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?


1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .