Beal's conjecture is a generalization of Fermat's Last Theorem. It states: If $A^x + B^y = C^z$, where $A, B, C, x, y$ and $z$ are positive integers and $x, y$ and $z$ are all greater than $2$, then $A, B$ and $C$ must have a common prime factor.
I was trying to understand what makes $x, y$ or $z$ equal to $2$ not needing $A, B$ or $C$ to have a common prime factor (such: as $13^2 + 7^3 = 8^3$), and I came up with the following:
When $A^2$:
$1^2, 2^2, 3^2, 4^2, 5^2, 6^2, 7^2 ...$
$(1) , 1(1+3) , 1(1+3+5) , 1(1+3+5+7) , 1(1+3+5+7+9) , 1(1+3+5+7+9+11) , 1(1+3+5+7+9+11+13)...$
When $A^3$:
$1^3, 2^3, 3^3, 4^3, 5^3, 6^3, 7^3 ...$
$(1) , 2(1+3) , 3(1+3+5) , 4(1+3+5+7) , 5(1+3+5+7+9) , 6(1+3+5+7+9+11) , 7(1+3+5+7+9+11+13)...$
When $A^4$:
$1^4, 2^4, 3^4, 4^4, 5^4, 6^4, 7^4 ...$
$(1) , 4(1+3) , 9(1+3+5) , 16(1+3+5+7) , 25(1+3+5+7+9) , 36(1+3+5+7+9+11) , 49(1+3+5+7+9+11+13)...$
So any $X^y$ can be translated as: $n(1+3+..)$ and for $X^2$ are the only ones that can also be translated as $1(1+3+...)$.
*Note $(1+3+..)$ is the beginning of the consecutive list: $(1+3+5+7+9+11+13+15+17+19+21...)$
So I can suppose that for any $X^y$ as ${A(1+3+...),B(1+3+...),C(1+3+...)}$:
$A(1+3+...) + B(1+3+...) = C(1+3+...)$, $A,B,C$ must have a common prime factor unless at least one of either $A or B$ or $C$ equal $1$ (which ironically is a common prime factor of all numbers)?
Further more, the consecutive list $(1+3+5+7+9+11+13+15+17+19+21...)$ itself can be seen as $1+3 + 3(1+3) +4(1+3) +5(1+3)...$
(1+3 = 1+3, 5+7 = 3(1+3), 7+9 = 4(1+3)...)
I am curious to know whether someone can point any benefits in studying the conjecture in such an approach?
Any references for such already made studies are appreciated.