$\therefore$ Given the equation $2 x^{2}+x y+y^{2}-x+2 y+1=0$, the number of integer solutions $(x, y)$ is $\quad($ )
(A) 1
(B) 2
(C) 3
(D) 1
5. (C).

The equation $2 x^{2}+x y+y^{2}-x+2 y+1=0$ can be viewed as a quadratic equation in $x$: $2 x^{2}+(y-1) x + \left(y^{2}+2 y+1\right)=0$. Since it has real solutions, its discriminant must be $(y-1)^{2}-2 \times 4\left(y^{2}+2 y+1\right) \geqslant 0$, which simplifies to $7 y^{2}+18 y+7 \leqslant 0$, or $\frac{1}{7}(-9-4 \sqrt{2}) \leqslant y \leqslant \frac{1}{7}(-9+4 \sqrt{2})$. Therefore, we know that $y=-1,-2$.
Thus, we have $x=0, y=-1 ; x=1, y=-1 ; x=1, y=-2$.