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(n^4) + 4 is composite ?by Gokul43201
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#1
Apr1904, 10:03 PM

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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
Apr1904, 10:06 PM

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Are there any factors that tend to occur frequently? Maybe you can solve the problem by cases.



#3
Apr2004, 02:10 AM

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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. 


#4
Apr2004, 08:19 AM

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(n^4) + 4 is composite ?
thanks crag,
I must be blind ! 


#5
Apr2004, 07:39 PM

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On no! The blind leading the blind!



#6
May2704, 10:06 AM

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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... 


#7
May2704, 10:41 AM

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I am almost tempted not to welcome you 


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