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Prove that 3^(2n+1) + 4^(2n+1)is divided by 7

  1. May 8, 2005 #1
    prove that 3^(2n+1) + 4^(2n+1)
    is divided by 7 for every natural n
     
  2. jcsd
  3. May 8, 2005 #2

    Zurtex

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    What have you done so far?
     
  4. May 8, 2005 #3
    this isn't that hard. It is just induction as you suspect. The only fact you need to know is that [tex]c_1n + c_2m \equiv b_1n + b_2m\, (mod\, k)[/tex] if [tex] c_1\equiv b_1,\,c_2\equiv b_2\,(mod\, k)[/tex].

    So just write down your sequence [tex]a_n = 3^{2n+1}+4^{2n+1}[/tex]. When [tex]a_n[/tex] is in that form can you write down [tex]a_{n+1}[/tex] without out modifying the exponent.

    Hope those hints help.
    Steven
     
  5. May 8, 2005 #4
    well assuming that 3^(2n+1) + 4^(2n+1) is divisible by 7
    im writing down what needs to be proven for n=n+1

    3^(2n+3) + 4^(2n+3) / 7 = some integer = Z

    9*3^(2n+1)+16*4^(2n+1) / 7 = Z

    3^(2n+1)+4^(2n+1) / 7 + 8*3^(2n+1)+15*4^(2n+1) / 7 = Z

    the first term is divisible because of the induction assumption so that leaves me with
    8*3^(2n+1)+15*4^(2n+1) / 7

    here im stuck i can't find a way to show that this term is divisible by 7
    i tried proving that with another induction but that led me no where.
    i also tried transforming 8*3^(2n+1)+15*4^(2n+1) / 7 to
    7*3^(2n+1)+7*4^(2n+1) / 7 + 3^(2n+1)+8*4^(2n+1) / 7

    3^(2n+1)+4^(2n+1) + 3^(2n+1)+8*4^(2n+1) / 7
    and that leaves me with the term 3^(2n+1)+8*4^(2n+1) / 7
    that does not help me i can't find a way to show that its divisible by 7
    any help/tips would be appreciated

    sorry for not using latex by the way it would just take me too much time to type all this in latex :smile:
     
  6. May 8, 2005 #5
    Oh you're so close!

    (3^(2n+1)+4^(2n+1)) / 7 + (8*3^(2n+1)+15*4^(2n+1)) / 7
    = (3^(2n+1)+4^(2n+1)) / 7 + (3^(2n+1)+4^(2n+1)) / 7 + (7*3^(2n+1)+14*4^(2n+1)) / 7

    You just needed to do your trick one more time over

    Hope that helps
    Steven
     
  7. May 9, 2005 #6
    of course! i can't believe i didn't notice this thank you very much! :smile:
     
  8. May 9, 2005 #7
    We can easily prove it without induction:

    [tex]a_n = 3^{2n+1}+4^{2n+1}\equiv 3^{2n+1} + (-3)^{2n+1} \equiv 3^{2n+1}\cdot(1 + (-1)^{2n+1}) \equiv 0 (mod 7)[/tex].
     
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