Super Congruences Involving Alternating Harmonic Sums

Zhongyan Shen¹, Tianxin Cai²

¹Department of Mathematics, Zhejiang International Studies University, Hangzhou, China.
²Department of Mathematics, Zhejiang University, Hangzhou, China.

DOI: 10.4236/apm.2020.1010037                

Abstract

Let p be an odd prime, the harmonic congruence such as , and many different variations and generalizations have been studied intensively. In this note, we consider the congruences involving the combination of alternating harmonic sums,  where P_P denotes the set of positive integers which are prime to p. And we establish the combinational congruences involving alternating harmonic sums for positive integer n=3,4,5.

Keywords

Bernoulli Numbers, Alternating Harmonic Sums, Congruences, Modulo Prime Powers

Keywords:

1. Introduction

At the beginning of the 21^th century, Zhao (Cf. [1] ) first announced the following curious congruence involving multiple harmonic sums for any odd prime $p>3$,

${\displaystyle \underset{i+j+k=p}{\sum }}\frac{1}{ijk}\equiv -2B_{p-3}\mathrm{<mi>\; </mi>}\left(\mathrm{mod}\mathrm{<mi>\; </mi>}p\right),$ (1)

which holds when $p=3$ evidently. Here, Bernoulli numbers $B_k$ are defined by the recursive relation:

${\displaystyle \underset{i=0}{\overset{n}{\sum }}}\left(\begin{array}{c}n+1\\ i\end{array}\right)B_i=0,n\ge 1.$

A simple proof of (1) was presented in [2]. This congruence has been generalized along several directions. First, Zhou and Cai [3] established the following harmonic congruence for prime $p>3$ and integer $n\le p-2$

${\displaystyle \underset{l_1+l_2+\cdots +l_n=p}{\sum }}\frac{1}{l_1{l}_2\cdots l_n}\equiv (\begin{array}{l}-\left(n-1\right)!B_{p-n}\left(\mathrm{mod}\mathrm{<mi>\; </mi>}p\right),\mathrm{<mi>\; </mi>\; <mi>\; </mi>\; <mi>\; </mi>}\text{if}\mathrm{<mi>\; </mi>2}\cancel{|}n,\\ -\frac{n\left(n!\right)}{2\left(n+1\right)}p{B}_{p-n-1}\left(\mathrm{mod}\mathrm{<mi>\; </mi>}{p}^2\right),\mathrm{<mi>\; </mi>\; <mi>\; </mi>\; <mi>\; </mi>}\text{if}\mathrm{<mi>\; </mi>2}|n.\end{array}$ (2)

Later, Xia and Cai [4] generalized (1) to

${\displaystyle \underset{i+j+k=p}{\sum }}\frac{1}{ijk}\equiv \frac{12B_{p-3}}{p-3}-\frac{3B_{2p-4}}{p-4}\left(\mathrm{mod}{p}^2\right),$

where $p>5$ is a prime.

Recently, Wang and Cai [5] proved for every prime $p\ge 3$ and positive integer r,

${\displaystyle \underset{\begin{array}{c}i+j+k=p^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{ijk}\equiv -2p^{r-1}{B}_{p-3}\mathrm{<mi>\; </mi>}\left(\mathrm{mod}{p}^r\right),$ (3)

where ${\mathcal{P}}_p$ denotes the set of positive integers which are prime to p.

Let $n=2$ or 4, for every positive integer $r\ge \frac{n}{2}$ and prime $p>n$, Zhao [6] extended (3) to

${\displaystyle \underset{\begin{array}{c}{i}_1+i_2+\cdots +i_n=p^r\\ i_1,i_2,\cdots ,i_n\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{i_1{i}_2\cdots i_n}\equiv -\frac{n!}{n+1}{p}^r{B}_{p-n-1}\mathrm{<mi>\; </mi>}\left(\mathrm{mod}{p}^{r+1}\right).$ (4)

For any prime $p>5$ and integer $r>1$, Wang [7] proved that

${\displaystyle \underset{\begin{array}{c}{i}_1+i_2+\cdots +i_5=p^r\\ i_1,i_2,\cdots ,i_5\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{i_1{i}_2\cdots i_5}\equiv -\frac{5!}{6}{p}^{r-1}{B}_{p-5}\left(\mathrm{mod}\mathrm{<mi>\; </mi>}{p}^r\right).$

We consider the following alternating harmonic sums

${\displaystyle \underset{\begin{array}{c}{i}_1+i_2+\cdots +i_n=p^r\\ i_1,i_2,\cdots ,i_n\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\sigma }_1^{i_1}{\sigma }_2^{i_2}\cdots {\sigma }_n^{i_n}}{i_1{i}_2\cdots i_n},$

where ${\sigma }_i\in \left\{1,-1\right\},\mathrm{<mi>\; </mi>}i=1,\mathrm{<mi>\; </mi>2},\mathrm{<mi>\; </mi>}\cdots ,\mathrm{<mi>\; </mi>}n$. Given n, we only need to consider the following alternating harmonic sums,

${\displaystyle \underset{\begin{array}{c}{i}_1+i_2+\cdots +i_n=p^r\\ i_1,i_2,\cdots ,i_n\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1}}{i_1{i}_2\cdots i_n},{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+\cdots +i_n=p^r\\ i_1,i_2,\cdots ,i_n\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1+i_2}}{i_1{i}_2\cdots i_n},\cdots ,{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+\cdots +i_n=p^r\\ i_1,i_2,\cdots ,i_n\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1+i_2+\cdots +i_{\left[\frac{n}{2}\right]}}}{i_1{i}_2\cdots i_n}$

where $\left[x\right]$ denotes the largest integer less than or equal to x.

In this paper, we consider the congruences involving the combination of alternating harmonic sums,

${\displaystyle \underset{\begin{array}{c}{i}_1+i_2+\cdots +i_n=p^r\\ i_1,i_2,\cdots ,i_n\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1}}{i_1{i}_2\cdots i_n},{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+\cdots +i_n=p^r\\ i_1,i_2,\cdots ,i_n\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1+i_2}}{i_1{i}_2\cdots i_n},\cdots ,{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+\cdots +i_n=p^r\\ i_1,i_2,\cdots ,i_n\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1+i_2+\cdots +i_{\left[\frac{n}{2}\right]}}}{i_1{i}_2\cdots i_n}.$

We obtain the following theorems. Among them, Theorem 1 and Theorem 2 have been proved by Wang [8] using different method.

Theorem 1. Let p be an odd prime and r a positive integer, then

${\displaystyle \underset{\begin{array}{c}i+j+k=2p^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}\equiv p^{r-1}{B}_{p-3}\mathrm{<mi>\; </mi>}\left(\mathrm{mod}\mathrm{<mi>\; </mi>}{p}^r\right).$

Remark 1. There is no solution $\left(i\mathrm{,\; <mi>\; </mi>}j\mathrm{,\; <mi>\; </mi>}k\right)$ for the equation $i+j+k=2p^r$ with $i\mathrm{,}j\mathrm{,}k\in {\mathcal{P}}_{2p}$.

