1. Find all functions $f: \mathbf{Z}^{*} \rightarrow \mathbf{R}$ (where $\mathbf{Z}^{*}$ is the set of non-negative integers) that satisfy $f(n+m)+f(n-m)=f(3 n), m, n \in \mathbf{Z}^{*}, n \geqslant m$.
(1979 Austrian-Polish Competition)
1. In the given condition, let $m=0$ to get $2 f(n)=f(3 n), n \in \mathbf{Z}^{*}$. When $n=m=0$, we have $f(0)=0$. Also, in the given condition, let $n=m$, to get $f(2 n)+f(0) \equiv f(3 n)$, i.e., $f(2 n)=f(3 n)$. Therefore, on one hand, for any $m \in \mathbf{Z}^{*}$, we have $f(4 m)=f(6 m)=f(9 m)$. On the other hand, in the given condition, take $n=3 m$, to get $f(4 m)+f(2 m)=$ $f(9 m)$. Thus, for any $m \in \mathbf{Z}^{*}$, we have $f(2 m)=0$. Hence, for any $n \in \mathbf{Z}^{*}$, we have $f(n)=\frac{1}{2} f(3 n)=$ $\frac{1}{2} f(2 n)=0$, so $f(n)=0$. This shows that the required $f(n)$ is only $f(n)=0$ to satisfy the original condition.