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Modular arith 2

  1. Oct 2, 2008 #1
    prove that for any integer n, n[tex]^{2}[/tex] [tex]\cong[/tex] 0 or 1 (mod 3), and n[tex]^{2}[/tex] [tex]\cong[/tex] 0,1,4(mod 5)
     
  2. jcsd
  3. Oct 2, 2008 #2

    morphism

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    And what have you tried...?
     
  4. Oct 3, 2008 #3
    The only thing i found was that if you can prove n[tex]\cong[/tex]m mod 3 than n[tex]^{2}[/tex] [tex]\cong[/tex] m[tex]^{2}[/tex] mod 3

    but i couldnt prove n [tex]\cong[/tex] 0 mod 3 so i gave up
     
  5. Oct 4, 2008 #4

    morphism

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    You only have to consider n^2 (mod 3) when n=0,1,2. A same type of comment applies mod 5. (Why?)
     
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