Factorial Equation

I posted this problem on Brilliant a few months back here so -_-.

Firstly, $a,b > 1$ , which can be proven by trivially checking these cases. Now, we will prove that $c \geq a, b$ . We have the following

$\begin{aligned} a!+b!+c!&=a!b!\\ \implies c!+1&=a!b!-a!-b!+1\\ \implies c!+1&=(a!-1)(b!-1)\\ \implies a!-1 &\mid c!+1\\ \implies a!-1 &\leq c!+1\\ \implies a!-2 &\leq c!\\ \implies a &\leq c \quad \forall a\geq 3 \end{aligned}$

Similarly, this can be proved for $b$ . Thus, $c \geq a, b$ if they are both $\geq 3$ . If $a=2$ , we have $b!-1 = c!+1$ , which isn't satisfied for any integral $b, c$ . Same applies if $b=2$ . Thus, $a, b > 2$ . We will now prove $a=b$ . We have

$\begin{aligned}a!+b!+c!&=a!b!\\ \implies b!+c!&=a!b!-a!\\ \implies b!(1+\frac{c!}{b!})&=a!(b!-1)\\ \gcd (b!, b!-1)=1 \implies b! &\mid a! \end{aligned}$

Similarly, we can prove $a! \mid b!$ . Thus, $a!=b!$ , so $a=b$ . We now have

$\begin{aligned} 2a! + c! &= a!^2\\ c! &= a!(a!-2) \end{aligned}$

Firstly, note that $a < c$ . Next, we will prove that the product of $n$ consecutive numbers is divisible by $n!$ . But this is almost trivially true by looking at binomial coefficients:

$\begin{aligned} \dbinom{k+n}{k} &= \dfrac {(k+n)!}{k!n!}\\ &= \dfrac {(k+1)(k+2)\ldots(k+n)}{n!} \end{aligned}$

Since all binomial coefficients are integers, it follows that the product of $n$ consecutive numbers is divisible by $n!$ .

Using this, we have

$\begin{aligned} \dfrac {c!}{a!} &= a! - 2\\ (a+1)(a+2)\ldots (c) &= a!-2\\ \implies (c-a)! &\mid a!-2 \end{aligned}$

However, we have that $a! - 2 \equiv 2 \pmod{4}$ if $a \geq 4$ . Thus, $4 \not \mid (c-a)!$ , so $c-a \leq 3$ . We will now consider each of the cases:

Case 1: $c=a+1$

We have

$\begin{aligned} a+1 &= a!-2\\ 3 &= a((a-1)!-1)\\ \implies a &= 3 \end{aligned}$

The value of $a$ we pertain from the factorisation of the LHS, using $a \geq 3$ . Checking, we find this to be true, so we get $c=4$ . Thus, one such solution is $(3, 3, 4)$ .

Case 2: $c=a+2$

We have

$\begin{aligned} (a+1)(a+2) &= a! - 2\\ 4 &= a((a-1)! - a - 3)\\ \implies a&= 4 \end{aligned}$

However, when we check the inside of the bracket in the RHS, we find that the value of $a$ doesn't satisfy. Thus, no solutions in this case.

Case 3: $c=a+3$

We have

$\begin{aligned} (a+1)(a+2)(a+3) &= a! - 2\\ 8 &= a((a-1)! - a^2 - 6a - 11)\\ \implies a&= 8, 4 \end{aligned}$

However, when we check the inside of the bracket in the RHS, we find that neither values of $a$ doesn't satisfy. Thus, no solutions in this case.

Thus, the only solution is $(3, 3, 4)$ . Therefore, the answer is $3+3+4=10$ .

Note: We do not consider the alternate case when $a < 4$ , since it implies $a=3$ , which is covered above.