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arXiv · 2608.18608

On the Diophantine Equation $p^x+ (2p+1)^y =z^2$ with Consecutive Exponents

Subhasis Panda
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Abstract

We study the Diophantine equation $p^x+ (2p+1)^y =z^2$ over positive integers $x$, $y$ and $z$ for every odd prime $p$. We prove that $(x,y,z)=(2,1,p+1)$ is the unique solution except possibly when $2p+1$ is composite. In that case, it reduces to a family depending on one parameter, and we show that the parameter must be odd and satisfies an explicit upper bound, reducing the problem to finitely many cases.

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Subhasis Panda. 2026-08-19. On the Diophantine Equation $p^x+ (2p+1)^y =z^2$ with Consecutive Exponents. https://arxiv.org/abs/2608.18608

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