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Gaussian integers and norms

  1. Feb 22, 2012 #1
    2q3bw2w.png

    Show that N(a+bi) = even => a+bi divisible by 1+i

    So, N(a+bi) = a2+b2 = even

    so 2 divides a2+b2

    Write 2 = (1+i)(1-i)

    so we have 1+i divides a2+b2

    so 1+i divides either (a+bi) or (a-bi)

    if 1+i divides a+bi we are done

    what if 1+i divides a-bi though?

    Thats where I'm stuck!
     
  2. jcsd
  3. Feb 22, 2012 #2

    tiny-tim

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    hint: what is (1 - i)/(1 + i) ? :wink:

    (or generally (a + ib)/(1 + i) ?)
     
  4. Feb 22, 2012 #3
    hmm (1 - i)/(1 + i) = -i and (a + ib)/(1 + i) = (1-i)(a+bi)/2

    Not sure where you're going with it though!
     
  5. Feb 22, 2012 #4

    tiny-tim

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    and which of those factors are in Z ? :wink:
     
  6. Feb 22, 2012 #5


    so only 1+i is a factor?
     
  7. Feb 22, 2012 #6

    tiny-tim

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    which of the two factors (1-i)(a+bi)/2 are in Z ?
     
  8. Feb 22, 2012 #7


    Well (1-i) if (a+bi)/2 is the other factor

    Or (a+bi) if (1-i)/2 is the other factor

    Which should I take?
     
    Last edited: Feb 22, 2012
  9. Feb 22, 2012 #8

    tiny-tim

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    sorry, i've got confused :redface:

    i meant, is (1-i)(a+bi)/2 in Z ?
     
  10. Feb 22, 2012 #9


    nope!
     
  11. Feb 22, 2012 #10

    tiny-tim

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    why not?
     
  12. Feb 22, 2012 #11
    Since (a+b)+(bi - ai)/2

    So a=b mod 2

    i.e can only be a gaussian integer if both a&b are either even or odd?

    So it can be a Gaussian integer providing that, and if a&b are even or odd then a2 + b2 = even, which is the norm of a+bi?
     
  13. Feb 22, 2012 #12

    tiny-tim

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    perfect! :smile:
     
  14. Feb 22, 2012 #13

    Dick

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    I you want just a bit different take on it, then if (1+i) divides (a-bi) take the complex conjugate. So (1-i) divides (a+bi). But (1-i)=(-i)(1+i).
     
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