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?

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.)