Is new $n$-conjecture as follows correct?

Question

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: $$\min\{h(P_1), h(P_2), h(P_1+P_2)\} \leq 3$$

If the conjecture is right. Then there are no positive integer $x, y, z$, $n, m, k>3$ satisfy the equation $x^n+y^m=z^k$ — Đào Thanh Oai, Jul 11 '20 at 14:55
But one has to formulate it in a way to exclude cases like $2^n+2^n=2^{n+1}$ — Pietro Majer, Apr 11 '21 at 04:18
Here's a near miss for $n=3$: $2^{14}+7^5+13^6=2^5\cdot3^5\cdot5^4$. — Thomas Browning, Apr 11 '21 at 19:49
Draw attention because the nice problem was check with $P_1+P_2< 10^{18}$ — Đào Thanh Oai, Apr 24 '21 at 11:47

score 14 · Answer 1 · edited Apr 11 '21 at 06:15

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.)

a = 1022288301691921314835532892967014277786302791344455107816139963763145069687359424810667270039489345929029301393007247303344511065237 
b = 4005821025365458069945118311017282675402206671149976403498624129498574702167905733126870212117037684063261425637225699359421949547271 
c =   10621830276852061412855232703108032231130723932854745057900981539571749281974534306702514113168069346943754838515856358759614674721
d = 4676830625123658957500070687744472849236478810555581279857582626862200039312130562436302012081022720213179152015505627679021327325170

Prof. Elkies, I hyperlinked the formula: however, if you do not like it, please feel free to revert. — Daniele Tampieri, Apr 11 '21 at 06:19
The "n=2 case" (no solution of A+B=C in coprime 4th-powerful numbers) is plausible -- and it would be reduced to a finite computation if we knew an effective ABC conjecture in the form N >> C^r with any r > 3/4 -- but it seems very difficult and I don't think a proof is known. — Noam D. Elkies, Jul 26 '22 at 22:05
@NoamD.Elkies Thank you so much, I hope you enjoy this issue. Or introduce it with your student — Đào Thanh Oai, Aug 09 '22 at 04:31

Is new $n$-conjecture as follows correct?

1 Answers1