## Equation : (x + m)^3 = n*x

Consider,     $(x + m)^3 \; = \; n \, x$    ……….   (1)

where   $m$   and   $n$   are positive integers, and the equation (1) has exactly three distinct integer solutions in   $x$ .

Let’s take, for example,   $m = 30$

$(x + 30)^3 \; = \; n \, x$

possible solutions:
$n \; = \; 6859 \; ( \,= 19^3 \,)$ ,
$n \; = \; 8575 \; ( \,= 5^2\times \; 7^3 \,)$ ,    and
$n \; = \; 29791 \; ( \,= 31^3 \,)$

$n \; = \; 6859$    —–>    $(x + 30)^3 \; = \; 6859 \, x$

then,

$x \; = \; -125 \; = \; -5^3$
$x \; = \; 8 \; = \; 2^3$
$x \; = \; 27 \; = \; 3^3$

$n \; = \; 8575$    —–>    $(x + 30)^3 \; = \; 8575 \, x$

$x = -135 = (-5)\times \; 3^3$
$x = 5 = 5\times \; 1^3$
$x = 40 = 5\times \; 2^3$

$n \; = \; 29791$    —–>    $(x + 30)^3 \; = \; 31^3 \, x$

$x \; = \; -216 \; = \; -6^3$
$x \; = \; 1 \; = \; 1^3$
$x \; = \; 125 \; = \; 5^3$

$gcd(30, 6859) \; = \; 1$
$gcd(30, 8575) \; = \; 5$
$gcd(30, 29791) \; = \; 1$

If we add the condition that $gcd(m, n) \; = \; 1$ ,   we end up with two solutions.

Can we say that there exist infinitely many pairs   $( \,m, n \,)$   satisfying the given conditions?