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)?

 A: @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
$$(a,b,c)=\left(\dfrac{p^2-4p+1}{p^2+4p+1},\dfrac{p^2+1}{2p^2-2},\dfrac{3p^2-1}{p^3-3p}\right).$$
A: 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.
A: Call (the projective completion of) your surface $S$. It admits three double covers of $\mathbb P^2$, namely $$\pi_1(a,b,c)=(a,b), \quad\pi_2(a,b,c)=(a,c),\quad\pi_3(a,b,c)=(b,c).$$ Each double cover induces an involution, so we get
three involutions $$\sigma_1,\sigma_2,\sigma_3:S\to S.$$ These involutations don't commute, and if you form $f=\sigma_1\circ\sigma_2$ and $g=\sigma_1\circ\sigma_3$, then $f$ and $g$ generate a subgroup $G$ of $\text{Aut}(S)$ that is a free group on two generators. And presumably for most starting points $P\in S(\mathbb{Q})$, repeated application of $f$ and $g$ will give you a tree of rational solutions. (You might compare this with the generation of all positive integer solutions to the Hurwitz equation $x^2+y^2+z^2=3xyz$ starting from $(1,1,1)$.) Here are some references to papers that have studied rational points on K3 surfaces admitting 3 involutions:


*

*Baragar, A. Rational points on K3 surfaces in $\mathbb P^1\times\mathbb P^1\times\mathbb P^1$. Math. Ann.
305 (1996), no. 3, 541–558.

*Wang, L., Rational Points and Canonical Heights on K3-surfaces in $\mathbb P^1\times\mathbb P^1\times\mathbb P^1$, Contemporary Math. 186 (1995), 273 – 289
A: 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 $(c+3)(3c+1)=\Box$.
