Super Congruences Involving Alternating Harmonic Sums
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 PP 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: 1. Introduction

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

$\underset{i+j+k=p}{\sum }\frac{1}{ijk}\equiv -2{B}_{p-3}\left(\mathrm{mod}p\right),$ (1)

which holds when $p=3$ evidently. Here, Bernoulli numbers ${B}_{k}$ are defined by the recursive relation:

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

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

$\underset{{l}_{1}+{l}_{2}+\cdots +{l}_{n}=p}{\sum }\frac{1}{{l}_{1}{l}_{2}\cdots {l}_{n}}\equiv \left(\begin{array}{l}-\left(n-1\right)!{B}_{p-n}\left(\mathrm{mod}p\right),\text{if}2\overline{)|}n,\\ -\frac{n\left(n!\right)}{2\left(n+1\right)}p{B}_{p-n-1}\left(\mathrm{mod}{p}^{2}\right),\text{if}2|n.\end{array}$ (2)

Later, Xia and Cai  generalized (1) to

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

where $p>5$ is a prime.

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

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

$\underset{\begin{array}{c}{i}_{1}+{i}_{2}+\cdots +{i}_{n}={p}^{r}\\ {i}_{1},{i}_{2},\cdots ,{i}_{n}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{i}_{1}{i}_{2}\cdots {i}_{n}}\equiv -\frac{n!}{n+1}{p}^{r}{B}_{p-n-1}\left(\mathrm{mod}{p}^{r+1}\right).$ (4)

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

$\underset{\begin{array}{c}{i}_{1}+{i}_{2}+\cdots +{i}_{5}={p}^{r}\\ {i}_{1},{i}_{2},\cdots ,{i}_{5}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{i}_{1}{i}_{2}\cdots {i}_{5}}\equiv -\frac{5!}{6}{p}^{r-1}{B}_{p-5}\left(\mathrm{mod}{p}^{r}\right).$

We consider the following alternating harmonic sums

$\underset{\begin{array}{c}{i}_{1}+{i}_{2}+\cdots +{i}_{n}={p}^{r}\\ {i}_{1},{i}_{2},\cdots ,{i}_{n}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\sigma }_{1}^{{i}_{1}}{\sigma }_{2}^{{i}_{2}}\cdots {\sigma }_{n}^{{i}_{n}}}{{i}_{1}{i}_{2}\cdots {i}_{n}},$

where ${\sigma }_{i}\in \left\{1,-1\right\},i=1,2,\cdots ,n$. Given n, we only need to consider the following alternating harmonic sums,

$\underset{\begin{array}{c}{i}_{1}+{i}_{2}+\cdots +{i}_{n}={p}^{r}\\ {i}_{1},{i}_{2},\cdots ,{i}_{n}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}}}{{i}_{1}{i}_{2}\cdots {i}_{n}},\underset{\begin{array}{c}{i}_{1}+{i}_{2}+\cdots +{i}_{n}={p}^{r}\\ {i}_{1},{i}_{2},\cdots ,{i}_{n}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}+{i}_{2}}}{{i}_{1}{i}_{2}\cdots {i}_{n}},\cdots ,\underset{\begin{array}{c}{i}_{1}+{i}_{2}+\cdots +{i}_{n}={p}^{r}\\ {i}_{1},{i}_{2},\cdots ,{i}_{n}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}+{i}_{2}+\cdots +{i}_{\left[\frac{n}{2}\right]}}}{{i}_{1}{i}_{2}\cdots {i}_{n}}$

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,

$\underset{\begin{array}{c}{i}_{1}+{i}_{2}+\cdots +{i}_{n}={p}^{r}\\ {i}_{1},{i}_{2},\cdots ,{i}_{n}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}}}{{i}_{1}{i}_{2}\cdots {i}_{n}},\underset{\begin{array}{c}{i}_{1}+{i}_{2}+\cdots +{i}_{n}={p}^{r}\\ {i}_{1},{i}_{2},\cdots ,{i}_{n}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}+{i}_{2}}}{{i}_{1}{i}_{2}\cdots {i}_{n}},\cdots ,\underset{\begin{array}{c}{i}_{1}+{i}_{2}+\cdots +{i}_{n}={p}^{r}\\ {i}_{1},{i}_{2},\cdots ,{i}_{n}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}+{i}_{2}+\cdots +{i}_{\left[\frac{n}{2}\right]}}}{{i}_{1}{i}_{2}\cdots {i}_{n}}.$

