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  1. #1
    Junior Member darence's Avatar
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    equation and her solutions...

    If a and b are solutions of equation x^2-6x+1=0, proove that \forall n \in N number a^n+b^n is integer not divisible with 5.

    Please help, no idea
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  2. #2
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    Re: equation and her solutions...

    since x2 - 6x + 1 = (x - a)(x - b) = x2 - (a+b)x + ab,

    we have a + b = 6, which is clearly an integer. note 5 does not divide 6. note also that ab = 1, as we will need this later.

    also (a + b)2 = (a + b)(a + b) - 2ab = 36 - 2 = 34, which is also an integer.

    now let's prove an + bn is an integer, by induction on n.

    suppose ak-1 + bk-1 is an integer, for all k < n as our induction hypothesis.

    then an + bn = (an-1 + bn-1)(a + b) - (ab)(ak-2 + bk-2)

    since the RHS side is all integers, the LHS must be, as well.
    Last edited by Deveno; January 29th 2013 at 02:42 PM.
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  3. #3
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    Re: equation and her solutions...

    Yes, but the question is about a^n + b^n and not (a+b)^n.
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    Re: equation and her solutions...

    Quote Originally Posted by emakarov View Post
    Yes, but the question is about a^n + b^n and not (a+b)^n.
    its funny how your mind plays tricks on you. i noticed this just as i posted my answer, and amended it accordingly.
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  5. #5
    Junior Member darence's Avatar
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    Re: equation and her solutions...

    thank you very much.
    How to proove a^n+b^n is not divisible with 5?
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    Re: equation and her solutions...

    Quote Originally Posted by Deveno View Post
    an + bn = (an-1 + bn-1)(a + b) - (ab)(ak-2 + bk-2)
    And concerning divisibility by 5, it is sufficient to consider this recurrence relation modulo 5: this is a periodic sequence.

    Quote Originally Posted by Deveno View Post
    its funny how your mind plays tricks on you.
    I can relate. On the other forum, there was a discussion about formal proofs. I argued that verifying proofs on the computer ensures their correctness for all practical purposes, but agreed that there is no absolutely 100% guarantee, if only for this reason. Though typos can usually be eliminated by a community of researchers.
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