Given a positive integer $P>1$, let its prime factorization be written as$$P=p_1^{a_1}p_2^{a_2}p_3^{a_3}\cdots p_k^{a_k}.$$
Define the functions $h(P)$ by $h(1)=1$ and $h(P)=\min(a_1, a_2,\ldots,a_k).$
Is new the $n$-conjecture, formulated as follows, correct?
Conjecture: if ${P_1,P_2,…,P_n}$ are positive integer and pairwise coprime, then,
$$\min\{h(P_1), h(P_2),…,h(P_n), h(P_1+P_2+…+P_n)\} \leq n+1.$$
I proposed the case $n=2$ two years ago here (Is the conjecture A+B=C following correct?). Now I reformulate that question as follows:
Let ${P_1,P_2}$ are coprime, then:
$$\min\{h(P_1), h(P_2), h(P_1+P_2)\} \leq 3$$
Best Answer
Such attempted generalizations of ABC to four or more variables often fail to specializations of the identity $$ (x^2+xy-y^2)^3 + (x^2-xy-y^2)^3 = 2 (x^6 - y^6). \label{1}\tag{*} $$ One can use elliptic curves to make both $x^2 + xy - y^2$ and $x^2 - xy - y^2$ "powerful" (of the form $A^2 B^3$), which makes each of the four terms $(x^2+xy-y^2)^3$, $(x^2-xy-y^2)^3$, $2x^6$, $2y^6$ have $h=6$ but for a stray factor of $2$ which should not matter in the context of the ABC conjecture. For example, the pairwise prime numbers $a,b,c,d$ below satisfy $2a^6 + b^6 + 61^9 c^6 = 2d^6$. Here $d$ is even but $a$ is odd, so $2a^6$ has a "stray factor of $2$", and the expansion to $a^6 + a^6 + b^6 + 61^9 c^6 = 2d^6$ loses pairwise coprimality; so either way we don't quite get a counterexample. Still, this suggests that generalizations of ABC to four or more variables can run afoul of identities such as \eqref{1}. (It is "well known" that the Mason-Stothers theorem forbids the disproof of ABC itself by such an identity.)