A Common q-Analogue of Two Supercongruences

Abstract

We give a q-congruence whose specializations $q=-1$ and $q=1$ correspond to supercongruences (B.2) and (H.2) on Van Hamme’s list (in: p-Adic Functional Analysis (Nijmegen, 1996), Lecture Notes in Pure and Applied Mathematics, vol 192. Dekker, New York, pp 223–236, 1997):

$$\begin{aligned}&\sum _{k=0}^{(p-1)/2}(-1)^k(4k+1)A_k\equiv p(-1)^{(p-1)/2}\;({\text {mod}}p^3) \quad \text {and}\quad \\&\sum _{k=0}^{(p-1)/2}A_k\equiv a(p)\;({\text {mod}}p^2), \end{aligned}$$

where $p>2$ is prime,

$$\begin{aligned} A_k=\prod _{j=0}^{k-1}\biggl (\frac{1/2+j}{1+j}\biggr )^3=\frac{1}{2^{6k}}{\left( {\begin{array}{c}2k\\ k\end{array}}\right) }^3 \quad \text {for}\; k=0,1,2,\ldots , \end{aligned}$$

and a(p) is the pth coefficient of the modular form $q\prod _{j=1}^\infty (1-q^{4j})^6$ (of weight 3). We complement our result with a general common q-congruence for related hypergeometric sums.

Introduction

The formula of Bauer [1] from 1859,

$$\begin{aligned} \sum _{k=0}^\infty (-1)^k(4k+1)A_k=\frac{2}{\pi }, \quad \text {where}\; A_k=\frac{1}{2^{6k}}{\left( {\begin{array}{c}2k\\ k\end{array}}\right) }^3 \;\;\text {for}\; k=0,1,2,\ldots , \end{aligned}$$

(1.1)

is one of traditional targets for different methods of proofs of hypergeometric identities. Its special status is probably linked to the fact that it belongs to a family of series for $1/\pi$ of Ramanujan type, after Ramanujan [21] brought to life in 1914 a long list of similar looking equalities for the constant but with a faster convergence. Identity (1.1) is a particular instance of $_4F_3$ hypergeometric summation (known to Ramanujan) but there are several proofs of it, including the original one [1] of Bauer, that do not require any knowledge of hypergeometric functions. One notable—computer—proof of (1.1) was given in 1994 by Ekhad and Zeilberger [2] using the Wilf–Zeilberger (WZ) method of creative telescoping.

It was observed in 1997 by Van Hamme [28] that many Ramanujan’s and Ramanujan-like evaluations have nice p-adic analogues; for example, the congruence

$$\begin{aligned} \sum _{k=0}^{(p-1)/2}(-1)^k(4k+1)A_k\equiv p(-1)^{(p-1)/2}\;({\text {mod}}p^3) \end{aligned}$$

(1.2)

(tagged (B.2) on Van Hamme’s list) is valid for any prime $p>2$ and corresponds to the equality (1.1). The congruence (1.2) was first proved by Mortenson [19] using a $_6F_5$ hypergeometric transformation; it later received another proof by one of these authors [29] via the WZ method [in fact, using the very same ‘WZ certificate’ as in [2] for (1.1)]. Notice that (1.2) is an example of supercongruence meaning that it holds modulo a power of p greater than 1.

Another entry on Van Hamme’s 1997 list [28], tagged (H.2), is the congruence

$$\begin{aligned} \sum _{k=0}^{(p-1)/2}A_k \equiv {\left\{ \begin{array}{ll} -\Gamma _p(1/4)^4 \;({\text {mod}}p^2) &{}\quad \text { if } p\equiv 1\;({\text {mod}}4), \\ 0\;({\text {mod}}p^2) &{}\quad \text { if } p\equiv 3\;({\text {mod}}4), \end{array}\right. } \end{aligned}$$

(1.3)

again for any $p>2$ prime, and $\Gamma _p(x)$ is the p-adic Gamma function. Van Hamme not only observed but also proved (1.3) in [28], and it was later generalized by Sun [23, 24, Theorem 2.5], Guo and Zeng [12, Corollary 1.2], Long and Ramakrishna [17], Liu [15, 16, Theorem 1.5] in different ways. For example, Long and Ramakrishna [17, Theorem 3] gave the following generalization of (1.3):

$$\begin{aligned} \sum _{k=0}^{(p-1)/2} A_k \equiv {\left\{ \begin{array}{ll} -\Gamma _p(1/4)^4 \;({\text {mod}}p^3) &{}\quad \text { if } p\equiv 1\;({\text {mod}}4),\\ -\dfrac{p^2}{16}\,\Gamma _p(1/4)^4\;({\text {mod}}p^3) &{}\quad \text { if } p\equiv 3\;({\text {mod}}4). \end{array}\right. } \end{aligned}$$

(1.4)

Recently, these authors [14, Theorem 2] proved that, for any positive odd integer n, modulo $\Phi _n(q)^2$,

$$\begin{aligned} \sum _{k=0}^{(n-1)/2}\frac{(q;q^2)_k^2(q^2;q^4)_k}{(q^2;q^2)_k^2(q^4;q^4)_k}\,q^{2k} \equiv {\left\{ \begin{array}{ll} \dfrac{(q^2;q^4)_{(n-1)/4}^2}{(q^4;q^4)_{(n-1)/4}^2}\,q^{(n-1)/2} &{}\quad \text {if}\; n\equiv 1\;({\text {mod}}4), \\ 0 &{}\quad \text {if}\; n\equiv 3\;({\text {mod}}4). \end{array}\right. } \end{aligned}$$

(1.5)

Here and in what follows, $\Phi _n(q)$ denotes the nth cyclotomic polynomial; the q-shifted factorial is given by $(a;q)_0=1$ and $(a;q)_n=(1-a)(1-aq)\ldots (1-aq^{n-1})$ for $n\geqslant 1$ or $n=\infty$, while $[n]=[n]_q=1+q+\cdots +q^{n-1}$ stands for the q-integer. Van Hamme [27, Theorem 3] also proved that

$$\begin{aligned} \left( {\begin{array}{c}-1/2\\ (p-1)/4\end{array}}\right) \equiv -\frac{\Gamma _p(1/4)^2}{\Gamma _p(1/2)}\;({\text {mod}}p^2); \end{aligned}$$

in view of $\Gamma _p(1/2)^2=-1$ for $p\equiv 1\;({\text {mod}}4)$, by letting $q\rightarrow 1$ in (1.5) for $n=p$ we immediately obtain (1.3).

One feature of (1.3) (not highlighted in [28]) is its connection with the coefficients

$$\begin{aligned} a(p)={\left\{ \begin{array}{ll} 2(a^2-b^2) &{}\quad \text { if } p=a^2+b^2, a\, odd, \\ 0 &{}\quad \text { if } p\equiv 3\;({\text {mod}}4), \end{array}\right. } \end{aligned}$$

(1.6)

of CM modular form $q\prod _{j=1}^\infty (1-q^{4j})^6$ of weight 3, namely, the congruence

$$\begin{aligned} a(p)\equiv -\Gamma _p(1/4)^4\;({\text {mod}}p^2) \quad \text {for primes}\; p\equiv 1\;({\text {mod}}4). \end{aligned}$$

This served as a main motivation in [14] for not only establishing (1.5) but also speculating on possible q-deformation of modular forms.

For some other recent progress on q-analogues of supercongruences, the reader is referred to [4, 5, 7,8,9,10,11, 13, 20, 22, 26, 29]. In particular, the authors [13] introduced and executed a new method of creative microscoping to prove (and reprove) many q-analogues of classical supercongruences and also raised some problems on q-congruences. Using this method, the first author [6] gave a refinement of (1.5) modulo $\Phi _n(q)^3$ for $n\equiv 3\;({\text {mod}}4)$, in other words, a q-analogue of (1.4) for $p\equiv 3\;({\text {mod}}4)$.

A goal of this note is to present the following new q-analogue of Van Hamme’s supercongruence (1.3).

Theorem 1.1

Let n be a positive odd integer. Then

$$\begin{aligned}&\sum _{k=0}^{(n-1)/2}\frac{(1+q^{4k+1})\,(q^2;q^4)_k^3}{(1+q)\,(q^4;q^4)_k^3}\,q^{k} \nonumber \\&\quad \equiv \dfrac{[n]_{q^2}(q^3;q^4)_{(n-1)/2}}{(q^5;q^4)_{(n-1)/2}}\,q^{(1-n)/2} {\left\{ \begin{array}{ll} \;({\text {mod}}\Phi _n(q)^2 \Phi _n(-q)^3) &{}\quad \text {if}\; n\equiv 1\;({\text {mod}}4), \\ \;({\text {mod}}\Phi _n(q)^3 \Phi _n(-q)^3) &{}\quad \text {if}\; n\equiv 3\;({\text {mod}}4). \end{array}\right. } \end{aligned}$$

(1.7)

Note that $\Phi _n(q)\Phi _n(-q)=\Phi _n(q^2)$ for odd indices n.

The $n\equiv 3\;({\text {mod}}4)$ case of Theorem 1.1 confirms a conjecture of these authors [13, Conjecture 4.13], which states that, for $n\equiv 3\;({\text {mod}}4)$,

$$\begin{aligned} \sum _{k=0}^{(n-1)/2}\frac{(1+q^{4k+1})\,(q^2;q^4)_k^3}{(1+q)\,(q^4;q^4)_k^3}\,q^{k} \equiv 0\;({\text {mod}}\Phi _n(q)^2 \Phi _n(-q)). \end{aligned}$$

It is not difficult to verify that

$$\begin{aligned} \frac{(3/4)_{(p-1)/2}}{(5/4)_{(p-1)/2}} \equiv -\frac{p}{16}\Gamma _p\left( 1/4\right) ^4 \;({\text {mod}}p^2) \end{aligned}$$

for $p\equiv 3\;({\text {mod}}4)$, where $(a)_n=a(a+1)\ldots (a+n-1)$ denotes the rising factorial (also known as Pochhammer’s symbol). Therefore, the q-congruence (1.7) reduces to (1.4) for $p\equiv 3\;({\text {mod}}4)$ when $n=p$ and $q\rightarrow 1$, and it reduces to (1.3) for $p\equiv 1\;({\text {mod}}4)$ when $n=p$ and $q\rightarrow 1$. Moreover, letting $n=p$ and $q\rightarrow -1$ in (1.7), we immediately get (1.2). Thus, Theorem 1.1 presents a common q-analogue of supercongruences (1.2) and (1.3). We point out that other different q-analogues of (1.2) have been given in [7, 8].

Recently, Mao and Pan [18] (see also Sun [25, Theorem 1.3]) proved that, if $p\equiv 1\;({\text {mod}}4)$ is a prime, then

$$\begin{aligned} \sum _{k=0}^{(p+1)/2}\frac{(-1/2)_k^3}{k!^3}\equiv 0\;({\text {mod}}p^2). \end{aligned}$$

(1.8)

In this note, we prove the following q-analogue of (1.8).

Theorem 1.2

Let $n>1$ be an odd integer. Then

$$\begin{aligned}&\sum _{k=0}^{(n+1)/2}\frac{(1+q^{4k-1})\,(q^{-2};q^4)_k^3}{(1+q)\,(q^4;q^4)_k^3}\,q^{7k} \\&\quad \equiv \dfrac{[n]_{q^2}(q;q^4)_{(n-1)/2}}{(q^7;q^4)_{(n-1)/2}}\,q^{(n-3)/2} {\left\{ \begin{array}{ll} \;({\text {mod}}\Phi _n(q)^3 \Phi _n(-q)^3) &{}\quad \text {if}\; n\equiv 1\;({\text {mod}}4), \\ \;({\text {mod}}\Phi _n(q)^2 \Phi _n(-q)^3) &{}\quad \text {if}\; n\equiv 3\;({\text {mod}}4). \end{array}\right. } \nonumber \end{aligned}$$

The $n\equiv 1\;({\text {mod}}4)$ case of Theorem 1.2 also confirms a conjecture of the first author and Schlosser [11, Conjecture 10.2].

For n prime, letting $q\rightarrow 1$ in Theorem 1.2 we obtain the following generalization of (1.8).

