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Find $(x, a, b, c)$ if $$x! = a! + b! + c!$$

I want to know if there are more solutions to this apart from $(x, a, b, c) = (3, 2, 2, 2)$.

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Not a mathematical comment, but my (weak) intuition says that it is difficult (or impossible), since the factorial function grows really fast. –  Ian Mateus Jun 30 '12 at 18:10
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up vote 12 down vote accepted

First note that $a,b,c < x$, since $1 \leq n!$. This means that $a,b,c \leq x-1$. This implies that $$x! = a! + b! + c! \leq 3 (x-1)! \implies x \leq 3$$ Further, $x! = a! + b! + c! \geq 3 \implies x >2$. Hence, $x=3$ is the only possibility left.

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If $x\ge 4$ then $a \le x-1$, and we have $a!\le x!/4$, and similarly for $b$ and $c$. Then $a!+b!+c!\le \frac{3}{4}x!$, so there are no solutions with $x\ge 4$, and the other cases can be eliminated quickly.

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