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Many thanks for your replies. Cracked I thought I'd cracked it, I thinkbut 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:

$c = \dfrac{(r - q) d}{p d - 1} = \dfrac{1}{p dfrac{d}{p q r}$

Giving:

$r = \dfrac{1}{c p q}$

$d = \dfrac{c}{c p + q - \dfrac{1}{c p q}}$

But this, although it looks tantalizingly simple, is the end of the line for the present attempt!

1

Many thanks for your replies. Cracked it, I think.

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:

$c = \dfrac{(r - q) d}{p d - 1} = \dfrac{1}{p q r}$

Giving:

$r = \dfrac{1}{c p q}$

$d = \dfrac{c}{c p + q - \dfrac{1}{c p q}}$