Cool Integers

We can assume $a \leq b$

$a = 1$ : $b! = 1 + b! + c!$ - no solutions

$a = 2$ : $2b! = 2 + b! + c! \Rightarrow b! = 2 + c!$ - no solutions

For $a \geq 3$ : if $c \leq b$ then $a! + b! + c! \leq 3b! < a!b!$ - therefore $b < c$

Let $p(n) = k$ be the integer value $n = 2^k(2m-1)$ for some positive integer $m$

when $p(a!) < p(b!)$ we have

$p(a!b!) \geq 2p(a!)$ and $p(a! + b! + c!) = p(a!) \Rightarrow$ no solutions for $b > a + 1$

when $b = a + 1$ :

$a!(a+1)! = a! + (a + 1)! + c! \Rightarrow a!$ divided by $(a+1)!$ - no solutions

finally when $b = a$ : $(a!)^2 = 2a! + c! \Rightarrow a! = 2 + \frac{c!}{a!}$

if $c \geq a + 3$ , $a! \equiv 0 \pmod{3}$ and $2 + \frac{c!}{a!} \equiv 2 \pmod{3}$ - no solutions

if $c = a + 2$ - either $2 + \frac{c!}{a!} \equiv 2 \pmod{3}$ or $2 + \frac{c!}{a!} \equiv 1 \pmod{3}$ - no solutions

the only possibility left: $a = b, c = a + 1$

$a! \cdot a! = 2a! + (a + 1)! \Rightarrow a! = 2 + a + 1 = a + 3 \Rightarrow$

$a = b = 3, c = 4 \Rightarrow a + b + c = \boxed{10}$

hm... was it author's solution BTW? :-)

a!b! - a! - b! + 1 = c! + 1

(a! - 1)(b! - 1) = c! + 1

a = 3, b = 3, c = 4