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Divisibility question

  1. Sep 8, 2009 #1
    How can I show that (a^n -b^n) doesn't divide (a^n + b ^n) for all integers a,b, and n?

    I have that if (a^n -b^n) did divide (a^n + b ^n), then (a^n +b^n) = q (a^n -b^n) which implies b^n = -q*b^n (mod a^n). Then 1 = -q (mod a^n), meaning gcd(b, a^n) = 1. I am unsure of what more I can deduce. Any help is appreciated.
     
  2. jcsd
  3. Sep 8, 2009 #2
    I think you need to assume that a, b and n are all > 1. Otherwise there are simple counterexamples.
     
  4. Sep 8, 2009 #3
    Yes, of course.
     
  5. Sep 9, 2009 #4

    CRGreathouse

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    a^n - b^n | a^n + b^n
    implies
    a^n - b^n | 2b^n
    and so, if a >= b,
    ka^n = (2 - k)b^n
    for some positive k. It must be the case that k = 1 (why?), so
    a^n = b^n
    and hence
    a = b.
     
  6. Sep 9, 2009 #5
    Shouldn't the equation

    ka^n = (2 - k)b^n

    instead be

    ka^n = (2 + k)b^n?
     
  7. Sep 9, 2009 #6
    I think I have a solution. Here's a hint: Prove that you can reduce the problem to the case in which a and b are relatively prime.

    HTH

    Petek
     
  8. Sep 10, 2009 #7
    We assume a and b have no common factor, because if they did, it could be removed.

    To avoid the case of a^n-b^n = 0, we need only consider a prime p, such that p divides a^n-b^n. We eliminate the trivial cases where n=1, or where a,b or n =0. Then [tex]a^n \equiv b^n Mod p [/tex]; [tex] a^n + b^n \equiv 2b^n Mod p [/tex]

    Thus p divides 2 or p divides b^n. but 2 can be a factor of b since then it would also divide a, and consequently we can chose p >2. But then it follows both a and b are divisible by p contrary to assumption.
     
    Last edited: Sep 11, 2009
  9. Sep 11, 2009 #8
    That is easy.
    a=3,b=2,n=2 implies
    a^n+b^n=13 which is not divisible by a^n-b^n=5.
    So, if it is not valid for 3,2,and 2, then it is not valid for all integers.
     
  10. Sep 11, 2009 #9

    Borek

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    Hm, if a^n+b^n=2b^n doesn't it mean a^n-b^n=0?
     
  11. Sep 11, 2009 #10
    This proof seems to be correct (with "but 2 can be a factor of b" an obvious typo for "but 2 cannot be a factor of b"), and is simpler than the one I had in mind. Nice!

    Petek
     
  12. Sep 11, 2009 #11
    Yes, thank you: "But 2 CANNOT be a factor of b since then it would also divide a...."
     
  13. Sep 12, 2009 #12

    Office_Shredder

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    an-bn|an+bn

    Then

    an - bn|an+bn-(an-bn)
    Which gives

    an-bn|2bn

    Nothing here says they're equal though
     
  14. Sep 12, 2009 #13

    Borek

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    Thanks :smile:
     
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