Studying the Beal's conjecture, it states: $A^x + B^y = C^z$, 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.
It made me think of why the rule doesn't apply if $x$ or $y$ or $z$ equal $2$, and so I collected some data to try and find any difference.
The only thing that seems different is that:
$\textrm{For } A^3\!: \ (A^3 - A) \bmod 3 = 0$
For $A^4$: $(A^4 - A^2) modulo 3 = 0$
For $A^5$: $(A^5 - A) modulo 3 = 0$ | $(A^5 - A^3) modulo 3 = 0$ *
So when it comes to $A^x$ where $x$ is bigger than 2, the rule is always $(A^x - A^{x-2}) modulo 3 = 0$
And the same doesn't exist for $A^2$ where it is *always ($modulo 3 = 0$)
- I mention *always as visualized in the image below
My question is if there might be a connection to Beal's Conjecture (or at least partially, while considering that $(A^x - A^{x-2}) + (B^y - B^{y-2}) = (C^z - C^{z-2})$ where $x,y,z$ are bigger than 2
(Please see the image as a reference)
I am well aware that this doesn't solve the Beal's conjecture, and my intent is solely to seek any references, opinions or an answer for whether there might be a connection.
