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Number theorem

  1. Nov 3, 2013 #1
    1. The problem statement, all variables and given/known data

    How to prove the following:

    Let p be a prime p=3,5 (mod8). Show that the
    sequence n!+n^p-n+2 contains at most finitely many squares.

    Should I build a contardiction or prove it directly? I really need some help


    2. The attempt at a solution
    Use Fermats Little we have n^p-n=0 (mod p) then n!+n^p-n+2=n!+2(mod p)
    how should i keep going??
     
  2. jcsd
  3. Nov 3, 2013 #2

    Dick

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    To keep going you should notice that if n>=p then n!=0(mod p) so n!+n^p-n+2=2(mod p). Now if you can show 2 is NOT a quadratic residue of p, then for n>=p, that expression cannot be a square.
     
    Last edited: Nov 3, 2013
  4. Nov 3, 2013 #3
    Thank u so much!!!
     
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