How do I show that if Dm2−n2D2 is a perfect square for some integers m and n (n≠0), D is the sum of two (non-zero) perfect squares? I tried solving for D but that only gives me D=m22n2±√m4−4n2k22n2 for integers m, n, and k, which doesn't seem easier.
EDIT: D itself should not be a perfect square.
Answer
If Dm2−D2n2=a2, then Dm2 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.
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