Solve $ (x^2+y^2)^3=(x^2-y^2)^2$ in rational numbers

Question

Everyone talks about Pythagoras equations but here is something similar:

$$ (x^2+y^2)^3=(x^2-y^2)^2$$

This is the shape of a rose curve. It does have a peculiar 8-fold intersection at the origin. Two self tangencies at the origin: $y = x$ and $y = -x$.

Wikipedia says this curve is genus 0, meaning there should be a "map" to projective space. We can find a point on the curve e.g. $(x,y)=(0,0)$ or $(1,0)$ and intersect the curve with various lines of rational slope $y=mx+b$ with $m\in \mathbb{Q}$.

There could also be an integer equation by writing the equation in homogenous coordinates $[x:y:z]\in P^1(\mathbb{Q})$.

$$ (x^2+y^2)^3=(x^2-y^2)^2 \, z^2$$

The $z$ coordinate looks extra, but in this part $x,y,z \in \mathbb{Z}$.

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Answer 1

$$(x^2+y^2)^3= (x^2-y^2)^2$$ Let $x= r \cos (t)$, y = r $\sin(t)$ $$r^6= r^4 (\cos^2(t)-sin^2(t))^2$$ $$r^2= \cos^2(2t)$$ If $(a,b,c)$ is a Pythagorean triple, then $\cos(t) =\frac{a}{c}$, $\sin(t) =\frac{b}{c}$ should give a rational solution. Because then $r$ is rational, hence $x$ and $y$ are.

Answer 2

Let $x=tu, y=tv$. Then, we have $$t^6\,(u^2+v^2)^3=t^4\,(u^2-v^2)^2,$$ meaning $$u^2+v^2=\frac{(u^2-v^2)^2}{t^2\,(u^2+v^2)^2}=w^2.$$ The solution of this equation in rationals is known: $$u=w\,\frac{2s}{1+s^2}, v=w\,\frac{1-s^2}{1+s^2}, t=\frac{u^2-v^2}{w^3}\tag1$$ So $$(x,y)=\left(t\,u,t\,v\right)$$ with $u,v,t$ from (1) are the non-trivial solutions of the equation: we can reconstruct every rational solution $\neq(0,0)$ from it by letting $u=x$ and $v=y$, then we get $t=1$ because the equation is satisfied.
With $w=1, s=1/2$, we get $$(x,y)=\left(\frac{28}{125}, \frac{21}{125}\right).$$