If $p$, $q$, and $r$ are nonzero integers satisfying \[p^2+q^2 = r^2,\] compute the smallest possible value of $(p+q+r)^2$.

[i]Proposed by David Altizio[/i]
1. We start with the given equation \( p^2 + q^2 = r^2 \). This is a Pythagorean triple, meaning \( p \), \( q \), and \( r \) are integers that satisfy the Pythagorean theorem.

2. The smallest Pythagorean triple is \( (3, 4, 5) \). However, we need to consider all possible signs for \( p \), \( q \), and \( r \) to find the smallest value of \( (p + q + r)^2 \).

3. Let's consider the following cases:
   - \( p = 3 \), \( q = 4 \), \( r = 5 \)
   - \( p = 3 \), \( q = 4 \), \( r = -5 \)
   - \( p = 3 \), \( q = -4 \), \( r = 5 \)
   - \( p = 3 \), \( q = -4 \), \( r = -5 \)
   - \( p = -3 \), \( q = 4 \), \( r = 5 \)
   - \( p = -3 \), \( q = 4 \), \( r = -5 \)
   - \( p = -3 \), \( q = -4 \), \( r = 5 \)
   - \( p = -3 \), \( q = -4 \), \( r = -5 \)

4. Calculate \( (p + q + r)^2 \) for each case:
   - For \( p = 3 \), \( q = 4 \), \( r = 5 \):
     \[
     (p + q + r)^2 = (3 + 4 + 5)^2 = 12^2 = 144
     \]
   - For \( p = 3 \), \( q = 4 \), \( r = -5 \):
     \[
     (p + q + r)^2 = (3 + 4 - 5)^2 = 2^2 = 4
     \]
   - For \( p = 3 \), \( q = -4 \), \( r = 5 \):
     \[
     (p + q + r)^2 = (3 - 4 + 5)^2 = 4^2 = 16
     \]
   - For \( p = 3 \), \( q = -4 \), \( r = -5 \):
     \[
     (p + q + r)^2 = (3 - 4 - 5)^2 = (-6)^2 = 36
     \]
   - For \( p = -3 \), \( q = 4 \), \( r = 5 \):
     \[
     (p + q + r)^2 = (-3 + 4 + 5)^2 = 6^2 = 36
     \]
   - For \( p = -3 \), \( q = 4 \), \( r = -5 \):
     \[
     (p + q + r)^2 = (-3 + 4 - 5)^2 = (-4)^2 = 16
     \]
   - For \( p = -3 \), \( q = -4 \), \( r = 5 \):
     \[
     (p + q + r)^2 = (-3 - 4 + 5)^2 = (-2)^2 = 4
     \]
   - For \( p = -3 \), \( q = -4 \), \( r = -5 \):
     \[
     (p + q + r)^2 = (-3 - 4 - 5)^2 = (-12)^2 = 144
     \]

5. The smallest value of \( (p + q + r)^2 \) from all the cases is \( 4 \).

The final answer is \(\boxed{4}\).