$x^2 + y^2 + xy = x^2y^2$ has no solutions

Question

Prove that there do not exist positive integers $x$ and $y$ such that $x^2 + y^2 + xy = x^2y^2$ has no solutions.

Answer 1

Divide through by $x^2y^2$. Then we get, $$\frac{1}{y^2} + \frac{1}{xy} + \frac{1}{x^2} = 1$$ One of the denominators must be less than or equal to three. x = 1 and y = 1 are ruled out. Hence, xy = 2 or 3. But then x or y needs to be 1, which is not possible. Hence it has no solutions.

I think another way to do this would be by infinite descent, which would be pretty straightforward.

Answer 2

Add $xy$ to both sides, and you get:

$$(x+y)^2=xy(xy+1)$$

Show this means $xy$ and $xy+1$ must be squares.

Answer 3

Hint:

The discriminant of $$x^2(1-y^2)+xy+y^2=0$$ is

$$y^2-4y^2(1-y^2)=y^2(4y^2-3)$$

So we need $4y^2-3$ to be perfect square $=z^2$(say) where $z\ge0$

$3=(2y-z)(2y+z)\implies2y+z=3,2y-z=1\implies y=z=1$

Can you take it from here?