## Factorial And Power Of 2

#### 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^{c1}, which leads to `a`!/2 = 1 + 2^{c1}.

If `a` 3, `a`!/2 is even, so 2^{c1} = 1. But then we get `a`!/2 = 2: no solution.

If `a` = 3, 3 = 1 + 2^{c1} `c` = 2.

Hence we obtain the only solution: 3!2! = 3! + 2! + 2^{2}.

Related problems:

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

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

Factorial And Square: `a`!`b`! = `a`! + `b`! + `c`^{2}