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