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Math Help - Divisibility

  1. #1
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    Divisibility

    I am so sorry for misleading you. I needed to solve this problem, because it occurred to me whilst I was solving another one. Indeed, what I had written is not true. In such case, could you help me find all integers n >= 1 for which
    1+2^{n+1} + 2^{2n+2}

    i divisible by

    1+2^{n} + 2^{2n}?
    Last edited by gollum; September 21st 2011 at 10:55 AM.
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  2. #2
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    Re: Divisibility

    Quote Originally Posted by gollum View Post
    Could somebody help me prove that for any integer n >= 1
    1+2^{n+1} + 2^{2n+2}

    i divisible by

    1+2^{n} + 2^{2n}?

    I have already tried proving it using induction but I got stuck in the final step where n=k+1. Please, help.
    Have you checked what happens if you try putting n=2, or n=3?
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  3. #3
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    Re: Divisibility

    It is impossible to prove something that is NOT true! What reason do you have to believe that this statement is true?
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  4. #4
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    Re: Divisibility

    Hi! I have already corrected the post.
    Last edited by gollum; September 21st 2011 at 12:03 PM.
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  5. #5
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    Re: Divisibility

    You could try to show that the ratio \frac{1+2^{n+1} + 2^{2n+2}}{1+2^{n} + 2^{2n}} increases towards a limiting value 4. The ratio is equal to 3 when n=1. For n>1, it will lie between 3 and 4, so can never again be an integer.
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  6. #6
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    Re: Divisibility

    Thanks, but how do I find maximum and minimum of a function like this - I mean it is exponential as well as rational.
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  7. #7
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    Re: Divisibility

    Probably the easiest approach is to write 4 -\frac{1+2^{n+1} + 2^{2n+2}}{1+2^{n} + 2^{2n}} in the form \frac{???}{1+2^{n} + 2^{2n}}. Then show that that fraction is positive and less than 1 when n>1.
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