# Is new $n$-conjecture as follows correct?

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 at 4:18
• I want mean $x,y$ are positive integer and pairwise coprime – Đào Thanh Oai Apr 11 at 4:19
• Here's a near miss for $n=3$: $2^{14}+7^5+13^6=2^5\cdot3^5\cdot5^4$. – Thomas Browning Apr 11 at 19:49
• Thanks to dear @ThomasBrowning – Đào Thanh Oai Apr 11 at 20:49

## 1 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.)

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

• Thank to Dear Professor @Noam D.Elkies, I really want mention case $n=2$, and I want more people attend the case $(n=2)$. But I can't ask question with $n=2$ because I mention earlier, so If I want more people attend the problem, I must change with $n>2$ to ask a new question. I am very happy if you like case $n=2$ – Đào Thanh Oai Apr 11 at 5:33
• Prof. Elkies, I hyperlinked the formula: however, if you do not like it, please feel free to revert. – Daniele Tampieri Apr 11 at 6:19