A NEW $\boldsymbol {q}$-ANALOGUE OF VAN HAMME’S (A.2) SUPERCONGRUENCE

Abstract

We give a new q-analogue of the (A.2) supercongruence of Van Hamme. Our proof employs Andrews’ multiseries generalisation of Watson’s $_{8}\phi _{7}$ transformation, Andrews’ terminating q-analogue of Watson’s $_{3}F_{2}$ summation, a q-Watson-type summation due to Wei–Gong–Li and the creative microscoping method, developed by the author and Zudilin [‘A q-microscope for supercongruences’, Adv. Math. 346 (2019), 329–358]. As a conclusion, we confirm a weaker form of Conjecture 4.5 by the author [‘Some generalizations of a supercongruence of van Hamme’, Integral Transforms Spec. Funct. 28 (2017), 888–899].

1 Introduction

India’s great mathematician Ramanujan mentioned the formula

(1.1) $$ \begin{align} \sum_{k=0}^{\infty} (-1)^{k}(4k+1)\frac{(\tfrac{1}{2})_{k}^{5}}{k!^{5}}= \frac{2}{\Gamma(\tfrac{3}{4})^{4}} \end{align} $$

in his second letter to Hardy on February 27, 1913. Here $\Gamma (x)$ stands for the Gamma function and $(a)_{k}=a(a+1)\cdots (a+k-1)$ is the rising factorial. In 1997, Van Hamme [15] observed that thirteen Ramanujan-type formulae possess neat p-adic analogues. For instance, the formula (1.1) corresponds to the supercongruence

(1.2) $$ \begin{align} \sum_{k=0}^{(p-1)/2} (-1)^{k}(4k+1)\frac{(\tfrac{1}{2})_{k}^{5}}{k!^{5}} \equiv \begin{cases} -\dfrac{p}{\Gamma_{p}(\tfrac{3}{4})^{4}} \pmod{p^{3}} &\text{if }p\equiv 1\pmod 4\text{,}\\ 0\pmod{p^{3}} &\text{if }p\equiv 3\pmod 4\text{,} \end{cases} \end{align} $$

(tagged as (A.2) in [15]). Here and in what follows, p always denotes an odd prime and $\Gamma _{p}(x)$ is Morita’s p-adic Gamma function (see, for example, [12, Ch. 7]). The supercongruence (1.2) was first confirmed by McCarthy and Osburn [11]. Swisher [13] further proved that (1.2) is true modulo $p^{5}$ for $p\equiv 1\pmod {4}$ and $p>5$ . Liu [10] extended (1.2) for $p\equiv 3\pmod {4}$ to a congruence modulo $p^{4}$ . Recently, among other things, Wei [18] gave the following generalisation of the second case of (1.2):

$$ \begin{align*} \sum_{k=0}^{(p-1)/2} (-1)^{k}(4k+1)\frac{(\tfrac{1}{2})_{k}^{5}}{k!^{5}} \equiv p^{2}\frac{(\tfrac{3}{4})_{(p-1)/2}}{(\tfrac{5}{4})_{(p-1)/2}}\pmod{p^{5}} \quad\text{for }p\equiv 3\pmod 4\text{.} \end{align*} $$

During the past few years, there has been considerable interest in q-analogues of supercongruences. In particular, using the creative microscoping method introduced by the author and Zudilin [7], Wang and Yue [16], together with the author [5], gave a q-analogue of (1.2): modulo $[n]\Phi _{n}(q)^{2}$ ,

(1.3) $$ \begin{align} \sum_{k=0}^{M} (-1)^{k}[4k+1]\frac{(q;q^{2})_{k}^{4}(q^{2};q^{4})_{k}}{(q^{2};q^{2})_{k}^{4}(q^{4};q^{4})_{k}}q^{k} \equiv \begin{cases} \dfrac{(q^{2};q^{4})_{(n-1)/4}^{2}}{(q^{4};q^{4})_{(n-1)/4}^{2}}[n] &\text{if }n\equiv 1\pmod 4\text{,}\\ 0 &\text{if }n\equiv 3\pmod 4\text{,} \end{cases} \end{align} $$

where $M=(n-1)/2$ or $n-1$ . Wei [17, 18] further generalised (1.3) to the moduli $[n]\Phi _{n}(q)^{3}$ and $[n]\Phi _{n}(q)^{4}$ .