Corollary 1.3

Let p be an odd prime. Then

$$\begin{aligned} \sum _{k=0}^{(p+1)/2} \frac{(-1/2)_k^3}{k!^3} \equiv p\,\dfrac{(1/4)_{(p-1)/2}}{(7/4)_{(p-1)/2}} {\left\{ \begin{array}{ll} \;({\text {mod}}p^3) &{}\quad \text {if } p\equiv 1\;({\text {mod}}4),\\ \;({\text {mod}}p^2) &{}\quad \text {if } p\equiv 3\;({\text {mod}}4). \end{array}\right. } \end{aligned}$$

On the other hand, for n prime and $q\rightarrow -1$ in Theorem 1.2, we are led to the following result:

$$\begin{aligned} \sum _{k=0}^{(p+1)/2} (-1)^k (4k-1)\frac{(-1/2)_k^3}{k!^3} \equiv p(-1)^{(p+1)/2} \;({\text {mod}}p^3). \end{aligned}$$

(1.9)

It should be mentioned that a different q-analogue of (1.9) was given in [13, Theorem 4.9] with $r=-1$, $d=2$ and $a=1$ (see also [11, Section 5]).

Moreover, for the summation formula

$$\begin{aligned} \sum _{k=0}^\infty \frac{(-\frac{1}{2})_k^3}{k!^3}=12\frac{\Gamma (3/4)^4}{\pi ^3}, \end{aligned}$$

we have the following q-analogue.

Theorem 1.4

We have

$$\begin{aligned} \sum _{k=0}^\infty \frac{(1+q^{4k-1})\,(q^{-2};q^4)_k^3}{(1+q^{-1})\,(q^4;q^4)_k^3}\,q^{7k} =\frac{(q^2;q^4)_\infty (q^5;q^4)_\infty ^2 (q^6;q^4)_\infty }{(q^3;q^4)_\infty (q^4;q^4)_\infty ^2 (q^7;q^4)_\infty }. \end{aligned}$$

Both Theorems 1.1 and 1.2 are particular cases of a more general result, which we state and prove in the next section, while Theorem 1.4 follows from a classical q-identity.

A Family of q-Congruences from the q-Dixon Sum

In this section we establish the following family of one-parameter q-congruences.

Theorem 2.1

Let $n\geqslant 1$ be an odd integer and $\ell$ an integer with $0\leqslant \ell \leqslant (n-1)/2$. Then

$$\begin{aligned}&\sum _{k=0}^{n-1}\frac{(1+q^{4k-2\ell +1})\,(q^{2-4\ell };q^4)_k^3}{(1+q^{1-2\ell })\,(q^4;q^4)_k^3}\,q^{(6\ell +1)k} \nonumber \\&\equiv \frac{(1-q^{2n})\,(q^{3-6\ell };q^4)_{(n-1)/2+\ell }}{(1-q^{2-4\ell })\,(q^{5-2\ell };q^4)_{(n-1)/2+\ell }}\,q^{(2\ell -1)((n-1)/2+\ell )} {\left\{ \begin{array}{ll} \;({\text {mod}}\Phi _n(q)^2 \Phi _n(-q)^3) \\ \quad \text {if}\; n+2\ell \equiv 1\;({\text {mod}}4), \\ \;({\text {mod}}\Phi _n(q)^3 \Phi _n(-q)^3) \\ \quad \text {if}\; n+2\ell \equiv 3\;({\text {mod}}4). \end{array}\right. } \end{aligned}$$

(2.1)

Note that the q-congruence (2.1) remains true when the sum is over k from 0 to $(n-1)/2+\ell$, since $(q^{2-4\ell };q^4)_k/(q^4;q^4)_k\equiv 0\;({\text {mod}}\Phi _n(q^2))$ for $(n-1)/2+\ell <k\leqslant n-1$. Furthermore, when $\ell =0$ and $\ell =1$ (hence $n\ge 3$) the theorem reduces to Theorems 1.1 and 1.2, respectively.

The following easily proved q-congruence (see [11, Lemma 3.1]) is necessary in our derivation of Theorem 2.1.

Lemma 2.2

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

$$\begin{aligned} \frac{(aq;q^2)_{(n-1)/2-k}}{(q^2/a;q^2)_{(n-1)/2-k}} \equiv (-a)^{(n-1)/2-2k}\frac{(aq;q^2)_k}{(q^2/a;q^2)_k}\,q^{(n-1)^2/4+k} \;({\text {mod}}\Phi _n(q)). \end{aligned}$$

Like the proofs given in [13], we start with the following generalization of (1.7) with an extra parameter a.

