Find two incongruent solutions to $x^{2}\equiv-1 (\text{mod}37)$ (use Wilson's theorem) Here is some of the proof:
Let $p=37$ which is an odd prime. Then, by Wilson's Theorem, $x^{2}\equiv -1 (\text{mod}p)$ has a solution if and only if $p \equiv 1 (\text{mod}4).$ Furthermore, that solution is $(\frac{p-1}{2}!)^{2} \equiv -1 (\text{mod}p).$ (still according to the theorem). 
And, we can see by inspection that $37 \equiv 1 (\text{mod}4)$, and so the solutions fulfiill $(\frac{36}{2}!)^{2}\equiv -1 (\text{mod}37)$, or $(18!)^{2} \equiv -1(\text{mod}37).$ But
$18!=18(17)(16)(15)(14)(13)(12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)$
...? But I can't figure out how to finish it. I forget how to simplify this kind of situation, or how I can simplify it in a way that will be helpful.
 A: It seems like the two incongruent results implied by Wilson's Theorem are $18!$ and $(18!)^3$. You already have $18!$ and since obviously $(18!)^4\equiv 1 \bmod 37$, the other root must be $(18!)^3$ since $(18!)^6 \equiv (18!)^2 \equiv -1 \bmod 37$.
It's not necessary to reduce these values down to their remainders, but for information, $18!\equiv 31 \bmod 37$.
A: I don't know of a quick and dirty way to reduce $18!$ modulo $37$.  We can observe that 
$$
2 \times 18 = 4 \times 9 = 36 \equiv -1 \bmod 37
$$
and then
$$
3 \times 5 \times 6 \times 7 = 630 \equiv 1 \bmod 37
$$
so that leaves
$$
18! \equiv 8 \times 10 \times 11 \times 12 \times 13 \times 14 \times 15 \times 16 \times 17 \bmod 37
$$
But beyond that, I don't know a better way than successive multiplication and division:
$$
8 \times 10 = 80 \equiv 6 \bmod 37
$$
$$
6 \times 11 = 66 \equiv -8 \bmod 37
$$
$$
-8 \times 12 = -96 \equiv 15 \bmod 37
$$
and so forth.  This process eventually yields $31 \equiv -6 \bmod 37$, and then Joffan's answer yields the other answer, $6$.
A: (1). Remark: In this Q observe that $37$ is $1$ more than a square : $37=6^2+1.$ So $-1\equiv 6^2\equiv (37-6)^2\equiv 31^2 \pmod {37}.$
(2). For reduction of $(p-1)! \pmod p$ : 
For a prime $p$ that is not large, it is often easy and quick, for small $n$, to find $m$ such that $nm\equiv \pm 1 \pmod p. $ And some shortcuts are available: E.g., given $(7)(16)\equiv 1\pmod {37}$ we have $(8)(14)=(7)(16)\equiv 1\pmod {37}.$
Writing $a.b$ for $ab:\;$ Modulo $37 $we have $$18!\equiv (2.18).(3.12).(4.9). (5.15). (6).(7.16). (8.14). (10.11).( 13.17)\equiv$$ $$\equiv (-1)(-1)(-1)(+1)(6)(+1)(+1)(-1)(-1)\equiv -6\equiv 31.$$ 
Alternately, calculating only the terms in the above product that precede $(10.11)(13.17)$ we could have obtained $18!\equiv (-6).(10.11.13.17)$ which can be easily reduced without prior knowledge that $10.11 \equiv 13.17 \equiv -1.$
Also notice that  the first few terms (e.g $2.18$) are obtained by factoring $pk\pm 1 $ for  small $k$. 
A computer programmer would likely ask why you didn't stop this process when you got to $6$ and found $6^2\equiv -1.$