Theorem 2. Let p be an odd prime and r a positive integer, then

${\displaystyle \underset{\begin{array}{c}i+j+k=p^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}\equiv \frac{1}{2}{p}^{r-1}{B}_{p-3}\mathrm{<mi>\; </mi>}\left(\mathrm{mod}\mathrm{<mi>\; </mi>}{p}^r\right).$

Theorem 3. Let $p\ge 5$ be a prime and r a positive integer, then

$\begin{array}{l}4{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1}}{i_1{i}_2{i}_3{i}_4}+3{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1+i_2}}{i_1{i}_2{i}_3{i}_4}\\ \equiv (\begin{array}{l}\frac{216}{5}p{B}_{p-5}\left(mod{p}^2\right)\mathrm{,\; <mi>\; </mi>}\text{if}\mathrm{<mi>\; </mi>}r=\mathrm{1,}\\ \frac{36}{5}{p}^r{B}_{p-5}\left(mod{p}^{r+1}\right)\mathrm{,\; <mi>\; </mi>}\text{if}\mathrm{<mi>\; </mi>}r>1.\end{array}\end{array}$

Theorem 4. Let $p>5$ be a prime and r a positive integer, then

$\begin{array}{l}{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4+i_5=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4,\mathrm{<mi>\; </mi>}{i}_5\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1}}{i_1{i}_2{i}_3{i}_4{i}_5}+2{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4+i_5=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4,\mathrm{<mi>\; </mi>}{i}_5\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1+i_2}}{i_1{i}_2{i}_3{i}_4{i}_5}\\ \equiv (\begin{array}{l}12B_{p-5}\left(modp\right)\mathrm{,\; <mi>\; </mi>}\text{if}\mathrm{<mi>\; </mi>}r=\mathrm{1,}\\ 6p^{r-1}{B}_{p-5}\left(mod{p}^r\right)\mathrm{,\; <mi>\; </mi>}\text{if}\mathrm{<mi>\; </mi>}r>1.\end{array}\end{array}$

2. Preliminaries

In order to prove the theorems, we need the following lemmas.

Lemma 1 ( [5] ) Let p be an odd prime and r, m positive integers, then

${\displaystyle \underset{\begin{array}{c}i+j+k=m{p}^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{ijk}\equiv -2m{p}^{r-1}{B}_{p-3}\mathrm{<mi>\; </mi>}\left(\mathrm{mod}\mathrm{<mi>\; </mi>}{p}^r\right).$

Lemma 2. Let p be an odd prime and r, m positive integers, then

${\displaystyle \underset{\begin{array}{c}i+j+k=m{p}^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{ijk}=\frac{6}{m{p}^r}{\displaystyle \underset{\begin{array}{c}1\le j<l\le m{p}^r\\ j,\mathrm{<mi>\; </mi>}l,\mathrm{<mi>\; </mi>}l-j\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{jl}.$

Proof. It is easy to see that

${\displaystyle \underset{\begin{array}{c}i+j+k=m{p}^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{ijk}=\frac{1}{m{p}^r}{\displaystyle \underset{\begin{array}{c}i+j+k=m{p}^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{i+j+k}{ijk}=\frac{3}{m{p}^r}{\displaystyle \underset{\begin{array}{c}i+j+k=m{p}^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{ij}.$

Let $l=j+k$, then $1\le j<l\le m{p}^r$ and $j\mathrm{,\; <mi>\; </mi>}l\mathrm{,\; <mi>\; </mi>}l-j\in {\mathcal{P}}_p$. By symmetry, we have

$\frac{3}{m{p}^r}{\displaystyle \underset{\begin{array}{c}i+j+k=m{p}^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{ij}=\frac{3}{m{p}^r}{\displaystyle \underset{\begin{array}{c}i+j<m{p}^r\\ i,\mathrm{<mi>\; </mi>}j,\mathrm{<mi>\; </mi>}l\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{l}\frac{i+j}{ij}=\frac{6}{m{p}^r}{\displaystyle \underset{\begin{array}{c}1\le j<l\le m{p}^r\\ j,\mathrm{<mi>\; </mi>}l,\mathrm{<mi>\; </mi>}l-j\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{jl}.$

This completes the proof of Lemma 2. ¨

Lemma 3. Let $p>4$ be a prime and r, m positive integers, then

${\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4=m{p}^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{i_1{i}_2{i}_3{i}_4}=\frac{24}{m{p}^r}{\displaystyle \underset{\begin{array}{c}1\le u_1<u_2<u_3\le m{p}^r\\ u_1,\mathrm{<mi>\; </mi>}{u}_3,\mathrm{<mi>\; </mi>}{u}_2-u_1,\mathrm{<mi>\; </mi>}{u}_3-u_2\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1{u}_2{u}_3}.$

Proof. The proof of Lemma 3 is similar to the proof of Lemma 2. ¨

Lemma 4 ( [3] ) Let $r\mathrm{,\; <mi>\; </mi>}{\alpha }_1\mathrm{,\; <mi>\; </mi>}\cdots \mathrm{,\; <mi>\; </mi>}{\alpha }_n$ be positive integers, $r={\alpha }_1+\cdots +{\alpha }_n\le p-3$ , then

${\displaystyle \underset{\begin{array}{c}1\le l_1\mathrm{,\; <mi>\; </mi>}\cdots \mathrm{,\; <mi>\; </mi>}{l}_n\le p-1\\ l_i\ne l_j\mathrm{,\; <mi>\; </mi>}\forall i\ne j\end{array}}{\sum }}\frac{1}{l_1^{{\alpha }_1}{l}_2^{{\alpha }_2}\cdots l_n^{{\alpha }_n}}\equiv (\begin{array}{l}{\left(-1\right)}^n\left(n-1\right)\mathrm{<mo>\; !\; </mo>}\frac{r\left(r+1\right)}{\mathrm{2\; <mo\; stretchy="false">\; (\; </mo>}r+\mathrm{2\; <mo\; stretchy="false">\; )\; </mo>}}{B}_{p-r-2}{p}^2\mathrm{<mi>\; </mi>}\left(mod\mathrm{<mi>\; </mi>}{p}^3\right)\mathrm{,\; <mi>\; </mi>}\text{if}\mathrm{<mi>\; </mi>2}\cancel{|}r\mathrm{,}\\ {\left(-1\right)}^{n-1}\left(n-1\right)\mathrm{<mo>\; !\; </mo>}\frac{r}{r+1}{B}_{p-r-1}p\mathrm{<mi>\; </mi>}\left(mod\mathrm{<mi>\; </mi>}{p}^2\right)\mathrm{,\; <mi>\; </mi>}\text{if}\mathrm{<mi>\; </mi>2}|r\mathrm{.}\end{array}$

Lemma 5 ( [7] ). Let p be an odd prime, and ${\alpha }_1\mathrm{,\; <mi>\; </mi>}\cdots \mathrm{,\; <mi>\; </mi>}{\alpha }_n$ positive integers, where $r={\alpha }_1+\cdots +{\alpha }_n\le p-3$ , then

