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, why is the "no vanishing subsums" condition necessary? Are there examples of $n$-tuples violating only this hypothesis that violate the conclusion? It seemed to me that (assuming the truth of the $m$-variable version for $m<n$) having vanishing subsums only improves the bound.

Second, 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 $$ R(|a_1a_2a_3a_4|)^{2\cdot4-5} \le (3R(abc))^3 \le 27c^3 = 27\max\{|a_1|,|a_2|,|a_3|,|a_4|\}. $$ But note that these $4$-tuples are only relatively prime, not pairwise relatively prime. Has anyone mulled over whether the exponent $2n-5$ can be reduced if one strengthens the hypothesis to pairwise coprimality? A probabilistic argument suggests (I believe) that the exponent $1+\varepsilon$ should suffice. Or, looking at these examples another way, is the exponent $2n-5$ there 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?