Diophantine equation: (x – y – z)(x – y + z)(x + y – z) = 8 xyz

Does the Diophantine equation

$(x - y - z) \,(x - y + z) \,(x + y - z) \; = \; 8 \, x \, y \, z$

have an infinite number of relatively prime solutions?

Solution:

we have the trivial solutions   $(x, \; y, \; z) \; = \; (\pm 1, \; \pm 1, \; 0)$   and permutations thereof.

For other solutions, note that each of the three equations

$x \; - \; y \; - \; z \; = \; 2 \, \sqrt {y z}$
$x \; - \; y \; + \; z \; = \; 2 \, \sqrt {x z}$
$x \; + \; y \; - \; z \; = \; 2 \, \sqrt {x y}$

is satisfied by   $\sqrt {x} \; = \; \sqrt {y} \; + \; \sqrt {z}$

consequently,   we have the infinite set of solutions

$y \; = \; m^2$
$z \; = \; n^2$
$x \; = \; (m + n)^2$

where   $(m, \; n) \; = \; 1$

I don’t know whether or not there are any other infinite sets of relatively prime solutions.

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

2 Responses to Diophantine equation: (x – y – z)(x – y + z)(x + y – z) = 8 xyz

1. Paul says:

Yes, if you look at these few solutions you can see the pattern.

{1,4,9}
{1,9,16}
{1,16,25}
{1,25,36}
{1,36,49}
{1,49,64}
{1,64,81}
{4,9,25}
{4,25,49}
{4,49,81}
{9,16,49}
{9,25,64}
{16,25,81}