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Thread: Determine solvablity

  1. #1
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    Determine solvablity

    Prove that $\displaystyle x^{2}+2\equiv0 \mod p$ is unsolvable if $\displaystyle p\equiv 5,7\mod 8$ .
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  2. #2
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    Quote Originally Posted by chipai View Post
    Prove that $\displaystyle x^{2}+2\equiv0 \mod p$ is unsolvable if $\displaystyle p\equiv 5,7\mod 8$ .
    Thanks
    Can we use Gauss's Lemma ?

    To determine whether the given number $\displaystyle a \neq 0 $ is quadratic residue , consider the list :

    $\displaystyle a , 2a , 3a , ..., Pa ~,~ P = \frac{p-1}{2} $

    If the number of the elements in the above list lying between $\displaystyle P+1 $ and $\displaystyle 2P $ inclusively is even , then $\displaystyle a $ is quadratic residue , otherwise , it is quadratic non-residue .

    In this case , $\displaystyle a \equiv -2 \bmod{p} $ and $\displaystyle P $ is either $\displaystyle 2\bmod{4} $ or $\displaystyle 3 \bmod{4} $

    The elements lying between the range should be :

    $\displaystyle -2 , -4 , .... , (-2) \frac{P}{2} $ for $\displaystyle P \equiv 2 \bmod{4} $ so the number is $\displaystyle \frac{P}{2} $ an odd number .

    Or

    $\displaystyle -2 , -4 , ..., (-2) \frac{P-1}{2} $ for $\displaystyle P \equiv 3 \bmod{4} $ so the number is $\displaystyle \frac{P-1}{2} $ also an odd .

    Therefore , $\displaystyle - 2 $ is not the quadratic residue of modulo $\displaystyle p \equiv 5,7 \bmod{8} $ .

    Remarks : so we can see that $\displaystyle - 2 $ is quadratic residue of modulo $\displaystyle p \equiv 1,3 \bmod{8} $ .
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