1. (16 points) Given that $m, n$ are integers, and satisfy $2 m^{2}+n^{2}+$ $3 m+n-1=0$. Find $m, n$.


Three, 1. Taking $m$ as the variable, we get the quadratic equation in $m$: $2 m^{2}+3 m+n^{2}+n-1=0$.
Since $m$ has integer solutions, we have
$$
\Delta=9-8\left(n^{2}+n-1\right)=17-8 n(n+1) \geqslant 0
$$

and it must be a perfect square.
Also, since $n$ is an integer, we consider the following cases:
(1) When $n=0$ or $n=-1$, $\Delta=17$, which is not a perfect square.
(2) When $n=1$ or $n=-2$, $\Delta=1$, substituting gives $2 m^{2}+3 m+1=0$.
Solving, we get $m_{1}=-1, m_{2}=-\frac{1}{2}$ (discard).
(3) When $n \geqslant 2$ or $n \leqslant-3$, $\Delta<0$, the equation has no solution.
In summary, $m=-1, n=1$ or $m=-1, n=-2$.