6. 126 Find all functions $f: Z^{*} \rightarrow R$ that satisfy
$f(n+m)+f(n-m) \equiv f(3 n), n, m \in Z^{*}\left(Z^{*}\right.$ is the set of non-negative integers), $n \geqslant m$.
[Solution] In the identity concerning $f(x)$, let $m=0$, we get $2 f(n) \equiv f(3 n)$, $n \in Z^{*}$. When $n=m=0$, we have $f(0)=0$. Next, in the identity, let $n=m$, we get $f(2 n)+f(0) \equiv f(3 n)$, that is, $f(2 n) \equiv f(3 n)$. Therefore, on one hand, for any $m \in Z^{*}$, we have
$$f(4 m)=f(6 m)=f(9 m)$$

On the other hand, in the original identity, taking $n=3 m$, we get
$$f(4 m)+f(2 m)=f(9 m)$$

Thus, for any $m \in Z^{*}$, we have $f(2 m)=0$. Therefore, for any $n \in Z^{*}$, we have
$$f(n)=\frac{1}{2} f(3 n)=\frac{1}{2} f(2 n)=0 \text {. }$$

In other words, only when $f(n) \equiv 0$ can (and actually does) the original identity be satisfied.