7. All ordered integer pairs $(x, y)$ that satisfy the equation $x^{2}+2 x y+3 y^{2}-2 x$ $+y+1=0$ are
7. The original equation is $x^{2}+2(y-1) x+\left(3 y^{2}\right.$ $+y+1)=0 . \because x \in R, \quad \therefore \Delta=4\left[(y-1)^{2}\right.$ $\left.-\left(3 y^{2}+y+1\right)\right] \geqslant 0, -\frac{3}{2} \leqslant y \leqslant 0$, but $y$ is an integer, so $y$ can only take the values $0, -1$.

When $y=0$, $x^{2}-2 x+1=0, x=1$; when $y=-1$, $x^{2}-4 x+3=0, x=1$ or 3.
$\therefore$ All the integer pairs $(x, y)$ that satisfy the given equation are $(1,0),(1,-1),(3,-1)$.