# For which $n$ is $\frac{n!}{4}$ equal to $\left\lfloor \sqrt{\frac{n!}{4}}\right\rfloor\left(\left\lfloor\sqrt{\frac{n!}{4}}\right\rfloor + 1\right)$?

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!

• Google "Brocard's problem". It works for $n= 4,5,$ and $7$, it is conjectured those are the only such values. If you accept the proof of the abc conjecture then it is known that there are only finitely many such values of $n$. – Nate Jun 7 at 21:12

## 1 Answer

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.