Below is perhaps a possible proof of Brocard’s Problem, which is a famous problem stating that for $n, m \in \mathbb{Z^+}$, there only exists finitely many solutions to the equation $n! + 1 = m^2$ such that only three ordered pairs $(m, n)$ satisfy this equation. Each pair are known as the Brown Numbers, and a notable mathematician named Paul Erdós had also conjectured this problem.
$\longrightarrow$ A possible proof of Brocard’s Problem $-$ Joseph Marx $\longleftarrow$
I read through it and it does look correct. It mentions that $\sqrt{n!}$ is irrational, even though $\sqrt{0!} = 1^2$ but that case is not important. Since $n$ and $m$ are positive integers, this implies that $n > 0$, so we do not consider this case. Thus, for all the rest of the positive integers $n$, we know that $\sqrt{n!}$ is irrational, therefore $\sqrt{n!} \neq m$ because $m$ is also a positive integer, and all positive integers are rational.
So the question is as follows: Is the supposed proof of Brocard’s Problem actually valid? I can be naive sometimes...
The "proof" is not very sophisticated and falls apart at the end through a misunderstanding of the basic properties of limits. Starting from
$$\sqrt n! = m O(\sqrt n!),$$
which is just a rearrangement of the initial equation $n!=m^2+1$, the author takes the limit on one side to get:
$$\lim_{n\to\infty} \sqrt n! = m O(\sqrt n!).$$
This makes little sense. The author then uses the fact that $\lim_{n\to\infty}O(\sqrt n!)=1$ to claim that there is a contradiction, since $\sqrt n! \neq m$. This is not how limits work.