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.
On
The proposition is false. Suppose every prime divisor (if any) of $D$ is congruent to $3$ modulo $4$ and that $D$ is a square. Then $D$ is not the sum of two non-zero squares. Suppose $(p,q,r)$ is any Pythagorean triplet with $p^2=q^2+r^2$. Let $D=E^2$. We have $D(Ep)^2-q^2D^2=(rD)^2.$ Examples:(I)D=1,m=5,n=4. (II).D=9,m=15,n=4.
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.