Find all triples of non-negative Integers $a,b,c$ such that $a!b!=a!+b!+c!$

Question

Exactly what it says in the Title; not much development from there :/

Answer 1

I assume you mean you've determined that without loss of generality $b\geq a$. We then have $$b!=1+\frac{b!}{a!}+\frac{c!}{a!},\,c\geq a.$$ Since $a!b!>a!+b!$, we have $a!,\,b!\geq 2$ so $b\geq a\geq 2$. In fact if $a=2$ then $2\times b! = 2+b!+c!,\,b!=2+c!$, which has no solutions. So $b\geq a\geq 3$ and $3|b!$, so $\frac{b!}{a!},\,\frac{c!}{a!}$ are not both multiples of $3$, and $b,\,c$ are not both $\geq a+3$. But $a!=1+\frac{a!}{b!}+\frac{c!}{b!}$, so either $a=b\leq c$ or $a<b>c$. The latter case precludes $c\geq a+3$, and so does the former since $\frac{c!}{a!}=a!-2\notin 3\mathbb{Z}$. So $c\in \left\{a,\,a+1,\,a+2\right\}$.

Case I: $c=a$ so $3|b!=2+\frac{b!}{a!}$ so $b\geq a+2$. If $b=a$ then $a!=3$; if $b=a+1$ then $\left(a+1\right)!=a+3$; if $b=a+2$ then $\left(a+2\right)!=a^2+3a+4$. One can easily check none of these is solved by integers $a \geq 3$.

Case II: $c=a+1$ so $b!=a+2+\frac{b!}{a!}$ and $a!-1|b!-\frac{b!}{a!}=a+2$. Since $a!-1 > a+2$ for $a\geq 4$ we have $a=3,\,c=4$, giving one solution with $b=3$. It turns out there are no others (see the next paragraph).

Case III: $c=a+2$ so $b!=a^2+3a+3+\frac{b!}{a!}$, so $a!-1|a^2+3a+3$, which is consistent with neither $a=3$ nor $a=4$. But $a!-1>a^2+3a+3$ whenever $a\geq 5$, so there are no more solutions.

Answer 2

First assume $a = b,$ so the equation becomes $a!(a!-2) = c!.$ In particular this implies that $a!-2 = 0$ or $(a+1)\cdots(a+k)$ for some $k > 0.$ The first case is impossible, and in the second, either $a+1\mid a!$ or $a!\equiv -1\mod (a+1)$ and $a+1$ is prime (Wilson's theorem). Actually there is a remaining case, in which $a! - 2 = a+1,$ which is when $a = b =3,$ giving a solution with $c = 4.$ Evidently when $a > 3,$ this is no longer possible, so we reduce to the cases above.

If $a+1\mid a!,$ then $a+1$ does not divide $a!-2$ unless $a+1\mid 2,$ whence $a = 1$ or $0.$ This is impossible, so $a+1$ is prime and $a! - 2\equiv -3\mod (a+1).$ But $a+1\mid a! - 2,$ so $a+1 = 3\Rightarrow a = b = 2,$ which again is impossible. Hence $a\not= b.$

Without loss of generality let $1 < a < b.$ Note then that $c > b,$ since $b!\mid RHS \Rightarrow b! \mid a! + c! < b! + c!.$ Divide through by $a!$ to get $$b! = 1 + b!/a! + c!/a!.$$ Then mod $a+1,$ the LHS is $0,$ while the RHS is $1,$ so the only solution is $(a,b,c) = (3,3,4).$