${\displaystyle \underset{\begin{array}{l}1\le l_1\mathrm{,\; <mi>\; </mi>}\cdots \mathrm{,\; <mi>\; </mi>}{l}_n\le 2p\\ l_i\ne l_j\mathrm{,\; <mi>\; </mi>}{l}_i\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{l_1^{{\alpha }_1}{l}_2^{{\alpha }_2}\cdots l_n^{{\alpha }_n}}\equiv (\begin{array}{l}{\left(-1\right)}^n\left(n-1\right)\mathrm{<mo>\; !\; </mo>}\frac{2r\left(r+1\right)}{r+2}{B}_{p-r-2}{p}^2\mathrm{<mi>\; </mi>}\left(mod\mathrm{<mi>\; </mi>}{p}^3\right)\mathrm{,\; <mi>\; </mi>}\text{if}\mathrm{<mi>\; </mi>2}\cancel{|}r\mathrm{,}\\ {\left(-1\right)}^{n-1}\left(n-1\right)\mathrm{<mo>\; !\; </mo>}\frac{2r}{r+1}{B}_{p-r-1}p\mathrm{<mi>\; </mi>}\left(mod\mathrm{<mi>\; </mi>}{p}^2\right)\mathrm{,\; <mi>\; </mi>}\text{if}\mathrm{<mi>\; </mi>2}|r\mathrm{.}\end{array}$

Lemma 6. Let $p>4$ be a prime, then

${\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4\mathrm{<mo>\; =\; </mo>2}p\\ i_1\mathrm{,\; <mi>\; </mi>}{i}_2\mathrm{,\; <mi>\; </mi>}{i}_3\mathrm{,\; <mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{i_1{i}_2{i}_3{i}_4}\equiv -\frac{240}{5}p{B}_{p-5}\mathrm{<mi>\; </mi>}\left(mod\mathrm{<mi>\; </mi>}{p}^2\right)\mathrm{.}$

Proof. By Lemma 3, we have

${\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4=2p\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{i_1{i}_2{i}_3{i}_4}=\frac{24}{2p}{\displaystyle \underset{\begin{array}{c}1\le u_1<u_2<u_3\le 2p\\ u_1,\mathrm{<mi>\; </mi>}{u}_3,\mathrm{<mi>\; </mi>}{u}_2-u_1,\mathrm{<mi>\; </mi>}{u}_3-u_2\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1{u}_2{u}_3}.$ (5)

It is easy to see that

${\displaystyle \underset{\begin{array}{c}1\le u_1<u_2<u_3\le 2p\\ u_1,\mathrm{<mi>\; </mi>}{u}_3,\mathrm{<mi>\; </mi>}{u}_2-u_1,\mathrm{<mi>\; </mi>}{u}_3-u_2\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1{u}_2{u}_3}={\displaystyle \underset{\begin{array}{c}1\le u_1<u_2<u_3\le 2p\\ u_1,\mathrm{<mi>\; </mi>}{u}_2,\mathrm{<mi>\; </mi>}{u}_3,\mathrm{<mi>\; </mi>}{u}_2-u_1,\mathrm{<mi>\; </mi>}{u}_3-u_2\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1{u}_2{u}_3}+{\displaystyle \underset{\begin{array}{c}1\le u_1<p<u_3\le 2p\\ u_1,\mathrm{<mi>\; </mi>}{u}_3\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_{1}p{u}_3}.$

By Lemma 4, we have

${\displaystyle \underset{\begin{array}{c}1\le u_1<p<u_3\le 2p\\ u_1,\mathrm{<mi>\; </mi>}{u}_3\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_{1}p{u}_3}=\frac{1}{p}{\displaystyle \underset{1\le u_1<p}{\sum }}\frac{1}{u_1}{\displaystyle \underset{p<u_3<2p}{\sum }}\frac{1}{u_3}\equiv 0\left(\mathrm{mod}{p}^3\right).$ (6)

Hence

$\begin{array}{l}{\displaystyle \underset{\begin{array}{c}1\le u_1<u_2<u_3\le 2p\\ u_1,\mathrm{<mi>\; </mi>}{u}_3,\mathrm{<mi>\; </mi>}{u}_2-u_1,\mathrm{<mi>\; </mi>}{u}_3-u_2\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1{u}_2{u}_3}\equiv {\displaystyle \underset{\begin{array}{c}1\le u_1<u_2<u_3\le 2p\\ u_1,\mathrm{<mi>\; </mi>}{u}_2,\mathrm{<mi>\; </mi>}{u}_3,\mathrm{<mi>\; </mi>}{u}_2-u_1,\mathrm{<mi>\; </mi>}{u}_3-u_2\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1{u}_2{u}_3}\\ \equiv {\displaystyle \underset{\begin{array}{c}1\le u_1<u_2<u_3\le 2p\\ u_1,\mathrm{<mi>\; </mi>}{u}_2,\mathrm{<mi>\; </mi>}{u}_3\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1{u}_2{u}_3}-{\displaystyle \underset{\begin{array}{c}1\le u_1<u_1+p<u_3\le 2p\\ u_1,\mathrm{<mi>\; </mi>}{u}_3\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1(u_1+p)u_3}\\ -{\displaystyle \underset{\begin{array}{c}1\le u_1<u_2<u_2+p\le 2p\\ u_1,\mathrm{<mi>\; </mi>}{u}_2\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1{u}_2\left(u_2+p\right)}\left(\mathrm{mod}{p}^3\right).\end{array}$

Replace $u_3=u_2+p$, then

$\begin{array}{l}{\displaystyle \underset{\begin{array}{c}1\le u_1<u_1+p<u_3\le 2p\\ u_1,\mathrm{<mi>\; </mi>}{u}_3\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1\left(u_1+p\right)u_3}={\displaystyle \underset{\begin{array}{c}1\le u_1<u_1+p<u_2+p\le 2p\\ u_1,\mathrm{<mi>\; </mi>}{u}_2\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1\left(u_1+p\right)\left(u_2+p\right)}\\ \equiv {\displaystyle \underset{1\le u_1<u_2<p}{\sum }}\frac{1}{u_1^2{u}_2}\left(1-\frac{p}{u_1}+\frac{p^2}{u_1^2}\right)\left(1-\frac{p}{u_2}+\frac{p^2}{u_2^2}\right)\left(\mathrm{mod}{p}^3\right).\end{array}$

and

${\displaystyle \underset{\begin{array}{c}1\le u_1<u_2<u_2+p\le 2p\\ u_1,\mathrm{<mi>\; </mi>}{u}_2\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1{u}_2\left(u_2+p\right)}\equiv {\displaystyle \underset{1\le u_1<u_2<p}{\sum }}\frac{1}{u_1{u}_2^2}\left(1-\frac{p}{u_2}+\frac{p^2}{u_2^2}\right)\left(\mathrm{mod}{p}^3\right).$

