6. Given the function $f: Z \rightarrow R$ satisfies for any integers $x, y$ that
$$
f(x+y) f(x-y)=f(x)+f(y) .
$$

Then $f(2011)=$ $\qquad$ .
6.0 or 2 or -1.

From equation (1) we get
$$
\begin{array}{l}
f(x)+f(-y)=f(x+y) f(x-y)=f(x)+f(y) \\
\Rightarrow f(y)=f(-y) .
\end{array}
$$

In equation (1), let $x=y$, we get
$$
f(2 x) f(0)=2 f(x) \text {. }
$$

In equation (2), let $x=0$, we get $f(0)=0$ or 2.
If $f(0)=0$, then from equation (2) we get
$$
f(x)=0(\forall x \in \mathbf{Z}) \text {. }
$$

In this case, $f(2011)=0$.
If $f(0)=2$, then from equation (2) we get
$$
f(2 x)=f(x)(\forall x \in \mathbf{Z}) \text {. }
$$

In equation (1), let $y=0$, we get
$$
f^{2}(x)=f(x)+2 \Rightarrow f(x)=2 \text { or }-1(\forall x \in \mathbf{Z}) \text {. }
$$

In equation (1), let $y=1$, we get
$$
f(x+1) f(x-1)=f(x)+f(1) \text {. }
$$

If $f(1)=-1$, then from equation (3) we get $f(2)=-1$.
In equation (4), let $x=2$, we get $f(3)=2$.
By mathematical induction and equation (4), it is easy to prove: for any integer $k$, we have $f(3 k)=2, f(3 k+1)=f(3 k+2)=-1$, and it satisfies equation (1). In this case, $f(2011)=-1$.
If $f(1)=2$, then from equation (3) we get $f(2)=2$.
By mathematical induction and equation (4), it is easy to know
$$
f(x)=2(\forall x \in \mathbf{Z}) \text {. }
$$

In this case, $f(2011)=2$.