TITLE: Integer solutions of $ z^3 y^2 = x(x-1)(x+1)$
QUESTION [5 upvotes]: According to a conjecture there are no three
consecutive powerful numbers.
Necessary condition for this is integer solution of
$$ z^3 y^2 = x(x-1)(x+1) \qquad (1) $$

What are integer solutions of (1)?

For fixed $z$ Weierstrass model is
$$ v^2 = u^3 - z^6 u$$
$x = u/z^3, y= v/z^6$. Since $z$ is integer $u,v$ must be integers
too.

REPLY [3 votes]: According to the conjecture of Schinzel and Tijdeman (Schinzel, Α., Tijdeman, R., On the equation $y^n = P(x)$, Acta Arith. 31 (1976), 199-204), if a polynomial $P(x)$ with rational coefficients has at least three simple zeros, then the equation
$$
P(x) = y^2z^3
$$
has only finitely many solutions inintegers $x,y,z$ with $yz\neq 0$.
See also P. G. Walsh, On a conjecture of Schinzel and Tijdeman, in: Number Theory in Progress, Edited by: Kálmán Györy, Henryk Iwaniec and Jerzy Urbanowicz. De Gruyter 1999.