1. Given $p$ is a prime number, and satisfies $p \equiv 3(\bmod 8)$. Find all integer solutions $(x, y)$ of the equation
$$
y^{2}=x^{3}-p^{2} x
$$
If $p \nmid x$, then $x, x-p, x+p$ are pairwise coprime, so $x, x-p, x+p$ are all perfect squares. Then
$$
2 p=(x+p)-(x-p) \equiv 0(\bmod 8),
$$
which is a contradiction.
If $p \mid x$, let $x=a p$. Then
$$
(x, x-p, x+p)=p \Rightarrow p^{2} \mid y \text {. }
$$

Let $y=p^{2} b$. Then from equation (1) $\Rightarrow p b^{2}=a\left(a^{2}-1\right)$.
If $p \mid a$, let $a=p c^{2}, a^{2}-1=d^{2}$.
Then $a=1, d=0$.
Thus $(a, b)=(0,0),(-1,0),(1,0)$.
If $p \nmid a$, let
$$
a=c^{2}, a^{2}-1=p d^{2}
$$
$\Rightarrow c^{4}-1=p d^{2} \equiv 0$ or $3(\bmod 8)$
$\Rightarrow c$ is odd
$$
\Rightarrow\left\{\begin{array}{l}
c^{2}+1=2 e^{2}, \\
c^{2}-1=8 p f .
\end{array}\right.
$$

From a solution $(c, f)(c>1)$ of the system of equations (2), we can obtain another solution $\left(\sqrt{\frac{c+1}{2}}, \sqrt{\frac{c-1}{16 p}}\right)$.

Since $1<\sqrt{\frac{c+1}{2}}<c$, continuing this process, we obtain an infinite descending sequence of positive integers, which is a contradiction.