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)? [on stackexchange]:https://math.stackexchange.com/questions/334448/1-a21-b21-c2-8abc-a-b-c-in-mathbbq-has-infinitely-many-sol/336803#336803