TITLE: For which $n$ is $\frac{n!}{4}$ equal to $\left\lfloor \sqrt{\frac{n!}{4}}\right\rfloor\left(\left\lfloor\sqrt{\frac{n!}{4}}\right\rfloor + 1\right)$?
QUESTION [2 upvotes]: For how many values of $n$, is $\frac{n!}{4}$ equal to  $\left\lfloor \sqrt{\frac{n!}{4}}\right\rfloor\left(\left\lfloor\sqrt{\frac{n!}{4}}\right\rfloor + 1\right)$?
Further more, is there a way to approximate (or maybe even find the precise answer to) $\left\lfloor \sqrt{\frac{n!}{4}}\right\rfloor\left(\left\lfloor\sqrt{\frac{n!}{4}}\right\rfloor + 1\right)$? I tried approximating $n!$ but had no luck.
Thanks in advance!

REPLY [2 votes]: It's not a full solution, but maybe it will help:
Let $x=\sqrt{\frac{n!}{4}}$.
$$ x^2 = \lfloor x\rfloor \cdot (\lfloor x\rfloor + 1)$$
$$ x^2 + \frac14 =  (\lfloor x\rfloor + \frac12)^2$$
$$ -\frac12 + \sqrt{x^2+ \frac14} = \lfloor x\rfloor$$
$$ -1 + \sqrt{n!+ 1} = 2\lfloor x\rfloor \in 2\mathbb N$$
So we get a necessary condition
$$ \exists m\in\mathbb N : n! + 1 = (2m+1)^2$$
It's easy to see that it is also a sufficient condition, because
$$ \sqrt{n!} < \sqrt{n! + 1} < \sqrt{n!} + 1$$
$$ \sqrt{n!} < 2m+1 < \sqrt{n!} + 1$$
$$ m < \sqrt{\frac{n!}{4}} < m + \frac12$$
so $\lfloor \sqrt{\frac{n!}{4}}\rfloor = m$.
For low $n$, $n=4$ and $n=5$ satisfy this condition, but I have no proof there is no more solutions.
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