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Math Help - 2 to even power as a sum of squares

  1. #1
    Junior Member
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    2 to even power as a sum of squares

    Problem: Show that if  2^{2k} = a^{2} + b^{2} then either a or b is zero for some integers a, b and a non-negative integer k.

    My solution attempt outline:

    > Assume a>0, b>0
    > Consider a Pythagorean Triple of form (a, b, 2^{k})
    > Show that we cannot obtain a primitive P.T. from above ( 2^{k} is not odd) so the equation  (2^{k})^{2} = a^{2} + b^{2} does not have a solution for any k
    > So, we cannot have a>0, b>0
    > Show that 2^{2k} = (2^{k})^{2} + 0^{2} so (a>0 and b=0) OR (a=0 and b>0)

    Anything wrong with this? Does anyone have a simpler solution? Thanks.
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  2. #2
    MHF Contributor
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    Re: 2 to even power as a sum of squares

    Looks good.

    Last line is a bit confusing. I'd just say that if a=0 then clearly b=2k, and similarly if b=0
    Thanks from director
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