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Is the conjecture on A+B=C following correct ?

Conjecture: Let $A, B, C$ be three positive integer numbers such that $A+B=C$ with $\gcd(A, B, C) = 1$. By Fundamental theorem of arithmetic we write:

$A=a_1^{x_1}a_2^{x_2}...a_n^{x_n}$,

$B=b_1^{y_1}b_2^{y_2}...b_m^{y_m}$,

$C=c_1^{z_1}c_2^{z_2}...c_k^{z_k}$

Let $d = \min\{x_i, y_j, z_h \}$ where $1 \le i \le n, 1\ \le j \le m, 1\le h \le k$ then $$d \le 5$$

PS: I read above one hunded papers, I observed that in any case $\min\{x_i, y_j, z_h \} \le 3$

Example 1: Ten solutions of Catalan-Fermat equation

Example 2:

$2^4.3^5.7^6+5^9.11^8=19^1.23^1.47^1.6679^1.3051977^1$

See also:

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    $\begingroup$ If $\min\{x_i,y_j,z_h\}\geq 6$, then $rad(ABC)\leq \sqrt[6]{ABC}\leq C^{1/2}$, so $C\geq rad(ABC)^2$. The truth of the abc conjecture would imply there are only finitely-many triples $A,B,C$ with $\min\{x_i,y_j,z_h\}\geq 6$. $\endgroup$ Jun 19, 2018 at 4:11
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    $\begingroup$ @JulianRosen: In fact Baker conjectured that $C<rad(ABC)^{7/4}$ which would mean that there are no solutions (for coprime $A,B,C$). $\endgroup$
    – GH from MO
    Jun 19, 2018 at 4:12
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    $\begingroup$ Of course, if you believe ABC (alas, it looks like there is a flaw in Mochizuki's proof, if I understand the latest news right), all such questions become simple exercises. But can we say something unconditional (or, at least, based on weaker conjectures)? $\endgroup$
    – fedja
    Jun 19, 2018 at 13:06
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    $\begingroup$ The number of trivial edits is getting ridiculous. Please don’t edit the question just to get it back on the front page. $\endgroup$ Jun 26, 2018 at 0:22
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    $\begingroup$ There has been another completely trivial edit. Can you please stop fiddling with this question? If knowledgeable people are interested and have something to say, they will say it. It may be the case that people do not have anything useful to say, in which case you should accept this and move on $\endgroup$
    – Yemon Choi
    Jul 8, 2018 at 15:43

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