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)?


@Allan methods it's nice! here is my answer:
since
$$\left(\dfrac{1-a^2}{2a}\right)\left(\dfrac{1-b^2}{2b}\right)\left(\dfrac{1-c^2}{2c}\right)=1$$
so let
$$\dfrac{x}{y}=\dfrac{1-a^2}{2a},\;\dfrac{y}{z}=\dfrac{1-b^2}{2b},\;\dfrac{z}{x}=\dfrac{1-c^2}{2c}$$
and solving for $a,b,c$,
$$a = \frac{-x+\sqrt{x^2+y^2}}{y},\;\;b = \frac{-y+\sqrt{y^2+z^2}}{z},\;\;
c = \frac{-z+\sqrt{x^2+z^2}}{x}$$
it is easy to see
$x^2+y^2,y^2+z^2,z^2+x^2$ must be square. so we use  Euler bricks solution
$$x=u|4v^2-w^2|,y=v|4u^2-w^2|,z=4uvw$$
then it is not hard to find to give solution
$$(a,b,c)=\left(\dfrac{p^2-4p+1}{p^2+4p+1},\dfrac{p^2+1}{2p^2-2},\dfrac{3p^2-1}{p^3-3p}\right).$$
