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Math Help - Sum of two squares

  1. #1
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    Sum of two squares

    Suppose that p=x^2+y^2 p is prime with x odd, y even. Show that (\frac{x}{p})=1
    I know that p \equiv 1(4), so I can use LQR for Jacobi symbol
    (\frac{x}{p})=(\frac{p}{x})=(\frac{x^2+y^2}{x})=(\  frac{y^2}{x})=1, but for the last equality holds I need to have x,y are relatively prime. Are they relatively prime in this case? Can someone help me argue that they are relatively prime?
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  2. #2
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    Quote Originally Posted by namelessguy View Post
    Are they relatively prime in this case? Can someone help me argue that they are relatively prime?
    Let (x,y) = d then d|x^2 + y^2 = p . Thus d = 1 or p .
    But d = p \Rightarrow x,y  \geq p but thats a contradiction
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