<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] It's quite easy to make that polynomial a square. Let,
$$P = (4/3)b^2c^2+(4/3)a^2d^2-(1/3)a^2c^2-(4/3)b^2d^2$$
Then,
$$P = (a d)^2,\quad \text{if}\, a = 2b\tag{1}$$
and all other variables are free. Similarly,
$$P = (b c)^2,\quad \text{if}\, c = 2d\tag{2}$$
$$P = (2b d)^2,\quad \text{if}\, a = 2b,\, \text{or}\, c = 2d\tag{3}$$
where the last one is equivalent to restrest's solution. (Note that $a = 6$ and $b = 3$ is $a=2b$.)
Are there any other conditions on $P$? 
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