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Browkin and Brzezinski, in "Some remarks on the $abc$-conjecture", Math. Comp. 62 (1994), no. 206, 931–939, state the following generalization of the $abc$ conjecture to more than three variables:

Given $n\ge3$, let $a_1,\dots,a_n \in \Bbb Z$ satisfy $\gcd(a_1,a_2,\dots,a_n)=1$ and $a_1+\dots+a_n=0$, while no proper subsum of the $a_j$ equals $0$. Then for every $\varepsilon>0$, $$ \max\{|a_1|,\dots,|a_n|\} \ll_{n,\varepsilon} R(|a_1\cdots a_n|)^{2n-5+\varepsilon}, $$ where $R(m)$ denotes the radical of $m$ (the product of the distinct primes dividing $m$).

I have two questions about this conjecture.

First, the authors give constructions of $n$-tuples that show that the exponent $2n-5$ on the right-hand side cannot be improved. For example, when $n=4$, one takes any $abc$ triple $(a,b,c)$ and chooses $(a_1,a_2,a_3,a_4)=(a^3,b^3,3abc,-c^3)$, so that $$ R(|a_1a_2a_3a_4|)^{2\cdot4-5} \le (3R(abc))^3 \le 27c^3 = 27\max\{|a_1|,|a_2|,|a_3|,|a_4|\}. $$ On the other hand, a probabilistic argument suggests (I believe) that the exponent $1+\varepsilon$ should suffice. Is the exponent $2n-5$ present only because of integer points on certain lower-dimension varieties like $y^3=-27wxz$, on which the examples $(a^3,b^3,3abc,-c^3)$ all live?

Second, note that the $4$-tuples given above are only relatively prime, not pairwise relatively prime. Has anyone mulled over whether the exponent $2n-5$ can be reduced if one simply strengthens the hypothesis to pairwise coprimality?

(An earlier version of this question wondered why the "no vanishing subsums" condition was present, but that has been answered to my satisfaction.)

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    $\begingroup$ This is a somewhat degenerate example, but $(2^n, -2^n, 3^m, -3^{m})$ violates the conjecture for large $n,m$ if the vanishing subsums condition is removed. $\endgroup$ – Terry Tao Aug 29 '14 at 1:56
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In A more general abc conjecture, p. 7 Paul Vojta conjectures exponent $1 + \epsilon$ outside a proper Zariski-closed subset


Your question appears in The abc-conjecture and the n-conjecture,Coen Ramaekers p. 23

5.1 Strong $n$-conjecture. Adding pairwise coprimality, the exponent is $1$ without additional restrictions. Searching the web for it returns only 3 hits.

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Vojta's conjecture predicts $1+\epsilon$ for $n$-tuples outside a closed subvariety.

I think the case of subsums vanishing can lead to things like $a_i+b_i+c_i=0,i=1,2$ and $sa_1+sb_1+sc_1+ta_2+tb_2+tc_2=0$ with $(s,t)=1$ and $s,t$ as big as you like causing trouble.

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  • $\begingroup$ Thanks Felipe! Can you provide a reference to a version of Vojta's conjecture (presumably one where we ignore multiplicity in the local heights) that implies a $1+\epsilon$ version of $n$-tuple $abc$? or better yet, somewhere in the literature where this implication has been recorded? $\endgroup$ – Greg Martin Aug 29 '14 at 3:49
  • $\begingroup$ I thought the argument would be in his book, but there he only does $n=3$, i.e., the abc conjecture. The general argument is the same. He does need the more general conjecture with the ramification term by passing to $\sum x_i^m=0$ with the algebraic point $x_i = a_i^{1/m}$ and letting $m \to \infty$. $\endgroup$ – Felipe Voloch Aug 29 '14 at 4:09
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The relative primality certainly plays an important role. One can formulate a conjecture that accommodates the extent to which a particular summand is pairwise coprime, 3-wise coprime, etc, with other elements. My memory is that the paper of Brownawell and Masser:

Vanishing sums in function fields. Math. Proc. Cambridge Philos. Soc. 100 (1986), no. 3, 427–434.

provides a good function field analogue from which to formulate such a conjecture, though I do not remember the output (except that if one rules out vanishing subsums ... as plainly one should ... then for $n$-wise coprime $n$-tuples the appropriate exponent from such work is $(n-1)(n-2)/2$). Felipe has a paper at the same time (maybe preceded BM?) that likely is just as useful in this regard.

Another "special subvariety" example is:

$$x^4+y^4+(x+y)^4=2(x^2+xy+y^2)^2,$$

where if one takes $x+y+z=0$ as an extremal $abc$-example, then the summands can be made almost pairwise coprime, and the relevant $abcd$-exponent is $4/3$. I think that I first heard about this example from Andrew Granville in the early 1990's.

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    $\begingroup$ In the function field case, Mason proved a bound for which the exponent depended exponentially in $n$ and then, Brownawell-Masser and I more or less at the same time proved the result with an exponent quadratic in $n$. In any case, that exponent hasn't moved in nearly 30 years. Browkin and Brzezinski show that it can't be lower than $2n-5$ (unless one excludes a Zariski-closed set, perhaps). It would be very interesting to settle the exponent in the function field case. $\endgroup$ – Felipe Voloch Aug 29 '14 at 13:04
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    $\begingroup$ Isn't it easier to make $x^2+xy+y^2=a^n$ for $n$ sufficiently large and this will give quality close to $4/3$? The gcd will be bounded by $2$. Looks like $n>4$ already gives fixed good quality. $\endgroup$ – joro Aug 29 '14 at 13:09

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