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

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

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

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

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

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

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

$\begin{array}{l}4\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}}+3\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}+{i}_{2}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}}\\ \equiv \left(\begin{array}{l}\frac{216}{5}p{B}_{p-5}\left(mod{p}^{2}\right),\text{if}r=1,\\ \frac{36}{5}{p}^{r}{B}_{p-5}\left(mod{p}^{r+1}\right),\text{if}r>1.\end{array}\end{array}$

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

$\begin{array}{l}\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}+{i}_{5}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4},{i}_{5}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}{i}_{5}}+2\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}+{i}_{5}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4},{i}_{5}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}+{i}_{2}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}{i}_{5}}\\ \equiv \left(\begin{array}{l}12{B}_{p-5}\left(modp\right),\text{if}r=1,\\ 6{p}^{r-1}{B}_{p-5}\left(mod{p}^{r}\right),\text{if}r>1.\end{array}\end{array}$

2. Preliminaries

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

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

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

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

$\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}}\underset{\begin{array}{c}1\le j

Proof. It is easy to see that

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

Let $l=j+k$, then $1\le j and $j,l,l-j\in {\mathcal{P}}_{p}$. By symmetry, we have

$\frac{3}{m{p}^{r}}\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}}\underset{\begin{array}{c}i+j

This completes the proof of Lemma 2. ¨

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

$\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=m{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}}=\frac{24}{m{p}^{r}}\underset{\begin{array}{c}1\le {u}_{1}<{u}_{2}<{u}_{3}\le m{p}^{r}\\ {u}_{1},{u}_{3},{u}_{2}-{u}_{1},{u}_{3}-{u}_{2}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}{u}_{2}{u}_{3}}.$

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

Lemma 4 (  ) Let $r,{\alpha }_{1},\cdots ,{\alpha }_{n}$ be positive integers, $r={\alpha }_{1}+\cdots +{\alpha }_{n}\le p-3$ , then

$\underset{\begin{array}{c}1\le {l}_{1},\cdots ,{l}_{n}\le p-1\\ {l}_{i}\ne {l}_{j},\forall i\ne j\end{array}}{\sum }\frac{1}{{l}_{1}^{{\alpha }_{1}}{l}_{2}^{{\alpha }_{2}}\cdots {l}_{n}^{{\alpha }_{n}}}\equiv \left(\begin{array}{l}{\left(-1\right)}^{n}\left(n-1\right)!\frac{r\left(r+1\right)}{2\left(r+2\right)}{B}_{p-r-2}{p}^{2}\left(mod{p}^{3}\right),\text{if}2\overline{)|}r,\\ {\left(-1\right)}^{n-1}\left(n-1\right)!\frac{r}{r+1}{B}_{p-r-1}p\left(mod{p}^{2}\right),\text{if}2|r.\end{array}$

Lemma 5 (  ). Let p be an odd prime, and ${\alpha }_{1},\cdots ,{\alpha }_{n}$ positive integers, where $r={\alpha }_{1}+\cdots +{\alpha }_{n}\le p-3$ , then

$\underset{\begin{array}{l}1\le {l}_{1},\cdots ,{l}_{n}\le 2p\\ {l}_{i}\ne {l}_{j},{l}_{i}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{l}_{1}^{{\alpha }_{1}}{l}_{2}^{{\alpha }_{2}}\cdots {l}_{n}^{{\alpha }_{n}}}\equiv \left(\begin{array}{l}{\left(-1\right)}^{n}\left(n-1\right)!\frac{2r\left(r+1\right)}{r+2}{B}_{p-r-2}{p}^{2}\left(mod{p}^{3}\right),\text{if}2\overline{)|}r,\\ {\left(-1\right)}^{n-1}\left(n-1\right)!\frac{2r}{r+1}{B}_{p-r-1}p\left(mod{p}^{2}\right),\text{if}2|r.\end{array}$