We now need to familiarise ourselves with the standard q-notation. The q-shifted factorial is defined by $(a;q)_{n}=(1-a)(1-aq)\cdots (1-aq^{n-1})$ for $n\geqslant 1$ and $(a;q)_{0}=1$ . For simplicity, we also use the abbreviated notation $(a_{1},a_{2},\ldots ,a_{m};q)_{n}=(a_{1};q)_{n} (a_{2};q)_{n}\cdots (a_{m};q)_{n}$ for $n\geqslant 0$ . The q-integer is $[n]=[n]_{q}=(1-q^{n})/(1-q^{n})$ . The nth cyclotomic polynomial $\Phi _{n}(q)$ is given by

$$ \begin{align*} \Phi_{n}(q)=\prod_{\substack{1\leqslant k\leqslant n\\ \gcd(k,n)=1}}(q-\zeta^{k}), \end{align*} $$

where $\zeta $ is a primitive nth root of unity.

Letting $n=p\equiv 1\pmod 4$ and taking $q\to 1$ in (1.3), we obtain

(1.4) $$ \begin{align} \sum_{k=0}^{(p-1)/2} (-1)^{k}(4k+1)\frac{(\tfrac{1}{2})_{k}^{5}}{k!^{5}} \equiv \frac{(\tfrac{1}{2})_{(p-1)/4}^{2}}{(1)_{(p-1)/4}^{2}}p={-1/2 \choose (p-1)/4}^{2} p \pmod{p^{3}}. \end{align} $$

From [14, Theorem 3], we know that

$$ \begin{align*}{-1/2 \choose (p-1)/4}\equiv \frac{\Gamma_{p}(\tfrac{1}{4})^{2}}{\Gamma_{p}(\tfrac{1}{2})}\pmod{p^{2}}. \end{align*} $$

Since $\Gamma _{p}(\tfrac 12)^{2}=-1$ for $p\equiv 1\pmod {4}$ , by the identity $\Gamma _{p}(\tfrac 14)^{4} \Gamma _{p}(\tfrac 34)^{4}=1$ , we see that the supercongruence (1.4) is just (1.2) for $p\equiv 1\pmod {4}$ . This implies that (1.3) for $M=(n-1)/2$ really is a q-analogue of the (A.2) supercongruence of Van Hamme.

Note that supercongruences may have different q-analogues. See [8] for such examples. In this note, we shall give the following new q-analogue of (1.2).

Theorem 1.1. Let $n>1$ be an odd integer. Then, modulo $[n]_{q^{2}}\Phi _{n}(q^{2})^{2}$ ,

(1.5) $$ \begin{align} &\sum_{k=0}^{M} (-1)^{k}[4k+1]_{q^{2}}[4k+1]^{2}\frac{(q^{2};q^{4})_{k}^{4}(q^{4};q^{8})_{k}} {(q^{4};q^{4})_{k}^{4}(q^{8};q^{8})_{k}}q^{-2k} \notag\\ &\quad\equiv \begin{cases} -\dfrac{2q(q^{4};q^{8})_{(n-1)/4}^{2}}{(1+q^{2})(q^{8};q^{8})_{(n-1)/4}^{2}}[n]_{q^{2}}&\text{if }n\equiv 1\pmod{4}\text{,}\\[10pt] \dfrac{(1+q)^{2}(q^{4},q^{12}; q^{8})_{(n-3)/4}} {(1+q^{2})(1+q^{4})(q^{8},q^{16}; q^{8})_{(n-3)/4}}[n]_{q^{2}}& \text{if }n\equiv 3\pmod{4}\text{,} \end{cases} \end{align} $$

where $M=(n-1)/2$ or $n-1$ .

