On some conjectural congruences

TL;DR

This paper proves several conjectured congruences involving prime numbers and binomial coefficients, confirming recent hypotheses in number theory.

Contribution

It verifies conjectured congruences proposed by Guo and Schlosser, advancing understanding of prime-related binomial sum congruences.

Findings

Confirmed specific congruences modulo p^5 for primes p>3.

Validated conjectures involving binomial coefficients and prime moduli.

Enhanced the theoretical framework for congruences in number theory.

Abstract

In this paper, we confirm some congruences conjectured by V.J.W. Guo and M.J. Schlosser recently. For example, we show that for primes $p>3$, $$ \sum_{k=0}^{p-1}(2pk-2k-1)\frac{\left(\frac{-1}{p-1}\right)_k^{2p-2}}{(k!)^{2p-2}}\equiv0\pmod{p^5}. $$

Equations

$$\sum _{k=0}^{p-1}(2pk-2k-1)\frac{{\left(\frac{-1}{p-1}\right)}_k^{2p-2}}{(k!{)}^{2p-2}}\equiv 0(\mathrm{mod}{p}^5).$$

$$\underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^n=e,$$

$$\sum _{k=0}^{p-1}{\left(\genfrac{}{}{0ex}{}{\frac{-1}{p+1}}{k}\right)}^{p+1}\equiv 0(\mathrm{mod}{p}^5)$$

$$\sum _{k=0}^{p-1}{\left(\genfrac{}{}{0ex}{}{\frac{1}{p-1}}{k}\right)}^{p-1}\equiv \frac{2}{3}{p}^4{B}_{p-3}(\mathrm{mod}{p}^5),$$

$$\sum _{k=0}^{p-1}k\cdot \frac{{\left(\frac{1}{p+1}\right)}_k^{p+1}}{(k!{)}^{p+1}}\equiv 0(\mathrm{mod}{p}^3),$$

$$\frac{{\left(\frac{1}{p+1}\right)}_k}{k!}=(-1{)}^k(\genfrac{}{}{0ex}{}{\frac{-1}{p+1}}{k}).$$

$$\sum _{k=0}^{p-1}k\cdot \frac{{\left(\frac{1}{p+1}\right)}_k^{p+1}}{(k!{)}^{p+1}}\equiv \frac{p^3}{4}(\mathrm{mod}{p}^4).$$

$$\sum _{k=0}^{p-1}k\cdot \frac{{\left(\frac{1}{p+1}\right)}_k^{p+1}}{(k!{)}^{p+1}}\equiv \frac{p^3}{4}-\frac{p^4}{8}+\frac{3}{16}{p}^5-\frac{1}{36}p{B}_{p-3}(\mathrm{mod}{p}^6).$$

$$\sum _{k=0}^{p-1}{k}^r{\left(k+\frac{1}{p+1}\right)}^r\cdot \frac{{\left(\frac{1}{p+1}\right)}_k^{p+1}}{(k!{)}^{p+1}}\equiv 0(\mathrm{mod}{p}^4).$$

$$\sum _{k=0}^{p-1}(2pk-2k-1)\frac{{\left(\frac{-1}{p-1}\right)}_k^{2p-2}}{(k!{)}^{2p-2}}\equiv 0(\mathrm{mod}{p}^5).$$

$$\frac{{\left(\frac{1}{p+1}\right)}_k^{p+1}}{(k!{)}^{p+1}}-\prod _{j=1}^k\left(1-\frac{p}{j}\right)$$

$$\equiv$$

$$-\frac{p^5}{6}\sum _{1\le i<j\le k}\left(\frac{1}{i^{3}j}+\frac{1}{i{j}^3}\right)+\frac{p^5}{2}\sum _{1\le i_1<i_2\le k}\frac{1}{i_1{i}_2}\left(\sum _{j=1}^k\frac{1}{j^2}-\frac{1}{i_1^2}-\frac{1}{i_2^2}\right)(\mathrm{mod}{p}^6),$$

$$\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)(-1{)}^k{k}^m=0forallm=0,1,\dots ,n-1.$$

