Diophantine equation $(8n)!-4n+1=k^2$. Solve or prove that there are no integer solutions. 
Solve the Diophantine equation or prove there are no solutions. $$(8n)!-4n+1=k^2$$

I try to use that $k^2 \equiv 0\text{ or }1 \pmod 3$ and $k^2 \equiv 0\text{ or }1 \pmod 4$ but I can not come to a contradiction. Thankful for any ideas.
 A: Rewrite the equation as 
$$ (8n)(8n-1)(8n-2)(8n-3)! - (4n-1) = k^2 $$
Factor out $(4n-1)$ from the left side:
$$ (4n-1)[2(8n)(8n-1)(8n-3)! - 1] = k^2 $$
Now $(4n-1)|(8n-3)!$, so the term in brackets is coprime to $(4n-1)$. Thus $(4n-1)$ has to be a square number. But, as you noted, $(4n-1)$ can't be a square number.
A: The number $4n-1$ is not an exact square $4n-1\equiv  3 (mod 4)$. 
Therefore, in the canonical decomposition, it has a factor of the form $p^k$, where $p$ is prime and $k$ is odd. The factorial has factors $4n-1$ and $8n-2$, therefore it is divisible by $p^{2k}$. 
The difference is divided by $p^k$, but not by $p^{k+1}$, that is, in the canonical decomposition of the number from the condition, the exponent of $p$ is odd.
So $(8n)!-4n+1$ is not an exact square.
A: Let's prove more generally that $M!-m=k^2$ has no solutions in positive integers if $M\gt2m$. (The OP's result follows by setting $m=4n-1$ and $M=8n=2m+2$.)
If $p^r\mid\mid m$, then $p^{2r}\mid M!$, hence $p^r\mid\mid k^2$, which implies $r$ is even, hence $m$ must be a perfect square, say $m=n^2$. Since we now have $M\ge2n^2+1$, it follows that $M!=3n^2N$ for some $N\ge1$.  (The role of the $3$ will soon be apparent.) Consequently $(3N-1)n^2=k^2$, which implies $n\mid k$.  If we write $k=hn$, then we have $3N-1=h^2$, which is impossible, since $-1$ is not a square mod $3$.
