Determine all positive integers $a$, $b$ and $c$ which satisfy the equation
$$a^2+b^2+1=c!.$$

[i]Proposed by Nikola Velov[/i]
To determine all positive integers \(a\), \(b\), and \(c\) that satisfy the equation 
\[ a^2 + b^2 + 1 = c!, \]
we can proceed as follows:

1. **Consider the properties of \(a^2 + b^2 + 1\) modulo 4:**
   - Note that \(a^2\) and \(b^2\) can only be congruent to 0 or 1 modulo 4 because the square of any integer is either 0 or 1 modulo 4.
   - Therefore, the possible values of \(a^2 + b^2\) modulo 4 are:
     \[
     \begin{aligned}
     &0 + 0 \equiv 0 \pmod{4}, \\
     &0 + 1 \equiv 1 \pmod{4}, \\
     &1 + 0 \equiv 1 \pmod{4}, \\
     &1 + 1 \equiv 2 \pmod{4}.
     \end{aligned}
     \]
   - Adding 1 to these values, we get:
     \[
     \begin{aligned}
     &0 + 1 \equiv 1 \pmod{4}, \\
     &1 + 1 \equiv 2 \pmod{4}, \\
     &2 + 1 \equiv 3 \pmod{4}.
     \end{aligned}
     \]
   - Therefore, \(a^2 + b^2 + 1\) can be congruent to 1, 2, or 3 modulo 4.

2. **Consider the factorial \(c!\) modulo 4:**
   - For \(c \geq 4\), \(c!\) is divisible by 4, so \(c! \equiv 0 \pmod{4}\).
   - For \(c = 3\), \(3! = 6 \equiv 2 \pmod{4}\).
   - For \(c = 2\), \(2! = 2 \equiv 2 \pmod{4}\).
   - For \(c = 1\), \(1! = 1 \equiv 1 \pmod{4}\).

3. **Match the possible values of \(a^2 + b^2 + 1\) with \(c!\):**
   - Since \(a^2 + b^2 + 1\) can be 1, 2, or 3 modulo 4, and \(c!\) can be 0, 1, or 2 modulo 4, we need to check the cases where these values match.

4. **Check specific values of \(c\):**
   - For \(c \geq 4\), \(c! \equiv 0 \pmod{4}\), which does not match any possible value of \(a^2 + b^2 + 1\). Therefore, \(c < 4\).
   - For \(c = 3\), \(3! = 6\):
     \[
     a^2 + b^2 + 1 = 6 \implies a^2 + b^2 = 5.
     \]
     The pairs \((a, b)\) that satisfy this are \((2, 1)\) and \((1, 2)\).
   - For \(c = 2\), \(2! = 2\):
     \[
     a^2 + b^2 + 1 = 2 \implies a^2 + b^2 = 1.
     \]
     The pairs \((a, b)\) that satisfy this are \((1, 0)\) and \((0, 1)\). However, since we are looking for positive integers, these pairs are not valid.
   - For \(c = 1\), \(1! = 1\):
     \[
     a^2 + b^2 + 1 = 1 \implies a^2 + b^2 = 0.
     \]
     The only pair \((a, b)\) that satisfies this is \((0, 0)\), which is not valid since we are looking for positive integers.

Therefore, the only solutions are for \(c = 3\) with pairs \((a, b) = (2, 1)\) and \((1, 2)\).

The final answer is \( \boxed{ (a, b, c) = (2, 1, 3) } \) and \((1, 2, 3)\).