According to wikipedia's page on 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.
Per @Henri's comment I am attempting a special case of Beal's conjecture where $=−1$ and $=$.
Examples :
$3^2 + 4 ^2= 5^2$
$13 ^ 2 + 7^3 = 8^3$
It is way beyond my level to try and solve a generalized case, so I will focus on a case where $z$ and $y$ equal $= 3$, so we have:
$A^x + (C-1) ^3 = C^3$
Or we can see it as
$C^3 - (C-1) ^3 = A^x$
In the case of $C^3$, we can construct the following consecutive list:
$1 , 8 , 27 , 64 , 125 , 216 , 343 , 512 , 729, 1000 ...$
The increments can be seen as the following:
A) $7 , 19 , 37 , 61 , 91 , 127 , 169 , 217 , 271 ...$
The Increments in A) can be seen as:
B) $(1^2 + 2^2 + 1*2)$ , $(2^2 + 3^2 + 2*3)$ , $(3^2 + 4^2 + 3*4)$ , $(4^2 + 5^2 + 4*5)$ , $(5^2 + 6^2 + 5*6)$ , $(6^2 + 7^2 + 6*7) ...$
As you can see the list is growing by: $(q^2 + (q +1)^2 + q*(q+1))$ Which is $(q^2 + (q +1)(q + 1) + q^2 + q))$ which is equal to $(q^2 + q^2 +2q + 1 + q^2 + q))$ which is equal to $(3q^2 + 3q +1)$.
OR
C) $(1 * 6 +1), (3 *6 +1), (6 *6 +1) , (10 *6 +1) , (15 *6 +1) , (21 *6 +1) , (28 *6 +1) , (36 *6 +1) , (45 *6 +1)...$
which is growing by $r * 6 + 1$
and $r$ being the following: $1, 3, 6, 10, 15, 21, 28, 36, 45..$ which are increments of $+2, +3, +4 ,+5, +6, +7, +8, +9...$
According to B) $(3q^2 + 3q +1)) = A^x$ only if $x =2$?
I personally don't see how it can be solved with that approach.
According to C) I can conclude that if $r * 6 + 1 = A^x$, than for the possibility for $x > 2$ and $x$ being odd, could only result with $A$ possibly being $1, 7 , 13, 19, 25...$. Since for all $(A^x - A)/6 \mod 6 = 0$ When $x$ id odd and $x > 2$.
Am I correct so far? Is there a way to further cross out the possibilities for $A$ to be $1 , 7 , 13, 19 , 25...$ when $x > 2$ and $x$ is odd ?
(I am aware that it still maintains the possibility for $x>2$ while $x$ is even, but that in enough for my question)