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Chinese remainder theorem (Gaussian ints)

  1. Nov 23, 2009 #1
    Figured it would be a quick review of ideas that turned into a nightmare...

    [tex] x \equiv i ([/tex]mod [tex]1 + i)[/tex]
    [tex] x \equiv 1 ([/tex]mod [tex]2-i)[/tex]
    [tex] x \equiv 1+i ([/tex]mod [tex]3+4i)[/tex]

    we find [tex]s \equiv 0 [/tex](mod [tex]1+i)[/tex] and [tex]s \equiv 1 [/tex](mod [tex]5+10i)[/tex] but this is easy.

    So now find [tex]s' \equiv 0 [/tex](mod [tex]2-i)[/tex] and [tex]s' \equiv 1 [/tex](mod [tex]-1+7i)[/tex] is complicated...Using the euclidean algorithm I got that there is a gcd of [tex]\pm i[/tex] for [tex]2-i[/tex] and [tex]-1+7i[/tex]

    I think i'm kind of stuck, I also can't seem to find
    [tex]s'' \equiv 0 [/tex](mod [tex]3+4i)[/tex] and [tex]s'' \equiv 1 [/tex](mod [tex]3+i)[/tex].
     
  2. jcsd
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