rational solutions of $x^4+x^2y^2+y^4=x^2$ I am trying to find rational solutions of $x^4+x^2y^2+y^4=x^2$, except for $(\pm 1,0), (0,0) $. I guess these three solutions are the only ones, but I failed to prove it. It seems that $x^4+x^2y^2+y^4=(x^2+xy+y^2)(x^2-xy+y^2) $ may be useful, but I do not know how to proceed.

Start with the HINT by Piquito: $$(x^2+y^2)^2=x^4+x^2y^2+y^4+x^2y^2=x^2+x^2y^2$$ Rearranging yields $$(x^2+y^2)^2=x^2(y^2+1)$$ Since the LHS is a perfect square, and $x^2$ is a perfect square, it follows that $(y^2+1)$ must be a perfect square, and the only solution in the integers for that is $y=0$. The implications of $y=0$ have been elaborated in a previous answer.
