$p$ is a prime number such that its remainder divided by 8 is 3. Find all pairs of rational numbers $(x,y)$ that satisfy the following equation.
$$p^2 x^4-6px^2+1=y^2$$
To solve the given equation \( p^2 x^4 - 6px^2 + 1 = y^2 \) for rational numbers \((x, y)\), we will first prove the claim that if integers \(a, b, c\) satisfy \( p^2 a^4 - 6pa^2b^2 + b^4 = c^2 \), then \(abc = 0\).

1. **Assume \(a, b, c\) are natural numbers and \(abc \neq 0\)**:
   - Suppose there exist natural numbers \(a, b, c\) such that \( p^2 a^4 - 6pa^2b^2 + b^4 = c^2 \) and \(abc \neq 0\).
   - Let \(d = pa^2 - 3b^2\). Then, we have \(d^2 - 8b^4 = c^2\).

2. **Factorize the equation**:
   - The equation \(d^2 - 8b^4 = c^2\) can be rewritten as:
     \[
     8b^4 = (d + c)(d - c)
     \]

3. **Introduce gcd and new variables**:
   - Let \(x = \gcd(c, d)\), \(e = \frac{c}{x}\), and \(f = \frac{d}{x}\). Since \(\gcd(e, f) = 1\), we have \(x \mid b\).
   - Let \(g = \frac{b}{x}\). Then:
     \[
     8x^2g^4 = (f + e)(f - e)
     \]

4. **Analyze the factors**:
   - Since \(2 \mid (f + e)\) and \(2 \mid (f - e)\), we can write:
     \[
     2x^2g^4 = \left(\frac{f + e}{2}\right) \left(\frac{f - e}{2}\right)
     \]
   - Given that \(\gcd\left(\frac{f + e}{2}, \frac{f - e}{2}\right) = 1\), we have:
     \[
     \left\{\frac{f + e}{2}, \frac{f - e}{2}\right\} = \{2h^2, j^2\}
     \]
     for some natural numbers \(h, j\) with \(\gcd(h, j) = 1\) and \(2 \nmid j\).

5. **Express \(f\) and \(d\) in terms of \(h\) and \(j\)**:
   - We have:
     \[
     f = 2h^2 + j^2 \quad \text{and} \quad hj = xg^2
     \]
   - By the definition of \(d\):
     \[
     pa^2 = d + 3b^2 = xf + 3x^2g^2 = x(2h^2 + j^2 + 3hj)
     \]
   - Thus:
     \[
     pa^2g^2 = hj(2h + j)(h + j)
     \]

6. **Case analysis on the factors**:
   - Since \(j\) is an odd number, \(\gcd(j, 2h + j) = 1\). Therefore, \(h, j, h + j, 2h + j\) are pairwise coprime.
   - One of these must be of the form \(p \times (\text{square})\), while the others are squares. We analyze each case:

   **Case I: \(h + j\) is \(p \times (\text{square})\)**:
   - This implies \((\text{square}) + (\text{square}) = h + j = p \times (\text{square})\). By modulo 4, \(h, j, h + j\) must be even, which is a contradiction.

   **Case II: \(2h + j\) is \(p \times (\text{square})\)**:
   - Similarly, \((\text{square}) + (\text{square}) = 2h + j = p \times (\text{square})\). This also leads to a contradiction.

   **Case III: \(j\) is \(p \times (\text{square})\)**:
   - Since \(j\) is odd, \(j \equiv 3 \pmod{8}\). This implies \(h \equiv 1 \pmod{8}\), \(h + j \equiv 4 \pmod{8}\), and \(2h + j \equiv 5 \pmod{8}\), leading to a contradiction.

   **Case IV: \(h\) is \(p \times (\text{square})\)**:
   - By similar modulo work as in Case III, \(h\) must be even, while \(j, h + j, 2h + j\) are odd, leading to a contradiction.

7. **Conclude the claim**:
   - Since all cases lead to contradictions, the initial assumption that \(abc \neq 0\) must be false. Therefore, \(abc = 0\).

8. **Apply the claim to the original equation**:
   - By the claim, the solutions to \( p^2 x^4 - 6px^2 + 1 = y^2 \) are \((0, \pm 1)\).

The final answer is \(\boxed{(0, \pm 1)}\).