"Surprise!" said c

Without loss of generality, let's assume that $b \leq a$ .

If $b = a$ , then $\displaystyle (a!)^2 = 2a! + c^2$ .

$\displaystyle \begin{aligned} \therefore (a!)^2 - 2a! + 1 & = c^2 + 1 \\ (a! - 1)^2 & = c^2 + 1 \end{aligned}$

Note that $c > 0 \implies c^2 > 0$ . Since there exists no positive square which is one more than another positive square, we have $b < a$ .

Also, $b \neq 1$ - otherwise, $a! = a! + 1 + c^2 \implies c^2 = -1$ .

Dividing $a!b! = a! + b! + c^2$ by $b!$ we have

$\displaystyle a! = \frac{a!}{b!} + 1 + \frac{c^2}{b!}$

Clearly the LHS is an integer, so since $\displaystyle \frac{a!}{b!}$ is an integer, $\displaystyle \frac{c^2}{b!}$ must also be an integer. Furthermore, the RHS $\geq 3$ , which means that $a! \geq 3 \implies a \geq 3$ .

From $a!b! = a! + b! + c^2$ , we have $a!b! - a! - b! + 1 = c^2 + 1$

$\displaystyle \therefore (a! - 1)(b! - 1) = c^2 + 1$

Let us assume that a prime $p \equiv 3 \pmod{4}$ divides the RHS.

If $c^2 + 1 \equiv 0 \pmod{p}$ , it follows that $c^2 \equiv -1 \pmod{p}$ , and it is clear that $p \ \not \mid\ c$ .

$\displaystyle \therefore (c^2)^{\frac{p - 1}{2}} = c^{p - 1} \equiv (- 1)^{\frac{p - 1}{2}} \pmod{p}$

But we are given that $p \equiv 3 \pmod{4}$ , so $p - 1 \equiv 2 \pmod{4}$ , which means that $\displaystyle \frac{p - 1}{2}$ will be odd, and $\displaystyle (- 1)^{\frac{p - 1}{2}} = - 1$ .

So $c^{p - 1} \equiv - 1 \pmod{p}$ , but that is a contradiction since $\gcd(c, p) = 1$ , and by Fermat's Little Theorem , $c^{p - 1} \equiv 1 \pmod{p}$ .

In other words, there exists no prime $p \equiv 3 \pmod{4} \ | \ c^2 + 1$ .

However, for $a \geq 4$ , $a! \equiv 0 \pmod{4}$ , and $a! - 1 \equiv 3 \pmod{4}$ ; that is, the LHS is divisible by a prime $p \equiv 3 \pmod{4}$ .

Hence, $1 < b < a \leq 3$ , but we have already established that $a \geq 3$ , so we have $a = 3$ and $b = 2$ , and from the original equation, $c = 2$ .

$\displaystyle \implies \frac{a^b}{c} = \frac 92$

$\implies m + n = 11$

$We~ have~~~ f=a!*b! - (a!+!b) -c^2=0\\ Since~ a>b, ~~minimum~ a=2~ and~ b=1.\\ But~ if~ b=1,~~~ f=-c^2 , ~~so~c=0.~But~ given~ c>0. \\ So~ minimum ~b=2,~and~~ a>2. \\ If~ a=3,~~and ~b=2,~~~ f=6*2~-~(6~+~2)~-~c^2~=~4~-~c^2=0.\\ So~ ~c=\pm 2.~~ But ~c>0.~~~ So~ c=2.\\ \dfrac m n~=~\dfrac {3^2} 2.\\ \therefore~~ m~+~n~=~9~+~2=\color{#D61F06}{11}.$