Find all solution $a,b,c$ with $(1-a^2)(1-b^2)(1-c^2)=8abc$ Two years ago, I made a conjecture on stackexchange:
Today, I tried to find all solutions in integers $a,b,c$ to
$$(1-a^2)(1-b^2)(1-c^2)=8abc,\quad a,b,c\in \mathbb{Q}^{+}.$$  
I have found some solutions, such as
$$(a,b,c)=(5/17,1/11,8/9),(1/7,5/16,9/11),(3/4,11/21,1/10),\cdots$$
$$(a,b,c)=\left(\dfrac{4p}{p^2+1},\dfrac{p^2-3}{3p^2-1},\dfrac{(p+1)(p^2-4p+1)}{(p-1)(p^2+4p+1)}\right),\quad\text{for $p>2+\sqrt{3}$ and $p\in\mathbb {Q}^{+}$}.$$
Here is another simple solution:
$$(a,b,c)=\left(\dfrac{p^2-4p+1}{p^2+4p+1},\dfrac{p^2+1}{2p^2-2},\dfrac{3p^2-1}{p^3-3p}\right).$$

My question is: are there solutions of another form (or have we found all solutions)?


I'm late for this party, but using math110's method employing Euler bricks, couldn't resist giving some simple rational solutions to,
$$(1-a^2)(1-b^2)(1-c^2) = 8abc$$
Solution 1:
$$a,\,b,\,c = \frac{-(x-z)(2x+z)}{(2x-z)y},\;\frac{z}{2x},\;\frac{-2y+z}{2y+z}\tag1$$
where $x^2+y^2=z^2.$
Solution 2:
$$a,\,b,\,c = \frac{2z^2}{xy},\;\frac{x-z}{x+z},\;-\frac{y+z}{y-z}\tag2$$
where $x^2+y^2=5z^2$, and which may also be solved as a Pell equation.
