Find all ordered pairs of integers $(x, y)$ with $y \ge 0$ such that $x^2 + 2xy +   y! = 131$.
To find all ordered pairs of integers \((x, y)\) with \(y \ge 0\) such that \(x^2 + 2xy + y! = 131\), we can proceed as follows:

1. **Rewrite the given equation:**
   \[
   x^2 + 2xy + y! = 131
   \]

2. **Complete the square on the left-hand side:**
   \[
   x^2 + 2xy + y! = 131 \implies (x + y)^2 - y^2 + y! = 131
   \]

3. **Rearrange the equation:**
   \[
   (x + y)^2 = y^2 - y! + 131
   \]

4. **Analyze the right-hand side (RHS) for different values of \(y\):**
   - For \(y \geq 6\), \(y!\) grows very quickly, making \(y^2 - y! + 131\) negative. Since the square of a real number cannot be negative, we conclude that \(y \leq 5\).

5. **Check each possible value of \(y\) from 0 to 5:**

   - **For \(y = 0\):**
     \[
     (x + 0)^2 = 0^2 - 0! + 131 \implies x^2 = 131
     \]
     Since 131 is not a perfect square, there are no integer solutions for \(x\).

   - **For \(y = 1\):**
     \[
     (x + 1)^2 = 1^2 - 1! + 131 \implies (x + 1)^2 = 1 - 1 + 131 \implies (x + 1)^2 = 131
     \]
     Again, 131 is not a perfect square, so there are no integer solutions for \(x\).

   - **For \(y = 2\):**
     \[
     (x + 2)^2 = 2^2 - 2! + 131 \implies (x + 2)^2 = 4 - 2 + 131 \implies (x + 2)^2 = 133
     \]
     Since 133 is not a perfect square, there are no integer solutions for \(x\).

   - **For \(y = 3\):**
     \[
     (x + 3)^2 = 3^2 - 3! + 131 \implies (x + 3)^2 = 9 - 6 + 131 \implies (x + 3)^2 = 134
     \]
     Since 134 is not a perfect square, there are no integer solutions for \(x\).

   - **For \(y = 4\):**
     \[
     (x + 4)^2 = 4^2 - 4! + 131 \implies (x + 4)^2 = 16 - 24 + 131 \implies (x + 4)^2 = 123
     \]
     Since 123 is not a perfect square, there are no integer solutions for \(x\).

   - **For \(y = 5\):**
     \[
     (x + 5)^2 = 5^2 - 5! + 131 \implies (x + 5)^2 = 25 - 120 + 131 \implies (x + 5)^2 = 36
     \]
     Since 36 is a perfect square, we have:
     \[
     x + 5 = 6 \implies x = 1 \quad \text{or} \quad x + 5 = -6 \implies x = -11
     \]
     Therefore, the solutions for \(y = 5\) are \((x, y) = (1, 5)\) and \((x, y) = (-11, 5)\).

Conclusion:
The only ordered pairs of integers \((x, y)\) with \(y \ge 0\) that satisfy the equation \(x^2 + 2xy + y! = 131\) are \((1, 5)\) and \((-11, 5)\).

The final answer is \(\boxed{(1, 5), (-11, 5)}\).