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

$\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=2p\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}}\equiv -\frac{240}{5}p{B}_{p-5}\left(mod{p}^{2}\right).$

Proof. By Lemma 3, we have

$\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=2p\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}}=\frac{24}{2p}\underset{\begin{array}{c}1\le {u}_{1}<{u}_{2}<{u}_{3}\le 2p\\ {u}_{1},{u}_{3},{u}_{2}-{u}_{1},{u}_{3}-{u}_{2}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}{u}_{2}{u}_{3}}.$ (5)

It is easy to see that

$\underset{\begin{array}{c}1\le {u}_{1}<{u}_{2}<{u}_{3}\le 2p\\ {u}_{1},{u}_{3},{u}_{2}-{u}_{1},{u}_{3}-{u}_{2}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}{u}_{2}{u}_{3}}=\underset{\begin{array}{c}1\le {u}_{1}<{u}_{2}<{u}_{3}\le 2p\\ {u}_{1},{u}_{2},{u}_{3},{u}_{2}-{u}_{1},{u}_{3}-{u}_{2}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}{u}_{2}{u}_{3}}+\underset{\begin{array}{c}1\le {u}_{1}

By Lemma 4, we have

$\underset{\begin{array}{c}1\le {u}_{1} (6)

Hence

$\begin{array}{l}\underset{\begin{array}{c}1\le {u}_{1}<{u}_{2}<{u}_{3}\le 2p\\ {u}_{1},{u}_{3},{u}_{2}-{u}_{1},{u}_{3}-{u}_{2}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}{u}_{2}{u}_{3}}\equiv \underset{\begin{array}{c}1\le {u}_{1}<{u}_{2}<{u}_{3}\le 2p\\ {u}_{1},{u}_{2},{u}_{3},{u}_{2}-{u}_{1},{u}_{3}-{u}_{2}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}{u}_{2}{u}_{3}}\\ \equiv \underset{\begin{array}{c}1\le {u}_{1}<{u}_{2}<{u}_{3}\le 2p\\ {u}_{1},{u}_{2},{u}_{3}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}{u}_{2}{u}_{3}}-\underset{\begin{array}{c}1\le {u}_{1}<{u}_{1}+p<{u}_{3}\le 2p\\ {u}_{1},{u}_{3}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}\left({u}_{1}+p\right){u}_{3}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }-\underset{\begin{array}{c}1\le {u}_{1}<{u}_{2}<{u}_{2}+p\le 2p\\ {u}_{1},{u}_{2}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}{u}_{2}\left({u}_{2}+p\right)}\left(\mathrm{mod}{p}^{3}\right).\end{array}$

Replace ${u}_{3}={u}_{2}+p$, then

$\begin{array}{l}\underset{\begin{array}{c}1\le {u}_{1}<{u}_{1}+p<{u}_{3}\le 2p\\ {u}_{1},{u}_{3}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}\left({u}_{1}+p\right){u}_{3}}=\underset{\begin{array}{c}1\le {u}_{1}<{u}_{1}+p<{u}_{2}+p\le 2p\\ {u}_{1},{u}_{2}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}\left({u}_{1}+p\right)\left({u}_{2}+p\right)}\\ \equiv \underset{1\le {u}_{1}<{u}_{2}

and

$\underset{\begin{array}{c}1\le {u}_{1}<{u}_{2}<{u}_{2}+p\le 2p\\ {u}_{1},{u}_{2}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}{u}_{2}\left({u}_{2}+p\right)}\equiv \underset{1\le {u}_{1}<{u}_{2}

Thus

$\begin{array}{l}\underset{\begin{array}{c}1\le {u}_{1}<{u}_{2}<{u}_{3}\le 2p\\ {u}_{1},{u}_{3},{u}_{2}-{u}_{1},{u}_{3}-{u}_{2}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}{u}_{2}{u}_{3}}\\ \equiv \underset{\begin{array}{c}1\le {u}_{1}<{u}_{2}<{u}_{3}\le 2p\\ {u}_{1},{u}_{2},{u}_{3}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}{u}_{2}{u}_{3}}-\underset{1\le {u}_{1}<{u}_{2} (7)

