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Number Theory

  1. Nov 16, 2009 #1
    Find all odd primes [tex]p[/tex], if any, so that [tex]p[/tex] divides [tex]\sum_{n=1}^{103} n^{p-1} [/tex]
     
  2. jcsd
  3. Nov 16, 2009 #2
    By Fermat's Little Theorem,
    [tex]n^{p-1} \equiv 1 (mod p) [/tex]

    [tex] \sum_{n=1}^{103} n^{p-1} \equiv 103 (mod p)[/tex] ,

    Whence [tex]p \mid 103[/tex]

    Since 103 is prime , therefore 103 it is the only prime.

    It seems like my proof is wrong please correct me. (=
     
  4. Nov 16, 2009 #3
    Your proof is valid for p>103. If p<=103 then you can have p | n and [itex] n^{p-1} (modp)[/itex] will be 0
     
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