6. Let $M=\left\{a \mid a=x^{2}-y^{2}, x, y \in \mathbf{Z}\right\}$. Then for any integer $n$, among the numbers of the form $4 n, 4 n+1, 4 n+2, 4 n+3$, the number that is not an element of the set $M$ is ( ).
(A) $4 n$
(B) $4 n+1$
(C) $4 n+2$
(D) $4 n+3$
6. C.

Notice that for any integer $n$,
$$
\begin{array}{l}
4 n=(n+1)^{2}-(n-1)^{2}, \\
4 n+1=(2 n+1)^{2}-(2 n)^{2}, \\
4 n+3=(2 n+2)^{2}-(2 n+1)^{2},
\end{array}
$$

and $n-1, n+1, 2 n, 2 n+1, 2 n+2 \in \mathbf{Z}$.
Thus, $4 n, 4 n+1, 4 n+3 \in M$.
If $4 n+2$ is an element of set $M$, then there exist $x, y \in \mathbf{Z}$ such that $4 n+2=x^{2}-y^{2} (x, y \in \mathbf{Z})$.

However, since $x^{2}-y^{2}=(x+y)(x-y)$, and $x+y$ and $x-y$ have the same parity, $(x+y)(x-y)$ is either odd or a multiple of 4, meaning $4 n+2$ is not an element of set $M$.