Find all integer solutions $(x,y)$ of the equation $y^2=x^3-p^2x,$ where $p$ is a prime such that $p\equiv 3 \mod 4.$
We are given the equation \( y^2 = x^3 - p^2 x \) where \( p \) is a prime such that \( p \equiv 3 \pmod{4} \). We need to find all integer solutions \((x, y)\).

### Case 1: \( p \mid x \)

1. Let \( x = kp \) for some integer \( k \). Then the equation becomes:
   \[
   y^2 = (kp)^3 - p^2 (kp) = p^3 k^3 - p^3 k = p^3 (k^3 - k)
   \]
   Dividing both sides by \( p^2 \), we get:
   \[
   \left( \frac{y}{p} \right)^2 = p (k^3 - k)
   \]
   Let \( y = p^2 q \) for some integer \( q \). Then:
   \[
   q^2 = k^3 - k
   \]
   We need to solve \( q^2 = k(k-1)(k+1) \).

2. Notice that \( k, k-1, k+1 \) are three consecutive integers, so their greatest common divisor is at most 2. Therefore, at least one of them must be divisible by 2, and at least one of them must be divisible by 3.

3. Since \( k, k-1, k+1 \) are consecutive integers, one of them must be a multiple of 2 and one of them must be a multiple of 3. We analyze the possible values of \( k \):
   - If \( k = 0 \), then \( q^2 = 0 \), so \( q = 0 \) and \( y = 0 \). Thus, \( x = 0 \).
   - If \( k = 1 \), then \( q^2 = 0 \), so \( q = 0 \) and \( y = 0 \). Thus, \( x = p \).
   - If \( k = -1 \), then \( q^2 = 0 \), so \( q = 0 \) and \( y = 0 \). Thus, \( x = -p \).

4. Therefore, the solutions in this case are \( (x, y) = (0, 0), (p, 0), (-p, 0) \).

### Case 2: \( \gcd(p, x) = 1 \)

1. In this case, the equation \( y^2 = x(x^2 - p^2) \) can be factored as:
   \[
   y^2 = x(x - p)(x + p)
   \]

2. If \( x \) is odd, then \( x, x-p, x+p \) are pairwise coprime. Therefore, each of these factors must be a perfect square. Let:
   \[
   x = a^2, \quad x - p = b^2, \quad x + p = c^2
   \]
   Then:
   \[
   a^2 - b^2 = p \quad \text{and} \quad c^2 - a^2 = p
   \]
   This implies:
   \[
   (a - b)(a + b) = p \quad \text{and} \quad (c - a)(c + a) = p
   \]
   Since \( p \) is a prime, the only factor pairs are \( (1, p) \) and \( (-1, -p) \).

3. Solving \( a - b = 1 \) and \( a + b = p \), we get:
   \[
   2a = p + 1 \quad \Rightarrow \quad a = \frac{p + 1}{2}
   \]
   \[
   2b = p - 1 \quad \Rightarrow \quad b = \frac{p - 1}{2}
   \]
   Therefore:
   \[
   x = a^2 = \left( \frac{p + 1}{2} \right)^2
   \]

4. Since \( p \equiv 3 \pmod{4} \), \( p + 1 \equiv 0 \pmod{4} \), so \( \frac{p + 1}{2} \) is an integer. Thus, \( x = \left( \frac{p + 1}{2} \right)^2 \) is an integer.

5. If \( x \) is even, then \( x = 2m \) for some integer \( m \). Then:
   \[
   y^2 = 2m(2m - p)(2m + p)
   \]
   Since \( 2m \) is even, \( 2m - p \) and \( 2m + p \) must be odd. This implies that \( y^2 \) is even, which is a contradiction since \( y^2 \) must be a perfect square.

Therefore, the only solutions are when \( x \) is odd.

The final answer is \( \boxed{ (x, y) = (0, 0), (p, 0), (-p, 0), \left( \left( \frac{p^2 + 1}{2} \right), 0 \right) } \).