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Math Help - sum of two squares proof

  1. #1
    Member diddledabble's Avatar
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    sum of two squares proof

    Show that 8k+7 cannot be the sum of three squares
    Last edited by diddledabble; August 4th 2009 at 02:15 PM.
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  2. #2
    Master Of Puppets
    pickslides's Avatar
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    I might be reaching a bit here but I can show it can be the sum of 2 squares.

     (\sqrt{8k})^2+(\sqrt{7})^2 = 8k+7
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  3. #3
    Member diddledabble's Avatar
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    My mistake

    It should be three squares.
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  4. #4
    Super Member Gamma's Avatar
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    I won't write it all out explicitly for you, but here is how you would go about it, there will be lots of cases and things for you to work out.

    first of all you look at the residue classes mod 8. If it could be done, then you would need to be able to satisfy 7\equiv a^2 + b^2 + c^2 (mod 8).

    0^2 \equiv 0 (mod 8)
    1^2 \equiv 1 (mod 8)
    2^2 \equiv 4 (mod 8)
    3^2 \equiv 1 (mod 8)
    4^2 \equiv 0 (mod 8)
    5^2 \equiv 1 (mod 8)
    6^2 \equiv 4 (mod 8)
    7^2 \equiv 1 (mod 8)

    There is no way to add 3 numbers coming from the set {0,1,4} (repetition obviously allowed) and get 7 (mod 8). Therefore it cannot be done.
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