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Prove a^5 congruent to a (mod 15)

  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove that a5 [tex]\equiv[/tex]a (mod 15) for every integer a.


    2. Relevant equations



    3. The attempt at a solution
    I do not know how to show a5-a is divisible by 15
     
  2. jcsd
  3. Feb 4, 2010 #2
    [tex]a^5 - a = (a-1)a(a+1)(a^2 + 1)[/tex] can you show that this expression is divisible by 3 and by 5 individually? This would imply it's divisible by 15.
     
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