Find all $(n,x,y,z,p)$ of positive integers with $p$ prime and 
$$(x^2 + 4y^2)(y^2 + 4z^2)(z^2 + 4x^2) = p^n$$
Given the equation:
\[
(x^2 + 4y^2)(y^2 + 4z^2)(z^2 + 4x^2) = p^n
\]
where \( p \) is a prime number and \( n, x, y, z \) are positive integers, we need to find all possible solutions.

1. **Assume \((x, y, z) = 1\)**:
   - If there exists a common divisor \( q \) of \( x, y, z \), then \( q = p \). By dividing both sides by \( p^2 \), we get the same equation. If \((x, y, z)\) is a solution, then \((xp, yp, zp)\) will also satisfy the equation.
   - Assume \( p \) does not divide any of \( x, y, z \).

2. **Analyze the factors**:
   - Since \( x^2 + 4y^2, y^2 + 4z^2, z^2 + 4x^2 > 1 \), we can write:
     \[
     x^2 + 4y^2 = p^{a+b+c}, \quad y^2 + 4z^2 = p^{b+c}, \quad z^2 + 4x^2 = p^c
     \]
     where \( a, b \geq 0 \) and \( c \geq 1 \).

3. **Case \( p \neq 2, 5, 13 \)**:
   - Consider the congruences:
     \[
     x^2 \equiv -4y^2 \pmod{p} \implies \frac{x^2}{y^2} \equiv \frac{y^2}{z^2} \equiv \frac{z^2}{x^2} \pmod{p}
     \]
   - This implies \( p \mid (x^2 - yz)(x^2 + yz) \) and similarly for the other terms.

4. **Claim: \( p \not\equiv 3 \pmod{4} \)**:
   - If \( p \equiv 3 \pmod{4} \), then \( p \mid x^2 + 4y^2 \implies p \mid x, 2y \), which is a contradiction.

5. **If \( p \mid x^2 + yz \)**:
   - Then:
     \[
     -4 \equiv -\frac{z}{y} \equiv \frac{y^2}{z^2} \pmod{p} \implies p \mid y^3 + z^3
     \]
   - If \( p \mid y + z \), then \( p \mid x^2 + yz \equiv x^2 - y^2 \). Since \( p \) does not divide \( x - y \), we get \( p \mid x + y \) and \( p \mid y + z \), leading to \( p \mid 5z^2 \), which is a contradiction.

6. **If \( p \mid y^2 - yz + z^2 \)**:
   - Then:
     \[
     -4 \equiv -\frac{z}{y} \equiv \frac{y^2}{yz - y^2} \equiv \frac{y}{z - y} \pmod{p}
     \]
   - So \( p \mid 3y - 4z \implies \frac{16}{9}z^2 - \frac{4}{3}z^2 + z^2 = \frac{13}{16}z^2 \implies p = 13 \), which is a contradiction.

7. **Case \( p = 2 \)**:
   - If \( p = 2 \):
     \[
     x^2 + 4y^2 = 2^{a+b+c}, \quad y^2 + 4z^2 = 2^{b+c}, \quad z^2 + 4x^2 = 2^c
     \]
   - This implies \( 2 \mid x, y, z \), which is a contradiction.

8. **Case \( p = 5 \)**:
   - If \( p = 5 \):
     \[
     (x^2 + 4y^2)(y^2 + 4z^2)(z^2 + 4x^2) = 5^n
     \]
     \[
     x^2 + 4y^2 = 5^{a+b+c}, \quad y^2 + 4z^2 = 5^{b+c}, \quad z^2 + 4x^2 = 5^c
     \]
   - This leads to:
     \[
     5(x^2 + y^2 + z^2) = 5^c(5^{b+c} + 5^c + 1)
     \]
     \[
     x^2 + y^2 + z^2 = 5^{c-1}(5^{b+c} + 5^c + 1)
     \]
   - If \( c \geq 2 \), then \( 5 \mid x^2 + y^2 + z^2 \), which is not possible since \( x, y, z \not\equiv 0 \pmod{5} \). So \( c = 1 \implies z^2 + 4x^2 = 5 \implies x = z = 1 \).
   - Both \( 4y^2 + 1 \) and \( y^2 + 4 \) are powers of 5. Thus:
     \[
     y^2 + 4 = 5^{b+1}, \quad 5^{b+2} > 5^{b+1} \cdot 4 > 4y^2 + 1 = 5^{a+b+1} \implies a = 0 \implies 4y^2 + 1 = y^2 + 4 \implies y = 1
     \]
   - Therefore, \(\boxed{(1, 1, 1)}\) is the solution for this case. This gives \(\boxed{(5^k, 5^k, 5^k)}\).

9. **Case \( p = 13 \)**:
   - If \( p = 13 \):
     \[
     x^2 + 4y^2 = 13^{a+b+c}, \quad y^2 + 4z^2 = 13^{b+c}, \quad z^2 + 4x^2 = 13^c
     \]
   - This leads to:
     \[
     x^2 + y^2 \equiv y^2 + z^2 \equiv z^2 + x^2 \equiv 1 \pmod{3}
     \]
   - This is a contradiction.