Proof of a basic hypergeometric supercongruence modulo the fifth power of a cyclotomic polynomial

TL;DR

This paper proves a novel supercongruence involving basic hypergeometric series modulo the fifth power of a cyclotomic polynomial, extending previous results limited to the fourth power.

Contribution

It introduces the first proof of a hypergeometric supercongruence modulo the fifth power of a cyclotomic polynomial, using the $q$-Zeilberger algorithm.

Findings

Proved a supercongruence modulo the fifth power of a cyclotomic polynomial

Established related parametric supercongruences

Extended the known bounds of hypergeometric supercongruences

Abstract

By means of the $q$-Zeilberger algorithm, we prove a basic hypergeometric supercongruence modulo the fifth power of the cyclotomic polynomial $\Phi_n(q)$. This result appears to be quite unique, as in the existing literature so far no basic hypergeometric supercongruences modulo a power greater than the fourth of a cyclotomic polynomial have been proved. We also establish a couple of related results, including a parametric supercongruence.

Equations

$$\sum _{k=0}^{\infty }(-1{)}^k(4k+1)\frac{(\frac{1}{2}{)}_k^3}{k{!}^3}=\frac{2}{\pi },$$

$$\sum _{k=0}^{\frac{p-1}{2}}(-1{)}^k(4k+1)\frac{(\frac{1}{2}{)}_k^3}{k{!}^3}\equiv p(-1{)}^{\frac{p-1}{2}}(\mathrm{mod}{p}^3),$$

$$\sum _{k=0}^{\frac{n-1}{2}}[4k+1]\frac{(q;q^2{)}_k^4}{(q^2;q^2{)}_k^4}\equiv q^{\frac{1-n}{2}}[n]+\frac{(n^2-1)(1-q{)}^2}{24}{q}^{\frac{1-n}{2}}[n{]}^3(\mathrm{mod}[n]{\Phi }_n(q{)}^3),$$

$$\sum _{k=0}^{\frac{n+1}{2}}[4k-1]\frac{(q^{-1};q^2{)}_k^4}{(q^2;q^2{)}_k^4}{q}^{4k}$$

$$\sum _{k=0}^{n-1}[4k-1]\frac{(q^{-1};q^2{)}_k^4}{(q^2;q^2{)}_k^4}{q}^{4k}$$

$$\sum _{k=0}^{\frac{n+1}{2}}[4k-1]\frac{(a{q}^{-1};q^2{)}_k(q^{-1}/a;q^2{)}_k(q^{-1};q^2{)}_k^2}{(a{q}^2;q^2{)}_k(q^2/a;q^2{)}_k(q^2;q^2{)}_k^2}{q}^{4k}$$

$$\sum _{k=0}^{n-1}[4k-1]\frac{(a{q}^{-1};q^2{)}_k(q^{-1}/a;q^2{)}_k(q^{-1};q^2{)}_k^2}{(a{q}^2;q^2{)}_k(q^2/a;q^2{)}_k(q^2;q^2{)}_k^2}{q}^{4k}$$

$$\sum _{k=0}^{\frac{n+1}{2}}[4k-1]\frac{(q^{-2};q^4{)}_k^2}{(q^4;q^4{)}_k^2}{q}^{4k}$$

$$\sum _{k=0}^{n-1}[4k-1]\frac{(q^{-2};q^4{)}_k^2}{(q^4;q^4{)}_k^2}{q}^{4k}$$

$$(a;q{)}_n=(1-a)(1-aq)\cdots (1-a{q}^{n-1})$$

$$[n]=[n{]}_q=1+q+\cdots +q^{n-1}$$

$$\left[\begin{array}{c}n\\ k\end{array}\right]={\left[\begin{array}{c}n\\ k\end{array}\right]}_q:=\frac{(q;q{)}_n}{(q;q{)}_k(q;q{)}_{n-k}}$$

