Does $(x^2 - 1)(y^2 - 1) = c z^4$ have a rational point, with z non-zero, for any given rational c? I need this result for something else. It seems fairly hard, but I may be missing something obvious.
Just one non-trivial solution for any given $c$ would be fine (for my application).
 A: Many thanks for your replies. I thought I'd cracked it, but after posting noticed a mistake at the end. However, I'll leave the following, as far as it goes, in case it suggests any alternative angles to others.
Firstly, note that the equation can be expressed as:
$\dfrac{x - 1}{z} \dfrac{x + 1}{z} \dfrac{y - 1}{z} \dfrac{y + 1}{z} = c$
So taking:
$X, Y, Z, T = \dfrac{x - 1}{z}, \dfrac{y + 1}{z}, \dfrac{x + 1}{z}, \dfrac{y - 1}{z}$
(which is obviously unirational, i.e. "reversible") we can express it as:
$X + Y = Z + T$
$X Y Z T = c$
Now (reusing the original x, y, z for convenience) take:
$X, Y, Z, T = \dfrac{c x}{d}, \dfrac{y}{d}, \dfrac{z}{d}, \dfrac{t}{d}$
Then the preceding pair becomes:
$c x + y = z + t$
$x y z t = d^4$
Now as a special case assume t = c, so the first of this pair gives:
$t = c = \dfrac{z - y}{x - 1}$
and the second then becomes:
$x y z (z - y) = (x - 1) d^4$
Finally, letting:
$x, y, z = p d, q d, r d$
we obtain:
$c = \dfrac{(r - q) d}{p d - 1} = \dfrac{d}{p q r}$
But this, although it looks tantalizingly simple, is the end of the line for the present attempt!
A: [edited again mainly to add the Euler link, see last paragraph]
Yes, and indeed there are infinitely many rational points:
the birationally equivalent Diophantine equation
given by J.Ramsden in his partial answer to his own question,
$$
X + Y = Z + T,
\phantom{and}
XYZT = c,
$$
was already studied by Euler (in the equivalent form $xyz(x+y+z)=a$),
who in 1749 obtained the rational curve of solutions
$$
X = 6\frac{cr(2r^4+c)^2}{(r^4-4c)(r^8+10cr^4-2c^2)},
\phantom{and}
Y = -2\frac{r^8+10cr^4-2c^2}{3r^3(r^4-4c)},
$$ $$
Z = -3\frac{r^5(r^4-4c)^2}{2(2r^4+c)(r^8+10cr^4-2c^2)},
\phantom{and}
T = \frac{r^8+10cr^4-2c^2}{r^3(2r^4+c)}.
$$
Inverting J.Ramsden's birational transformation
$$
(X,Y,Z,T) = \left(
  \frac{x-1}{z},
  \frac{y+1}{z},
  \frac{x+1}{z},
  \frac{y-1}{z}
\right)
$$
then yields
$$
x = \frac{(r^8+2c^2)(r^8-44cr^4-2c^2)}{(r^8+10cr^4-2c^2)^2},
\phantom{and}
y = \frac{7r^4+8c}{9r^4},
\phantom{and}
z = \frac{-4(r^4-4c)(2r^4+c)}{3r(r^8+10cr^4-2c^2)}.
$$
Euler's solution is unpublished and somewhat mysterious;
he gave the formulas in a letter to Goldbach but didn't explain
how he found them, writing only that he obtained the solution
"endlich nach vieler angewandter Mühe" [at last, after applying much effort].
The curve is singular, and finding such a curve requires
going somewhat beyond the usual manipulation of
elliptic fibrations on K3 surfaces, which for these surfaces will not work unless
$c$ is a square (the case in which R.Kloosterman already found a nonsingular
rational curve of solutions).  I've lectured on Euler's surface and solution
a number of times over the past few years;
here's
the latest iteration.  Thanks to Franz Lemmermeyer for bringing
Euler's letter to my attention.
