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Number Theory: Wilson's Theorem

  1. Nov 2, 2008 #1
    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]

  2. jcsd
  3. Nov 3, 2008 #2
    Update: I got #2, I still don't know how to do #1.
  4. Nov 4, 2008 #3
    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 ...
  5. Nov 5, 2008 #4
    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.
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