## Factorial And Square

#### Problem

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

`a`!`b`! = `a`! + `b`! + `c`^{2}

#### Solution

Without loss of generality let us assume that `b` `a`.

If `b` = `a`, then, (`a`!)^{2} = 2(`a`!) + `c`^{2}.

(`a`!)^{2} 2(`a`!) + 1 = `c`^{2} + 1

(`a`! 1)^{2} = `c`^{2} + 1

However, as there exists no square which is one more than another square we deduce that `b` `a`.

In addition, `b` 1, otherwise, `a`! = `a`! + 1 + `c`^{2} `c`^{2} = -1.

Dividing `a`!`b`! = `a`! + `b`! + `c`^{2} by `b`! we get, `a`! = `a`!/`b`! + 1 + `c`^{2}/`b`!.

Clearly LHS is integer, and as `a`!/`b`! is integer, `c`^{2}/`b`! must also be integer. Furthermore, RHS 3, which means that `a`! 3 `a` 3.

From `a`!`b`! = `a`! + `b`! + `c`^{2}, we get `a`!`b`! `a`! `b`! + 1 = `c`^{2} + 1.

(`a`! 1)(`b`! 1) = `c`^{2} + 1.

Let us assume that a prime, `p` 3 mod 4 divides the RHS.

If `c`^{2} + 1 0 mod `p`, it follows that `c`^{2} -1 mod `p`, and it is clear that `p` does not divide `c`.

(`c`^{2})^{(p1)/2} = `c`^{p1} (-1)^{(p1)/2} mod `p`.

But we are given that `p` 3 mod 4, so `p`1 2 mod 4, which means that (`p`1)/2 will be odd, and (-1)^{(p1)/2} = -1.

Hence `c`^{p1} -1 mod `p`, which is a contradiction, because HCF(`c`, `p`) = 1 and by Fermat's Little Theorem, `c`^{p1} 1 mod `p`.

In other words, there exists no prime, `p` 3 mod 4, which divides `c`^{2} + 1.

However, for `a` 4, `a`! 0 mod 4, and `a`! 1 3 mod 4; that is, LHS is divisible by a prime, `p` 3 mod 4.

Hence 1 `b` `a` 3, but we have already established that `a` 3, so we deduce that `a` = 3 and `b` = 2.

From `a`!`b`! = `a`! + `b`! + `c`^{2}, we get, 12 = 6 + 2 + `c`^{2} `c` = 2.

That is, the original equation has a unique solution: `a` = 3, `b` = `c` = 2.

Related problems:

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

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

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