Monday, 5 February 2018

diophantine equations - Dm2n2D2 is a perfect square then D is the sum of two squares




How do I show that if Dm2n2D2 is a perfect square for some integers m and n (n0), D is the sum of two (non-zero) perfect squares? I tried solving for D but that only gives me D=m22n2±m44n2k22n2 for integers m, n, and k, which doesn't seem easier.



EDIT: D itself should not be a perfect square.


Answer



If Dm2D2n2=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.


No comments:

Post a Comment

real analysis - How to find lim_{hrightarrow 0}frac{sin(ha)}{h}

How to find \lim_{h\rightarrow 0}\frac{\sin(ha)}{h} without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...