20.4. (Austria-Poland, 79). Find all functions $f: \mathbf{Z}^{+} \rightarrow \mathbf{R}$, satisfying the identity

$$
f(n+m)+f(n-m) \equiv f(3 n), \quad n, m \in \mathbf{Z}^{+}, \quad n \geqslant m
$$
20.4. Setting $m=0$ in the original identity for the function $f(n)$, we get $2 f(n) \equiv f(3 n)\left(n \in \mathbb{Z}^{+}\right)$, and for $n=m=0$ we have $f(0)=0$. Further, 
setting $n=m$ in the identity, we get

$$
f(2 n)+f(0) \equiv f(3 n), \text{ i.e. } f(2 n) \equiv f(3 n)
$$

From this, on the one hand, for any value of $m \in \mathbb{Z}^{+}$ we have the equalities

$$
f(4 m)=f(6 m)=f(9 m)
$$

and on the other hand, from the identity for $n=3 m$ we get

$$
f(4 m)+f(2 m) \equiv f(9 m)
$$

which is possible only if $f(2 m) \equiv 0$. Therefore, for any value of $n \in \mathbb{Z}^{+}$ we have

$$
f(n)=(1 / 2) f(3 n)=(1 / 2) f(2 n)=0
$$

i.e., the original identity can only be satisfied (and indeed is satisfied) by the function $f(n)$ that is identically equal to 0.