Further solutions of the auxiliary Diophantine equation

Determine whether the auxiliary equation \[(2k+1)^{2b+1}-(2k-1)^{2b+1}=2(2k^2-1)^{b+1}, \] with integers \(k\ge 2\), odd integers \(b\ge 3\), and \(p=2k^2-1\) prime, admits any solutions beyond the known solution \(k=2\), \(b=1\), corresponding to \(p=7\) and \((x,y,z)=(4,3,76)\).

Background

The paper studies positive-integer solutions of $p^x+(2p+1)^y=z^2$ for an odd prime $p$ under the consecutive-exponent condition $|x-y|=1$. Apart from the universal solution $(x,y,z)=(2,1,p+1)$, the authors reduce any exceptional solution to the case where $2p+1$ is composite and has the factorization $(2k-1)(2k+1)$, with $p=2k^2-1$.

In the remaining nontrivial factorization case, the exponents satisfy $x=2(b+1)$ and $y=2b+1$, and the problem becomes the displayed auxiliary equation. The paper proves an upper bound on $b$ for each fixed $k$, hence reducing the search to finitely many exponents for each prime, but it does not establish whether additional solutions exist. It records the example $k=2,b=1$, yielding $7^4+15^3=76^2$, and reports that no solutions were found computationally for odd $b\ge3$.