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).



Does $(x^2 - 1)(y^2 - 1) = c z^4$ have a rational point, with z non-zero, for any given rational c?
*[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Euler's solution is unpublished and somewhat mysterious;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](http://math.harvard.edu/~elkies/euler_12u.pdf)'s
the latest iteration. Thanks to Franz Lemmermeyer for bringing
Euler's letter to my attention.


*Added later:*
[The same lecture notes](http://math.harvard.edu/~elkies/euler_12u.pdf)
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](http://math.dartmouth.edu/~euler/correspondence/letters/OO0853.pdf). See the last page.