Thus

$\begin{array}{l}{\displaystyle \underset{\begin{array}{c}1\le u_1<u_2<u_3\le 2p\\ u_1,\mathrm{<mi>\; </mi>}{u}_3,\mathrm{<mi>\; </mi>}{u}_2-u_1,\mathrm{<mi>\; </mi>}{u}_3-u_2\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1{u}_2{u}_3}\\ \equiv {\displaystyle \underset{\begin{array}{c}1\le u_1<u_2<u_3\le 2p\\ u_1,\mathrm{<mi>\; </mi>}{u}_2,\mathrm{<mi>\; </mi>}{u}_3\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1{u}_2{u}_3}-{\displaystyle \underset{1\le u_1<u_2<p}{\sum }}(\frac{1}{u_1^2{u}_2}+\frac{1}{u_1{u}_2^2}-p\left(\frac{1}{u_1^3{u}_2}+\frac{1}{u_1^2{u}_2^2}+\frac{1}{u_1{u}_2^3}\right)\\ +p^2\left(\frac{1}{u_1^4{u}_2}+\frac{1}{u_1^3{u}_2}+\frac{1}{u_1^2{u}_2^3}+\frac{1}{u_1^1{u}_2^4}\right)\\ \equiv \frac{1}{3!}{\displaystyle \underset{\begin{array}{c}1\le u_1,\mathrm{<mi>\; </mi>}{u}_2,\mathrm{<mi>\; </mi>}{u}_3\le 2p\\ u_i\ne u_j,\mathrm{<mi>\; </mi>}{u}_i\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1{u}_2{u}_3}-{\displaystyle \underset{1\le u_1,u_2<p}{\sum }}(\frac{1}{u_1^2{u}_2}-p\left(\frac{1}{u_1^3{u}_2}+\frac{1}{2}\frac{1}{u_1^2{u}_2^2}\right)\\ +p^2\left(\frac{1}{u_1^4{u}_2}+\frac{1}{u_1^3{u}_2}\right))\left(\mathrm{mod}{p}^3\right).\end{array}$ (7)

Using Lemma 5 in the first sum of the right hand in (7) and using Lemma 4 in the second sum, we have

$\begin{array}{c}{\displaystyle \underset{\begin{array}{c}1\le u_1<u_2<u_3\le 2p\\ u_1,\mathrm{<mi>\; </mi>}{u}_3,\mathrm{<mi>\; </mi>}{u}_2-u_1,\mathrm{<mi>\; </mi>}{u}_3-u_2\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1{u}_2{u}_3}\equiv \frac{1}{3!}{\left(-1\right)}^3\left(3-1\right)!\frac{24}{5}{B}_{p-5}{p}^2-{\left(-1\right)}^2\frac{12}{10}{B}_{p-5}{p}^2\\ +\frac{3p}{2}\left(-\frac{4}{5}{B}_{p-5}p\right)-p^2\frac{30}{14}{B}_{p-7}{p}^2\\ \equiv -\frac{20}{5}{p}^2{B}_{p-5}\left(mod{p}^3\right)\mathrm{.}\end{array}$ (8)

Combining (5) with (8), we complete the proof of Lemma 6. ¨

Lemma 7. Let $p>4$ be a prime and $r>1$ a positive integer, then

${\displaystyle \underset{\begin{array}{l}{i}_1+i_2+i_3+i_4=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{i_1{i}_2{i}_3{i}_4}\equiv -\frac{48}{5}{p}^r{B}_{p-5}\mathrm{<mi>\; </mi>}\left(\mathrm{mod}\mathrm{<mi>\; </mi>}{p}^{r+1}\right).$

Proof. The proof of Lemma 7 is similar to the proof method of (4) in [6]. ¨

Lemma 8 ( [7] ). Let $p>5$ be a prime and $r\mathrm{,\; <mi>\; </mi>}m$ positive integers, $\left(m,\mathrm{<mi>\; </mi>}p\right)=1$ , then

${\displaystyle \underset{\begin{array}{c}{i}_1+i_2+\cdots +i_5\mathrm{<mo>\; =\; </mo>}mp\\ i_1\mathrm{,\; <mi>\; </mi>}{i}_2\mathrm{,}\cdots \mathrm{,\; <mi>\; </mi>}{i}_5\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{i_1{i}_2{i}_3{i}_4{i}_5}\equiv (\begin{array}{l}-4\left(5m+m^3\right)B_{p-5}\mathrm{<mi>\; </mi>}\left(mod\mathrm{<mi>\; </mi>}p\right)\mathrm{,\; <mi>\; </mi>}\text{if}\mathrm{<mi>\; </mi>}r=\mathrm{1,}\\ -20m{p}^{r-1}{B}_{p-5}\mathrm{<mi>\; </mi>}\left(mod\mathrm{<mi>\; </mi>}{p}^r\right)\mathrm{,\; <mi>\; </mi>}\text{if}\mathrm{<mi>\; </mi>}r>1.\end{array}$

Lemma 9. Let $p>5$ be a prime and $r\mathrm{,\; <mi>\; </mi>}m$ positive integers, then

${\displaystyle \underset{\begin{array}{c}{i}_1+i_2+\cdots +i_5=m{p}^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}\cdots ,\mathrm{<mi>\; </mi>}{i}_5\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{i_1{i}_2{i}_3{i}_4{i}_5}=\frac{120}{m{p}^r}{\displaystyle \underset{\begin{array}{c}1\le u_1<u_2<u_3<u_4\le m{p}^r\\ u_1,\mathrm{<mi>\; </mi>}{u}_4,\mathrm{<mi>\; </mi>}{u}_2-u_1,\mathrm{<mi>\; </mi>}{u}_3-u_2,\mathrm{<mi>\; </mi>}{u}_4-u_3\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1{u}_2{u}_3{u}_4}.$

Proof. The proof of Lemma 9 is similar to the proof of Lemma 2. ¨

3. Proofs of the Theorems

Proof of Theorem 1. It is easy to see that

$\begin{array}{c}{\displaystyle \underset{\begin{array}{c}i+j+k=2p^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}=\frac{1}{2p^r}{\displaystyle \underset{\begin{array}{c}i+j+k=2p^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^i\left(i+j+k\right)}{ijk}\\ =\frac{1}{2p^r}{\displaystyle \underset{\begin{array}{c}i+j+k=2p^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\left(\frac{{\left(-1\right)}^i}{jk}+\frac{2{\left(-1\right)}^i}{ij}\right).\end{array}$ (9)

Let $l=j+k$,then $1\le j<l\le 2p^r$ and $j\mathrm{,\; <mi>\; </mi>}l\mathrm{,\; <mi>\; </mi>}l-j\in {\mathcal{P}}_p$, hence

${\displaystyle \underset{\begin{array}{c}i+j+k=2p^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{jk}={\displaystyle \underset{\begin{array}{c}1\le j<l\le 2p^r\\ j,\mathrm{<mi>\; </mi>}l,\mathrm{<mi>\; </mi>}l-j\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{l}\frac{{\left(-1\right)}^l\left(j+k\right)}{jk}={\displaystyle \underset{\begin{array}{c}1\le j<l\le 2p^r\\ j,\mathrm{<mi>\; </mi>}l,\mathrm{<mi>\; </mi>}l-j\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{2{\left(-1\right)}^l}{jl}.$ (10)

Let $l^{\prime }=i+j$, then $1\le i<l^{\prime }\le 2p^r$ and $i\mathrm{,\; <mi>\; </mi>}{l}^{\prime }\mathrm{,\; <mi>\; </mi>}{l}^{\prime }-i\in {\mathcal{P}}_p$, hence

${\displaystyle \underset{\begin{array}{c}i+j+k=2p^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{2{\left(-1\right)}^i}{ij}={\displaystyle \underset{\begin{array}{c}1\le i<l^{\prime }\le 2p^r\\ i,\mathrm{<mi>\; </mi>}{l}^{\prime },\mathrm{<mi>\; </mi>}{l}^{\prime }-i\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{l^{\prime }}\frac{2{\left(-1\right)}^i\left(i+j\right)}{ij}={\displaystyle \underset{\begin{array}{c}1\le i<l^{\prime }\le 2p^r\\ i,\mathrm{<mi>\; </mi>}{l}^{\prime },\mathrm{<mi>\; </mi>}{l}^{\prime }-i\in {\mathcal{P}}_p\end{array}}{\sum }}\left(\frac{2{\left(-1\right)}^i}{j{l}^{\prime }}+\frac{2{\left(-1\right)}^i}{i{l}^{\prime }}\right).$ (11)

Noting that $i=l^{\prime }-j$, ${\left(-1\right)}^{l^{\prime }-j}={\left(-1\right)}^{l^{\prime }+j}$ and we rename $l^{\prime }$ to $l$, then

${\displaystyle \underset{\begin{array}{c}1\le i<l^{\prime }\le 2p^r\\ i,\mathrm{<mi>\; </mi>}{l}^{\prime },\mathrm{<mi>\; </mi>}{l}^{\prime }-i\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{2{\left(-1\right)}^i}{j{l}^{\prime }}={\displaystyle \underset{\begin{array}{c}1\le j<l\le 2p^r\\ j,\mathrm{<mi>\; </mi>}l,\mathrm{<mi>\; </mi>}l-j\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{2{\left(-1\right)}^{j+l}}{jl}.$ (12)

Rename i to j and $l^{\prime }$ to $l$, then

${\displaystyle \underset{\begin{array}{c}1\le i<l^{\prime }\le 2p^r\\ i,\mathrm{<mi>\; </mi>}{l}^{\prime },\mathrm{<mi>\; </mi>}{l}^{\prime }-i\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{2{\left(-1\right)}^i}{i{l}^{\prime }}={\displaystyle \underset{\begin{array}{c}1\le j<l\le 2p^r\\ j,\mathrm{<mi>\; </mi>}l,\mathrm{<mi>\; </mi>}l-j\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{2{\left(-1\right)}^j}{jl}.$ (13)

Combining (9)-(13), we have

$\begin{array}{c}{\displaystyle \underset{\begin{array}{c}i+j+k=2p^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}=\frac{1}{p^r}{\displaystyle \underset{\begin{array}{c}1\le j<l\le 2p^r\\ j,\mathrm{<mi>\; </mi>}l,\mathrm{<mi>\; </mi>}l-j\in {\mathcal{P}}_p\end{array}}{\sum }}\left(\frac{{\left(-1\right)}^l}{jl}+\frac{{\left(-1\right)}^{j+l}}{jl}+\frac{{\left(-1\right)}^j}{jl}\right)\\ =\frac{1}{p^r}{\displaystyle \underset{\begin{array}{c}1\le j<l\le 2p^r\\ j,\mathrm{<mi>\; </mi>}l,\mathrm{<mi>\; </mi>}l-j\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{\left(1+{\left(-1\right)}^l\right)\left(1+{\left(-1\right)}^j\right)}{jl}-\frac{1}{p^r}{\displaystyle \underset{\begin{array}{c}1\le j<l\le 2p^r\\ j,\mathrm{<mi>\; </mi>}l,\mathrm{<mi>\; </mi>}l-j\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{jl}\\ =\frac{1}{p^r}{\displaystyle \underset{\begin{array}{c}1\le j<l\le 2p^r\\ j,\mathrm{<mi>\; </mi>}l,\mathrm{<mi>\; </mi>}l-j\in {\mathcal{P}}_p,j\text{even},\mathrm{<mi>\; </mi>}l\text{even}\end{array}}{\sum }}\frac{4}{jl}-\frac{1}{p^r}{\displaystyle \underset{\begin{array}{c}1\le j<l\le 2p^r\\ j,\mathrm{<mi>\; </mi>}l,\mathrm{<mi>\; </mi>}l-j\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{jl}.\end{array}$ (14)

Let $j=2j^{\prime },\mathrm{<mi>\; </mi>}l=2l^{\prime }$ in the first sum of (14) and noting that

${\displaystyle \underset{\begin{array}{c}1\le j^{\prime }<l^{\prime }\le p^r\\ j^{\prime },\mathrm{<mi>\; </mi>}{l}^{\prime },\mathrm{<mi>\; </mi>}{l}^{\prime }-j^{\prime }\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{j^{\prime }{l}^{\prime }}={\displaystyle \underset{\begin{array}{c}1\le j<l\le p^r\\ j,\mathrm{<mi>\; </mi>}l,\mathrm{<mi>\; </mi>}l-j\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{jl},$

(14) is equal to

${\displaystyle \underset{\begin{array}{c}i+j+k=2p^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}=\frac{1}{p^r}{\displaystyle \underset{\begin{array}{c}1\le j<l\le p^r\\ j,\mathrm{<mi>\; </mi>}l,\mathrm{<mi>\; </mi>}l-j\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{jl}-\frac{1}{p^r}{\displaystyle \underset{\begin{array}{c}1\le j<l\le 2p^r\\ j,\mathrm{<mi>\; </mi>}l,\mathrm{<mi>\; </mi>}l-j\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{jl}.$ (15)

By Lemma 1, Lemma 2 and (15), we obtain

${\displaystyle \underset{\begin{array}{c}i+j+k=2p^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}=\frac{1}{p^r}{\displaystyle \underset{\begin{array}{c}1\le j<l\le p^r\\ j,\mathrm{<mi>\; </mi>}l,\mathrm{<mi>\; </mi>}l-j\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{jl}-\frac{1}{p^r}{\displaystyle \underset{\begin{array}{c}1\le j<l\le 2p^r\\ j,\mathrm{<mi>\; </mi>}l,\mathrm{<mi>\; </mi>}l-j\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{jl}\equiv p^{r-1}{B}_{p-3}\left(\mathrm{mod}\mathrm{<mi>\; </mi>}{p}^r\right).$

This completes the proof of Theorem 1. ¨

Proof of Theorem 2. For every triple $\left(i\mathrm{,\; <mi>\; </mi>}j\mathrm{,\; <mi>\; </mi>}k\right)$ of positive integers which satisfies $i+j+k=2p^r,\mathrm{<mi>\; </mi>}i,j,k\in {\mathcal{P}}_p$, we take it to 3 cases.

Cases 1. Let $A\left(p^r\right)=\left\{\left(i\mathrm{,}j\mathrm{,}k\right)\mathrm{<mo>\; |\; </mo>\; <mi>\; </mi>1}\le i\mathrm{,\; <mi>\; </mi>}j\mathrm{,\; <mi>\; </mi>}k\le p^r-\mathrm{1\; <mi>\; </mi>}\text{and}\mathrm{<mi>\; </mi>}i\mathrm{,}j\mathrm{,}k\in {\mathcal{P}}_p\right\}$. $\left(i\mathrm{,\; <mi>\; </mi>}j\mathrm{,\; <mi>\; </mi>}k\right)\leftrightarrow \left(p^r-i\mathrm{,\; <mi>\; </mi>}{p}^r-j\mathrm{,\; <mi>\; </mi>}{p}^r-k\right)$ is a bijection between the solutions of $i+j+k=2p^r,\mathrm{<mi>\; </mi>}\left(i,j,k\right)\in A\left(p^r\right)$ and $i+j+k=p^r,\mathrm{<mi>\; </mi>}i,j,k\in {\mathcal{P}}_p$, we have

