1. Given $x^{2}=y^{2}+4 y+20$. Then the number of all integer pairs satisfying the equation is ( ).
(A) 4
(B) 6
(C) 8
(D) infinitely many
- 1. B.

From the given equation, we have $x^{2}-\left(y^{2}+4 y+4\right)=16$, which simplifies to $(x+y+2)(x-y-2)=16$.
Since $x+y+2$ and $x-y-2$ have the same parity, we have the system of equations
$\left\{\begin{array}{l}x+y+2= \pm 2, \\ x-y-2= \pm 8\end{array}\right.$ or $\left\{\begin{array}{l}x+y+2= \pm 8, \\ x-y-2= \pm 2\end{array}\right.$ or $\left\{\begin{array}{l}x+y+2= \pm 4, \\ x-y-2= \pm 4 .\end{array}\right.$

Solving these, we get $(x, y)=(5,-5),(-5,1),(-5,-5),(5, 1),(4,-2),(-4,-2)$.