Perfect squares in reciprocal-sum sequences and primes that are inert in quadratic fields
Abstract
Let k be a positive integer, and consider the sequences of positive rationals with x0∈N and xn+1=k/(x0+x1+…+xn). Write xn=an/bn in lowest terms. We show that there is a rational constant c>0 such that c an+bn is a perfect power for every such sequence and every n≥2 if and only if k=h2 and every prime factor of h is congruent to $3$ modulo $4$; in that case c=2/h and the powers are squares. The proof rests on the observation that tn=(x0+…+xn)/h satisfies tn+1=tn+1/tn, a recursion under which reduced fractions never cancel. This yields an exact formula for c an+bn, shows that along any single sequence each prime factor of h spoils at most one term, and leads to two generalizations. For arbitrary k the invariant bn2−4/k an2 is always a rational square, and it is always an integer square exactly when the primes dividing the square part of k satisfy an inertness condition in Q(√(−k)). For the recursions xn+1=h2/(x0+…+xn+nμh) the role of the Gaussian integers is played by the quadratic fields Q(√(μ2−4)), which include Q(√(−3)) and every real quadratic field.
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BibTeX
@misc{matijasevick2026perfect,
title = {{Perfect squares in reciprocal-sum sequences and primes that are inert in quadratic fields}},
author = {Mateo Matijasevick and Santiago Rodríguez and Gregorio Salazar},
howpublished = {arXiv},
year = {2026},
month = oct,
doi = {10.48550/arxiv.2610.00765},
eprint = {2610.00765},
archivePrefix = {arXiv},
url = {https://arxiv.org/abs/2610.00765},
}
RIS
TY - UNPB
TI - Perfect squares in reciprocal-sum sequences and primes that are inert in quadratic fields
AU - Mateo Matijasevick
AU - Santiago Rodríguez
AU - Gregorio Salazar
PY - 2026
DA - 2026/10/01
PB - arXiv
DO - 10.48550/arxiv.2610.00765
UR - https://arxiv.org/abs/2610.00765
AB - Let k be a positive integer, and consider the sequences of positive rationals with x_0∈N and x_n+1=k/(x_0+x_1+…+x_n). Write x_n=a_n/b_n in lowest terms. We show that there is a rational constant c>0 such that c a_n+b_n is a perfect power for every such sequence and every n≥2 if and only if k=h^2 and every prime factor of h is congruent to $3$ modulo $4$; in that case c=2/h and the powers are squares. The proof rests on the observation that t_n=(x_0+…+x_n)/h satisfies t_n+1=t_n+1/t_n, a recursion under which reduced fractions never cancel. This yields an exact formula for c a_n+b_n, shows that along any single sequence each prime factor of h spoils at most one term, and leads to two generalizations. For arbitrary k the invariant b_n^2−4/k a_n^2 is always a rational square, and it is always an integer square exactly when the primes dividing the square part of k satisfy an inertness condition in Q(√(−k)). For the recursions x_n+1=h^2/(x_0+…+x_n+nμh) the role of the Gaussian integers is played by the quadratic fields Q(√(μ^2−4)), which include Q(√(−3)) and every real quadratic field.
ER -