The Beal Conjecture states that if $A^x + B^y = C^z$, where $A, B, C, x, y, z$ are positive integers and $x$, $y$, $z$ are $≥ 3$, then $A$, $B$, and $C$ have a common prime factor.

There is some reward for a proof or a counterexample.

After a few minutes of searching, I found:

$9998500104995450136496997050049356506434949950300298635004549989500014
99999$

$$ +$$
$9998400119994400181995632080078856012869885600800795632018199944000119
9998400001$
$$ = $$

$9998500104995450136496997050049356506434949950300298635004549989500014
9999900000.$

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Note that

* $9998500...499999 = 99999^{15}$,
* $9998400...400001 = 9999800001^8$, and
* $9998500...900000 = 9999700002999990^5$.

So is the conjecture solved? Unfortunately, no. It turns out that the greatest common divisor of these numbers is $99999$, so this cannot be a counterexample to Beal's conjecture according to the definition above.

Now the question arises, is there a list (maybe in the OEIS) about ''near misses'' to counterexamples of Beal's conjecture? Or perhaps a list of known general solutions to $A^x + B^y = C^z$ where I can add this term?