$$f(n,k)$$

$$g(n,k)$$

$$(2k-3)f(n,k-1)-(2k-4)f(n,k)=g(n+1,k)-g(n,k).$$

$$F(n,k)$$

$$G(n,k)$$

$$[2k-3]F(n,k-1)-[2k-4]F(n,k)=G(n+1,k)-G(n,k).$$

$$\frac{F(n,k-1)}{G(n,k)}$$

$$\frac{F(n,k)}{G(n,k)}$$

$$\frac{G(n+1,k)}{G(n,k)}$$

$$\frac{q^{2n-4k+6}(1-q^{4n-1})(1-q^{2k-3}{)}^3}{(1-q^{2n-2k+2})(1-q^{2n}{)}^3}+\frac{q^{4-2k}(1-q^{2k-4})(1-q^{4n-1})(1-q^{2n+2k-3})}{(1-q^{2n}{)}^3}$$

$$=\frac{q^{4-2k}(1-q^{2n-1}{)}^3(1-q^{2n+2k-3})}{(1-q^{2n}{)}^3(1-q^{2n-2k+2})}-1,$$

$$[2k-3]\sum _{n=0}^{\frac{m+1}{2}}F(n,k-1)-[2k-4]\sum _{n=0}^{\frac{m+1}{2}}F(n,k)$$

$$=G\left(\frac{m+3}{2},k\right).$$

$$G\left(\frac{m+3}{2},1\right)$$

$$=\frac{q^{m-3}[m{]}^4}{[m+1{]}^4(-q;q{)}_{(m-1)/2}^8}{\left[\begin{array}{c}m-1\\ (m-1)/2\end{array}\right]}^4,$$

$$G\left(\frac{m+3}{2},2\right)$$

$$=-\frac{q^{-2}[m{]}^4[m+2]}{[m+1{]}^3(-q;q{)}_{(m-1)/2}^8}{\left[\begin{array}{c}m-1\\ (m-1)/2\end{array}\right]}^4.$$

$$\sum _{n=0}^{\frac{m+1}{2}}F(n,0)$$

$$=\frac{1+q}{q}G\left(\frac{m+3}{2},2\right)-qG\left(\frac{m+3}{2},1\right)$$

$$=-\frac{(1+q)[m{]}^4[m+1][m+2]+q^{m+1}[m{]}^4}{q^3[m+1{]}^4(-q;q{)}_{(m-1)/2}^8}{\left[\begin{array}{c}m-1\\ (m-1)/2\end{array}\right]}^4,$$

$$\sum _{n=0}^{\frac{m+1}{2}}[4n-1]\frac{(q^{-1};q^2{)}_n^4}{(q^2;q^2{)}_n^4}{q}^{4n}=-\frac{(1+q)[m{]}^4[m+1][m+2]+q^{m+1}[m{]}^4}{q[m+1{]}^4(-q;q{)}_{(m-1)/2}^8}{\left[\begin{array}{c}m-1\\ (m-1)/2\end{array}\right]}^4.$$

$$\left[\begin{array}{c}m-1\\ (m-1)/2\end{array}\right]=\prod _{k=1}^{(m-1)/2}\frac{1-q^{m-k}}{1-q^k}\equiv \prod _{k=1}^{(m-1)/2}\frac{1-q^{-k}}{1-q^k}=(-1{)}^{(m-1)/2}{q}^{(1-m^2)/8}(\mathrm{mod}{\Phi }_m(q)),$$

$$\sum _{n=0}^{\frac{m+1}{2}}[4n-1]\frac{(q^{-1};q^2{)}_n^4}{(q^2;q^2{)}_n^4}{q}^{4n}\equiv -((1+q{)}^2+q)[m{]}^4(\mathrm{mod}[m{]}^4{\Phi }_m(q)).$$

$$[2k-3]\sum _{n=0}^{m-1}F(n,k-1)-[2k-4]\sum _{n=0}^{m-1}F(n,k)=G(m,k),$$

$$\sum _{n=0}^{m-1}[4n-1]\frac{(q^{-1};q^2{)}_n^4}{(q^2;q^2{)}_n^4}{q}^{4n}$$

Full text

The first author was partially supported by the National Natural Science Foundation of China (grant 11771175).

