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.
 A: Let $a=x^2,b=y^2$. Then the equation simplifies to $a^2 + ab+b^2 = a$, which on grouping becomes $a^2 + a(b-1) + b^2=0$. Solving for $a$,
$$
a= \frac{1-b + \sqrt{1+b^2-2b-4b^2}}{2} = \frac{1-b + \sqrt{1-2b-3b^2}}{2} 
$$
(The other root would be negative, but a is positive).
Therefore, we would like $1-2b-3b^2$ to be a perfect square, however, this is clearly only possible when $b=0$, because the expression is negative for $b \geq 1$. 
Therefore, $y^2=0 \implies y=0 \implies x^4=x^2$, and :
$$
x^4-x^2=0 \implies x^2(x-1)(x+1)=0
$$
Hence there are three possible solutions, namely $(x,y)=(1,0),(0,0),(-1,0)$.

For rational non-zero $b$, suppose that $b = \frac{p^2}{q^2}$ in simplest form, where $p,q\neq 0$, so  that $\gcd(p,q) = 1$. We are basically asking whether $1 - 2b - 3b^2$ can be a perfect square. First of all, note that : $$
(3b^2-2b-1) =(3b+1)(b-1) = \left(\frac{3p^2}{q^2}+1\right)\left(\frac{p^2}{q^2} -1\right) = \left(\frac{(3p^2+q^2)(p^2-q^2)}{q^2}\right)
$$
We claim that $\gcd(3p^2 + q^2,p^2 - q^2) = 1$ or $4$. To see this,let $p^2  = a$ and $q^2 = b$. Then, if $t | 3a+b$ and $t | a-b$, by summing $t | 4a$  and by taking $t | 3a-3b$ and subtracting we get $t | 4b$. Therefore, $t | \gcd(4a,4b) = 4$ as $a,b$ are coprime. However, $3a+b \equiv a-b \mod 4$, and in our case $a-b = p^2- q^2$ cannot take the value $2$ mod $4$, therefore the $\gcd$ cannot be two, since $p^2-q^2$ cannot divide two without dividing $4$, in which case the other must too.
$(3b+1)(b-1)$ is a perfect square if and only if $(3p^2 + q^2)(p^2 - q^2)$ is. If the two terms are coprime, then they are individually squares. If they are not, then $\frac{3p^2+q^2}{4},\frac{p^2-q^2}{4}$ are coprime whose product is a perfect square, and now $4$ is a perfect square so these terms are perfect squares implies that $3p^2+q^2$ and $p^2-q^2$ are both perfect squares.
That is, for co-prime $p$ and $q$, we want $p^2-q^2=c^2$ and $3p^2+q^2=d^2$ to both be squares, which are co prime.
However, $c^2 + d^2 = 4p^2$. Since $c^2$ and $d^2$ cannot both divide $4$,  $c^2+d^2$ cannot be a multiple of $4$ from what we know about squares modulo $4$.
Therefore, no non-zero rational value of $b$ produces a square value of the expression. Hence, we have only $b= 0$ to consider, which of course forms part of the integral possibilities we discussed earlier.
A: HINT.-A better way could be the following:$$(x^2+y^2)^2=x^4+x^2y^2+y^4+x^2y^2=x^2+x^2y^2$$ Then by Pythagorean triples you do have two possibilities
$$\begin{cases}1)\space\space x=2xy\space\space\text{ and }\space\space xy=x^2-y^2\\2)\space\space x=x^2-y^2\space\space\text{ and }\space\space xy=2xy\end{cases}$$ Can you end now?
A: 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.