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

$\begin{array}{c}\underset{\begin{array}{c}1\le {u}_{1}<{u}_{2}<{u}_{3}\le 2p\\ {u}_{1},{u}_{3},{u}_{2}-{u}_{1},{u}_{3}-{u}_{2}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}{u}_{2}{u}_{3}}\equiv \frac{1}{3!}{\left(-1\right)}^{3}\left(3-1\right)!\frac{24}{5}{B}_{p-5}{p}^{2}-{\left(-1\right)}^{2}\frac{12}{10}{B}_{p-5}{p}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{3p}{2}\left(-\frac{4}{5}{B}_{p-5}p\right)-{p}^{2}\frac{30}{14}{B}_{p-7}{p}^{2}\\ \equiv -\frac{20}{5}{p}^{2}{B}_{p-5}\left(mod{p}^{3}\right).\end{array}$ (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

$\underset{\begin{array}{l}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}}\equiv -\frac{48}{5}{p}^{r}{B}_{p-5}\left(\mathrm{mod}{p}^{r+1}\right).$

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

Lemma 8 (  ). Let $p>5$ be a prime and $r,m$ positive integers, $\left(m,p\right)=1$ , then

$\underset{\begin{array}{c}{i}_{1}+{i}_{2}+\cdots +{i}_{5}=mp\\ {i}_{1},{i}_{2},\cdots ,{i}_{5}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}{i}_{5}}\equiv \left(\begin{array}{l}-4\left(5m+{m}^{3}\right){B}_{p-5}\left(modp\right),\text{if}r=1,\\ -20m{p}^{r-1}{B}_{p-5}\left(mod{p}^{r}\right),\text{if}r>1.\end{array}$

Lemma 9. Let $p>5$ be a prime and $r,m$ positive integers, then

$\underset{\begin{array}{c}{i}_{1}+{i}_{2}+\cdots +{i}_{5}=m{p}^{r}\\ {i}_{1},{i}_{2},\cdots ,{i}_{5}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}{i}_{5}}=\frac{120}{m{p}^{r}}\underset{\begin{array}{c}1\le {u}_{1}<{u}_{2}<{u}_{3}<{u}_{4}\le m{p}^{r}\\ {u}_{1},{u}_{4},{u}_{2}-{u}_{1},{u}_{3}-{u}_{2},{u}_{4}-{u}_{3}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}{u}_{2}{u}_{3}{u}_{4}}.$

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

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

Let $l=j+k$,then $1\le j and $j,l,l-j\in {\mathcal{P}}_{p}$, hence

$\underset{\begin{array}{c}i+j+k=2{p}^{r}\\ i,j,k\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{i}}{jk}=\underset{\begin{array}{c}1\le j (10)

Let ${l}^{\prime }=i+j$, then $1\le i<{l}^{\prime }\le 2{p}^{r}$ and $i,{l}^{\prime },{l}^{\prime }-i\in {\mathcal{P}}_{p}$, hence

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

$\underset{\begin{array}{c}1\le i<{l}^{\prime }\le 2{p}^{r}\\ i,{l}^{\prime },{l}^{\prime }-i\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{2{\left(-1\right)}^{i}}{j{l}^{\prime }}=\underset{\begin{array}{c}1\le j (12)

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

$\underset{\begin{array}{c}1\le i<{l}^{\prime }\le 2{p}^{r}\\ i,{l}^{\prime },{l}^{\prime }-i\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{2{\left(-1\right)}^{i}}{i{l}^{\prime }}=\underset{\begin{array}{c}1\le j (13)

Combining (9)-(13), we have

$\begin{array}{c}\underset{\begin{array}{c}i+j+k=2{p}^{r}\\ i,j,k\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{i}}{ijk}=\frac{1}{{p}^{r}}\underset{\begin{array}{c}1\le j (14)

Let $j=2{j}^{\prime },l=2{l}^{\prime }$ in the first sum of (14) and noting that

$\underset{\begin{array}{c}1\le {j}^{\prime }<{l}^{\prime }\le {p}^{r}\\ {j}^{\prime },{l}^{\prime },{l}^{\prime }-{j}^{\prime }\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{j}^{\prime }{l}^{\prime }}=\underset{\begin{array}{c}1\le j