For n prime, letting $q\to -1$ in Theorem 1.1, we get (1.2). However, for n prime and $q\to 1$ in Theorem 1.1, we arrive at

(1.6) $$ \begin{align} \sum_{k=0}^{(p-1)/2} (-1)^{k}(4k+1)^{3}\frac{(\tfrac{1}{2})_{k}^{5}}{k!^{5}} \equiv \begin{cases} \dfrac{p}{\Gamma_{p}(\tfrac{3}{4})^{4}} \pmod{p^{3}} &\text{if }p\equiv 1\pmod 4\text{,}\\[10pt] \dfrac{(\tfrac{1}{2})_{(p-3)/4}(\tfrac{3}{2})_{(p-3)/4}p}{(\frac{p-3}{4})!(\frac{p+1}{4})!}\pmod{p^{3}} &\text{if }p\equiv 3\pmod 4\text{.} \end{cases} \end{align} $$

Thus, Theorem 1.1 may be considered as a common q-analogue of (1.2) and (1.6).

Letting n be an odd prime power and $q\to 1$ in (1.3) and (1.5), we are led to the following results. If $p^{r}\equiv 1\pmod {4}$ , then

(1.7) $$ \begin{align} \sum_{k=0}^{(p^{r}-1)/d} (-1)^{k}(4k+1)\frac{(\tfrac{1}{2})_{k}^{5}}{k!^{5}} &\equiv {(p^{r}-1)/2\choose (p^{r}-1)/4}^{2}\frac{p^{r}}{2^{p^{r}-1}} \pmod{p^{r+2}}, \end{align} $$

(1.8) $$ \begin{align} \sum_{k=0}^{(p^{r}-1)/d} (-1)^{k}(4k+1)^{3}\frac{(\tfrac{1}{2})_{k}^{5}}{k!^{5}} &\equiv -{(p^{r}-1)/2\choose (p^{r}-1)/4}^{2}\frac{p^{r}}{2^{p^{r}-1}} \pmod{p^{r+2}}, \end{align} $$

where $d=1$ or $2$ . Since $4+1+(4k+1)^{3}=2(4k+1)(8k^{2}+4k+1)$ , combining (1.7) and (1.8), we obtain the following conclusion.

Corollary 1.2. If $p^{r}\equiv 1\pmod {4}$ , then

(1.9) $$ \begin{align} \sum_{k=0}^{(p^{r}-1)/d} (-1)^{k}(4k+1)(8k^{2}+4k+1) \frac{(\tfrac{1}{2})_{k}^{5}}{k!^{5}} &\equiv 0\pmod{p^{r+2}}, \end{align} $$

where $d=1$ or $2$ .

Note that the author [4, Conjecture 4.5] conjectured that (1.9) is true modulo $p^{3r}$ for $p\equiv 1\pmod {4}$ .

We shall prove Theorem 1.1 in the next section. In Section 3, we raise two related conjectures on supercongruences.

2 Proof of Theorem 1.1

We first give the following q-congruence. See [6, Lemma 3.1] for a short proof.

Lemma 2.1. Let n be a positive odd integer. Then, for $0\leqslant k\leqslant (n-1)/2$ ,

$$ \begin{align*} \frac{(aq;q^{2})_{(n-1)/2-k}}{(q^{2}/a;q^{2})_{(n-1)/2-k}} \equiv (-a)^{(n-1)/2-2k}\frac{(aq;q^{2})_{k}}{(q^{2}/a;q^{2})_{k}} q^{(n-1)^{2}/4+k} \pmod{\Phi_{n}(q)}. \end{align*} $$

We will use a powerful transformation of Andrews (see [1, Theorem 4]), which can be stated as follows:

