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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.

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  • $\begingroup$ Why are $u$ and $v$ integers? Can you give some reference or the proof? $\endgroup$ – GH from MO Apr 15 '14 at 16:46
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    $\begingroup$ If $x$ is even, then the three integers $x-1$, $x$ and $x+1$ are pairwise coprime, and the equation (1) is equivalent to wanting all of these to be square-full. If $x$ is odd, then $(x-1)$ and $x+1$ have a common factor $2$. Removing this common factor, we are led to ask if it is possible for $n$, $2n+1$, and $n+1$ all to be square-full. The same heuristic that suggests that there are no three consecutive square-full numbers, also suggests that there are at most finitely many square-full triples $n$, $2n+1$, $n+1$ (and maybe there are none). $\endgroup$ – Lucia Apr 15 '14 at 17:03
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    $\begingroup$ @GHfromMO I think they are integers because of the map from the Weierstrass model. x=u/z^3 <=> x z^3 = u $\endgroup$ – joro Apr 16 '14 at 5:15

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