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?

suspectinfinitely 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 somepropervariety. I am pretty sure I can findmanydegree $2$ curves with $x$ square or constant times square. $\endgroup$ – joro Oct 16 '14 at 17:38