1. Introduction

In 1997, Van Hamme [27] conjectured that 13 Ramanujan-type series including

[TABLE]

admit nice $p$-adic analogues, such as

[TABLE]

where $(a)_{n}=a(a+1)\cdots(a+n-1)$ denotes the Pochhammer symbol and $p$ is an odd prime. Up to present, all of the 13 supercongruences have been confirmed. See [21, 24] for historic remarks on these supercongruences. Recently, $q$-analogues of congruences and supercongruences have caught the interests of many authors (see, for example, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 25, 26, 29]). In particular, the first author and Zudilin [16] devised a method, called ‘creative microscoping’, to prove quite a few $q$-supercongruences by introducing an additional parameter $a$. In [13], the authors of the present paper proved many additional $q$-supercongruences by the creative microscoping method. Supercongruences modulo a higher integer power of a prime, or, in the $q$-case, of a cyclotomic polynomial, are very special and usually difficult to prove. As far as we know, until now the result

[TABLE]

for an odd positive integer $n$, due to the first author and Wang [15], is the unique $q$-supercongruence modulo $[n]\Phi_{n}(q)^{3}$ in the literature that was completely proved. (Several similar conjectural $q$-supercongruences are stated in [13] and in [16].) The purpose of this paper is to establish an even higher $q$-congruence, namely modulo a fifth power of a cyclotomic polynomial. Specifically, we prove the following three theorems. (The first two together confirm a conjecture by the authors [13, Conjecture 5.4]).

Theorem 1.1**.**

Let $n>1$ be a positive odd integer. Then

[TABLE]

Theorem 1.2**.**

Let $n>1$ be a positive odd integer. Then

[TABLE]

The $a=-1$ case of Theorem 1.2 admits an even stronger $q$-congruence.

Theorem 1.3**.**

Let $n>1$ be a positive odd integer. Then

[TABLE]

In the above $q$-supercongruences and in what follows,

[TABLE]

is the $q$-shifted factorial,

[TABLE]

is the $q$-number,

[TABLE]

is the $q$-binomial coefficient, and $\Phi_{n}(q)$ is the $n$-th cyclotomic polynomial of $q$. Note that the congruences in Theorem 1.1 modulo $[n]\Phi_{n}(q)^{2}$ and the congruences in Theorem 1.2 modulo $[n](1-aq^{n})(a-q^{n})$ have already been proved by the authors in [13, eqs. (5.5) and (5.10)].

2. Proof of Theorem 1.1 by the Zeilberger algorithm

The Zeilberger algorithm (cf. [22]) can be used to find that the functions

[TABLE]

satisfy the relation

[TABLE]

Of course, given this relation, it is not difficult to verify by hand that it is satisfied by the above pair of doubly-indexed sequences $f(n,k)$ and $g(n,k)$.

Here we use the convention $1/(1)_{m}=0$ for all negative integers $m$. We now define the $q$-analogues of $f(n,k)$ and $g(n,k)$ as follows:

[TABLE]

where we have used the convention that $1/(q^{2};q^{2})_{m}=0$ for $m=-1,-2,\ldots.$ Then the functions $F(n,k)$ and $G(n,k)$ satisfy the relation

[TABLE]

Indeed, it is straightforward to obtain the following expressions:

[TABLE]

It is easy to verify the identity

[TABLE]

which is equivalent to (4). (Alternatively, we could have established (4) by only guessing $F(n,k)$ and invoking the $q$-Zeilberger algorithm [28].)

Let $m>1$ be an odd integer. Summing (4) over $n$ from [math] to $(m+1)/2$, we get

[TABLE]

We readily compute

[TABLE]

Combining (5) and (6), we have

[TABLE]

i.e.,

[TABLE]

By [4, Lemma 2.1] (or [3, Lemma 2.1]), we have $(-q;q)_{(m-1)/2}^{2}\equiv q^{(m^{2}-1)/8}$ $\pmod{\Phi_{m}(q)}$. Moreover, it is easy to see that

[TABLE]

and $[m]$ is relatively prime to $(-q;q)_{(m-1)/2}$. It follows from (7) that

[TABLE]

Concluding, the congruence (2a) holds.

