Number Theory: Wilson's Theorem

mathsss2

I posted this question but I am not getting anywhere with this question, any help would be very appreciated:

1. let [tex]p[/tex] be odd prime explain why: [tex]2*4*...*(p-1)\equiv (2-p)(4-p)*...*(p-1-p)\equiv(-1)^{(p-1)/2}*1*3*...*(p-2)[/tex] mod [tex]p[/tex].

2. Using number 2 and wilson's thereom [[tex](p-1)!\equiv-1[/tex] mod p] prove [tex]1^23^25^2*....*(p-2)^2\equiv(-1)^{(p-1)/2}[/tex] mod [tex]p[/tex]

Thanks.

mathsss2

Update: I got #2, I still don't know how to do #1.

dodo

Let me give names to the individual expressions in question 1: call them, from left to right, A(p), B(p), C(p), so that your question 1 becomes
[tex]A(p) \equiv B(p) \equiv C(p) \mbox{ (mod p)}[/tex]

Now, B(p) and C(p) are the same expression. If you multiply by -1 each of the factors in B(p), (that's (p-1)/2 factors), they become (p-2)(p-4)...(p-(p-1)), that is, 1*3*...(p-2).

Now you only need to prove that either of B(p) or C(p) is congruent to A(p) modulo p. Just observe that [tex]-1 \equiv p-1 \mbox{ (mod p)}[/tex], and also [tex]-3 \equiv p-3 \mbox{ (mod p)}[/tex], and also ...

robert Ihnot

The fact of the matter is that [tex] 2-p \equiv 2 Mod p [/tex] Thus the first two expressions are of equal value. As Dodo has already explained.