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:
$9998500104995450136496997050049356506434949950300298635004549989500014 99999$
$$ +$$ $9998400119994400181995632080078856012869885600800795632018199944000119 9998400001$ $$ = $$
$9998500104995450136496997050049356506434949950300298635004549989500014 9999900000.$
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?
I don't think that there is a list in the oeis for equations of this type. Even if it does contain sequences of this type the oeis isn't really the place for this kind of problem. The more variables in an equation the harder is becomes to catalog. For example these three sequences A050787, A050788 and A050789 catalog solutions to the equation $A^3+B^3=C^3-1$. (one sequence for each variable.) These sequences are organized by increasing values of $C$. In the $A^x+B^y=C^z$ case there would be six sequences which would probably be organized by increasing values $C^z$. Having one equation spread over six sequences is bad enough but there are other problems. Unlike the Cube sequences mentioned earlier there would be whole bunch of duplicates. For example $256+256=512$ can be represented eight different ways assuming that it takes the form $A^x+B^y=C^z$ where $A,B,C,x,y,z\in\Bbb{N}\space |\space x,y,z\ge 3$.
$2^8+2^8=2^9$, $\quad 4^4+2^8=2^9$, $2^8+4^4=2^9$, $\quad 4^4+4^4=2^9$, $2^8+2^8=8^3$, $\quad 4^4+2^8=8^3$, $2^8+4^4=8^3$, $\quad 4^4+4^4=8^3$
The number of duplicates gets larger and larger as the numbers get lager. Also unlike the cube sequences it is fairly trivial to construct an infinite number solutions to $A^x+B^y=C^z$ when $A,B,C$ share a factor. I can even do so using the very large example in the original post. For the sake not typing large numbers that nobody is going to read including myself. I will represent the bases of the power numbers in the example of the original post as $p,q,r$ respectively. Therefore $p^{15}+q^8=r^5$. One way I could create an infinite number of solutions is to multiply by $d^{120}$ where $d\in\Bbb{N}\space|\space d>1$ Then we would have $\left(pd^8\right)^{15}+\left(qd^{15}\right)^{8}=\left(rd^{24}\right)^{5}$. Another way I could do this is to multiply by $p^{40d}$. Then we would have $p^{40d+15}+\left(qp^{5d}\right)^{8}=\left(rp^{8d}\right)^{5}$. When looking at a particular sequence in the oeis it displays around 30 terms. Sometimes there can be links to text files with 100,000 terms. Even with 100,000 terms it wouldn't even come remotely close to the large example found by the op. There is no point in adding that example to the oeis.