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Compute $(\frac{3}{379})$.

So what I did is as follows:

$(\frac{3}{379})= (\frac{379}{3})=1$ because using Euler's Criterion $\displaystyle 1^\frac{3-1}{2}\equiv 1$. Hence $3$ is a quadratic residue mod $379$ but the back of my book says $3 $ is a quadratic non-residue mod $379$. Could someone please explain what I did wrong.

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    $\begingroup$ $379 \equiv 3 \pmod 4$, so $(\frac{3}{379})= (\frac{379}{3})$ is false. $\endgroup$
    – sharding4
    Mar 31, 2018 at 16:45
  • $\begingroup$ I see where I went wrong. $\endgroup$
    – Rose
    Mar 31, 2018 at 16:47

1 Answer 1

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$$\left(\frac{3}{379}\right)=(-1)^{\frac{3-1}{2}\cdot\frac{379-1}{2}}\left(\frac{379}{3}\right)=-1\cdot\left(\frac{379}{3}\right)=-1$$

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