Find all pairs $(p,n)$ of integers so that $p$ is a prime and there exists $x,y\not\equiv0\pmod p$ with
$$x^2+y^2\equiv n\pmod p.$$
1. **Understanding the problem**: We need to find all pairs \((p, n)\) of integers such that \(p\) is a prime and there exist \(x, y \not\equiv 0 \pmod{p}\) with \(x^2 + y^2 \equiv n \pmod{p}\).

2. **Using properties of quadratic residues**: Let \(a\) be an integer coprime to \(p\) and \(p > 2\). We use the sum of Legendre symbols:
   \[
   \sum_{j=1}^{p-1} \left( \frac{a - j^2}{p} \right) = -\left( \frac{a}{p} \right) - \left( \frac{-1}{p} \right)
   \]
   Here, \(\left( \frac{a}{p} \right)\) is the Legendre symbol, which is \(+1\) if \(a\) is a quadratic residue modulo \(p\), \(-1\) if \(a\) is a non-residue, and \(0\) if \(a \equiv 0 \pmod{p}\).

3. **Existence of \(y\)**: For \(p > 5\), there exists an integer \(y\) such that \(0 < y < p\) and \(\left( \frac{a - y^2}{p} \right) = +1\). This implies that \(a - y^2\) is a quadratic residue modulo \(p\).

4. **Finding \(x\)**: Since \(a - y^2\) is a quadratic residue, there exists an \(x\) such that \(x^2 \equiv a - y^2 \pmod{p}\). Therefore, \(x^2 + y^2 \equiv a \pmod{p}\).

5. **Special cases for \(p \equiv 1 \pmod{4}\) and \(p \equiv 3 \pmod{4}\)**:
   - If \(p \equiv 1 \pmod{4}\), then \(-1\) is a quadratic residue modulo \(p\). This means there exist \(x, y \not\equiv 0 \pmod{p}\) such that \(x^2 + y^2 \equiv 0 \pmod{p}\).
   - If \(p \equiv 3 \pmod{4}\), then \(-1\) is not a quadratic residue modulo \(p\). Hence, there do not exist \(x, y \not\equiv 0 \pmod{p}\) such that \(x^2 + y^2 \equiv 0 \pmod{p}\).

6. **Conclusion**: For \(p > 5\), any \(n\) can be expressed as \(x^2 + y^2 \pmod{p}\) for some \(x, y \not\equiv 0 \pmod{p}\). For \(p \equiv 1 \pmod{4}\), \(n\) can be \(0\). For \(p \equiv 3 \pmod{4}\), \(n\) cannot be \(0\).

The final answer is \( \boxed{ (p, n) } \) such that \(p\) is a prime and \(n\) can be any integer modulo \(p\) for \(p > 5\). For \(p \equiv 1 \pmod{4}\), \(n\) can be \(0\). For \(p \equiv 3 \pmod{4}\), \(n\) cannot be \(0\).