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Proving mod relations

  1. Apr 12, 2013 #1
    [itex]\forall[/itex] b[itex]\in[/itex] [itex]Z[/itex] b [itex]\equiv[/itex] 1 (mod 2) [itex]\Rightarrow[/itex] b[itex]^{2}[/itex] [itex]\equiv[/itex] 1 (mod 8)

    How do I go about proving this? Can the Chinese Remainder Theorem be used to prove this or is there something easier?
     
  2. jcsd
  3. Apr 12, 2013 #2

    HallsofIvy

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    Directly from the definition, [tex]b\conguent 1 (mod 2)[/tex] means that b= 1+ 2k[/tex] for some integer k. Then [itex]b^2= 4k^2+ 4k+ 1= 4(k^2+ k)+ 1[/itex]

    Now consider two cases:
    1) k is even: k= 2n for some integer n. What is [itex]4(k^2+ k)+ 1[/itex] in this case?

    2) k is odd: k= 2n+ 1 for some integer n. What is [itex]4(k^2+ k)+ 1[/itex] in this case?
     
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