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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
    Commented Apr 15, 2014 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
    Commented Apr 15, 2014 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
    Commented Apr 16, 2014 at 5:15
  • $\begingroup$ Since $x^2-1$ and $x$ are coprime, both would need to be powerful. Then the $abc$ triple $(1, x^2-1, x^2)$ would have to have quality $q\ge 2\log(x)/(\log(x^2-1)/2+\log(x)/2)>4/3$. $\endgroup$
    – Yaakov Baruch
    Commented May 15, 2022 at 13:59
  • $\begingroup$ @YaakovBaruch Indeed. But abc allows finitely many solutions. $\endgroup$
    – joro
    Commented May 15, 2022 at 14:04

1 Answer 1

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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.

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  • $\begingroup$ Thanks. You mean simple zeros over the complex numbers, right? And does this imply finitely many three consecutive powerfull numbers? $\endgroup$
    – joro
    Commented May 15, 2022 at 8:55
  • $\begingroup$ Yes, simple zeros over $\mathbb{C}$ and, again, yes, this conjecture implies that there are only finitely many consecutive powerfull numbers. $\endgroup$
    – Maciej Ulas
    Commented May 15, 2022 at 9:09
  • $\begingroup$ @MaciejUlas - you must mean finitely may triples, right? (Is there any reason to even expect finitely many pairs?) $\endgroup$
    – Yaakov Baruch
    Commented May 15, 2022 at 23:55
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    $\begingroup$ @Yaakov, no, there is a reason to expect infinitely many pairs of consecutive powerful numbers: $8$ is the smaller element of such a pair, and, if $n$ is the smaller number of such a pair, then so is $4n(n+1)$. $\endgroup$
    – Gerry Myerson
    Commented May 16, 2022 at 3:09
  • $\begingroup$ Of course, I was thinking about three consecutive powerfull numbers but write consecutive powerfull numbers. Sorry for this. $\endgroup$
    – Maciej Ulas
    Commented May 16, 2022 at 4:53

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