(14) is equal to

$\underset{\begin{array}{c}i+j+k=2{p}^{r}\\ i,j,k\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{i}}{ijk}=\frac{1}{{p}^{r}}\underset{\begin{array}{c}1\le j (15)

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

$\underset{\begin{array}{c}i+j+k=2{p}^{r}\\ i,j,k\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{i}}{ijk}=\frac{1}{{p}^{r}}\underset{\begin{array}{c}1\le j

This completes the proof of Theorem 1. ¨

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

Cases 1. Let $A\left({p}^{r}\right)=\left\{\left(i,j,k\right)|1\le i,j,k\le {p}^{r}-1\text{and}i,j,k\in {\mathcal{P}}_{p}\right\}$. $\left(i,j,k\right)↔\left({p}^{r}-i,{p}^{r}-j,{p}^{r}-k\right)$ is a bijection between the solutions of $i+j+k=2{p}^{r},\left(i,j,k\right)\in A\left({p}^{r}\right)$ and $i+j+k={p}^{r},i,j,k\in {\mathcal{P}}_{p}$, we have

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

Cases 2. Let $B\left({p}^{r}\right)=\left\{\left(i,j,k\right)|{p}^{r}+1\le i\le 2{p}^{r}-1,1\le j,k\le {p}^{r}-1\text{and}i,j,k\in {\mathcal{P}}_{p}\right\}$. $\left(i,j,k\right)↔\left({p}^{r}+i,j,k\right)$ is a bijection between the solutions of $i+j+k=2{p}^{r},\left(i,j,k\right)\in B\left({p}^{r}\right)$ and $i+j+k={p}^{r},i,j,k\in {\mathcal{P}}_{p}$, we have

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

Cases 3. Let

$C\left({p}^{r}\right)=\left\{\left(i,j,k\right)|{p}^{r}+1\le j\le 2{p}^{r}-1,1\le i,k\le {p}^{r}-1\text{and}i,j,k\in {\mathcal{P}}_{p}\right\}$

and

$D\left({p}^{r}\right)=\left\{\left(i,j,k\right)|{p}^{r}+1\le k\le 2{p}^{r}-1,1\le i,j\le {p}^{r}-1\text{and}i,j,k\in {\mathcal{P}}_{p}\right\}.$

$\left(i,j,k\right)↔\left(i,{p}^{r}+j,k\right)$ in the former and $\left(i,j,k\right)↔\left(i,j,{p}^{r}+k\right)$ in the later are the bijections between the solutions of $i+j+k=2{p}^{r},\left(i,j,k\right)\in C\left({p}^{r}\right)\text{or}\left(i,j,k\right)\in D\left({p}^{r}\right)$ and $i+j+k={p}^{r},i,j,k\in {\mathcal{P}}_{p}$, we have

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

Combining (16)-(18), we have

$\begin{array}{c}\underset{\begin{array}{c}i+j+k=2{p}^{r}\\ i,j,k\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{i}}{ijk}=\underset{\begin{array}{c}i+j+k=2{p}^{r}\\ \left(i,j,k\right)\in A\left({p}^{r}\right)\end{array}}{\sum }\frac{{\left(-1\right)}^{i}}{ijk}+\underset{\begin{array}{c}i+j+k=2{p}^{r}\\ \left(i,j,k\right)\in B\left({p}^{r}\right)\end{array}}{\sum }\frac{{\left(-1\right)}^{i}}{ijk}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{\begin{array}{c}i+j+k=2{p}^{r}\\ \left(i,j,k\right)\in C\left({p}^{r}\right)\end{array}}{\sum }\frac{{\left(-1\right)}^{i}}{ijk}+\underset{\begin{array}{c}i+j+k=2{p}^{r}\\ \left(i,j,k\right)\in D\left({p}^{r}\right)\end{array}}{\sum }\frac{{\left(-1\right)}^{i}}{ijk}\\ \equiv 2\underset{\begin{array}{c}i+j+k={p}^{r}\\ i,j,k\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{i}}{ijk}\left(\mathrm{mod}{p}^{r}\right).\end{array}$

