Find the number of solutions to the given congruence$$x^{2}+y^{2}+z^{2} \equiv 2 a x y z \pmod p$$ where  $p$ is an odd prime and $x,y,z \in \mathbb{Z}$.

[i]Proposed by  math_and_me[/i]
1. **Lemma 1**: Let \( S_n = \sum_{x=0}^{p-1} x^n \). Then \( S_n \equiv -1 \pmod{p} \) if \( p-1 \mid n \) and \( S_n \equiv 0 \pmod{p} \) otherwise.

2. **Lemma 2**: Let \( a, b, c \in \mathbb{Z} \) and \( p \nmid a \). Then 
   \[
   \sum_{x=0}^{p-1} \left( \frac{ax^2 + bx + c}{p} \right) = 
   \begin{cases} 
   - \left( \frac{a}{p} \right) & \text{if } p \nmid b^2 - 4ac, \\
   (p-1) \left( \frac{a}{p} \right) & \text{if } p \mid b^2 - 4ac.
   \end{cases}
   \]

3. Consider the function \( f(x) = ax^2 + bx + c \). We have:
   \[
   \left( \frac{f(x)}{p} \right) \equiv f(x)^{\frac{p-1}{2}} \equiv a_0 + a_1 x + \dots + a_{p-1} x^{p-1} \pmod{p}.
   \]
   Summing over all \( x \) from 0 to \( p-1 \):
   \[
   \sum_{x=0}^{p-1} \left( \frac{f(x)}{p} \right) = \sum_{x=0}^{p-1} f(x)^{\frac{p-1}{2}} \equiv \sum_{x=0}^{p-1} \sum_{j=0}^{p-1} a_j x^j \pmod{p}.
   \]
   This simplifies to:
   \[
   \sum_{j=0}^{p-1} a_j \sum_{x=0}^{p-1} x^j \pmod{p}.
   \]
   Using Lemma 1, we get:
   \[
   \sum_{j=0}^{p-1} a_j S_j \pmod{p} \equiv -a_{p-1} \equiv -a^{\frac{p-1}{2}} \equiv - \left( \frac{a}{p} \right) \pmod{p}.
   \]
   Therefore,
   \[
   \sum_{x=0}^{p-1} \left( \frac{ax^2 + bx + c}{p} \right) \equiv - \left( \frac{a}{p} \right) \pmod{p}.
   \]

4. Given the congruence \( x^2 + y^2 + z^2 \equiv 2axyz \pmod{p} \), we rewrite it as:
   \[
   (z - axy)^2 \equiv (a^2x^2 - 1)y^2 - x^2 \pmod{p}.
   \]
   Fixing \( x \) and \( y \) in \( \{0, 1, 2, \dots, p-1\} \), the number of solutions for \( z \) is:
   \[
   1 + \left( \frac{(a^2x^2 - 1)y^2 - x^2}{p} \right),
   \]
   where \( \left( \frac{a}{p} \right) \) is the Legendre symbol.

5. The total number of solutions to the given equation is:
   \[
   \sum_{x=0}^{p-1} \sum_{y=0}^{p-1} \left( 1 + \left( \frac{(a^2x^2 - 1)y^2 - x^2}{p} \right) \right) = p^2 + \sum_{x=0}^{p-1} \sum_{y=0}^{p-1} \left( \frac{(a^2x^2 - 1)y^2 - x^2}{p} \right).
   \]

6. Using Lemma 2, we get:
   \[
   \sum_{y=0}^{p-1} \left( \frac{(a^2x^2 - 1)y^2 - x^2}{p} \right) = 
   \begin{cases} 
   - \left( \frac{a^2x^2 - 1}{p} \right) & \text{if } ax \not\equiv \pm 1 \pmod{p}, \\
   p \left( \frac{-1}{p} \right) & \text{if } ax \equiv \pm 1 \pmod{p}.
   \end{cases}
   \]

7. Summing over \( x \):
   \[
   \sum_{x=0}^{p-1} \sum_{y=0}^{p-1} \left( \frac{(a^2x^2 - 1)y^2 - x^2}{p} \right) = 2p \left( \frac{-1}{p} \right) - \sum_{x=0}^{p-1} \left( \frac{a^2x^2 - 1}{p} \right).
   \]
   Since \( \sum_{x=0}^{p-1} \left( \frac{a^2x^2}{p} \right) = \left( \frac{a^2}{p} \right) \sum_{x=0}^{p-1} \left( \frac{x^2}{p} \right) = \left( \frac{a^2}{p} \right) \cdot 0 = 0 \), we have:
   \[
   \sum_{x=0}^{p-1} \left( \frac{a^2x^2 - 1}{p} \right) = -1.
   \]
   Therefore,
   \[
   \sum_{x=0}^{p-1} \sum_{y=0}^{p-1} \left( \frac{(a^2x^2 - 1)y^2 - x^2}{p} \right) = 2p \left( \frac{-1}{p} \right) + 1.
   \]

8. The total number of solutions to the given congruence is:
   \[
   p^2 + 2p \left( \frac{-1}{p} \right) + 1 = p^2 + 2p \left( \frac{-1}{p} \right) + \left( \frac{-1}{p} \right)^2 = \left( p + \left( \frac{-1}{p} \right) \right)^2.
   \]

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