Does the equation
$$(a^{2}-b^{2})^{4}(a^{2} + b^{2}) = 4a^{2}b^{2} $$
Have any nonzero solutions, where $a^2$ and $b^2$ are both rational ?
If yes, i suspect there should be infinitely many of them, can they be parametrized ?
Does the equation
$$(a^{2}-b^{2})^{4}(a^{2} + b^{2}) = 4a^{2}b^{2} $$
Have any nonzero solutions, where $a^2$ and $b^2$ are both rational ?
If yes, i suspect there should be infinitely many of them, can they be parametrized ?
In any case there are only finitely many rational solutions. If I interpret your question correctly, you are looking for rational points on the curve $C$ given by $$ (x-y)^4(x+y)=4xy $$ where I have put $x=a^2$ and $y=b^2$. Taking $t = x/y$, we easily find that $y$ satisfies $$ y^3 = \frac{4t}{(t-1)^4(t+1)}, $$ so that $C$ is birational to the superelliptic curve $C'$ given by $$ \eta^3 = 4 \xi (\xi-1)^2 (\xi + 1)^2. $$ We thus see that $C'$ is a triple cover of $\mathbb{P}^1$ which is totally ramified in $4$ points (including one above infinity), so by Riemann--Hurwitz $C'$ has genus $2$, so has finitely many rational points.
Extending René's answer, the curve $C'$ is birational to $$C'' \colon y^2 = x^6 + 4,$$ whose Jacobian variety has finite Mordell-Weil group isomorphic to $({\mathbb Z}/3{\mathbb Z})^2$, from which it is easy to show that the only rational points on $C''$ are thoses with $x = 0$ and the two points at infinity. From this, it is easy to get all solutions to the original equation.
To get to this hyperelliptic equation, we rewrite René's equation as $$(\xi^2 - 1) \left(\frac{\eta}{\xi^2-1}\right)^3 - 4 \xi = 0$$ and set $u = \eta/(\xi^2-1)$. Taking the discriminant of $u^3\xi^2 - 4\xi - u^3$ as a polynomial in $\xi$ gives $$4 + u^6 = v^2$$ as claimed. The points with $u = 0$ give $\eta = 0$, so $y = 0, x, -x$, which leads to $x = y = 0$. The points at infinity have $\xi = \pm 1$, which gives the same result.