Likely a mistake, but got very large exceptional set in Vojta's
more general abc conjecture.


In [A more general abc conjecture](http://arxiv.org/abs/math/9806171), p. 7 Paul Vojta conjectures:


If $x\_0,\ldots x\_{n-1}$ are nonzero coprime integers satisfying $x\_0 + \cdots x\_{n-1}=0$


$$ \max\{|x\_0|,\ldots |x\_{n-1}|\} \le C \prod\_{p\mid x\_0 \cdots x\_{n-1}}p^{1+\epsilon}\qquad (1) $$


for all $x\_0 , \ldots, x\_{n-1}$ as above **outside a proper Zariski-closed subset**.


Define the quality $q(x\_0 \ldots x\_{n-1})=\frac{\log(\max(|x\_i|))}{\log\operatorname{rad}(x\_0 \cdots x\_{n-1})}$. The conjecture implies $\limsup q(x\_0 \ldots x\_{n-1})=1$ outside a proper Zariski-closed subset.


Consider the set of all genus $0$ curves parametrized by two quadratics.
$x=a t^2+b t + c,y=d t^2+ e t +f$. Assume $a,b$ are nonzero and $|a|\ge |d$
and $x,y$ are coprime over $\mathbb{Q}[t]$.


The explicit form of the curve is


$$ d^{2} x^{2} + (-2 a d) x y + a^{2} y^{2} + (-2 c d^{2} + b d e - a e^{2} + 2 a d f) x + (- b^{2} d + 2 a c d + a b e - 2 a^{2} f) y + c^{2} d^{2} - b c d e + a c e^{2} + b^{2} d f - 2 a c d f - a b e f + a^{2} f^{2}=0 \qquad(1)$$


Fix $a,b,c,d,e,f$. Take $x\_i$ the monomials in $x,y$ of (1), i.e. $x\_0 =d^{2} x^{2} ,x\_1= (-2 a d) x y$.
For infinitely many of them we get quality $4/3-o(1)$
infinitely often for $x,y$ on (1). 


$\max|x\_i|=a^2 y^2 \sim a^2 d t^4$.
The radical is $poly(a,b,c,d,e,f) x y \sim t^4$.


One can make the radical of the quadratic $x$ only $O(t)$ in at least
two ways: (a) there exist integers $k$ such that $a t^2+b t + c=k r^2$
has infinitely many integral solutions and (b) find integer $p$ such that for
arbitrary large $n, a t^2+b t + c \equiv 0 \pmod{p^n}$ via
Hensel lifting. The gcd of $x\_i$ is bounded by the resultants and
clearing bounded gcd doesn't decrease the quality.


Both (a) and (b) make the radical only $O(t^3)$ for arbitrary large $t$,
which gives quality $4/3-o(1)$ and we can get rid of the $o(1)$.


So (1) is in exceptional Vojta's set. By gross abuse of notation,
"almost all" $a,b,c,d,e$ give quality $4/3-o(1)$ so we can treat
them as variables in (1). This in general makes the dimension of 
(1) $7$, which is greater than the dimension of $x\_0 + \cdots + x\_5=0$.


I believe higher dimensional variety can't be subvariety of lower dimensional
variety.



> 
> Why isn't this construction counterexample?
> 
> 
> 



More on Vojta's exceptional set for a more general abc conjecture
Your construction is not a counterexample because your tuples $(x\_0,\ldots,x\_5)$
(obviously) satisfy the relation $x\_1^2 - 4x\_0x\_2 = 0$, so they all lie in the
proper Zariski closed subset defined by this equation.