$\begin{array}{c}{\displaystyle \underset{\begin{array}{c}i+j+k=2p^r\\ \left(i,j,k\right)\in A\left(p^r\right)\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}\equiv {\displaystyle \underset{\begin{array}{c}i+j+k=p^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{p^r-i}}{\left(p^r-i\right)\left(p^r-j\right)\left(p^r-k\right)}\\ \equiv {\displaystyle \underset{\begin{array}{c}i+j+k=p^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}\left(\mathrm{mod}{p}^r\right).\end{array}$ (16)

Cases 2. Let $B\left(p^r\right)=\left\{\left(i\mathrm{,}j\mathrm{,}k\right)\mathrm{<mo>\; |\; </mo>\; <mi>\; </mi>}{p}^r+1\le i\le 2p^r-\mathrm{1,1}\le j\mathrm{,\; <mi>\; </mi>}k\le p^r-\mathrm{1\; <mi>\; </mi>}\text{and}\mathrm{<mi>\; </mi>}i\mathrm{,}j\mathrm{,}k\in {\mathcal{P}}_p\right\}$. $\left(i\mathrm{,\; <mi>\; </mi>}j\mathrm{,\; <mi>\; </mi>}k\right)\leftrightarrow \left(p^r+i\mathrm{,\; <mi>\; </mi>}j\mathrm{,\; <mi>\; </mi>}k\right)$ is a bijection between the solutions of $i+j+k=2p^r,\mathrm{<mi>\; </mi>}\left(i,j,k\right)\in B\left(p^r\right)$ and $i+j+k=p^r,\mathrm{<mi>\; </mi>}i,j,k\in {\mathcal{P}}_p$, we have

${\displaystyle \underset{\begin{array}{c}i+j+k=2p^r\\ \left(i,j,k\right)\in B\left(p^r\right)\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}\equiv {\displaystyle \underset{\begin{array}{c}i+j+k=p^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{p^r+i}}{\left(p^r+i\right)jk}\equiv -{\displaystyle \underset{\begin{array}{c}i+j+k=p^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}\left(\mathrm{mod}{p}^r\right).$ (17)

Cases 3. Let

$C\left(p^r\right)=\left\{\left(i\mathrm{,}j\mathrm{,}k\right)\mathrm{<mo>\; |\; </mo>\; <mi>\; </mi>}{p}^r+1\le j\le 2p^r-\mathrm{1,1}\le i\mathrm{,\; <mi>\; </mi>}k\le p^r-\mathrm{1\; <mi>\; </mi>}\text{and}\mathrm{<mi>\; </mi>}i\mathrm{,}j\mathrm{,}k\in {\mathcal{P}}_p\right\}$

and

$D\left(p^r\right)=\left\{\left(i\mathrm{,}j\mathrm{,}k\right)\mathrm{<mo>\; |\; </mo>\; <mi>\; </mi>}{p}^r+1\le k\le 2p^r-\mathrm{1,1}\le i\mathrm{,\; <mi>\; </mi>}j\le p^r-\mathrm{1\; <mi>\; </mi>}\text{and}\mathrm{<mi>\; </mi>}i\mathrm{,}j\mathrm{,}k\in {\mathcal{P}}_p\right\}\mathrm{.}$

$\left(i\mathrm{,\; <mi>\; </mi>}j\mathrm{,\; <mi>\; </mi>}k\right)\leftrightarrow \left(i\mathrm{,\; <mi>\; </mi>}{p}^r+j\mathrm{,\; <mi>\; </mi>}k\right)$ in the former and $\left(i\mathrm{,\; <mi>\; </mi>}j\mathrm{,\; <mi>\; </mi>}k\right)\leftrightarrow \left(i\mathrm{,\; <mi>\; </mi>}j\mathrm{,\; <mi>\; </mi>}{p}^r+k\right)$ in the later are the bijections between the solutions of $i+j+k=2p^r\mathrm{,\; <mi>\; </mi>}\left(i\mathrm{,}j\mathrm{,}k\right)\in C\left(p^r\right)\mathrm{<mi>\; </mi>}\text{or}\mathrm{<mi>\; </mi>}\left(i\mathrm{,}j\mathrm{,}k\right)\in D\left(p^r\right)$ and $i+j+k=p^r\mathrm{,\; <mi>\; </mi>}i\mathrm{,}j\mathrm{,}k\in {\mathcal{P}}_p$, we have

$\begin{array}{l}{\displaystyle \underset{\begin{array}{c}i+j+k=2p^r\\ \left(i,j,k\right)\in C\left(p^r\right)\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}+{\displaystyle \underset{\begin{array}{c}i+j+k=2p^r\\ \left(i,j,k\right)\in D\left(p^r\right)\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}\\ \equiv {\displaystyle \underset{\begin{array}{c}i+j+k=p^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{i\left(p^r+j\right)k}+{\displaystyle \underset{\begin{array}{c}i+j+k=p^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ij\left(p^r+k\right)}\\ \equiv 2{\displaystyle \underset{\begin{array}{c}i+j+k=p^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}\left(\mathrm{mod}{p}^r\right).\end{array}$ (18)

Combining (16)-(18), we have

$\begin{array}{c}{\displaystyle \underset{\begin{array}{c}i+j+k=2p^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}={\displaystyle \underset{\begin{array}{c}i+j+k=2p^r\\ \left(i,j,k\right)\in A\left(p^r\right)\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}+{\displaystyle \underset{\begin{array}{c}i+j+k=2p^r\\ \left(i,j,k\right)\in B\left(p^r\right)\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}\\ +{\displaystyle \underset{\begin{array}{c}i+j+k=2p^r\\ \left(i,j,k\right)\in C\left(p^r\right)\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}+{\displaystyle \underset{\begin{array}{c}i+j+k=2p^r\\ \left(i,j,k\right)\in D\left(p^r\right)\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}\\ \equiv 2{\displaystyle \underset{\begin{array}{c}i+j+k=p^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}\left(\mathrm{mod}{p}^r\right).\end{array}$

By Theorem 1, we complete the proof of Theorem 2. ¨

Proof of Theorem 3. By symmetry, it is easy to see that

$\begin{array}{c}{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1}}{i_1{i}_2{i}_3{i}_4}=\frac{1}{2p^r}{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1}\left(i_1+i_2+i_3+i_4\right)}{i_1{i}_2{i}_3{i}_4}\\ =\frac{1}{2p^r}{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\left(\frac{{\left(-1\right)}^{i_1}}{i_2{i}_3{i}_4}+3\frac{{\left(-1\right)}^{i_1}}{i_1{i}_3{i}_4}\right).\end{array}$ (19)

Let $u_3=i_2+i_3+i_4$ in the first sum of the last equation in (19), then $i_1=2p^r-u_3$, (19) equals to