(2.1) $$ \begin{align} &\sum_{k\geqslant 0}\frac{(a,q\sqrt{a},-q\sqrt{a},b_{1},c_{1},\ldots,b_{m},c_{m},q^{-N};q)_{k}} {(q,\!\sqrt{a},-\sqrt{a},aq/b_{1},aq/c_{1},\ldots,aq/b_{m},aq/c_{m},aq^{N+1};q)_{k}} \bigg(\frac{a^{m}q^{m+N}}{b_{1}c_{1}\cdots b_{m}c_{m}}\bigg)^{k} \notag\\ &\quad=\frac{(aq,aq/b_{m}c_{m};q)_{N}}{(aq/b_{m},aq/c_{m};q)_{N}} \sum_{j_{1},\ldots,j_{m-1}\geqslant 0} \frac{(aq/b_{1}c_{1};q)_{j_{1}}\cdots(aq/b_{m-1}c_{m-1};q)_{j_{m-1}}} {(q;q)_{j_{1}}\cdots(q;q)_{j_{m-1}}} \notag\\ &\qquad\times\frac{(b_{2},c_{2};q)_{j_{1}}\cdots(b_{m},c_{m};q)_{j_{1}+\cdots+j_{m-1}}} {(aq/b_{1},aq/c_{1};q)_{j_{1}} \cdots(aq/b_{m-1},aq/c_{m-1};q)_{j_{1}+\cdots+j_{m-1}}} \notag\\ &\qquad\times\frac{(q^{-N};q)_{j_{1}+\cdots+j_{m-1}}} {(b_{m}c_{m}q^{-N}/a;q)_{j_{1}+\cdots+j_{m-1}}} \frac{(aq)^{\,j_{m-2}+\cdots+(m-2)j_{1}} q^{j_{1}+\cdots+j_{m-1}}} {(b_{2}c_{2})^{\,j_{1}}\cdots(b_{m-1}c_{m-1})^{j_{1}+\cdots+j_{m-2}}}. \end{align} $$

It should be pointed out that Andrews’ transformation is a multiseries generalisation of Watson’s $_{8}\phi _{7}$ transformation:

$$ \begin{align*} & {}_{8}\phi_{7}\!\left[\begin{array}{@{}cccccccc@{}} a,& qa^{{1}/{2}},& -qa^{{1}/{2}}, & b, & c, & d, & e, & q^{-n} \\ & a^{{1}/{2}}, & -a^{{1}/{2}}, & aq/b, & aq/c, & aq/d, & aq/e, & aq^{n+1} \end{array};\, q,\, \frac{a^{2}q^{n+2}}{bcde} \right] \notag\\[5pt] &\quad =\frac{(aq, aq/de;q)_{n}} {(aq/d, aq/e;q)_{n}} \,{}_{4}\phi_{3}\!\left[\begin{array}{@{}c@{}} aq/bc,\ d,\ e,\ q^{-n} \\ aq/b,\, aq/c,\, deq^{-n}/a \end{array};\, q,\, q \right] \end{align*} $$

(see [3, Appendix (III.18)]), where the basic hypergeometric series $_{r+1}\phi _{r}$ is defined as

$$ \begin{align*}_{r+1}\phi_{r}\left[\begin{array}{@{}c@{}} a_{1},a_{2},\ldots,a_{r+1}\\ b_{1},b_{2},\ldots,b_{r} \end{array};\,q,\, z \right] =\sum_{k=0}^{\infty}\frac{(a_{1},a_{2},\ldots, a_{r+1};q)_{k} } {(q,b_{1},\ldots,b_{r};q)_{k}} z^{k}. \end{align*} $$

We shall also use Andrews’ terminating q-analogue of Watson’s $_{3}F_{2}$ summation (see [2] or [3, (II.17)]):

(2.2) $$ \begin{align} {}_{4}\phi_{3}\left[\begin{array}{@{}c@{}} q^{-n},\, a^{2}q^{n+1},\,c,\, -c \\ aq,\, -aq,\, c^{2}\end{array};\,q,q\right] =\begin{cases} 0&\text{if }\textit{n}\text{ is odd},\\ \dfrac{c^{n}(q, a^{2}q^{2}/c^{2}; q^{2})_{n/2}} {(a^{2}q^{2},\, c^{2}q; q^{2})_{n/2}}& \text{if }\textit{n}\text{ is even,} \end{cases} \end{align} $$

and the following q-Watson-type summation due to Wei et al. [19, Corollary 5]:

(2.3) $$ \begin{align} {}_{4}\phi_{3}\bigg[\begin{array}{@{}c@{}} q^{-n},\, a^{2}q^{n+1},\,c,\, -cq \\ aq,\, -aq,\, c^{2}q\end{array};\,q,q\bigg] =\begin{cases} \dfrac{c^{n}(q;q^{2})_{(n+1)/2}(a^{2}q^{2}/c^{2}; q^{2})_{(n-1)/2}} {(a^{2}q^{2};q^{2})_{(n-1)/2}(c^{2}q; q^{2})_{(n+1)/2}}&\text{if }\textit{n}\text{ is odd},\\[10pt] \dfrac{c^{n}(q, a^{2}q^{2}/c^{2}; q^{2})_{n/2}} {(a^{2}q^{2},\, c^{2}q; q^{2})_{n/2}}& \text{if }\textit{n}\text{ is even.} \end{cases} \end{align} $$

We first prove the following parametric version of Theorem 1.1.

