Perfect squares in reciprocal-sum sequences and primes that are inert in quadratic fields

Source abstract

Let $k$ be a positive integer, and consider the sequences of positive rationals with $x_0\in\N$ and $x_{n+1}=k/(x_0+x_1+\dots+x_n)$. 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 $n\ge2$ 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 $t_n=(x_0+\dots+x_n)/h$ 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 $b_n^2-\frac4k a_n^2$ 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 $\Q(\sqrt{-k})$. For the recursions $x_{n+1}=h^2/(x_0+\dots+x_n+nμh)$ the role of the Gaussian integers is played by the quadratic fields $\Q(\sqrt{μ^2-4})$, which include $\Q(\sqrt{-3})$ and every real quadratic field.