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

Question

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:

Probably a trivial answer, but is this the only solution? If yes, how can we show it formally?
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

Mark Sapir · Accepted Answer · 2020-10-21 14:45:26Z

1

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

answered Oct 21 '20 at 14:43

Mark Sapir

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$\begingroup$ Wow nice! And how do I calculate it? $\endgroup$ – Eliyahu Abadi Oct 21 '20 at 14:45
$\begingroup$ There is no faster method them simple checking. $\endgroup$ – Mark Sapir Oct 21 '20 at 14:46

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