Theorem 2.2. Let $n>1$ be an odd integer. Then, modulo $\Phi _{n}(q^{2})(1-aq^{2n})(a-q^{2n})$ ,

(2.4) $$ \begin{align} &\sum_{k=0}^{(n-1)/2} (-1)^{k}[4k+1]_{q^{2}}[4k+1]^{2}\frac{(aq^{2},q^{2}/a;q^{4})_{k} (q^{2};q^{4})_{k}^{2}(q^{4};q^{8})_{k}} {(aq^{4},q^{4}/a;q^{4})_{k}(q^{4};q^{4})_{k}^{2}(q^{8};q^{8})_{k}}q^{-2k} \notag\\ &\quad\equiv \begin{cases} \bigg(1-\dfrac{(1+q)(1-aq^{2})(1-q^{2}/a)}{(1-q)(1-q^{4})}\bigg) \dfrac{(q^{4};q^{8})_{(n-1)/4}^{2}}{(q^{8};q^{8})_{(n-1)/4}^{2}}[n]_{q^{2}}&\text{if }n\equiv 1\pmod{4}\text{,}\\[10pt] \dfrac{(1+q)(1-aq^{2})(1-q^{2}/a)(q^{4},q^{12}; q^{8})_{(n-3)/4}} {(1-q)(1-q^{8})(q^{8},q^{16}; q^{8})_{(n-3)/4}}[n]_{q^{2}}& \text{if }n\equiv 3\pmod{4}\text{.} \end{cases} \end{align} $$

Proof. For $a=q^{-2n}$ or $a=q^{2n}$ , the left-hand side of (2.4) may be written as

$$ \begin{align*} &\sum_{k=0}^{(n-1)/2} (-1)^{k}[4k+1]_{q^{2}}[4k+1]^{2}\frac{(q^{2-2n},q^{2+2n};q^{4})_{k} (q^{2};q^{4})_{k}^{2}(q^{4};q^{8})_{k}} {(q^{4-2n},q^{4+2n};q^{4})_{k}(q^{4};q^{4})_{k}^{2}(q^{8};q^{8})_{k}}q^{-2k}. \end{align*} $$

Letting $m=3$ , $q\mapsto q^{4}$ , $a=q^{2}$ , $b_{1}=c_{1}=q^{5}$ , $b_{2}=c_{2}=q^{2}$ , $b_{3}=-q^{2}$ , $c_{3}=q^{2+2n}$ and $N=(n-1)/2$ in (2.1), we see that the above summation is equal to

(2.5) $$ \begin{align} &\frac{(q^{6},-q^{2-2n};q^{4})_{(n-1)/2}}{(-q^{4},q^{4-2n};q^{4})_{(n-1)/2}} \notag\\ &\quad\times \sum_{j_{1},j_{2}\geqslant 0} \frac{(q^{-4};q^{4})_{j_{1}}(q^{2};q^{4})_{j_{2}}(q^{2},q^{2};q^{4})_{j_{1}}(-q^{2},q^{2+2n},q^{2-2n};q^{4})_{j_{1}+j_{2}}}{(q^{4};q^{4})_{j_{1}} (q^{4};q^{4})_{j_{2}}(q,q;q^{4})_{j_{1}} (q^{4},q^{4},-q^{4};q^{4})_{j_{1}+j_{2}}}q^{6j_{1}+4j_{2}} \notag\\ &=(-1)^{(n-1)/2}q^{1-n}[n]_{q^{2}} \sum_{j_{2}=0}^{(n-1)/2}\frac{(q^{2},-q^{2},q^{2+2n},q^{2-2n};q^{4})_{j_{2}}}{(q^{4},q^{4},q^{4},-q^{4};q^{4})_{j_{2}}}q^{4j_{2}} \notag\\ &\quad+(-1)^{(n+1)/2}q^{3-n}[n]_{q^{2}} (1+q)^{2}\sum_{j_{2}=0}^{(n-3)/2}\frac{(q^{2};q^{4})_{j_{2}}(-q^{2},q^{2+2n},q^{2-2n};q^{4})_{j_{2}+1}}{(q^{4};q^{4}){j_{2}}(q^{4},q^{4},-q^{4};q^{4})_{j_{2}+1}}q^{4j_{2}}, \end{align} $$

where we have used the fact that $(q^{-4};q^{4})_{j_{1}}=0$ for $j_{1}>1$ .

