This was a tough problem. I'm not sure I can even explain how I got to my solution but here is the form of all solutions I found:
Consider the left and right sides of the equations as functions and shift them to the right by p:
x
^{3} = q*(xp)
this will make things a little simpler. Any solutions to this equation can just be shifted back, so the number of solutions is unchanged as is the number of pairs (p,q).
There are an infinite number pairs.
For any integer n, let p=n(n+1) = n²+n
and q = (n²+n+1)³
These are clearly relatively prime.
The equation becomes x=(n²+n+1)³(xn²n)
The three solutions are as follows:
If x=
(n²+n+1)(n²+n+1)³=(n²+n+1)³((n²+n+1)n²n)
If x=
n(n²+n+1)=(n³+n²+n)
n³(n²+n+1)³=(n²+n+1)³(n³+n²+nn²n)
If x=
(n+1)(n²+n+1)=n³2n²2n1
(n+1)³(n²+n+1)³=(n²+n+1)³(n³2n²2n1n²n)
For example if n=5
p=5*6=30
q=31³=29791
the equation
x³=29791(x30)
has solutions
x=31, x=155, x=186
Corresponding to the untranslated equation
(x+30)³=29791x
having solutions 1, 125, 216
Hmmm... This is the first I noticed those cubes so there may be something simpler going on than I found.

Posted by Jer
on 20150922 13:39:31 