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

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

$\begin{array}{c}\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}}=\frac{1}{2{p}^{r}}\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}}\left({i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}\right)}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}}\\ =\frac{1}{2{p}^{r}}\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\left(\frac{{\left(-1\right)}^{{i}_{1}}}{{i}_{2}{i}_{3}{i}_{4}}+3\frac{{\left(-1\right)}^{{i}_{1}}}{{i}_{1}{i}_{3}{i}_{4}}\right).\end{array}$ (19)

Let ${u}_{3}={i}_{2}+{i}_{3}+{i}_{4}$ in the first sum of the last equation in (19), then ${i}_{1}=2{p}^{r}-{u}_{3}$, (19) equals to

$\begin{array}{l}=\frac{1}{2{p}^{r}}\left[\underset{\begin{array}{c}{u}_{3}={i}_{2}+{i}_{3}+{i}_{4}<2{p}^{r}\\ {u}_{3},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{2{p}^{r}-{u}_{3}}\left({i}_{2}+{i}_{3}+{i}_{4}\right)}{{i}_{2}{i}_{3}{i}_{4}{u}_{3}}+3\underset{\begin{array}{c}{u}_{3}={i}_{1}+{i}_{3}+{i}_{4}<2{p}^{r}\\ {u}_{3},{i}_{1},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}}\left({i}_{1}+{i}_{3}+{i}_{4}\right)}{{i}_{1}{i}_{3}{i}_{4}{u}_{3}}\right]\\ =\frac{1}{2{p}^{r}}\left[3\underset{\begin{array}{c}{u}_{3}={i}_{2}+{i}_{3}+{i}_{4}<2{p}^{r}\\ {u}_{3},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{u}_{3}}}{{i}_{3}{i}_{4}{u}_{3}}+3\underset{\begin{array}{c}{u}_{3}={i}_{1}+{i}_{3}+{i}_{4}<2{p}^{r}\\ {u}_{3},{i}_{1},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}}}{{i}_{3}{i}_{4}{u}_{3}}+6\underset{\begin{array}{c}{u}_{3}={i}_{1}+{i}_{3}+{i}_{4}<2{p}^{r}\\ {u}_{3},{i}_{1},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}}}{{i}_{1}{i}_{3}{u}_{3}}\right].\end{array}$ (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

$\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}+{i}_{2}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}}=\frac{4}{{p}^{r}}\underset{\begin{array}{c}0<{u}_{1}<{u}_{2}<{u}_{3}<2{p}^{r}\\ {u}_{1},{u}_{3},{u}_{3}-{u}_{2},{u}_{2}-{u}_{1}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{u}_{2}}+{\left(-1\right)}^{{u}_{1}+{u}_{2}+{u}_{3}}+{\left(-1\right)}^{{u}_{1}+{u}_{3}}}{{u}_{1}{u}_{2}{u}_{3}}.$

Hence

$\begin{array}{l}4\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}}+3\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}+{i}_{2}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}}\\ =\frac{12}{{p}^{r}}\underset{\begin{array}{c}0<{u}_{1}<{u}_{2}<{u}_{3}<2{p}^{r}\\ {u}_{1},{u}_{3},{u}_{3}-{u}_{2},{u}_{2}-{u}_{1}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{\left[1+{\left(-1\right)}^{{u}_{1}}\right]\left[1+{\left(-1\right)}^{{u}_{2}}\right]\left[1+{\left(-1\right)}^{{u}_{3}}\right]-1}{{u}_{1}{u}_{2}{u}_{3}}\\ =\frac{12}{{p}^{r}}\underset{\begin{array}{c}0<{u}_{1}<{u}_{2}<{u}_{3}<{p}^{r}\\ {u}_{1},{u}_{3},{u}_{3}-{u}_{2},{u}_{2}-{u}_{1}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}{u}_{2}{u}_{3}}-\frac{12}{{p}^{r}}\underset{\begin{array}{c}0<{u}_{1}<{u}_{2}<{u}_{3}<2{p}^{r}\\ {u}_{1},{u}_{3},{u}_{3}-{u}_{2},{u}_{2}-{u}_{1}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}{u}_{2}{u}_{3}}.\end{array}$

