Prove that $-2, -3, -5, -7$ are quadratic non-residues modulo p

Question

Prove that if $p$ is prime and $p\equiv 173 \pmod{1680}$, then $-2, -3, -5, -7$ are quadratic non-residues modulo p.

Answer 1

Since $p\equiv1\pmod4,$ it follows that $\left(\dfrac{-1}{p}\right)=1.$ Then $p\equiv5\pmod8,$ so $\left(\dfrac{2}{p}\right)=-1.$
Also, $\left(\dfrac{-3}{p}\right)=(-1)^{(p-1)/2}(-1)^{(p-1)/2}\left(\dfrac{p}{3}\right)\equiv p\pmod3.$ But $p\equiv-1\pmod3,$ and hence $\left(\dfrac{-3}{p}\right)=-1.$
Now $\left(\dfrac{5}{p}\right)=\left(\dfrac{p}{5}\right)$ while $p\equiv3\pmod5.$ Thus $\left(\dfrac{-5}{p}\right)=\left(\dfrac{3}{5}\right)=\left(\dfrac{2}{3}\right)=-1.$
Finally, $\left(\dfrac{7}{p}\right)=(-1)^{(p-1)/2}\left(\dfrac{p}{7}\right)=\left(\dfrac{p}{7}\right).$ But $p\equiv5\pmod7.$ Therefore $\left(\dfrac{p}{7}\right)=\left(\dfrac{5}{7}\right)=\left(\dfrac{2}{5}\right)=-1.$

This uses a lot of quadratic reciprocity, so tell me if some steps are ambiguous.
Hope this helps.