=n$, we get $f(2 n)+f(0)=f(3 n)$, which means $f(2 n)=f(3 n)$. Therefore, on one hand, for every $m \in \mathbb{Z}^{+}$, it holds that $f(4 m)=f(6 m)=f(9 m)$, and on the other hand, if we set $n=3 m$ in the equation, we get $f(4 m)+f(2 m)=f(9 m)$, which is possible if and only if $f(2 m)=0$. Consequently, for any $n \in \mathbb{Z}^{+}$ we have

$$
f(n)=\frac{1}{2} f(3 n)=\frac{1}{2} f(2 n)=0
$$

which means the only function that satisfies the equation is $f(n)=0$, for every $n \in \mathbb{Z}^{+}$.