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Problem

Given that $a$, $b$, and $c$ are positive intgers, solve the following equation.

$a$!$b$! = $a$! + $b$! + 2^$c$

Solution

Without loss of generality let us assume that $a$ $b$ and divide through by $b$!: $a$! = $a$!/$b$! + 1 +2^$c$/$b$!, giving integers throughout.

As each term on the RHS is integer, RHS 3 $a$! 3 $a$ 3.

For $b$ 2, $b$! will contain a factor of 3, and so cannot divide 2^$c$. Thus $b$=1 or $b$=2.

If $b$=1, we get $a$! = $a$! + 1 + 2^$c$, which leads to 2^$c$ + 1 = 0: no solution.

If $b$=2, we get $a$! = $a$!/2 + 1 + 2^$c$1, which leads to $a$!/2 = 1 + 2^$c$1.

If $a$ 3, $a$!/2 is even, so 2^$c$1 = 1. But then we get $a$!/2 = 2: no solution.

If $a$ = 3, 3 = 1 + 2^$c$1 $c$ = 2.

Hence we obtain the only solution: 3!2! = 3! + 2! + 2².

Related problems:

Factorial Symmetry: $a$!$b$! = $a$! + $b$!

Factorial Equation: $a$!$b$! = $a$! + $b$! + $c$!

Factorial And Square: $a$!$b$! = $a$! + $b$! + $c$²