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Thread: Proove that...

  1. Sep 19th 2011, 01:24 PM #1
    Sorombo
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    Proove that...

    Proove that for every $n \in \mathbb{N}$

    n^3-9n+27\not\equiv 0 \mod 81
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  2. Sep 20th 2011, 11:15 PM #2
    CaptainBlack
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    Re: Proove that...

    Quote Originally Posted by Sorombo View Post
    Proove that for every $n \in \mathbb{N}$

    n^3-9n+27\not\equiv 0 \mod 81
    Suppose otherwise, then there exists a k\in \mathbb{Z} such that:

    n^3-9n+27=k\times 81=k\times 9^2

    which implies that 9|n^3 which in turn implies that 3|n.

    Hence there exists a \lambda \in \mathbb{N} such that:

    27 \lambda^3-27 \lambda +27=3 \times k \times 27

    or:

    \lambda^3- \lambda +1=3 \times k

    Now consider the left hand side modulo 3.

    CB
    Last edited by CaptainBlack; Sep 20th 2011 at 11:26 PM.
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