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Problem ; modulo p

  1. Oct 13, 2012 #1
    if p is prime and n is natural number, show that (p-n+1) . (p-2)! + n–1 =0 (mod p)

    i think i have to show that (p-n+1) . (p-2)! = 0 (mod p) and n – 1 =0 (mod p)
    using Wilson theorem
    (p-1)!=-1(p)
     
    Last edited: Oct 13, 2012
  2. jcsd
  3. Oct 13, 2012 #2


    Well, of course [itex]\,n-1\neq 0\pmod p\,[/itex] almost always. Working modulo p in the following:

    $$(p-2)!=\frac{(p-1)!}{p-1}=\frac{-1}{-1}=1\Longrightarrow (p-n+1)(p-2)!+n-1=(-n+1)\cdot 1+n-1=-n+1+n-1=0$$

    DonAntonio
     
  4. Oct 13, 2012 #3
    thanks a lot.
    very clever.
    also i solve it inductances
     
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