Confusion is possible, we got argument against Vojta's more general abc conjecture.

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.

Let $rad$ the radical of polynomial be the product of irreducible factors and the radical of integer is the product of the primes factors.

For natural $D > 2$ and variables $x,y$, define $A_D(x,y) : x_1=(x+y)^D,x_2=-x^D,x_3=-y^D,x_4= -(x_1+x_2+x_3)=xyF_{D-2}(x,y)$

with $\deg F_{D-2}=D-2$.

We have $\deg x_1=D$ and $\deg rad(x_1 x_2 x_3 x_4) \le D+1$ and in addition $xy$ divides $x_1 x_2 x_3 x_4$.

For small primes $p,q$ and large exponents $n,m$, set $X=p^n,Y=q^m, X>Y$, then we have the freedom to remove $xy$ from the radical.

Then in $A_D(X,Y)$ we have $x_1 \sim X^D$ and $rad(x_1 x_2 x_3 x_4)= O(pq X^{D-1})$.

For fixed $D$ we have infinitely many counterexamples which violate the hypothesis, so they must be in the exceptional set.

For fixed $D$, there is proper algebraic dependency between the $x_i$, but the algebraic dependencies are different for different $D$.

We believe this shows the exceptional set must be infinite.

Q1 What is wrong with this counterexample?


We asked Vojta and he kindly replied that this is not counterexample because the exceptional set is allowed to depend on epsilon and taking $\epsilon > 1/(D-1)$ kills the infinite family of quadruples.

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