diophantine equations - $Dm^2 - n^2D^2$ is a perfect square then $D$ is the sum of two squares

How do I show that if $$Dm^2 - n^2D^2$$ is a perfect square for some integers $m$ and $n$ ($n \neq 0$), $D$ is the sum of two (non-zero) perfect squares? I tried solving for $D$ but that only gives me $$D = \frac{m^2}{2n^2} \pm \frac{\sqrt{m^4 - 4n^2 k^2}}{2n^2}$$ for integers $m$, $n$, and $k$, which doesn't seem easier.

EDIT: $D$ itself should not be a perfect square.

Answer

If $Dm^2-D^2n^2= a^2$, then $Dm^2$ is a sum of two squares. Now an integer is a sum of two squares if and only if all primes $\equiv 3 \mod 4$ in its factorization occur with even multiplicities. The presence of the extra square $m^2$ doesn't affect this condition, so $D$ is a sum of two squares, also.