By Lemma 3, we have

$\begin{array}{l}4\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}}+3\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}+{i}_{2}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}}\\ =\frac{1}{2}\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}}-\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}}.\end{array}$

By (2) and Lemma 6, we have

$\begin{array}{l}4\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=2p\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}}+3\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=2p\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}+{i}_{2}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}}\\ \equiv -\frac{24}{5}p{B}_{p-5}+\frac{240}{5}p{B}_{p-5}\equiv \frac{216}{5}p{B}_{p-5}\left(\mathrm{mod}{p}^{2}\right).\end{array}$

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

$\begin{array}{l}4\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}}+3\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}+{i}_{2}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}}\\ \equiv -\frac{12}{5}{p}^{r}{B}_{p-5}+\frac{48}{5}{p}^{r}{B}_{p-5}\equiv \frac{36}{5}{p}^{r}{B}_{p-5}\left(\mathrm{mod}{p}^{r+1}\right).\end{array}$

This completes the proof of Theorem 3. ¨

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

$\begin{array}{c}\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}+{i}_{5}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4},{i}_{5}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}{i}_{5}}=\frac{12}{{p}^{r}}\underset{\begin{array}{c}0<{u}_{1}<{u}_{2}<{u}_{3}<{u}_{4}<2{p}^{r}\\ {u}_{1},{u}_{4},{u}_{2}-{u}_{1},{u}_{3}-{u}_{2},{u}_{4}-{u}_{3}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}{u}_{2}{u}_{3}{u}_{4}}\left[{\left(-1\right)}^{{u}_{1}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\left(-1\right)}^{{u}_{4}}+{\left(-1\right)}^{{u}_{1}+{u}_{2}}+\left(-1\right){\right)}^{{u}_{2}+{u}_{3}}+{\left(-1\right)}^{{u}_{3}+{u}_{4}}\right]\end{array}$

and

$\begin{array}{l}\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}+{i}_{5}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4},{i}_{5}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}+{i}_{2}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}{i}_{5}}\\ =\frac{6}{{p}^{r}}\underset{\begin{array}{c}0<{u}_{1}<{u}_{2}<{u}_{3}<{u}_{4}<2{p}^{r}\\ {u}_{1},{u}_{4},{u}_{2}-{u}_{1},{u}_{3}-{u}_{2},{u}_{4}-{u}_{3}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}{u}_{2}{u}_{3}{u}_{4}}\left[{\left(-1\right)}^{{u}_{2}}+{\left(-1\right)}^{{u}_{3}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\left(-1\right)}^{{u}_{1}+{u}_{3}}+{\left(-1\right)}^{{u}_{1}+{u}_{4}}+{\left(-1\right)}^{{u}_{2}+{u}_{4}}+{\left(-1\right)}^{{u}_{1}+{u}_{2}+{u}_{3}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\left(-1\right)}^{{u}_{1}+{u}_{2}+{u}_{4}}+{\left(-1\right)}^{{u}_{1}+{u}_{3}+{u}_{4}}+{\left(-1\right)}^{{u}_{2}+{u}_{3}+{u}_{4}}+{\left(-1\right)}^{{u}_{1}+{u}_{2}+{u}_{3}+{u}_{4}}\right]\end{array}$

