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Thread: solve in Nē

  1. #1
    Super Member dhiab's Avatar
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    solve in Nē

    solve in Nē:
    mē = 1+2ⁿ
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  2. #2
    MHF Contributor
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    Re: solve in Nē

    $$m^2-1 = 2^n$$

    $$(m+1)(m-1) = 2^n$$

    This implies that both $m-1$ and $m+1$ are both powers of $2$. However, there are no distinct powers of 2 that are a distance of 2 apart, so there are no solutions over the natural numbers.
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  3. #3
    Forum Admin topsquark's Avatar
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    Re: solve in Nē

    Quote Originally Posted by SlipEternal View Post
    $$m^2-1 = 2^n$$

    $$(m+1)(m-1) = 2^n$$

    This implies that both $m-1$ and $m+1$ are both powers of $2$. However, there are no distinct powers of 2 that are a distance of 2 apart, so there are no solutions over the natural numbers.
    m = 3, n = 3.

    -Dan
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  4. #4
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    Re: solve in Nē

    Quote Originally Posted by topsquark View Post
    m = 3, n = 3.

    -Dan
    Oops! Good point. I dunno what I was thinking.
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