(In reply to
Do any exist? by Gamer)
Nope.
After substituting x=z^2 and y=az^2 you can divide by z^2 and get
z^2(a^2 + 1) + 1 = 2kaz^2 which is possible only if z=1.
Then x=1 and y=a. Plugging in these values gives a quadratic in 'a' with discriminant 4(k^2  2).
But k^2  2 = 2 or 3 mod4, so will never be a square.
Thus, there are no solutions.

Posted by xdog
on 20110309 10:51:09 