115 Let the set $A=\left\{n \mid 2^{n}+3^{n}\right.$ is a multiple of 5. $\left.n \in \mathbf{N}\right\} \quad B=\left\{m \mid m=u^{2}-v^{2}, u\right.$, $v \in \mathbf{Z}\}$, then which of the following relations is correct?
A. $A \cap B=\emptyset$
B. $A \cup B=A$
C. $A \cup B=B$
D. $A=B$
$115 \mathrm{C}$.
$$
2^{n}+3^{n}=(5-3)^{n}+3^{n}=5 M+(-3)^{n}+3^{n}
$$

If $5 \mid 2^{n}+3^{n}$, then $5 \mid\left[(-3)^{n}+3^{n}\right]$.
When $n=2 k, k \in \mathbf{N}$, $(-3)^{n}+3^{n}=2 \cdot 3^{n} \quad 5 \mid 2 \cdot 3^{n}$,
i.e., $5 \mid \left[(-3)^{n}+3^{n}\right]$.
When $n=2 k-1, k \in \mathbf{N}$, $(-3)^{n}+3^{n}=0$, in this case $5 \mid\left[(-3)^{n}+3^{n}\right]$.
Therefore,
$$
\begin{aligned}
A & =\left\{n \mid 2^{n}+3^{n} \text { is a multiple of } 5 \right\} \\
& =\{n \mid n=2 k-1, k \in \mathbf{N}\} .
\end{aligned}
$$

Since $2 k-1=k^{2}-(k-1)^{2}$, we have $A \subseteq B$ i.e., $A \cup B=B$. Also, $8=3^{2}-1^{2}$, so $A \neq B$.