It is proved that if n > 0, D divides 2n ^ 2, then n ^ 2 + D is not a complete square number

It is proved that if n > 0, D divides 2n ^ 2, then n ^ 2 + D is not a complete square number

D = 2Kn ^ 2K is an integer
n^2+d=n^2(2k+1) k=4 d=8n^2
The proposition of n ^ 2 + D = 9N ^ 2 = (3n) ^ 2 complete square number
D = K (2n) ^ 2K is an integer
n^2+d=n^2(4k+1)
k=2 d=8n^2
The proposition of n ^ 2 + D = 9N ^ 2 = (3n) ^ 2 complete square number