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Thread: Another quadratic reciprocity problem

  1. #1
    MHF Contributor alexmahone's Avatar
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    Another quadratic reciprocity problem

    For which positive integers n do there exist integers x and y such with (x, n) = 1, (y, n) = 1, such that $\displaystyle x^2+y^2 \equiv 0\ (mod\ n)$?
    Last edited by alexmahone; Jul 21st 2011 at 04:56 PM.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: Another quadratic reciprocity problem

    It is easy to see that n = 2 satifies the given congruence. If n is odd, every prime p dividing n should satisfy $\displaystyle \left(\frac{-y^2}{p}\right)=1 \implies \left(\frac{-1}{p}\right)=1 \implies p \equiv 1\ (mod\ 4)$.

    The answer given in the book is: $\displaystyle n=2^{\alpha}\prod p^{\beta}$ where $\displaystyle \alpha$ = 0 or 1, the primes p in the product are all $\displaystyle \equiv 1\ (mod\ 4)$, and $\displaystyle \beta=\beta(p)=0,1,2,...$.

    How do they get the $\displaystyle 2^\alpha$ in the product?
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