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On the Diophantine Equation px+ (2p+1)y =z2 with Consecutive Exponents

  • Subhasis Panda
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Abstract

We determine all positive integer solutions to px+(2p+1)y=z2, where p is an odd prime and |x−y|=1. The unique solution is (x,y,z)=(2,1,p+1) except when p=7, for which there is one additional solution, (x,y,z)=(4,3,76). The proof reduces the nontrivial factorization case to an auxiliary equation, bounds its exponents using an AM--GM inequality, and isolates the exceptional solution with a divisibility argument modulo 2k2.

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Cite

BibTeXRIS
BibTeX
@misc{panda2026diophantine,
  title = {{On the Diophantine Equation $p^x+ (2p+1)^y =z^2$ with Consecutive Exponents}},
  author = {Subhasis Panda},
  howpublished = {arXiv},
  year = {2026},
  month = oct,
  doi = {10.48550/arxiv.2608.18608},
  eprint = {2608.18608},
  archivePrefix = {arXiv},
  url = {https://arxiv.org/abs/2608.18608},
}
RIS
TY  - UNPB
TI  - On the Diophantine Equation p^x+ (2p+1)^y =z^2 with Consecutive Exponents
AU  - Subhasis Panda
PY  - 2026
DA  - 2026/10/19
PB  - arXiv
DO  - 10.48550/arxiv.2608.18608
UR  - https://arxiv.org/abs/2608.18608
AB  - We determine all positive integer solutions to p^x+(2p+1)^y=z^2, where p is an odd prime and |x−y|=1. The unique solution is (x,y,z)=(2,1,p+1) except when p=7, for which there is one additional solution, (x,y,z)=(4,3,76). The proof reduces the nontrivial factorization case to an auxiliary equation, bounds its exponents using an AM--GM inequality, and isolates the exceptional solution with a divisibility argument modulo 2k^2.
ER  -