3. The integer solutions $(x, y)$ of the equation $2 x^{2}+x y+y^{2}-x+2 y+1=0$ are $\qquad$ _.
3. $(0,-1),(1,-1),(1,-2)$.

The original equation can be transformed into
$$
2 x^{2}+(y-1) x+y^{2}+2 y+1=0 \text {. }
$$

Considering $\Delta=(y-1)^{2}-8(y+1)^{2} \geqslant 0$.
Solving yields $\frac{-9-4 \sqrt{2}}{7} \leqslant y \leqslant \frac{-9+4 \sqrt{2}}{7}$.
Thus, $y=-1,-2$.
Therefore, when $y=-1$, $x=0,1$;
when $y=-2$, $x=1$.
Hence, the integer solutions are $(0,-1),(1,-1)$, $(1,-2)$.