## Factorial Symmetry

#### Problem

Given that `a` and `b` are positive integers, solve the following equation.

`a`!`b`! = `a`! + `b`!

#### Solution

If we divide through by `b`! we get, `a`! = `a`!/`b`! + 1.

As LHS is integer, `a`!/`b`! must be integer, and it follows that `b`! `a`!.

If, instead, we divide through by `a`! we get, `b`! = 1 + `b`!/`a`!, and in the same way we deduce that `a`! `b`!.

Hence `a`! = `b`!, giving the unique solution `a`! = 2 `a` = `b` = 2.

Alternatively, divide the original equation by `a`!`b`! to give 1 = 1/`b`! + 1/`a`!. Clearly `a`! = `b`! = 2.

Related problems:

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

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

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

Problem ID: 214 (09 Mar 2005) Difficulty: 2 Star