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Math Help - 2p as the sum of squares

  1. #1
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    2p as the sum of squares

    Suppose that the odd prime p = x^2 + y^2 with x > y > 0. Find integers u and v (in terms of x and y) such that 2p = u^2 + v^2 with u > v > 0.


    Okay, I know that this means 2p = 2x^2 + 2y^2, and then u^2 + v^2 = 2x^2 + 2y^2. I'm not sure it helps, but since p is prime and it can be written as the sum of two squares, then p ≡ 1 mod 4. I'm not sure what I should do next.
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  2. #2
    Super Member PaulRS's Avatar
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    What is (x+y)^2 + (x-y)^2 ?
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  3. #3
    Senior Member Shanks's Avatar
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    Apply the identity:
    if x=u^2+v^2, y=s^2+t^2, then xy=(us+vt)^2+(ut-vs)^2.
    since 2 and p both can be written as sum of two squares, so let x=2,y=p, the result follows immediately.
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