Similarly, summing (4) over $n$ from [math] to $m-1$, we get

[TABLE]

and so

[TABLE]

It is easy to see that

[TABLE]

and $(-q;q)_{m-1}\equiv 1\pmod{\Phi_{m}(q)}$ (see, for example, [4]). The proof of (2b) then follows easily from (8).

3. Proof of Theorems 1.2 and 1.3

Proof of Theorem 1.2.

It is easy to see by induction on $N$ that

[TABLE]

For $N=(n+1)/2$ or $N=n-1$, we see that $(aq;q^{2})_{N}(q/a;q^{2})_{N}$ contains the factor $(1-aq^{n})(1-q^{n}/a)$. Moreover,

[TABLE]

is a polynomial in $q$. Since $[(n+1)/2]$ and $[n]$ are relatively prime, we conclude that $\begin{bmatrix}\begin{smallmatrix}n\\ (n-1)/2\end{smallmatrix}\end{bmatrix}$ is divisible by $[n]$. Therefore, $\begin{bmatrix}\begin{smallmatrix}n+1\\ (n+1)/2\end{smallmatrix}\end{bmatrix}=(1+q^{(n+1)/2})\begin{bmatrix}\begin{smallmatrix}n\\ (n-1)/2\end{smallmatrix}\end{bmatrix}$ is also divisible by $[n]$. It is also well known that $\begin{bmatrix}\begin{smallmatrix}2n-2\\ n-1\end{smallmatrix}\end{bmatrix}$ is divisible by $[n]$. Moreover, it is easy to see that $[n]$ is relatively prime to $1+q^{m}$ for any non-negative integer $m$. The proof then follows from (9) by taking $N=(n+1)/2$ and $N=n-1$. ∎

Proof of Theorem 1.3.

For $a=-1$, the identity (9) reduces to

[TABLE]

Note that, in the proof of Theorem 1.2, we have proved that $\begin{bmatrix}\begin{smallmatrix}2N\\ N\end{smallmatrix}\end{bmatrix}_{q^{2}}$ is divisible by $[n]_{q^{2}}$ for both $N=(n+1)/2$ and $N=n-1$. Moreover, $[n]_{q^{2}}$ is relatively prime to $(-q^{2};q^{2})_{m}$ for $m\geqslant 0$. Hence the right-hand side of (10) is congruent to [math] modulo $[n]_{q^{2}}^{2}$ for $N=(n+1)/2$ or $N=n-1$. To further determine the right-hand side of (10) modulo $[n]_{q^{2}}^{2}\Phi_{n}(q^{2})$, we need only to use the same congruences (with $q\mapsto q^{2}$) used in the proof of Theorem 1.1. ∎

4. Immediate consequences

Notice that for $n=p^{r}$ being an odd prime power, $\Phi_{p^{r}}(q)=[p]_{q^{p^{r-1}}}$ holds. This observation was used in [15] to extend (1) to a supercongruence modulo $[p^{r}][p]_{q^{p^{r-1}}}^{3}$. In the same vein we immediately deduce from Theorem 1.1 the following result:

Corollary 4.1**.**

Let $p$ be an odd prime and $r$ a positive integer. Then

[TABLE]

The $q\to 1$ limiting cases of these two identities yield the following supercongruences:

Corollary 4.2**.**

Let $p$ be an odd prime and $r$ a positive integer. Then

[TABLE]

Similarly, we deduce from Theorem 1.3 the following result:

Corollary 4.3**.**

Let $p$ be an odd prime and $r$ a positive integer. Then

[TABLE]

The $q\to 1$ limiting cases of these two identities yield the following supercongruences:

Corollary 4.4**.**

Let $p$ be an odd prime and $r$ a positive integer. Then

[TABLE]

The supercongruences in Corollaries 4.2 and 4.4 are remarkable since they are valid for arbitrarily high prime powers. Swisher [24] had empirically observed several similar but different hypergeometric supercongruences and stated them without proof.