# A way to solve equations of the form $(\frac{n}{a})! = b$?

I found myself today needing the natural solutions for $$n$$ satisfying this equation:
$$(\frac{n}{5})! = 2$$

In this case I "got lucky" and was able to guess that 10 is a suitable solution.

Here are questions that popped in my mind:

1. Probably a trivial answer, but is this the only solution? If yes, how can we show it formally?
2. Perhaps a more interesting question:
given two constants $$a,b \in \Bbb{N}$$ , is there a way for calculating all Natural solutions for $$n$$ satisfying: $$(\frac{n}{a})!=b$$
?
• How do you define the factorial of a fraction? The usual technique is to use the Gamma function. It values at non-integers also tend to be non-integral. Given that, you are really solving for $\frac na$ in the integers. There can only be one solution because factorial is monotonically increasing. – Ross Millikan Oct 21 '20 at 14:40
• If $b$ is large, you can use the approx given by robjohn there math.stackexchange.com/a/2079043/399263, or the one by babler here math.stackexchange.com/a/432690/399263 based solely on log function (no W), else if $b$ small just use brute force. – zwim Oct 21 '20 at 14:58

$$(n/a)!$$ is (rapidly) increasing. So there can be at most one solution of $$(n/a)!=b$$ which can be easily found by checking $$n=a, 2a,...$$.