Taking $q\mapsto q^{4}$ , $a=1$ , $c=q^{2}$ and $n\mapsto (n-1)/2$ in (2.2), we have

$$ \begin{align*} \sum_{j_{2}=0}^{(n-1)/2}\frac{(q^{2},-q^{2},q^{2+2n},q^{2-2n};q^{4})_{j_{2}}}{(q^{4},q^{4},q^{4},-q^{4};q^{4})_{j_{2}}}q^{4j_{2}} =\begin{cases} q^{n-1}\dfrac{(q^{4}; q^{8})_{(n-1)/4}^{2}} {(q^{8}; q^{8})_{(n-1)/4}^{2}}&\text{if }n\equiv 1\pmod{4}\text{,}\\[5pt] 0& \text{if }n\equiv 3\pmod{4}\text{.} \end{cases} \end{align*} $$

Similarly, taking $q\mapsto q^{4}$ , $a=q^{4}$ , $c=q^{2}$ and $n\mapsto (n-3)/2$ in (2.3), we get

$$ \begin{align*} &\sum_{j_{2}=0}^{(n-3)/2}\frac{(q^{2};q^{4})_{j_{2}}(-q^{2},q^{2+2n},q^{2-2n};q^{4})_{j_{2}+1}}{(q^{4};q^{4}){j_{2}}(q^{4},q^{4},-q^{4};q^{4})_{j_{2}+1}}q^{4j_{2}} \\ &\quad=\frac{(1+q^{2})(1-q^{2+2n})(1-q^{2-2n})}{(1-q^{4})^{2}(1+q^{4})} \sum_{j_{2}=0}^{(n-3)/2}\frac{(q^{2},-q^{6},q^{6+2n},q^{6-2n};q^{4})_{j_{2}}}{(q^{4},q^{8},q^{8},-q^{8};q^{4})_{j_{2}}}q^{4j_{2}}\\ &\quad=\begin{cases} q^{n-3}\dfrac{(1+q^{2})(1-q^{2+2n})(1-q^{2-2n})(q^{4}; q^{8})_{(n-1)/4}^{2}} {(1-q^{4})^{2}(q^{8}; q^{8})_{(n-1)/4}^{2}}&\text{if }n\equiv 1\pmod{4}\text{,}\\[10pt] q^{n-3}\dfrac{(1+q^{2})(1-q^{2+2n})(1-q^{2-2n})(q^{4},q^{12}; q^{8})_{(n-3)/4}} {(1-q^{4})^{2}(1+q^{4})(q^{8},q^{16}; q^{8})_{(n-3)/4}}& \text{if }n\equiv 3\pmod{4}\text{.} \end{cases} \end{align*} $$

Substituting the above two identities into (2.5), we obtain

$$ \begin{align*} &\sum_{k=0}^{(n-1)/2} (-1)^{k}[4k+1]_{q^{2}}[4k+1]^{2}\frac{(q^{2-2n},q^{2+2n};q^{4})_{k} (q^{2};q^{4})_{k}^{2}(q^{4};q^{8})_{k}} {(q^{4-2n},q^{4+2n};q^{4})_{k}(q^{4};q^{4})_{k}^{2}(q^{8};q^{8})_{k}}q^{-2k}\\ &\quad= \begin{cases} \bigg(1-\dfrac{(1+q)(1-q^{2+2n})(1-q^{2-2n})}{(1-q)(1-q^{4})}\bigg) \dfrac{(q^{4};q^{8})_{(n-1)/4}^{2}}{(q^{8};q^{8})_{(n-1)/4}^{2}}[n]_{q^{2}}&\text{if }n\equiv 1\pmod{4}\text{,}\\[10pt] \dfrac{(1+q)(1-q^{2+2n})(1-q^{2-2n})(q^{4},q^{12}; q^{8})_{(n-3)/4}} {(1-q)(1-q^{8})(q^{8},q^{16}; q^{8})_{(n-3)/4}}[n]_{q^{2}}& \text{if }n\equiv 3\pmod{4}\text{.} \end{cases} \end{align*} $$

This proves that both sides of (2.4) are equal when $a=q^{\pm 2n}$ . Namely, the q-congruence (2.4) holds modulo $1-aq^{2n}$ or $a-q^{2n}$ .

