If $x,y\in\mathbb Z$, solve $x^2+xy+y^2=x^2y^2$.

Question

If $x,y\in\mathbb Z$, solve $$x^2+xy+y^2=x^2y^2$$

We could try some factorizations. $x^2+y^2=xy(xy-1)$. We may as well add $2xy$ to both sides: $(x+y)^2=xy(xy+1)$. Then we could subtract $x^2y^2$ from both sides: $(x+y+xy)(x+y-xy)=xy$

Or we could somehow use the fact that if an integer is between two consecutive integer squares, then it can't be a square itself.

I'd like to get some more help here, some other strategies if these can't be used. Thanks.

Answer 1

You could rewrite it like $$x^2+2xy+y^2=(xy)^2+xy.$$ When encountering things of the form $a^2+a$, always think of $\frac{(2a+1)^2-1}4$.

(Also, $a^2-a=\frac{(2a-1)^2-1}4$.) These identities often allow you to rewrite equations in a nice way, especially when doing number theory.

Once you know this, it becomes $$4(x+y)^2=(2xy+1)^2-1.$$

I bet you can continue from here.

Answer 2

$$ x^2+xy+y^2=x^2y^2 \quad \Longleftrightarrow \quad (1-y^2)x^2+yx+y^2=0$$

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

$$4y^2-3=p^2 \quad \implies \quad y^2=1\cdots $$