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Math Help - [SOLVED] Prove Composite

  1. #1
    Junior Member star_tenshi's Avatar
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    [SOLVED] Prove Composite

    Problem:
    Show that 8^{n}+1 is composite for all n\geq1.

    What I have so far:
    In general,

    a^{n}+1 = (a+1)(a^{(n-1)} - a^{(n-2)} + a^{(n-3)} - \cdots + 1)
    but this only holds for n odd.

    Is there a way to factor when n is even? Or would I have to find a solution specific to a=8?
    Last edited by star_tenshi; January 22nd 2009 at 12:23 PM. Reason: switched odd / even for n. (>.<)
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by star_tenshi View Post
    Problem:
    Show that 8^{n}+1 is composite for all n\geq1.

    What I have so far:
    In general,

    a^{n}+1 = (a+1)(a^{(n-1)} - a^{(n-2)} + a^{(n-3)} - \cdots + 1)
    but this only holds for n odd.

    Is there a way to factor when n is even? Or would I have to find a solution specific to a=8?
    Note: You may want to wait for a more NT minded member to answer this in case of error

    Have you tried induction?
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  3. #3
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by star_tenshi View Post
    Problem:
    Show that 8^{n}+1 is composite for all n\geq1.

    What I have so far:
    In general,

    a^{n}+1 = (a+1)(a^{(n-1)} - a^{(n-2)} + a^{(n-3)} - \cdots + 1)
    but this only holds for n odd.

    Is there a way to factor when n is even? Or would I have to find a solution specific to a=8?
    You can use that identity with n=3, if you write 8 as 2^3.
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  4. #4
    Junior Member star_tenshi's Avatar
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    Quote Originally Posted by Opalg View Post
    You can use that identity with n=3, if you write 8 as 2^3.
    Yes, I know that, but I am looking at the case where n is even. Even if I were to use 8^{n} + 1 = (2^{3})^m + 1 = 2^{3m} + 1, 3m is not always odd.
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  5. #5
    o_O
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    Make use of the identity: a^3 + b^3 = (a+b)(a^2 -ab + b^2)

    Here imagine: a = 2^n and b = 1

    For all n \geq 1, we have:
    \begin{aligned} 8^n + 1 & = \left(2^3\right)^n + 1 \\ & = \left(2^n\right)^3 + 1 \\ & = \left(2^n + 1\right)\left((2^n)^2 - 2^n + 1\right) \end{aligned}

    So what can we conclude?
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  6. #6
    Junior Member star_tenshi's Avatar
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    OHHH! Now I see what Opalg was trying to convey. Thanks again o_O for clearing that up!

    We can conclude that 8^{n} + 1 is composite.

    Then to generalize, any number that can be expressed as a^{3} + b^{3} is composite because it can be broken down into the factors (a+b)(a^{2}-ab+b^{2})!
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