Moreover, in view of Lemma 2.1, we can verify that the kth and $((n-1)/2-k)$ th summands cancel each other modulo $\Phi _{n}(q^{2})$ for any positive odd integer n. It follows that

(2.6) $$ \begin{align} \sum_{k=0}^{(n-1)/2} (-1)^{k}[4k+1]_{q^{2}}[4k+1]^{2}\frac{(aq^{2},q^{2}/a;q^{4})_{k} (q^{2};q^{4})_{k}^{2}(q^{4};q^{8})_{k}} {(aq^{4},q^{4}/a;q^{4})_{k}(q^{4};q^{4})_{k}^{2}(q^{8};q^{8})_{k}}q^{-2k} \equiv 0 \pmod{\Phi_{n}(q^{2})}. \end{align} $$

Noticing that $[n]_{q^{2}}\equiv 0\pmod {\Phi _{n}(q^{2})}$ for $n>1$ , we conclude that the q-congruence (2.4) also holds modulo $\Phi _{n}(q)$ .

Since $1-aq^{2n}$ , $a-q^{2n}$ and $\Phi _{n}(q^{2})$ are pairwise relatively prime polynomials in q, we complete the proof of the theorem.

Proof of Theorem 1.1.

It is easy to see that the denominators on both sides of (2.4) when $a=1$ are relatively prime to $\Phi _{n}(q^{2})$ . However, when $a=1$ , the polynomial $(1-aq^{2n})(a-q^{2n})$ contains the factor $\Phi _{n}(q^{2})^{2}$ . Therefore, the $a=1$ case of (2.4) implies that (1.5) is true modulo $\Phi _{n}(q^{2})^{3}$ for $M=(n-1)/2$ . Furthermore, since $(q^{2};q^{4})_{k}^{4}(q^{4};q^{8})_{k}/((q^{4};q^{4})_{k}^{4}(q^{8};q^{8})_{k})\equiv 0\pmod {\Phi _{n}(q^{2})^{5}}$ for $(n-1)/2<k\leqslant n-1$ , we see that (1.5) is also true modulo $\Phi _{n}(q^{2})^{3}$ for $M=n-1$ .

It remains to prove the following two q-congruences:

(2.7) $$ \begin{align} \sum_{k=0}^{(n-1)/2} (-1)^{k}[4k+1]_{q^{2}}[4k+1]^{2}\frac{(q^{2};q^{4})_{k}^{4}(q^{4};q^{8})_{k}}{(q^{4};q^{4})_{k}^{4}(q^{8};q^{8})_{k}}q^{-2k} \equiv 0\pmod{[n]_{q^{2}}}, \end{align} $$

(2.8) $$ \begin{align} \hspace{-10pt}\sum_{k=0}^{n-1} (-1)^{k}[4k+1]_{q^{2}}[4k+1]^{2}\frac{(q^{2};q^{4})_{k}^{4}(q^{4};q^{8})_{k}} {(q^{4};q^{4})_{k}^{4}(q^{8};q^{8})_{k}}q^{-2k} \equiv 0\pmod{[n]_{q^{2}}}. \end{align} $$

For $n>1$ , let $\zeta \ne 1$ be an nth root of unity, possibly not primitive. Suppose $\zeta $ is a primitive root of unity of odd degree d satisfying $d\mid n$ . Let $c_{q}(k)$ be the kth term on the left-hand side of the congruences (2.7) and (2.8). Then

$$ \begin{align*} c_{q}(k)=(-1)^{k}[4k+1]_{q^{2}}[4k+1]^{2}\frac{(q^{2};q^{4})_{k}^{4}(q^{4};q^{8})_{k}}{(q^{4};q^{4})_{k}^{4}(q^{8};q^{8})_{k}}q^{-2k}. \end{align*} $$

