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.)
Best Answer
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.