
Proof question
Recall the definition of n! (read n factorial"):
n! = (n)(n1)(n2) ….(2)(1) =∏(k)
In both (a) and (b) below, suppose p≥3 is prime.
(a) Prove that if x∈ Zpx is a solution to x square ≡1 (mod p), then x ≡±1 (mod p).
(b) Prove that (p1)!≡±1 (mod p)
Zpx x shoud be above p

Okay, Correct me if I'm wrong, but the two question are
Let p be prime.
1) If $\displaystyle x\in\mathbb{Z}_{p} $ and $\displaystyle x^{2}\equiv\\1$ (mod p) then $\displaystyle x\equiv\pm\\1$ (mod p)
2) $\displaystyle (p  1)!\equiv\pm\\1$ (mod p)
For 1 we get
$\displaystyle x^{2}\equiv\\1$ (mod p)
$\displaystyle \Rightarrow\\x^{2} 1\equiv\\0 $ (mod p)
$\displaystyle \Rightarrow\\(x1)(x+1)\equiv\\0$ (mod p)
$\displaystyle \Rightarrow\\p(x1)(x+1) $
$\displaystyle \Rightarrow\\p(x1) $ or $\displaystyle p(x+1) $
$\displaystyle \Rightarrow\\x\equiv\\1 $ (mod p) or $\displaystyle x\equiv\\1 $ (mod p)
$\displaystyle \Rightarrow\\x\equiv\pm\\1 $ (mod p)
2) is wrong. The correct version is $\displaystyle (p  1)!\equiv\\1$ (mod p), the result is known as Wilson's Theorem
$\displaystyle (p1)! = 1*2*3........(p2)(p1) $
Now 1 and p1 are the only numbers modulo p that are their own inverses.
For any other number x such that $\displaystyle 1<x<p1$ there exists $\displaystyle x^{1}$ such that $\displaystyle 1<x^{1}<p1$ and $\displaystyle x*x^{1}\equiv\\1$ (mod p).
So we match up each x with its inverse and we get:
$\displaystyle (p1)!\equiv\\1*2*3........(p2)(p1)\equiv\\1*1*1.......(p1)\equiv\\p1\equiv\\1$ (mod p)
Now we have the desired result.

Thank you for your answer
but for the second one
it is Wilson's theorem like you said
but the right one should be (p1)!≡1