Find all the integer solutions $(x,y,z)$ of the equation $(x + y + z)^5 = 80xyz(x^2 + y^2 + z^2)$,
1. We start with the given equation:
   \[
   (x + y + z)^5 = 80xyz(x^2 + y^2 + z^2)
   \]

2. Consider the expression:
   \[
   \sum_{\text{cyc}} (x + y - z)^5
   \]
   Expanding this cyclic sum, we get:
   \[
   (x + y - z)^5 + (y + z - x)^5 + (z + x - y)^5
   \]

3. Notice that:
   \[
   (x + y + z)^5 = (x + y - z + 2z)^5 = (x + y - z)^5 + 5(x + y - z)^4(2z) + \ldots + 32z^5
   \]
   Similarly, we can expand the other terms:
   \[
   (y + z - x)^5 = (y + z - x + 2x)^5 = (y + z - x)^5 + 5(y + z - x)^4(2x) + \ldots + 32x^5
   \]
   \[
   (z + x - y)^5 = (z + x - y + 2y)^5 = (z + x - y)^5 + 5(z + x - y)^4(2y) + \ldots + 32y^5
   \]

4. Summing these, we get:
   \[
   \sum_{\text{cyc}} (x + y - z)^5 = (x + y + z)^5 - 80xyz(x^2 + y^2 + z^2)
   \]

5. Therefore, the given equation can be rewritten as:
   \[
   (x + y + z)^5 - 80xyz(x^2 + y^2 + z^2) = 0
   \]
   This implies:
   \[
   \sum_{\text{cyc}} (x + y - z)^5 = 0
   \]

6. By Fermat's Last Theorem, which states that there are no three positive integers \(a, b, c\) that satisfy \(a^n + b^n = c^n\) for \(n > 2\), we infer that the only integer solutions to the equation \((x + y - z)^5 = 0\) are when one of the terms is zero or when the terms are equal in magnitude but opposite in sign.

7. Therefore, the solutions \((x, y, z)\) must be of the form \((a, -a, 0)\) and its permutations, where \(a\) is an integer.

The final answer is \( \boxed{ (a, -a, 0) } \) and its permutations, where \(a\) is an integer.