On the exponential Diophantine equation (P_n^(k))^x+(P_r^(k))^x=P_m^(k)

@misc{faye2026exponential,
  title = {{On the exponential Diophantine equation $\left(P_n^(k)\right)^x+\left(P_r^(k)\right)^x=P_m^(k)$}},
  author = {Fallou Faye and Bernadette Faye},
  howpublished = {arXiv},
  year = {2026},
  month = nov,
  doi = {10.48550/arxiv.2611.02295},
  eprint = {2611.02295},
  archivePrefix = {arXiv},
  url = {https://arxiv.org/abs/2611.02295},
  note = {CC0},
}

TY  - UNPB
TI  - On the exponential Diophantine equation (P_n^(k))^x+(P_r^(k))^x=P_m^(k)
AU  - Fallou Faye
AU  - Bernadette Faye
PY  - 2026
DA  - 2026/11/01
PB  - arXiv
DO  - 10.48550/arxiv.2611.02295
UR  - https://arxiv.org/abs/2611.02295
AB  - The Pell numbers satisfy P_n^2+P_n+1^2=P_2n+1, and Bravo, Luca and Pomeo recently found all the cases in which the sum of the x-th powers of two consecutive k-generalized Pell numbers is again a k-generalized Pell number. We ask what happens when the two indices are arbitrary, that is, we solve (P_n^(k))^x+(P_r^(k))^x=P_m^(k) in positive integers k≥2, n≥ r, m and x. The answer is that nothing new appears: apart from 1^x+1^x=2, every solution has consecutive indices. The proof combines linear forms in logarithms with an auxiliary Fibonacci equation which does not depend on k, and, for large k, with the fact that the dominant root of the sequence is very close to φ^2. For the reduction step, the method of Dujella and Pethő fails for large indices and has to be replaced by a criterion of Legendre type.
ER  -