Find the solutions of x^2 + xy + y^2 = x^2*y^2, where x and y are integers.

Question

anonymous · Answer

It's easy to show there are none when one is even and one is odd. When both are even, I've proved that the only solutions is (0,0), by infinite descent. But I'm having problems with the case where they are both odd. Then there is the solution (1,-1), but I can't prove that there aren't any other.

Or perhaps there's another way to solve it? Some clever factoring I didn't saw?

amistre64 · Answer

x^2 + xy + y^2 = x^2 y^2

add xy to each side?

x^2 + 2xy + y^2 = x^2 y^2 + xy

(x+y)^2 = x^2 y^2 + xy

if we knew a babylonian theyd prolly know for sure :)

amistre64 · Answer

not sure if I got the -4 part right but:  (x+y)^2 = (x-y)^2 + 4xy

anonymous · Answer

Yeah, and that solves it. A square can't, after all, be a product of two consecutive numbers, unless one of them is 1.

anonymous · Answer

I mean, (x+y)^2 = xy(1+xy).

amistre64 · Answer

yay!!

anonymous · Answer

Yay! Thanks. :)