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.
I am attempting a special case of Beal's conjecture where $=−1$ and $=$.
Example :
$13 ^ 2 + 7^3 = 8^3$
It is way beyond my level to try and solve a generalized case, so I will focus on an even more specific case where $z$ and $y$ both equal $= 3$, so we have:
$A^x + (C-1) ^3 = C^3$
Or we can see it as
$C^3 - (C-1) ^3 = A^x$
$1^3 - 0^3 = 1$
$2^3 - 1^3 = 7$
$3^3 - 2^3 = 19$
$4^3 - 3^3 = 37$
$5^3 - 4^3 = 61$
$6^3 - 5^3 = 91$
$7^3 - 6^3 = 127$
$8^3 - 7^3 = 169$ Here we have $169=13^2$
...
$1, 7 , 19, 37, 61, 91, 127 , 169, 217 ...$
I will denote an element in the above set as $n$
Growing at:
$1, 6, 12, 18, 24, 30 , 36 ...$
I will denote an element in the above set as $m$
*Note besides when $m=1$, for all other $m$: $m \bmod 3 =0$
So when $n=169=13^2$ we have $169 - 1 -6 -12-18 -24-30-36-42= 0$
But I am trying to prove that $n ≠ A^x$ if $x > 2$ and $x$ is odd
We know from Fermat's theorem that $(A^x - A) \bmod 3 = 0$ when $x$ is odd and $x>1$ (attention to the *Note)
So for any element $n$ (noted above) equal $(A^x - A)$, $A$ can only be $1,7,13,19,25...$ when $x$ is odd and $x>1$.
The problem is that:
Example: $7^3 = 343$
$343 - 1- 7 -19 -37 -61 -91 -127 =0$
But
$343 -1 -6 -12-18 -24 - 30 -36 -42 -48 -54 -60 = 24$ (before continuing on to negative)
Example: $7^5 = 16807$
For $A^5$ the set of elements $n$ is $1,31,211,781,2101,4651,9031...$
$16807 - 1-31 -211 -781 -2101 -4651-9031 =0$
But
$16807 -1 -6 -12 -18 -24 - 30 -36 -42 -48 -54 -60-.... = 630$ (before continuing on to negative)
Example: $7^7 = 823543$
For $A^7$ the set of elements $n$ is $1, 127 , 2059, 14197, 61741, 201811, 543607...$
$823543 - 1- 127 -2059 -14197 -61741 -201811 -543607-... =0$
But
$823543 -1 -6 -12 -18 -24 - 30 -36 -42 -48 -54 -60-.... = 4446$ (before continuing on to negative)
Example: $13^3 = 2197$
$2197 - 1- 7 -19 -37 -61 -91 -127 -... =0$
But
$2197 -1 -6 -12 -18 -24 - 30 -36 -42 -48 -54 -60 -... = 234$ (before continuing on to negative)
Example: $13^5 = 371293$
For $A^5$ the set of elements $n$ is $1,31,211,781,2101,4651,9031...$
$371293- 1-31 -211 -781 -2101 -4651-9031... =0$
But
$371293 -1 -6 -12-18 -24 - 30 -36 -42 -48 -54 -60-.... = 2982$ (before continuing on to negative)
...
As you can see:
It seems that for $A^x$ where $A = 7,13,19,25...$ and $x$ is odd and $x>1$, when ever applying:
$A^x - 1 -6 -12 -18 -24 - 30 -36 -42 -48 -54 -60-...$ Can not result in $0$ but only in an integer that is $ \bmod 3=0$ (examples above: $24,630,4446, 234,2982$)
Can it be proved? and if that could be proved, will it prove the specific case that for $A^x + (C-1) ^3 = C^3$ can occur only and only if $x ≤ 2$?