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Prove this is divisible by 15

  1. Jan 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove [itex] 4^{2n}-1 [/itex] is divisible by 15 for all positive integers
    3. The attempt at a solution
    Ok so I factor it into this
    [itex] (2^n+1)(2^n-1)(4^n+1) [/itex]
    I know how to get the factor of 3 because we know
    2^n is even so either 2^n+1 or 2^n-1 is divisible by 3 because we have 3 consecutive integers so at least one of them needs to be divisible by 3.
    But im not sure how to get the factor of 5.
    I thought about squaring some of their components and looking how close they
    are to 4^n+1.
     
  2. jcsd
  3. Jan 11, 2013 #2
    Consider the cases when n is odd and n is even. You should be able to obtain a factor of 5 from different terms in each case.
     
  4. Jan 11, 2013 #3

    pasmith

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    Did you not notice that [itex]4^2 = 16 = 15 + 1[/itex]?

    That suggests a proof, by induction on [itex]n[/itex], that [itex]4^{2n} = 15k_n + 1[/itex] for some integers [itex]k_1 = 1, k_2, k_3, \dots[/itex].
     
  5. Jan 11, 2013 #4
    You may also use the factorisation of:


    [itex]
    \left ( x^n - 1 \right ) = \left ( x - 1 \right ) \left ( x^{n-1} + \ldots + 1 \right )
    [/itex]
     
  6. Jan 11, 2013 #5

    Mentallic

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    Of course it would. Induction is a perfectly reasonable method of proof here.

    Assume [itex]4^{2k}-1=15N[/itex] for some integer N, and now prove that [itex]4^{2(k+1)}-1[/itex] is a multiple of 15.
     
  7. Jan 11, 2013 #6
    Well, I have found a solution :

    Let us assume that 42n-1 is divisible by 15.
    So, we have

    16n-1 is divisible by 16-1 ....

    Since 16n-1 = 16n - 1n .... it takes the form of an-bn is divisible by a-b. Just prove that it is possible and you have it ...
     
  8. Jan 11, 2013 #7

    Mentallic

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    That's not exactly induction, but rather the method that jfgobin detailed, except that assuming 42n-1 is divisible by 15 does absolutely nothing for your proof.
     
  9. Jan 11, 2013 #8
    Yeah it requires proof by induction. Its pretty straight foward. You have a base case. Let n=1 and show it works. Then for the inductive case basically what you assume that
    15|4^(2k)-1 where k ε Z. Moreover we want to show that 15|4^(2k+2)-1. Then rewrite the notation as such were 15r=4^(2k)-1 where rεZ and 15n=4^(2k+2)-1 where nεZ. Also notice that 15n=4^(2k+2)-1 is just 15n=4^(2k)*16. From there its pretty much easy.
     
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