Added later:
The same lecture notes
show on page 43 a somewhat simpler curve that I found on Euler's surface,
which also yields a somewhat simpler curve of solutions of
$(x^2-1)(y^2-1)=cz^4$:
$$
x = \frac{u^{12}+48cu^8-48c^2u^4+128c^3}{u^4(u^4-12c)^2},
\phantom{and}
y = \frac{5u^4+4c}{3u^4-4c},
\phantom{and}
z = \frac{4(u^4+4c)}{u(u^4-12c)}.
$$
For $c=2$, we get at $u=2$ the solution $(26, 11/5, 6)$ which is almost
as simple as the solution $(5/3,17,4)$ that Mark Sapir noted.
Added later yet: Here's a

transcription of Euler's 1749 letter.  See the last page.
A: Here are some thoughts, too long for a comment. We can assume that $c$ is not a rational square, otherwise we have many solutions with $x=y$ and $z\neq 0$.
We have $x^2-1=c_1 z^2$ and $y^2-1=c_2 z^2$ with $c_1 c_2=c$. Hence $(x,y,z)$ lies on the intersection of two space quadrics. As $c_1\neq c_2$, the intersection is an elliptic curve isomorphic to the plane curve
$$ Y^2=-8(X-1)(X+1)(-c_1X-2c_2+c_1),$$
see Pinch: Square values of quadratic polynomials here for details. Namely, apply Proposition 1.1 there with $(a,b,c,d,e,f)=(1,-c_1,1,1,-c_2,1)$ and $(p,q,r,s)=(1,0,1,1)$.
With the notation $2U=c_1(X-1)$ and $8V=c_1Y$ the last equation becomes
$$ V^2=U(U+c_1)(U+c_2). $$
A solution $z\neq 0$ corresponds to $V\neq 0$, hence the problem boils down to whether the last equation can be solved with $V\neq 0$ for some decomposition $c=c_1c_2$.
A: If you take the projective closure of your surface in $\mathbb{P}^3$ you find a singular quartic. This quartic has six singular points. Namely the two points found by Daniel and four points of the form $(\pm 1,\pm 1,0,1)$. Each of these points is an $A_1$ singularity. This implies that the resolution of the surface is a K3 surface.
This surfaces contains also several lines, namely $z=0,x=\pm w$ and $z=0, y=\pm w$.
Each line yields a genus one fibration by taking the pencil of planes through one of the lines.
As an example you can take (in the chart $w=1$) the pencil $z=T(x-1)$. If you substitute this in the affine equation for your surface you'll find 
$(x-1)*(-cT^4x^3+3cT^4x^2-x+xy^2-3cT^4x+cT^4-1+y^2)$
The second factor is a cubic curve in $\mathbb{P}^2_{\mathbb{Q}(T)}$.
which contains the rational point $(1,1,1)$. In Cassels' book on elliptic curves you can find an algorithm that puts this equation in Weierstrass form.
A trick that turns out to work is to move the point $(1,1,1)$ to $(0,1,0)$, you get then the following Weierstrass equation:
$y^2=x^3-(8cT^4+1/3)x+16c^2T^8+8/3cT^4+2/27$.
This is a degree 2 base change of a rational elliptic surface. (Set $u=T^2$.)
This rational elliptic surface has one fiber of type $I_6$, two of type $I_1$ and one of type $IV$. In particular, the Shioda-Tate formula tells us that the Mordell-Weil rank of this elliptic curve over $\overline{\mathbb{Q}}(u)$ is one.
By the work of Shioda we know that the Mordell-Weil group is generated by polynomials $x(u)$ and $y(u)$ of degree at most 2, resp. 3.
If you can find these polynomials then you can check whether this generator is defined over $\mathbb{Q}$ or not. If so then by multiplying this generator in the Mordell-Weil group and specializing $T$ you can find a bunch of points.
Edit: The Mordell-Weil group is generated by $x=1/3+4\sqrt{c}u$ and $y=4\sqrt{c}u(1+\sqrt{c}u)$. Hence this approach only works if $c$ is a square.
Edit 2: -Deleted- This remark was incorrect. There is a $\mathbb{Q}(T)$-rational point on the elliptic curve, with $x=1/3$. However this point correspond with one of the singular points on the K3 surface, and has order three in the Mordell-Weil group.
