Find all solutions of $a^2 + b^2 = n!$ for positive integers $a$, $b$, $n$ with $a \le b$ and  $n < 14$.
To find all solutions of \(a^2 + b^2 = n!\) for positive integers \(a\), \(b\), and \(n\) with \(a \le b\) and \(n < 14\), we can use the following steps:

1. **Understanding the Representation of Sums of Two Squares:**
   A number \(m\) can be represented as a sum of two squares if and only if every prime factor of the form \(4k+3\) in its prime factorization has an even exponent. This is derived from Fermat's theorem on sums of two squares.

2. **Checking Factorials for \(n < 14\):**
   We need to check the factorials \(n!\) for \(n = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13\) to see if they can be expressed as a sum of two squares.

3. **Eliminating Factorials with Prime Factors of the Form \(4k+3\):**
   For \(n > 6\), \(n!\) includes the prime number 7, which is of the form \(4k+3\). Since 7 appears with an odd exponent in \(n!\) for \(n \ge 7\), these factorials cannot be expressed as a sum of two squares. Therefore, we only need to check \(n = 1, 2, 3, 4, 5, 6\).

4. **Checking Each Case:**
   - \(n = 1\):
     \[
     1! = 1 = 1^2 + 0^2
     \]
     So, \(a = 1\) and \(b = 0\).

   - \(n = 2\):
     \[
     2! = 2 = 1^2 + 1^2
     \]
     So, \(a = 1\) and \(b = 1\).

   - \(n = 3\):
     \[
     3! = 6
     \]
     The prime factorization of 6 is \(2 \times 3\). Since 3 is of the form \(4k+3\) and appears with an odd exponent, 6 cannot be expressed as a sum of two squares.

   - \(n = 4\):
     \[
     4! = 24
     \]
     The prime factorization of 24 is \(2^3 \times 3\). Since 3 is of the form \(4k+3\) and appears with an odd exponent, 24 cannot be expressed as a sum of two squares.

   - \(n = 5\):
     \[
     5! = 120
     \]
     The prime factorization of 120 is \(2^3 \times 3 \times 5\). Since 3 is of the form \(4k+3\) and appears with an odd exponent, 120 cannot be expressed as a sum of two squares.

   - \(n = 6\):
     \[
     6! = 720 = 12^2 + 24^2
     \]
     So, \(a = 12\) and \(b = 24\).

Conclusion:
The solutions for \(a^2 + b^2 = n!\) for positive integers \(a\), \(b\), and \(n\) with \(a \le b\) and \(n < 14\) are:
- For \(n = 1\): \(a = 1\), \(b = 0\)
- For \(n = 2\): \(a = 1\), \(b = 1\)
- For \(n = 6\): \(a = 12\), \(b = 24\)

The final answer is \( \boxed{ (a, b, n) = (1, 0, 1), (1, 1, 2), (12, 24, 6) } \)