Show that $4(p-5)! + 1 \equiv 0 \pmod p$

Question

Show that $4(p-5)! + 1 \equiv 0 \pmod p$

I'm having trouble figuring out how to show this. The best I've come up with is : $$4(p-5)!\equiv (p-5)! \equiv (p-1)!\equiv -1$$ However, I'm not even sure if the congruences are true. I had just been going off of $(p-1)!+1\equiv 0 \pmod p$ and trying to get the LHS to look like that.

Thank you in advance for your help in this matter!

Answer 1

Wilson theorem

$$-1\equiv(p-1)!\pmod p$$

$(p-1)!\equiv(p-5)(-1)(-2)(-3)(-4)\pmod p$

$\equiv4!\cdot(p-5)!$

So, I believe there is a typo(missing $!$ symbol after $4$) in the question.

Answer 2

Starting with Wilson's Theorem:

$$ \begin{align} -1&\equiv(p-1)!\\ -1&\equiv(p-1)(p-2)(p-3)(p-4)(p-5)!\\ \end{align} $$

The claim is that $-1\equiv4(p-5)!$. So for it to be true:

$$ \begin{align} (p-1)(p-2)(p-3)(p-4)&\equiv4\\ (-1)(-2)(-3)(-4)&\equiv4\\ 24&\equiv4\\ 20&\equiv0 \end{align} $$

So this could only possibly be true for $p=2$ or $p=5$. The original relation makes no sense for $p=2$, and is true for $p=5$.

So it's true exactly when $p=5$.