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Wilson's Theorem

  1. Feb 24, 2010 #1
    Show that (p-1)(p-2)...(p-r)==(-1)^r*r!(mod p) for r=1, 2, ..., p-1. I am having trouble getting this proof started. Can you please give me some direction? Thank you.
  2. jcsd
  3. Feb 24, 2010 #2
    Well...first note that if you multiply out then every term in the resulting sum will be a multiple of p, and hence will disappear mod p. So all you really need to be concerned with is the constant term: (-1)(-2)...(-r).
  4. Feb 24, 2010 #3
    So something like
    -1 * -2 * ... * -r == (-1)(-2)...(-r)(mod p)
    == (-1)^r(1*28...*r)(mod p)
    == (-1)^r * r! (mod p)
    and we are done, right?

    Thank you.
  5. Feb 24, 2010 #4
    By Fermat's Little Theorem, [tex]a^{p-1}-1\equiv 0[/tex] mod p. Define a polynomial [tex]p(x)=x^{p-1}-1[/tex], and note that [tex]q[/tex] is satisfied for x=1,2,...,p-1, and that these are all the roots.

    So we can then write [tex]q(x)=(x-1)(x-2)\cdots(x-(p-1))[/tex]. By changing the order of subtraction in each factor, we can make a related polynomial [tex]q'(x)=(1-x)(2-x)\cdots((p-1)-x)=(-1)^{n-1}q(x)[/tex].

    What happens when we evaluate at zero? [tex]q'(0)=(1)(2)\cdots(p-1)=(-1)^{n-1}q(0)=(-1)^n[/tex]. I.e., [tex](p-1)!\equiv(-1)^n[/tex] mod p.

    It's a simple step from there to the result you need.
    Last edited: Feb 24, 2010
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