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Factorial And Square

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

$a$!$b$! = $a$! + $b$! + $c$²

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)^($p$1)/2 = $c$^$p$1 (-1)^($p$1)/2 mod $p$.

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

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