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(n^4) + 4 is composite ?

  1. Apr 19, 2004 #1

    Gokul43201

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    How do you prove that (n^4) +4 is composite for all n>1 ?

    I found this problem in a book I was reading. The even n part is obvious. The proof for odd n cannot be done by trying to factorize [(2k+1)^4]+4. I know this is true, because I've tried and moreover, the factors of the first few such numbers are : (2*2*5),(5*17),(2*2*5*13),(17*37),(2*2*5*5*13),(5*13*37),... The expected factors of 4 appear in the even numbers, but the factors for the odds seem patternless. So, I imagine that the proof must be done by some other means.

    Any ideas ?
     
  2. jcsd
  3. Apr 19, 2004 #2

    Hurkyl

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    Are there any factors that tend to occur frequently? Maybe you can solve the problem by cases.
     
  4. Apr 20, 2004 #3
    Are you sure they're patternless? Just look at the odd n for now:

    n=3: n^4+4 = 5*17
    n=5: n^4+4 = 17*37
    n=7: n^4+4 = 5*13*37 = 37*65

    And I'll add n=9 for you:

    n=9: n^4+4 = 5*13*101 = 65*101

    See any pattern yet? That should lead you to the general proof.
     
  5. Apr 20, 2004 #4

    Gokul43201

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    thanks

    thanks crag,
    I must be blind !
     
  6. Apr 20, 2004 #5
    On no! The blind leading the blind! :cool:
     
  7. May 27, 2004 #6
    how about:
    [tex]n^4 + 4 = n^4 + 4 - 4n^2 + 4n^2 = (n^2 + 2)^2 - (2n)^2 = (n^2 + 2n + 2)(n^2 - 2n + 2)[/tex]
    :)

    edit:
    sorry, i pulled up an old post... just realized that it's from April not May 20th...
     
  8. May 27, 2004 #7

    arildno

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    What a terrible misdemeanor on your part...
    I am almost tempted not to welcome you :smile:
     
  9. May 27, 2004 #8

    Gokul43201

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    thanks, yrch...that's nice.

    Now I really feel like $#!+.
     
    Last edited: May 27, 2004
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