In this page, I am not interested on the Beal conjecture itself. But I am interested on the following problem:
The author claimed:
Any solutions to the Beal conjecture will necessarily involve three terms all of which are 3-powerful numbers, i.e. numbers where the exponent of every prime factor is at least three.
My question is: Is there are a proof for this claim or it is just an assumption comes from the Beal conjecture.
Beal conjecture is
"If $A^x+B^y=C^z$ where $A$, $B$, $C$, $x$, $y$, and $z$ are positive integers with $x, y, z > 2$, then $A$, $B$, and $C$ have a common prime factor."
Thus in any case we are talking about an expression $A^x+B^y=C^z$ in which the exponents $x$, $y$ and $z$ are at least 3, so clearly $A^x$, $B^y$ and $C^z$ are at least 3-powerful numbers.
And this is independent from the fact that the current $A^x+B^y=C^z$ is a counterexample, i.e, $(A,B,C)=1$, or an example of Beal conjecture, i.e. $(A,B,C)>1$.
Based on this observation, the Wikipedia page shows a couple of examples, to demonstrate that, if you relax the conditions of Beal conjecture, i.e., instead of talking about perfect powers you talk about powerful numbers, then there are indeed triplets of relative prime 3-powerful numbers that add up.