Hence

$\begin{array}{l}\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}+{i}_{5}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4},{i}_{5}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}{i}_{5}}+2\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}+{i}_{5}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4},{i}_{5}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}+{i}_{2}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}{i}_{5}}\\ =\frac{12}{{p}^{r}}\underset{\begin{array}{c}0<{u}_{1}<{u}_{2}<{u}_{3}<{u}_{4}<2{p}^{r}\\ {u}_{1},{u}_{4},{u}_{2}-{u}_{1},{u}_{3}-{u}_{2},{u}_{4}-{u}_{3}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{\left[1+{\left(-1\right)}^{{u}_{1}}\right]\left[1+{\left(-1\right)}^{{u}_{2}}\right]\left[1+{\left(-1\right)}^{{u}_{3}}\right]\left[1+{\left(-1\right)}^{{u}_{4}}\right]-1}{{u}_{1}{u}_{2}{u}_{3}{u}_{4}}\\ =\frac{12}{{p}^{r}}\underset{\begin{array}{c}0<{u}_{1}<{u}_{2}<{u}_{3}<{u}_{4}<{p}^{r}\\ {u}_{1},{u}_{4},{u}_{2}-{u}_{1},{u}_{3}-{u}_{2},{u}_{4}-{u}_{3}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}{u}_{2}{u}_{3}{u}_{4}}-\frac{12}{{p}^{r}}\underset{\begin{array}{c}0<{u}_{1}<{u}_{2}<{u}_{3}<{u}_{4}<2{p}^{r}\\ {u}_{1},{u}_{4},{u}_{2}-{u}_{1},{u}_{3}-{u}_{2},{u}_{4}-{u}_{3}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{u}_{1}{u}_{2}{u}_{3}{u}_{4}}.\end{array}$

By Lemma 9, we have

$\begin{array}{l}\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}+{i}_{5}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4},{i}_{5}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}{i}_{5}}+2\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}+{i}_{5}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4},{i}_{5}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}+{i}_{2}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}{i}_{5}}\\ =\frac{1}{10}\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}+{i}_{5}={p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4},{i}_{5}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}{i}_{5}}-\frac{2}{10}\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}+{i}_{5}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4},{i}_{5}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{1}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}{i}_{5}}.\end{array}$

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

$\begin{array}{l}\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}+{i}_{5}=2p\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4},{i}_{5}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}{i}_{5}}+2\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}+{i}_{5}=2p\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4},{i}_{5}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}+{i}_{2}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}{i}_{5}}\\ \equiv -\frac{24}{10}{B}_{p-5}+\frac{144}{10}{B}_{p-5}\equiv 12{B}_{p-5}\left(\mathrm{mod}p\right).\end{array}$

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

$\begin{array}{l}\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}+{i}_{5}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4},{i}_{5}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}{i}_{5}}+2\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}+{i}_{5}=2{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4},{i}_{5}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}+{i}_{2}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}{i}_{5}}\\ \equiv -2{p}^{r-1}{B}_{p-5}+8{p}^{r-1}{B}_{p-5}\equiv 6{p}^{r-1}{B}_{p-5}\left(\mathrm{mod}{p}^{r}\right).\end{array}$

This completes the proof of Theorem 4. ¨

4. Conclusions

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

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

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

Let p be odd prime and $r,m$ positive integers, $\left(m,p\right)=1$, similar to the proof of Theorem 2, we can prove that

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

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

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

$4\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=2m{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}}+3\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}=2m{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}+{i}_{2}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}}.$

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

$\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}+{i}_{5}=2m{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4},{i}_{5}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}{i}_{5}}+2\underset{\begin{array}{c}{i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}+{i}_{5}=2m{p}^{r}\\ {i}_{1},{i}_{2},{i}_{3},{i}_{4},{i}_{5}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\left(-1\right)}^{{i}_{1}+{i}_{2}}}{{i}_{1}{i}_{2}{i}_{3}{i}_{4}{i}_{5}}.$

Similarly, we can consider the congruence $\left(mod{p}^{r+1}\right)$ for

$\underset{\begin{array}{c}{i}_{1}+{i}_{2}+\cdots +{i}_{n}={p}^{r}\\ {i}_{1},{i}_{2},\cdots ,{i}_{n}\in {\mathcal{P}}_{p}\end{array}}{\sum }\frac{{\sigma }_{1}^{{i}_{1}}{\sigma }_{2}^{{i}_{2}}\cdots {\sigma }_{n}^{{i}_{n}}}{{i}_{1}{i}_{2}\cdots {i}_{n}},$

where ${\sigma }_{i}\in \left\{1,-1\right\},i=1,2,\cdots ,n$, 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).

Cite this paper: Shen, Z. and Cai, T. (2020) Super Congruences Involving Alternating Harmonic Sums. Advances in Pure Mathematics, 10, 611-622. doi: 10.4236/apm.2020.1010037.
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