I was trying to study the Beal's Conjecture, which 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 was attempting a special case of the Beal's Conjecture where $=−1$ and $=$
So it means that $A^x + (C-1)^z = C^z$ or $C^z -(C-1)^z=A^x$ (where $C$ and $z$ are greater than $1$)
It means $A$ will always be a positive odd integer greater than $1$.
If I disobey the rules and allow both $z=2$ and $x=2$ ($z=x=2$), I will find many examples, such as:
$5^2 - 4^2=3^2 =9$
$13^2 - 12^2=5^2 =25$
$25^2 - 24^2=7^2 = 49$
So we can see that:
$A^2 = C^2 -(C-1)^2$: When $A$ is any positive odd integer greater than $1$, then $C^2 = (0.5(a^2+1))^2$ and $(C-1)^2 =(0.5(a^2-1))^2$
We know from Fermat's little theorem that the same is false for when $x=z$, $x$ is odd, and $x$ is greater than $1$ .
If I disobey the rules and allow only $z=2$, I will find even more examples because:
For all $C^2$:
$1^2=1$
$2^2=1+3$
$3^2=1+3+5$
$4^2=1+3+5+7$
$C^2=1+3+5+7+n...n+2$
So $C^2 - (C-1)^2) = n$ and $n$ can be any odd positive integer including $125=5^3$ ,$343=7^3$, $3125=5^5$ which can translate into any $A^x$ where $A$ is any odd positive integer and $x$ is any integer.
But here is the thing:
If I disobey the rules and allow only $x=2$ I have only found one example so far:
$8^3-7^3 =13^2$
At first I was trying to understand why is this case different which left me clueless, so I tried to search the web for other examples, and I couldn't find any so far.
Now I am wondering if there are at all any other examples of $C^z -(C-1)^z = A^2$ when $C$ and $z$ are positive integers greater than $1$ and $z$ is odd? If yes, is there a way detect them?
Any references are appreciated.
The equation $$C^z + B^z = A^2$$ for $z$ odd has been completely resolved by the work of many authors. You are looking at the case where $C + B = 1$ in particular.
First if $z$ has a prime factor $p>3$ we can consider the equation $$C^p + B^p = A^2$$ instead as any solution of the first equation gives a solution of the second.
If $p \ge 7$: then then only solutions are $(C,B,A) = (±1,∓1,0), (1,0,±1), (0,1,±1)$ by the paper H. Darmon and L. Merel, "Winding quotients and some variants of Fermat’s last theorem".
For $p = 5$: Poonen showed in "Some diophantine equations of the form $x^n + y^n = z^m$" that this has only the trivial solutions.
In the $z = 3$ case we have infinitely many solutions, which Cohen in his book "Number Theory Volume II: Analytic and Modern Tools" gives three parameterizations for these solutions: any solution must be given by picking some $s,t$ (satisfying some conditions) and then $A,B,C$ will be given by one of:
$$ \left\{\begin{array}{l} C=s(s+2 t)\left(s^{2}-2 t s+4 t^{2}\right) \\ B=-4 t(s-t)\left(s^{2}+t s+t^{2}\right) \\ A=\pm\left(s^{2}-2 t s-2 t^{2}\right)\left(s^{4}+2 t s^{3}+6 t^{2} s^{2}-4 t^{3} s+4 t^{4}\right) \end{array}\right. $$ $$ \left\{\begin{array}{l} C=s^{4}-4 t s^{3}-6 t^{2} s^{2}-4 t^{3} s+t^{4} \\ B=2\left(s^{4}+2 t s^{3}+2 t^{3} s+t^{4}\right) \\ A=3(s-t)(s+t)\left(s^{4}+2 s^{3} t+6 s^{2} t^{2}+2 s t^{3}+t^{4}\right) \end{array}\right. $$ $$ \left\{\begin{array}{l} C=-3 s^{4}+6 t^{2} s^{2}+t^{4} \\ B=3 s^{4}+6 t^{2} s^{2}-t^{4} \\ A=6 s t\left(3 s^{4}+t^{4}\right) \end{array}\right. $$
Where we may have to swap $C$ with $B$ to get all solutions.
Looking at the third parameterization we cannot have $C = 1 - B$ as $B \equiv C \pmod 2$ there.
In the second case $C + B$ is always a multiple of 3 so we do not get solutions.
In the first case $C + B$ factors as $(s^2 - 2st - 2t^2)^2$ so this is $1$ if and only if $s^2 - 2st - 2t^2 = \pm 1$ which after a change of variables is $s'^2 - 3t'^2 = \pm 1$ so this is a Pell equation (or rather two Pell equations) with infinitely many solutions for example the solution $$s' = 1351, t' = 780$$ gives us $$ s = 780 + 1351, t = 780$$ which gives the solution $$ 28712305723921^3 -28712305723920^3 = 49731172316281^2 $$ of the original equation. Here $s = 1, t = -1$ gives the solution $8^3 - 7^3 = 13^2$ that you found.
If $z = 3^n$ with $n \ge 2$ then Poonen shows in the same paper mentioned above that the equation $$C^9 + B^9 = A^2$$ has only trivial solutions.
Check out the survey by Michael Bennett, Preda Mihailescu and Samir Siksek at https://homepages.warwick.ac.uk/~maseap/papers/bealconj.pdf for more on some of the known cases of Beal.
In summary: To get a solution of $C^z - (C-1)^z = A^2$ for odd $z \ge3$ we must have $z=3$ (except for the trivial examples where $C=1$) and we can get all solutions by taking a solution $s',t'$ of the Pell equation $s'^2 - 3t'^2 = \pm1 $ and plug $s = s' + t', t =t'$ into
$$ \left\{\begin{array}{l} C=s(s+2 t)\left(s^{2}-2 t s+4 t^{2}\right) \\ A=\pm\left(s^{2}-2 t s-2 t^{2}\right)\left(s^{4}+2 t s^{3}+6 t^{2} s^{2}-4 t^{3} s+4 t^{4}\right) \end{array}\right. $$ And, more generally even without the condition that $B = C-1$ this equation has been completely resolved.