$$\sum _{k=0}^{p-1}k{H}_k^{(2)}\equiv -\frac{p^2}{2}{B}_{p-3}-\frac{p}{2}+\frac{1}{2}(\mathrm{mod}{p}^3),\sum _{k=0}^{p-1}k{H}_k^{(3)}\equiv \frac{1}{3}p{B}_{p-3}(\mathrm{mod}{p}^2),$$

$$\sum _{k=0}^{p-1}k{H}_k^{(4)}\equiv 0(\mathrm{mod}p).$$

$$\sum _{k=1}^{p-1}k{H}_k^{(m)}=$$

$$=$$

$$H_{p-1}\equiv -\frac{1}{3}{p}^2{B}_{p-3}(\mathrm{mod}{p}^3),H_{p-1}^{(2)}\equiv \frac{2}{3}p{B}_{p-3}(\mathrm{mod}{p}^3),H_{p-1}^{(3)}\equiv 0(\mathrm{mod}p).$$

$${\sigma }_1=\sum _{k=1}^{p-1}k\sum _{1\le i<j\le k}\left(\frac{1}{i{j}^2}+\frac{1}{i^{2}j}\right),{\sigma }_2=\sum _{k=1}^{p-1}k\sum _{1\le i<j\le k}\left(\frac{1}{i{j}^3}+\frac{1}{i^{3}j}\right),$$

$${\sigma }_3=\sum _{k=1}^{p-1}k\sum _{1\le i<j\le k}\frac{H_k^{(2)}}{ij}.$$

$${\sigma }_1\equiv -\frac{1}{2}p{B}_{p-3}+\frac{3}{4}(p-1)(\mathrm{mod}{p}^2),{\sigma }_2\equiv \frac{1}{2}{B}_{p-3}(\mathrm{mod}p),$$

$${\sigma }_3\equiv \frac{7}{8}+\frac{1}{2}{B}_{p-3}(\mathrm{mod}p).$$

$${\sigma }_1=$$

$$\equiv$$

$${\sigma }_3=$$

$$\equiv$$

$$=$$

$$\equiv$$

$$\sum _{k=0}^{p-1}k\frac{{\left(\frac{1}{p+1}\right)}_k^{p+1}}{(k!{)}^{p+1}}=\sum _{k=0}^{p-1}k\frac{{\left(\frac{1}{p+1}\right)}_k^{p+1}}{(k!{)}^{p+1}}-\sum _{k=1}^{p-1}(-1{)}^{k}k(\genfrac{}{}{0ex}{}{p-1}{k})$$

$$\equiv$$

$$+\frac{p(p^4-p^3)}{2}\left(-\frac{1}{2}p{B}_{p-3}+\frac{3}{4}(p-1)\right)-\frac{1}{12}{p}^5{B}_{p-3}+\frac{7}{16}{p}^5$$

$$\equiv$$

$$\sum _{k=0}^{p-1}{k}^r{\left(\frac{1}{p+1}+k\right)}^r(-1{)}^k(\genfrac{}{}{0ex}{}{p-1}{k})=0$$

$$\sum _{k=0}^{p-1}{k}^r{\left(k+\frac{1}{p+1}\right)}^r\cdot \frac{{\left(\frac{1}{p+1}\right)}_k^{p+1}}{(k!{)}^{p+1}}$$

Full text

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2010 Mathematics Subject Classification. Primary 11B65; Secondary 05A10, 11A07, 11B68.

Keywords. Congruences, binomial coeffients, Bernoulli numbers.

This work was supported by the National Natural Science Foundation of China (grant no. 11571162).

1. Introduction

In 2015, motivated by the well-known formula

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Z.-W. Sun [8] established some new congruences modulo prime powers. For example, he showed that for any prime $p>3$,

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and

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where $B_{0},B_{1},B_{2},\ldots$ are the well-known Bernoulli numbers (cf. [4]).

Recently, V.J.W. Guo and M.J. Schlosser [3] studied some interesting $q$-congruences for truncated basic hypergeometric series with the base being an even power of $q$. In their paper, as a corollary, they obtained that for any odd prime $p$,

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where $(x)_{k}=x(x+1)\cdots(x+k-1)$ denotes the Pochhammer symbol. Note that

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Thus this corollary is very similar to (1.1). To refine this corollary, Guo and Schlosser conjectured that under the same condition,

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We shall prove (1.3) by establishing the following extension.

Theorem 1.1**.**

Let $p>5$ be a prime. Then

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Remark**.**

We can directly verify that (1.3) holds for $p=5$.

Guo and Schlosser also posed some other conjectures which are similar to (1.3). Here we shall confirm two of these conjectures.

Theorem 1.2**.**

[3, Conjecture 2]** Let $r$ be a positive integer and let $p$ be a prime with $p>2r+1$. Then

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Theorem 1.3**.**

[3, Conjecture 4, (5.4)]** Let $p>3$ be a prime. Then

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Remark**.**

Via a similar discussion as the proof of Theorem 1.3, we can also prove that [3, Conjecture 3, (5.2)] holds modulo $p^{5}$. However, it is difficult to show this conjecture holds modulo $p^{7}$ by using the same method since the computation is very complicated.

In the next section, we shall prove Theorem 1.1 and Theorem 1.2. The proof of Theorem 1.3 will be given in the last section.

2. Proof of Theorems 1.1–1.2

To show theorems 1.1–1.2 we need the following lemmas.

Lemma 2.1**.**

[8]** For primes $p>5$ we have

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where $H_{k}^{(m)}=\sum_{i=1}^{k}1/i^{m}$ denotes the $k$-th harmonic number of order $m$.

Lemma 2.2**.**

[12, pp. 125–126]** For any positive integer $n$ we have

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Lemma 2.3**.**

Let $p>3$ be a prime. Then

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Proof.

For $m=2,3,4$, we have

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In view of [8, Lemma 2.1], we know that for primes $p>3$,

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Combining (2.2) with (2) we obtain Lemma 2.3. ∎

Lemma 2.4**.**

Let $p>3$ be a prime. Let

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Then we have

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Proof.

In light of (2.2) and noting that $\sum_{k=1}^{p-1}H_{k}/k^{2}\equiv-\sum_{k=1}^{p-1}H_{k}^{(2)}/k\equiv B_{p-3}\pmod{p}$ (cf. [11, (5.4)]) we obtain that

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The proof of $\sigma_{2}\pmod{p}$ is similar. Below we consider $\sigma_{3}\pmod{p}$. By Lemma 2.3, [8, (2.7)] and [10, Lemma 2.1] we arrive at

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Now the proof of Lemma 2.4 is complete. ∎

Proof of Theorem 1.1. In view of Lemmas 2.1–2.4, we have

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This proves Theorem 1.1.∎

Proof of Theorem 1.2. The case $p=5$ can be verified directly. Now we suppose that $p>5$. Since $p>2r+1$, it is easy to see that

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with the help of Lemma 2.2. Thus by Lemma 2.1 we arrive at

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Noting that

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we have

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This concludes the proof of Theorem 1.2.∎

3. Proof of Theorem 1.3

. We need the following preliminary results.

Lemma 3.1**.**

For any nonnegative integer $n$ we have the following identities.

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Proof.

Denote the left-hand side of (3.1) and (3.2) by $S_{1}(n)$ and $S_{2}(n)$ respectively. Via Zeilberger’s algorithm [5], we find that $S_{1}(n)$ satisfies

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and $S_{2}(n)$ satisfies

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Then we can easily show these two identities by induction on $n$. ∎

Lemma 3.2**.**

Let $p>3$ be a prime. Then

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Proof.

Noting (2.2) and [11, (5.4)] we have

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∎

Proof of Theorem 1.3. One can directly check that Theorem 1.3 holds for $p=5$. Below we assume that $p>5$. For each $k=\{1,\ldots,p-1\}$, we have

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and so

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Thus in view of Lemma 2.3–2.4, Lemma 3.2 and [8, Lemmas 2.1&2.2], we obtain that

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In 1990, Glaisher [1, 2] showed that

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This together with Lemma 3.1 gives that

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Combining (3.3) and (3) we immediately obtain Theorem 1.3.∎

Acknowledgments**.**

The author would like to thank Dr. Guo-Shuai Mao for his helpful comments.