<s>[INST] Parametric solutions to $(4/3)b^2c^2+(4/3)a^2d^2-(1/3)a^2c^2-(4/3)b^2d^2=\square$ Let $a,b,c$ and $d$ be rational.Find a rational parametric solutions for $a,b,c$ and $d$ so that 
$$(4/3)b^2c^2+(4/3)a^2d^2-(1/3)a^2c^2-(4/3)b^2d^2=\square.$$
 [/INST] Rewrite it as
$$4b^2(c^2-d^2) = a^2(4d^2-c^2) = a^2(4d^2-4c_0^2)$$
where $c_0 = c/2$. Then 
$$b^2(c^2-d^2) = a^2(d^2-c_0^2)$$
We can assign: $b^2 = d^2-c_0^2$ and $4d^2-c^2 = a^2$.
For the first equation we use the famous Pythagorean triple: $b = 3, d = 5, c_0 = 4 \rightarrow c = 8$.
Then it brings us to the last variable: $a = 6$.
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