AI 中文总结
研究倒数求和序列中完全平方数的条件，证明当且仅当k为平方数且其平方根的所有素因子≡3 mod 4时，存在常数c使c a_n+b_n为平方数，并推广到任意k及带参数μ的递推，涉及二次域中的惰性素数。
AI 中文摘要
设 $k$ 为正整数，考虑正有理数序列，其中 $x_0\in\N$ 且 $x_{n+1}=k/(x_0+x_1+\dots+x_n)$。将 $x_n$ 写为最简分数 $a_n/b_n$。我们证明，存在有理常数 $c>0$ 使得对每个这样的序列及每个 $n\ge2$，$c\\,a_n+b_n$ 都是完全幂，当且仅当 $k=h^2$ 且 $h$ 的每个素因子都同余于 $3$ 模 $4$；此时 $c=2/h$ 且这些幂是平方数。证明基于观察：$t_n=(x_0+\dots+x_n)/h$ 满足 $t_{n+1}=t_n+1/t_n$，这是一个在约分后分数永不消去的递推。这给出了 $c\\,a_n+b_n$ 的精确公式，表明沿任意单个序列，$h$ 的每个素因子至多破坏一项，并引出两个推广。对于任意 $k$，不变量 $b_n^2-\frac4k a_n^2$ 总是有理平方数，并且当且仅当整除 $k$ 的平方部分的素数在 $\Q(\sqrt{-k})$ 中满足惰性条件时，它总是整数平方数。对于递推 $x_{n+1}=h^2/(x_0+\dots+x_n+n\mu h)$，高斯整数的作用由二次域 $\Q(\sqrt{\mu^2-4})$ 扮演，这些域包括 $\Q(\sqrt{-3})$ 和每个实二次域。
英文摘要
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.
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