0
$\begingroup$

In A more general abc conjecture, p. 7 Paul Vojta conjectures:

If $x_0,\ldots x_{n-1}$ are nonzero coprime integers satisfing $x_0 + \cdots x_{n-1}=0$

$$ \max\{|x_0|,\ldots |x_{n-1}|\} \le C \prod_{p\mid x_0 \cdots x_{n-1}}p^{1+\epsilon}\qquad (1) $$

for all $x_0 , \ldots, x_{n-1}$ as above outside a proper Zariski-closed subset.

Similar claim here p.5.

For $F_n$ the fibonacci numbers and $L_n$ lucas numbers, $y,x=F_{2n+1}^2,F_{2n}^2$ satisfy $$ -x^2 + 3xy - y^2 + 2x - 2y - 1=0$$ and $x,y=L_{2n}^2,L_{2n+1}^2$ sastisfy $$ x^2 - 3xy + y^2 - 10x + 10y + 25=0$$

Take $x_i$ to be the monomials of any of the aboves.

In both cases $x,y$ are squares. $\max |x_i|=y^2$ and the radical is $O(y)$ because of the squares.

This gives abc quality of about $2$ and both identities are in Vojta's exceptional set.

In identities like this, one of $x,y$ being perfect power gives sufficiently large abc quality.

I conjecture that in general parametrizations of the forms $a_0 n^2,a_1n^2+a_2n+a_3$ give explicit genus $0$ curves of degree at most $2$ which give tuples violating Vojta's conjecture.

E.g. the parametrization $n^2,19n^2+n+1$ satisfies

$$ -y^2 + 38 y x - 361 x^2 + 2y - 37x - 1=0 $$

and the abc quality is about $1.5$.

Is the above true?

If this is true, there are infinitely many genus $0$ curves of degree $2$ where $x$ is perfect power and all of them violate the conjecture. Found some more binary recurrences for which $a_{2n}^2,a_{2n+1}^2$ are on degree $2$ genus $0$ curves (e.g. $a_n=2a_{n-1}+a_{n-2}$).

The requirement for perfect power can be strengthened. Let $f(x,y)=0$ be genus $0$ curve of degree $2$ with infinitely many integral points and $x,y$ are coprime infinitely often Assume $\log|x| \sim \log|y|$.

Fix $\epsilon > 0$

If $rad(x) < |x|^{1-\epsilon}$ or $rad(y) < |y|^{1-\epsilon}$ infinitely often the curves are in the exceptional set.

According the conjecture all them must be on some other varieties.

If yes, what is the proper Zariski-closed subset in the above examples?

$\endgroup$
  • $\begingroup$ The Zariski closed set in Vojta's conjecture depends on $n$ and $\epsilon$. Even if you fix those, say $n=2,\epsilon=1$, to describe the corresponding set amounts to proving the abc conjecture. Unless you are claiming that you are producing so many rational curves that they cover the whole ambient space and disproving Vojta's conjecture. Please clarify. $\endgroup$ – Felipe Voloch Oct 16 '14 at 16:59
  • $\begingroup$ @FelipeVoloch I fix degree $2$ and suspect infinitely many genus $0$ curves violate the conjecture via integral points with radical of $x$ or $y$ being sufficiently small while $x$ and $y$ are of the same size, might be wrong. I am asking if these curves with sufficiently small radical are on some proper variety. I am pretty sure I can find many degree $2$ curves with $x$ square or constant times square. $\endgroup$ – joro Oct 16 '14 at 17:38
  • $\begingroup$ @FelipeVoloch Re: fixing $n$ and $\epsilon$. I suppose I fix $n=6$ (degree=2) and $\epsilon=0.1$, which allegedly gives me infinitely curves with square $x$ of abc quality $1.5$ (possibly more) violating the epsilon. $\endgroup$ – joro Oct 16 '14 at 17:58
  • $\begingroup$ Infinitely many curves may all be contained in a Zariski closed set of higher dimension. $\endgroup$ – Felipe Voloch Oct 16 '14 at 18:31
  • $\begingroup$ In fact, if all your examples for $n=6$ are built starting with $x,y$ and producing $x_0,\ldots,x_5=1,x,y,x^2,xy,y^2$, they all satisfy $x_3 x_5=x_4^2$. $\endgroup$ – Felipe Voloch Oct 16 '14 at 18:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.