Observe that (2.6) is true for any odd $n>1$ . Thus, letting $a=1$ and $n=d$ in (2.6), we obtain

$$ \begin{align*} \sum_{k=0}^{(d-1)/2}c_{\zeta}(k)=\sum_{k=0}^{d-1}c_{\zeta}(k)=0 \quad\text{and}\quad\sum_{k=0}^{(d-1)/2}c_{-\zeta}(k)=\sum_{k=0}^{d-1}c_{-\zeta}(k)=0. \end{align*} $$

Noticing that

$$ \begin{align*} \frac{c_{\zeta}(\ell d+k)}{c_{\zeta}(\ell d)} =\lim_{q\to\zeta}\frac{c_{q}(\ell d+k)}{c_{q}(\ell d)} =c_{\zeta}(k), \end{align*} $$

we have

$$ \begin{align*} \sum_{k=0}^{n-1}c_{\zeta}(k)=\sum_{\ell=0}^{n/d-1}\sum_{k=0}^{d-1}c_{\zeta}(\ell d+k) =\sum_{\ell=0}^{n/d-1}c_{\zeta}(\ell d) \sum_{k=0}^{d-1}c_{\zeta}(k)=0, \end{align*} $$

and

$$ \begin{align*} \sum_{k=0}^{(n-1)/2}c_{\zeta}(k) =\sum_{\ell=0}^{(n/d-3)/2} c_{\zeta}(\ell d) \sum_{k=0}^{d-1}c_{\zeta}(k)+\sum_{k=0}^{(d-1)/2}c_{\zeta}((n-d)/2+k)=0. \end{align*} $$

This means that both the sums $\sum _{k=0}^{n-1}c_{q}(k)$ and $\sum _{k=0}^{(n-1)/2}c_{q}(k)$ are divisible by $\Phi _{d}(q)$ . Similarly, we can show that they are also divisible by $\Phi _{d}(-q)$ . Since d can be any divisor of n larger than $1$ , we deduce that each of them is congruent to $0$ modulo

$$ \begin{align*} \prod_{d\mid n,\, d>1}\Phi_{d}(q) \Phi_{d}(-q)=[n]_{q^{2}}, \end{align*} $$

thus establishing (2.7) and (2.8).

3 Two open problems

Swisher [13] proposed many interesting conjectures on generalisations of Van Hamme’s supercongruences (A.2)–(L.2). Recently, the author and Zudilin [9] have proved some conjectures of Swisher by establishing their q-analogues. Here we would like to propose a similar conjecture.

Conjecture 3.1. Let $p\equiv 1\pmod {4}$ and let $r,s\geqslant 1$ . Then

(3.1) $$ \begin{align} \sum_{k=0}^{(p^{r}-1)/d} (-1)^{k}(4k+1)^{2s+1}\frac{(\tfrac{1}{2})_{k}^{5}}{k!^{5}} \equiv -p\Gamma_{p}(\tfrac{1}{4})^{4}\, \sum_{k=0}^{(p^{r-1}-1)/d} (-1)^{k}(4k+1)^{2s+1}\frac{(\tfrac{1}{2})_{k}^{5}}{k!^{5}} \pmod{p^{3r-2}}, \end{align} $$

where $d=1$ or $2$ .

For $s=0$ , Swisher [13, (A.3)] and the author [5, Conjecture 4.1] conjectured that (3.1) holds modulo $p^{5r}$ for $p>5$ . From (1.8), we can easily see that (3.1) is true modulo $p^{r}$ for $s=1$ .

Finally, motivated by [4, Conjecture 4.5], we believe that the following generalisation of Corollary 1.2 for p of the form $4k+3$ should be true.

Conjecture 3.2. Let $p\equiv 3\pmod {4}$ and let $r\geqslant 2$ be even. Then

$$ \begin{align*} \sum_{k=0}^{(p^{r}-1)/d} (-1)^{k}(4k+1)(8k^{2}+4k+1) \frac{(\tfrac{1}{2})_{k}^{5}}{k!^{5}} &\equiv 0\pmod{p^{2r}}, \end{align*} $$

where $d=1$ or $2$ .