Is there a way to prove that $A^3 - (A-1)^3 = 13^x$ is false when $x$ is an odd positive integer greater than $1$?

Question

In the case of $A^3 - (A-1)^3 = B^x$ we can find some rare examples, such as:

$8^3-7^3 =13^2$

$28712305723921^3−28712305723920^3=49731172316281^2$

But according to the Beal's Conjecture $A^3 - (A-1)^3 = B^x$ should be false if $x$ is an odd positive integer greater than $1$.

So I was trying to prove the specific case of $A^3 - (A-1)^3 = 13^x$.

For $13^x$ when $x$ is an odd positive integer greater than $1$, we have:

$13^x = C$

$13^3=2197$

$13^5=371293$

$13^7=62748517$

$13^9=10604499373$

$13^{11}=1792160394037$

$13^{13}=302875106592253$

...

As you can see $C$ always ends in $7$ or $3$.

I have then listed the following:

$A^3 - (A-1)^3 = D$:

$1^3-0^3=1$

$2^3-1^3=7$

$3^3-2^3=19$

$4^3-3^3=37$

$5^3-4^3=61$

$6^3-5^3=91$

$7^3-6^3=127$

...

As you can see:

$D$ grows by $+6+12+18+24+30+36...$

$D$ never ends in $3$, therefor all the results of $C$ ending in $3$ are automatically excludes.

$C$ always ends in $7$ and $3$, therefor all the results in $D$ ending in $1$ or $9$ are automatically excluded.

The list resulting in $D$ is now left with only numbers ending in: $7$:

$7, 37 ,127 ,217,397, 547,817, 1027 ,1387, 1657, 2107...$

We have two growing sets here:

$7, 127 , 397 ,817,1387,2107..$ growing at $+120,+270,+420,+570,+720...$ (growing at $+150$)

$37,217,547,1027,1657...$ growing at $+180,+330,+480,+630...$ (growing at $+150$)

So we can now take any resulting $C$ and subtract either $-120-270-420-570-720-...$ to check whether it will land on $7$ OR either $-180-330-480-630-...$ to check whether it will land on $37$

So these are the results:

$13^3=2197$

$2197-120-270-420-570-720-...=97$

$2197-180-330-480-630-...=577$

$13^7=62748517$

$62748517-120-270-420-570-720-...=52687$

$62748517-180-330-480-630-...=134977$

$13 ^ {11} = 1792160394037$

$1792160394037-120-270-420-570-720-...=7020817$

$1792160394037-180-330-480-630-...=20933137$

...

As you can see the results always grow further and further away from possibly being equal to $7$ or $37$.

Assuming it will prove the specific case of $A^3 - (A-1)^3 = 13^x$, is there a way to prove that these results can only grow further away?

Answer 1

Actually, there are no integer solutions to $$ s^2 - 156 t^2 = -3$$

Which tells us that there are no integral solutions to $3m^2 - 3 m + 1 = 13 n^2$

The solutions to $$ u^2 - 12 v^2 = -3$$ have $v$ in the sequence $1, 13, 181, 2521, 35113, \ldots$ with $$ v_{k+2} =14 v_{k+1} - v_k$$ so that there are infinitely many solutions to $3m^2 - 3 m + 1 = n^2$

The bit about 156 amounts to this list

624    factored   2^4 * 3 *  13

    1.             1          24         -12   cycle length             2
    2.            -1          24          12   cycle length             2
    3.             3          24          -4   cycle length             2
    4.            -3          24           4   cycle length             2
    5.             5          22          -7   cycle length             4
    6.            -5          22           7   cycle length             4
    7.             7          22          -5   cycle length             4
    8.            -7          22           5   cycle length             4

  form class number is   8

or this cycle

0  form   1 24 -12   delta  -2
1  form   -12 24 1   delta  24
2  form   1 24 -12

or this presentation of the continued fraction for $\sqrt{156}$

$$ \sqrt { 156} = 12 + \frac{ \sqrt {156} - 12 }{ 1 } $$ $$ \frac{ 1 }{ \sqrt {156} - 12 } = \frac{ \sqrt {156} + 12 }{12 } = 2 + \frac{ \sqrt {156} - 12 }{12 } $$ $$ \frac{ 12 }{ \sqrt {156} - 12 } = \frac{ \sqrt {156} + 12 }{1 } = 24 + \frac{ \sqrt {156} - 12 }{1 } $$

Simple continued fraction tableau:
$$ \begin{array}{cccccccccc} & & 12 & & 2 & & 24 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 12 }{ 1 } & & \frac{ 25 }{ 2 } \\ \\ & 1 & & -12 & & 1 \end{array} $$

$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 156 \cdot 0^2 = 1 & \mbox{digit} & 12 \\ \frac{ 12 }{ 1 } & 12^2 - 156 \cdot 1^2 = -12 & \mbox{digit} & 2 \\ \frac{ 25 }{ 2 } & 25^2 - 156 \cdot 2^2 = 1 & \mbox{digit} & 24 \\ \end{array} $$