Let $m$ be an integer and $q$ be an odd prime factor of $m^2 + 1$. Is there an obvious reason that $\left(\frac{2m}{q}\right)$ always equals 1? From some numerics, this seems to be the case.

The last time I got stuck on something like this, it ended up just being because $-1 \equiv m^2 \pmod{m^2 + 1}$, so I'm wondering if there's something easy I'm missing.

This would be useful for an explicit 2-descent I'm doing for prime twists of elliptic curves defined in terms of $m$.