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)?


The original proposer asks for "simple methods". Simplicity, like beauty, is in the eye of the beholder. I am sure
that Noam Elkies and Joe Silverman feel their answers are extremely simple. The following discussion is, in my humble opinion,
simpler.
We can express the underlying equation as a quadratic in $a$,
\begin{equation*}
a^2+\frac{8bc}{(b^2-1)(c^2-1)}a-1
\end{equation*}
with the obvious condition that $|b| \ne 1$ and $|c| \ne 1$.
For $a$ to be rational, the discriminant must be a rational square, so there exists $D \in \mathbb{Q}$ such that
\begin{equation*}
D^2=(c^2-1)^2b^4-2(c^4-10c^2+1)b^2+(c^2-1)^2
\end{equation*}
This quartic has an obvious rational point when $b=0$, and so is birationally equivalent to an elliptic curve. We find the curve
\begin{equation*}
v^2=u(u+(c^2-1)^2)(u+4c^2)
\end{equation*}
with the reverse transformation
\begin{equation*}
b=\frac{v}{(c^2-1)(u+4c^2)}
\end{equation*}
The elliptic curve has $3$ points of order $2$, which give $b=0$ or $b$ undefined. There are also $4$ points of order $4$ at
\begin{equation*}
u=2c(c^2-1) \hspace{1cm} v= \pm 2c(c+1)(c-1)(c^2+2c-1)
\end{equation*}
and
\begin{equation*}
u=-2c(c^2-1) \hspace{1cm} v= \pm 2c(c+1)(c-1)(c^2-2c-1)
\end{equation*}
all of which give $|b|=1$. 
Thus, to get a non-trivial solution we need the elliptic curve to have rank at least $1$. Numerical investigations suggest that the rank is often $0$, so
solutions do not exist for all $c$.
We can derive parametric solutions by finding points of the curve, subject to certain conditions.
For example, $u=c^2-1$ would give a point if $5c^2-1=\Box$. We can parametrize this quadric using the solution when $c=1$, to give
\begin{equation*}
a=\frac{(p-2)(p-5)(3p-5)}{p(p-1)(p-3)(2p-5)} \hspace{1cm} b=\frac{p^2-4p+5}{2(p^2-5p+5)} \hspace{1cm} c=\frac{p^2-4p+5}{p^2-5}
\end{equation*}
which gives strictly positive solutions when $p > 5$.
Another simple point to consider could be $u=2c^2(c-1)(c+3)$ which gives a rational point when $(c+3)(3c+1)=\Box$.
