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 
$$(4/3)b^2c^2+(4/3)a^2d^2-(1/3)a^2c^2-(4/3)b^2d^2=\square.$$
 A: For the equation:
$$4c^2b^2+4d^2a^2-c^2a^2-4d^2b^2=3t^2$$
Can specify any number :  $d,c$ . Then decisions will be.
$$a=(d^2-c^2)p^2+3s^2$$
$$b=(d^2-c^2)p^2-3cps-3s^2$$
$$t=c(d^2-c^2)p^2+4(d^2-c^2)ps-3cs^2$$
$p,s$ - any integer asked us.
A: 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: $a = 6$.
A: 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 $P$? 