$\begin{array}{l}=\frac{1}{2p^r}\left[{\displaystyle \underset{\begin{array}{c}{u}_3=i_2+i_3+i_4<2p^r\\ u_3,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{2p^r-u_3}\left(i_2+i_3+i_4\right)}{i_2{i}_3{i}_4{u}_3}+3{\displaystyle \underset{\begin{array}{c}{u}_3=i_1+i_3+i_4<2p^r\\ u_3,\mathrm{<mi>\; </mi>}{i}_1,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1}\left(i_1+i_3+i_4\right)}{i_1{i}_3{i}_4{u}_3}\right]\\ =\frac{1}{2p^r}\left[3{\displaystyle \underset{\begin{array}{c}{u}_3=i_2+i_3+i_4<2p^r\\ u_3,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{u_3}}{i_3{i}_4{u}_3}+3{\displaystyle \underset{\begin{array}{c}{u}_3=i_1+i_3+i_4<2p^r\\ u_3,\mathrm{<mi>\; </mi>}{i}_1,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1}}{i_3{i}_4{u}_3}+6{\displaystyle \underset{\begin{array}{c}{u}_3=i_1+i_3+i_4<2p^r\\ u_3,\mathrm{<mi>\; </mi>}{i}_1,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1}}{i_1{i}_3{u}_3}\right].\end{array}$ (20)

Let $u_2=i_3+i_4$ in the second sum of the last equation in (20), since $u_3=i_1+i_3+i_4$, then $i_1=u_3-u_2$, (20) equals to

Similarly, we have

${\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1+i_2}}{i_1{i}_2{i}_3{i}_4}=\frac{4}{p^r}{\displaystyle \underset{\begin{array}{c}0<u_1<u_2<u_3<2p^r\\ u_1,\mathrm{<mi>\; </mi>}{u}_3,\mathrm{<mi>\; </mi>}{u}_3-u_2,\mathrm{<mi>\; </mi>}{u}_2-u_1\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{u_2}+{\left(-1\right)}^{u_1+u_2+u_3}+{\left(-1\right)}^{u_1+u_3}}{u_1{u}_2{u}_3}.$

Hence

$\begin{array}{l}4{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1}}{i_1{i}_2{i}_3{i}_4}+3{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1+i_2}}{i_1{i}_2{i}_3{i}_4}\\ =\frac{12}{p^r}{\displaystyle \underset{\begin{array}{c}0<u_1<u_2<u_3<2p^r\\ u_1,\mathrm{<mi>\; </mi>}{u}_3,\mathrm{<mi>\; </mi>}{u}_3-u_2,\mathrm{<mi>\; </mi>}{u}_2-u_1\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{\left[1+{\left(-1\right)}^{u_1}\right]\left[1+{\left(-1\right)}^{u_2}\right]\left[1+{\left(-1\right)}^{u_3}\right]-1}{u_1{u}_2{u}_3}\\ =\frac{12}{p^r}{\displaystyle \underset{\begin{array}{c}0<u_1<u_2<u_3<p^r\\ u_1,\mathrm{<mi>\; </mi>}{u}_3,\mathrm{<mi>\; </mi>}{u}_3-u_2,\mathrm{<mi>\; </mi>}{u}_2-u_1\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1{u}_2{u}_3}-\frac{12}{p^r}{\displaystyle \underset{\begin{array}{c}0<u_1<u_2<u_3<2p^r\\ u_1,\mathrm{<mi>\; </mi>}{u}_3,\mathrm{<mi>\; </mi>}{u}_3-u_2,\mathrm{<mi>\; </mi>}{u}_2-u_1\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1{u}_2{u}_3}.\end{array}$

By Lemma 3, we have

$\begin{array}{l}4{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1}}{i_1{i}_2{i}_3{i}_4}+3{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1+i_2}}{i_1{i}_2{i}_3{i}_4}\\ =\frac{1}{2}{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{i_1{i}_2{i}_3{i}_4}-{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{i_1{i}_2{i}_3{i}_4}.\end{array}$

By (2) and Lemma 6, we have

$\begin{array}{l}4{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4=2p\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1}}{i_1{i}_2{i}_3{i}_4}+3{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4=2p\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1+i_2}}{i_1{i}_2{i}_3{i}_4}\\ \equiv -\frac{24}{5}p{B}_{p-5}+\frac{240}{5}p{B}_{p-5}\equiv \frac{216}{5}p{B}_{p-5}\left(\mathrm{mod}{p}^2\right).\end{array}$

By (4) and Lemma 7, if $r\ge 2$, then

$\begin{array}{l}4{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1}}{i_1{i}_2{i}_3{i}_4}+3{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1+i_2}}{i_1{i}_2{i}_3{i}_4}\\ \equiv -\frac{12}{5}{p}^r{B}_{p-5}+\frac{48}{5}{p}^r{B}_{p-5}\equiv \frac{36}{5}{p}^r{B}_{p-5}\left(\mathrm{mod}{p}^{r+1}\right).\end{array}$

This completes the proof of Theorem 3. ¨

Proof of Theorem 4. Similar to the proofs of Theorem 1 and Theorem 3, we have

$\begin{array}{c}{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4+i_5=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4,\mathrm{<mi>\; </mi>}{i}_5\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1}}{i_1{i}_2{i}_3{i}_4{i}_5}=\frac{12}{p^r}{\displaystyle \underset{\begin{array}{c}0<u_1<u_2<u_3<u_4<2p^r\\ u_1,\mathrm{<mi>\; </mi>}{u}_4,\mathrm{<mi>\; </mi>}{u}_2-u_1,\mathrm{<mi>\; </mi>}{u}_3-u_2,\mathrm{<mi>\; </mi>}{u}_4-u_3\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1{u}_2{u}_3{u}_4}[{\left(-1\right)}^{u_1}\\ +{\left(-1\right)}^{u_4}+{\left(-1\right)}^{u_1+u_2}+\left(-1\right){)}^{u_2+u_3}+{\left(-1\right)}^{u_3+u_4}]\end{array}$

and

$\begin{array}{l}{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4+i_5=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4,\mathrm{<mi>\; </mi>}{i}_5\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1+i_2}}{i_1{i}_2{i}_3{i}_4{i}_5}\\ =\frac{6}{p^r}{\displaystyle \underset{\begin{array}{c}0<u_1<u_2<u_3<u_4<2p^r\\ u_1,\mathrm{<mi>\; </mi>}{u}_4,\mathrm{<mi>\; </mi>}{u}_2-u_1,\mathrm{<mi>\; </mi>}{u}_3-u_2,\mathrm{<mi>\; </mi>}{u}_4-u_3\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1{u}_2{u}_3{u}_4}[{\left(-1\right)}^{u_2}+{\left(-1\right)}^{u_3}\\ +{\left(-1\right)}^{u_1+u_3}+{\left(-1\right)}^{u_1+u_4}+{\left(-1\right)}^{u_2+u_4}+{\left(-1\right)}^{u_1+u_2+u_3}\\ +{\left(-1\right)}^{u_1+u_2+u_4}+{\left(-1\right)}^{u_1+u_3+u_4}+{\left(-1\right)}^{u_2+u_3+u_4}+{\left(-1\right)}^{u_1+u_2+u_3+u_4}]\end{array}$

