1. Given that $m$ and $n$ are constants, the sum of the three monomials $4 x^{2} y$, $m x^{3-n^{2}} y$, and $8 x^{3} y$ is still a monomial. Then the number of possible values for $m+n$ is $(\quad)$.
(A) 1
(B) 2
(C) 3
(D) 4
-、1.C.
Since $4 x^{2} y$ and $8 x^{3} y$ are not like terms, therefore, the sum of $4 x^{2} y$ and $m x^{3-n^{2}} y$ is 0 or the sum of $m x^{3-n^{2}} y$ and $8 x^{3} y$ is 0.
Thus, we have $\left\{\begin{array}{l}m=-4, \\ 3-n^{2}=2\end{array}\right.$ or $\left\{\begin{array}{l}m=-8, \\ 3-n^{2}=3 .\end{array}\right.$
Solving, we get $\left\{\begin{array}{l}m=-4, \\ n= \pm 1\end{array}\right.$ or $\left\{\begin{array}{l}m=-8, \\ n=0 .\end{array}\right.$
Hence, $m+n=-5,-3$ or -8.