## Equation : x^2 + y^2 + z^2 = (x – y)(y – z)(z – x)

Show that the equation

$x^2 \; + \; y^2 \; + \; z^2 \; = \; (x - y) \, (y - z) \, (z - x)$

has infinitely many solutions in integers   $x, \; y, \; z$

Solution :

Let’s find solutions   $(x, y, z)$   which are in arithmetic progression.

$y - x \; = \; z - y \; = \; d \; > \; 0$
so that the equation reduces to the form

$(y - d)^2 + y^2 + (y+d)^2$
$= 2 \, d^2 + 3 \, y^2$

$(y - x) \,(z - y) \,(z - x)$
$= \; (d) \,(d) \,(2 d)$
$= \; 2 \, d^3$

$2 \, d^2 + 3 \, y^2 \; = \; 2 \, d^3$

Thus we get    $3 \, y^2 \; = \; 2 \,(d - 1) \, d^2$

$2 \,(d - 1)$   is 3 times a square.

This is satisfied if   $d - 1 = 6 \, n^2$   for some   $n$

Thus

$d \; = \; 6 \, n^2 + 1$    and    $3 \, y^2 \; = \; d^2 \, 2 \,(6 \, n^2)$

giving us

$y^2 \; = \; 4 \, d^2 \, n^2$

Thus we can take

$y \; = \; 2 \,d \,n \; = \; 2 \,n \,(6 \,n^2 + 1)$

From this we obtain

$x \; = \; y - d \; = \; (2 \,n - 1) \,(6 \,n^2 + 1)$
$z \; = \; y + d \; = \; (2 \,n + 1) \,(6 \,n^2 + 1)$

It is easily verified that

$(x, y, z) \; = \; ((2 \,n - 1) \,(6 \,n^2 + 1), 2 \,n \,(6 \,n^2 + 1),(2 \,n + 1) \,(6 \,n^2 + 1))$

is indeed a solution for a fixed   $n$   and this gives an infinite set of solutions as   $n$   varies over the set of integers.