Can someone check my argument on this problem? If p is prime and $\displaystyle p\equiv 1 ( mod 4)$ and $\displaystyle m^2+n^2=p$ and $\displaystyle m>0$ odd. Then $\displaystyle (\frac{m}{p})=1$

By LQR,

$\displaystyle (\frac{m}{p})=(\frac{p}{m})=(\frac{m^2+n^2}{m})=(\ frac{n^2}{m})$. Here, do I need to check the cases when m is prime and when m is composite to figure out if $\displaystyle (\frac{n^2}{m})=1$?

In class, I think we have a property that $\displaystyle (\frac{n^2}{p})=1$ with p prime.