2021年4月根源杯初高数学衔接联考二试第四题

n为正整数. 设1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}=\dfrac{a_n}{b_n}, 其中a_n, b_n为互素的正整数. 对素数p, 令集合S_P=\left\{n\mid n\in\mathbf{N}^+, p\mid a_n\right\}. 对任何一个素数p\geqslant 5, 证明: 

(1) p-1\in S_p

(2) p^2-1\in S_p

(3) 除p-1, p^2-1外, 集合S_p中还至少有1个元素. 

引理: 设素数p\geqslant 5, k\in\mathbf{N}, 令

    \[\dfrac{1}{kp+1}+\dfrac{1}{kp+2}+\cdots+\dfrac{1}{kp+p-1}=\dfrac{t_k}{s_k}, \]

其中t_k, s_k为互素的正整数, 则p^2\mid t_k

引理的证明: \dfrac{t_k}{s_k}=\displaystyle\sum\limits_{i=1}^{p-1}\dfrac{1}{kp+i}=\dfrac{1}{2}\displaystyle\sum\limits_{i=1}^{p-1}\bigg(\dfrac{1}{kp+i}+\dfrac{1}{kp+p-i}\bigg)=\dfrac{\left(2k+1\right)p}{2}\sum\limits_{i=1}^{p-1}\dfrac{1}{\left(kp+i\right)\left(kp+p-i\right)}.

S=\displaystyle\sum\limits_{i=1}^{p-1}\dfrac{1}{\left(kp+i\right)\left(kp+p-i\right)}. 欲证p^2\mid t_k, 只要证p\mid S

注意到在1\sim p-1中, 恰好配成\dfrac{p-1}{2}对数论倒数, 所以

    \[S=\sum\limits_{i=1}^{p-1}\dfrac{1}{\left(kp+i\right)\left(kp+p-i\right)}\equiv \sum\limits_{i=1}^{p-1}\dfrac{1}{-i^2}\equiv -\sum\limits_{i=1}^{p-1}\dfrac{1}{i^2}\equiv -\sum\limits_{j=1}^{p-1} j^2=-\dfrac{\left(p-1\right)p\left(2p-1\right)}{6}\pmod{p}.\]

而素数p\geqslant 5, 故p\mid S. 引理得证. 

回到原题.

(1) 由引理, 知p^2\mid a_{p-1}. 由p\mid a_{p-1}, 得p-1\in S_p

(2) \dfrac{a_{p^2-1}}{b_{p^2-1}}=\displaystyle\sum\limits_{i=1}^{p^2-1}\dfrac{1}{i}=\dfrac{1}{p}\displaystyle\sum\limits_{i=1}^{p-1}\dfrac{1}{i}+\displaystyle\sum\limits_{k=0}^{p-1}\displaystyle\sum\limits_{i=1}^{p-1}\dfrac{1}{kp+i}=\dfrac{1}{p}\cdot \dfrac{a_{p-1}}{b_{p-1}}+\displaystyle\sum\limits_{k=0}^{p-1}\dfrac{t_k}{s_k}

p^2\mid a_{p-1}, p^2\mid t_k, 故p\mid a_{p^2-1}. 因此p^2-1\in S_p

(3) 下面证明p\left(p-1\right)\in S_p

\dfrac{a_{p\left(p-1\right)}}{b_{p\left(p-1\right)}}=\displaystyle\sum\limits_{i=1}^{p\left(p-1\right)}\dfrac{1}{i}=\dfrac{1}{p}\displaystyle\sum\limits_{i=1}^{p-1}\dfrac{1}{i}+\displaystyle\sum\limits_{k=0}^{p-2}\displaystyle\sum\limits_{i=1}^{p-1}\dfrac{1}{kp+i}=\dfrac{1}{p}\cdot \dfrac{a_{p-1}}{b_{p-1}}+\displaystyle\sum\limits_{k=0}^{p-2}\dfrac{t_k}{s_k}

p^2\mid a_{p-1}, p^2\mid t_k, 故p\mid a_{p\left(p-1\right)}. 因此p\left(p-1\right)\in S_p

p\left(p-1\right)不等于p-1, p^2-1, 故p\left(p-1\right)作为除p-1, p^2-1外的元素, 也属于S_p