Hence

$\begin{array}{l}{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4+i_5=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4,\mathrm{<mi>\; </mi>}{i}_5\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1}}{i_1{i}_2{i}_3{i}_4{i}_5}+2{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4+i_5=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4,\mathrm{<mi>\; </mi>}{i}_5\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1+i_2}}{i_1{i}_2{i}_3{i}_4{i}_5}\\ =\frac{12}{p^r}{\displaystyle \underset{\begin{array}{c}0<u_1<u_2<u_3<u_4<2p^r\\ u_1,\mathrm{<mi>\; </mi>}{u}_4,\mathrm{<mi>\; </mi>}{u}_2-u_1,\mathrm{<mi>\; </mi>}{u}_3-u_2,\mathrm{<mi>\; </mi>}{u}_4-u_3\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{\left[1+{\left(-1\right)}^{u_1}\right]\left[1+{\left(-1\right)}^{u_2}\right]\left[1+{\left(-1\right)}^{u_3}\right]\left[1+{\left(-1\right)}^{u_4}\right]-1}{u_1{u}_2{u}_3{u}_4}\\ =\frac{12}{p^r}{\displaystyle \underset{\begin{array}{c}0<u_1<u_2<u_3<u_4<p^r\\ u_1,\mathrm{<mi>\; </mi>}{u}_4,\mathrm{<mi>\; </mi>}{u}_2-u_1,\mathrm{<mi>\; </mi>}{u}_3-u_2,\mathrm{<mi>\; </mi>}{u}_4-u_3\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1{u}_2{u}_3{u}_4}-\frac{12}{p^r}{\displaystyle \underset{\begin{array}{c}0<u_1<u_2<u_3<u_4<2p^r\\ u_1,\mathrm{<mi>\; </mi>}{u}_4,\mathrm{<mi>\; </mi>}{u}_2-u_1,\mathrm{<mi>\; </mi>}{u}_3-u_2,\mathrm{<mi>\; </mi>}{u}_4-u_3\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{u_1{u}_2{u}_3{u}_4}.\end{array}$

By Lemma 9, we have

$\begin{array}{l}{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4+i_5=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4,\mathrm{<mi>\; </mi>}{i}_5\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1}}{i_1{i}_2{i}_3{i}_4{i}_5}+2{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4+i_5=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4,\mathrm{<mi>\; </mi>}{i}_5\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1+i_2}}{i_1{i}_2{i}_3{i}_4{i}_5}\\ =\frac{1}{10}{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4+i_5=p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4,\mathrm{<mi>\; </mi>}{i}_5\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{i_1{i}_2{i}_3{i}_4{i}_5}-\frac{2}{10}{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4+i_5=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4,\mathrm{<mi>\; </mi>}{i}_5\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{1}{i_1{i}_2{i}_3{i}_4{i}_5}.\end{array}$

By (2) and Lemma 8 (1), we have

$\begin{array}{l}{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4+i_5=2p\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4,\mathrm{<mi>\; </mi>}{i}_5\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1}}{i_1{i}_2{i}_3{i}_4{i}_5}+2{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4+i_5=2p\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4,\mathrm{<mi>\; </mi>}{i}_5\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1+i_2}}{i_1{i}_2{i}_3{i}_4{i}_5}\\ \equiv -\frac{24}{10}{B}_{p-5}+\frac{144}{10}{B}_{p-5}\equiv 12B_{p-5}\left(\mathrm{mod}p\right).\end{array}$

By Lemma 8 (2), if $r\ge 2$, then

$\begin{array}{l}{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4+i_5=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4,\mathrm{<mi>\; </mi>}{i}_5\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1}}{i_1{i}_2{i}_3{i}_4{i}_5}+2{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4+i_5=2p^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4,\mathrm{<mi>\; </mi>}{i}_5\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1+i_2}}{i_1{i}_2{i}_3{i}_4{i}_5}\\ \equiv -2p^{r-1}{B}_{p-5}+8p^{r-1}{B}_{p-5}\equiv 6p^{r-1}{B}_{p-5}\left(\mathrm{mod}{p}^r\right).\end{array}$

This completes the proof of Theorem 4. ¨

4. Conclusions

Let p be an odd prime and $r\mathrm{,\; <mi>\; </mi>}m$ positive integers, $\left(m,p\right)=1$, using Lemma 1 and Lemma 2, similar to the proof of Theorem 1, we can prove that

${\displaystyle \underset{\begin{array}{c}i+j+k=2m{p}^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}\equiv m{p}^{r-1}{B}_{p-3}\mathrm{<mi>\; </mi>}\left(\mathrm{mod}\mathrm{<mi>\; </mi>}{p}^r\right).$

In particular, if $m=1$, it becomes Theorem 1.

Let p be odd prime and $r\mathrm{,\; <mi>\; </mi>}m$ positive integers, $\left(m,p\right)=1$, similar to the proof of Theorem 2, we can prove that

${\displaystyle \underset{\begin{array}{c}i+j+k=m{p}^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}\equiv \frac{1}{2}{\displaystyle \underset{\begin{array}{c}i+j+k=2m{p}^r\\ i,j,k\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^i}{ijk}\equiv \frac{m}{2}{p}^{r-1}{B}_{p-3}\mathrm{<mi>\; </mi>}\left(\mathrm{mod}\mathrm{<mi>\; </mi>}{p}^r\right).$

In particular, if $m=1$, it becomes Theorem 2.

Let $p>4$ be a prime and $r\mathrm{,\; <mi>\; </mi>}m$ positive integers, $\left(m,p\right)=1$, we can deduce the congruence $\left(mod\mathrm{<mi>\; </mi>}{p}^{r+1}\right)$ for

$4{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4=2m{p}^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1}}{i_1{i}_2{i}_3{i}_4}+3{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4=2m{p}^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1+i_2}}{i_1{i}_2{i}_3{i}_4}.$

Let $p>5$ be a prime and $r\mathrm{,\; <mi>\; </mi>}m$ positive integers, $\left(m,p\right)=1$, we can deduce the congruence $\left(mod\mathrm{<mi>\; </mi>}{p}^r\right)$ for

${\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4+i_5=2m{p}^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4,\mathrm{<mi>\; </mi>}{i}_5\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1}}{i_1{i}_2{i}_3{i}_4{i}_5}+2{\displaystyle \underset{\begin{array}{c}{i}_1+i_2+i_3+i_4+i_5=2m{p}^r\\ i_1,\mathrm{<mi>\; </mi>}{i}_2,\mathrm{<mi>\; </mi>}{i}_3,\mathrm{<mi>\; </mi>}{i}_4,\mathrm{<mi>\; </mi>}{i}_5\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\left(-1\right)}^{i_1+i_2}}{i_1{i}_2{i}_3{i}_4{i}_5}.$

Similarly, we can consider the congruence $\mathrm{<mo\; stretchy="false">\; (\; </mo>}mod\mathrm{<mi>\; </mi>}{p}^{r+1}\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ for

${\displaystyle \underset{\begin{array}{c}{i}_1+i_2+\cdots +i_n=p^r\\ i_1,i_2,\cdots ,i_n\in {\mathcal{P}}_p\end{array}}{\sum }}\frac{{\sigma }_1^{i_1}{\sigma }_2^{i_2}\cdots {\sigma }_n^{i_n}}{i_1{i}_2\cdots i_n}\mathrm{<mi>\; </mi>},$

where ${\sigma }_i\in \left\{1,-1\right\},\mathrm{<mi>\; </mi>}i=1,\mathrm{<mi>\; </mi>2},\mathrm{<mi>\; </mi>}\cdots ,\mathrm{<mi>\; </mi>}n$, but it seems much more complicated.

Founding

This work is supported by the Natural Science Foundation of Zhejiang Province, Project (No. LY18A010016) and the National Natural Science Foundation of China, Project (No. 12071421).

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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