Numbers that are clearly NOT a Square Although I have never studied math very seriously, I have heard of Brocard's Problem, which asks for integer solutions for the following Diophantine Equation:$$n!+1=m^2$$
The only solutions are conjectured to be $(4,5),(5,11),(7,71)$, and there are no solutions for $n < 10^{9}$. 
However, it is clear there are an infinite amount of $n$ where it is simple to verify $n!+1$ is not a square. 
For example, take $n=81$. Assume that $81!+1$ is a square. 
This implies that $81!-1=a^2-2$ for some integer $a$. 
Note that by Wilson's Theorem, $82!\equiv -1 \pmod {83}$, implying $81! \equiv 1 \pmod {83}$. 
Thus, $a^2-2 \equiv 0 \pmod {83}$. A contradiction, since $2$ is not a quadratic residue of $83$. 
In a similar way, we can claim that for all $n>5$ where $n+2$ is a prime number $p$ such that $ p \equiv 3,5 \pmod 8$, then $n!+1$ is not a square. 
My question is, are there any other integers $n$ where it is simple to verify $n!+1$ is not a square? Any help would be appreciated.
 A: Your nice argument can be generalized. 
For example, take $n=80$. You deduced from Wilson's Theorem that $81!\equiv 1\pmod {83}$. Now $$80!\equiv \frac{1}{81}\equiv -\frac{1}{2}\pmod{83}.$$ Here $\frac{1}{2}$ means the inverse of $2$ modulo $83$. It follows that $80!+1\equiv -\frac{1}{2}+1=\frac{1}{2}\pmod{ 83}$. Because $2$ is not a quadratic residue modulo $83$, neither is its inverse, so $80!+1$ cannot be a square. Of course, this works for any $n$ for which $n+3$ is prime congruent to $3$ or $5$ modulo $8$.
More generally, for any prime $p$ and positive integer $k\leq p-1$,
$$
(p-k)!+1\equiv \frac{(-1)^k}{(k-1)!}+1=\frac{(k-1)! \big[(k-1)!+(-1)^k\big]}{((k-1)!)^2}\mod p.
$$
So $(p-k)!+1$ is a quadratic residue modulo $p$ if and only if $(k-1)! \big[(k-1)!+(-1)^k\big]$ is a quadratic residue. Via quadratic reciprocity this can be translated to a condition on $p$.
For each $k\geq 1$, there is a condition similar to yours. I'll list the first three (they start looking complicated very fast). The quantity $n!+1$ cannot be a square if any of the following conditions hold:


*

*$n+2$ is a prime congruent to $3$ or $5$ mod $8$

*$n+3$ is a prime congruent to $3$ or $5$ mod $8$

*$n+4$ is a prime  congruent to $5$, $9$, $21$, $23$, $31$, $37$, $43$, $45$, $49$, $51$, $55$, $57$, $59$, $65$, $67$, $69$, $71$, $73$, 
$77$, $81$, $83$, $85$, $87$, $91$, $95$, $97$, $99$, $101$, $103$, $109$, $111$, $113$, $117$, $119$, $123$, $125$, $131$, $137$, $145$, $147$, $159$, or $163$ mod $168$.


The primes in the third case are exactly those for which $42$ is a quadratic non-residue.