Theorem 2.3

Let $n>1$ be an odd integer and $0\leqslant \ell \leqslant (n-1)/2$. Then

$$\begin{aligned}&\sum _{k=0}^{n-1}\frac{(1+q^{4k-2\ell +1})\,(aq^{2-4\ell };q^4)_k (q^{2-4\ell }/a;q^4)_k (q^{2-4\ell };q^4)_k}{(1+q^{1-2\ell })\,(aq^4;q^4)_k (q^4/a;q^4)_k (q^4;q^4)_k}\,q^{(6\ell +1)k} \nonumber \\&\equiv \frac{(1-q^{2n})\,(q^{3-6\ell };q^4)_{(n-1)/2+\ell }}{(1-q^{2-4\ell })\,(q^{5-2\ell };q^4)_{(n-1)/2+\ell }}\,\nonumber \\&\qquad \times q^{(2\ell -1)((n-1)/2+\ell )}{\left\{ \begin{array}{ll} \;({\text {mod}}\Phi _n(-q) (1-aq^{2n})(a-q^{2n})) \\ \quad \text {if}\; n+2\ell \equiv 1\;({\text {mod}}4), \\ \;({\text {mod}}\Phi _n(q^2) (1-aq^{2n})(a-q^{2n})) \\ \quad \text {if}\; n+2\ell \equiv 3\;({\text {mod}}4). \end{array}\right. } \end{aligned}$$

(2.2)

Proof

Performing the parameter substitutions $q\mapsto q^4$, $a\mapsto q^{2-4\ell }$, $b\mapsto bq^{2-4\ell }$ and $c\mapsto cq^{2-4\ell }$ in the q-Dixon sum [3, Appendix (II.13)], we obtain

$$\begin{aligned}&\sum _{k=0}^\infty \frac{(1+q^{4k-2\ell +1})\,(q^{2-4\ell };q^4)_k (bq^{2-4\ell };q^4)_k (cq^{2-4\ell };q^4)_k}{(1+q^{1-2\ell })\,(q^4/b;q^4)_k (q^4/c;q^4)_k (q^4;q^4)_k} \biggl (\frac{q^{6\ell +1}}{bc}\biggr )^k \nonumber \\&\qquad =\frac{(q^{6-4\ell };q^4)_\infty (q^{2\ell +3}/b;q^4)_\infty (q^{2\ell +3}/c;q^4)_\infty (q^{4\ell +2}/bc;q^4)_\infty }{(q^4/b;q^4)_\infty (q^4/c;q^4)_\infty (q^{5-2\ell };q^4)_\infty (q^{6\ell +1}/bc;q^4)_\infty }. \end{aligned}$$

(2.3)

Since n is odd, putting $b=q^{-2n}$ and $c=q^{2n}$ in (2.3) we see that the left-hand side terminates and is equal to

$$\begin{aligned}&\sum _{k=0}^{(n-1)/2+\ell }\frac{(1+q^{4k-2\ell +1})\,(q^{2-4\ell -2n};q^4)_k (q^{2-4\ell +2n};q^4)_k (q^{2-4\ell };q^4)_k}{(1+q^{1-2\ell })\,(q^{4-2n};q^4)_k (q^{4+2n};q^4)_k (q^4;q^4)_k}\,q^{(6\ell +1)k} \\&\quad =\sum _{k=0}^{n-1}\frac{(1+q^{4k-2\ell +1})\,(q^{2-4\ell -2n};q^4)_k (q^{2-4\ell +2n};q^4)_k (q^{2-4\ell };q^4)_k}{(1+q^{1-2\ell })\,(q^{4-2n};q^4)_k (q^{4+2n};q^4)_k (q^4;q^4)_k}\,q^{(6\ell +1)k}, \end{aligned}$$

while the right-hand side becomes

$$\begin{aligned}&\frac{(q^{2\ell -2n+3};q^4)_{(n-1)/2+\ell } (q^{6-4\ell };q^4)_{(n-1)/2+\ell }}{(q^{4-2n};q^4)_{(n-1)/2+\ell } (q^{5-2\ell };q^4)_{(n-1)/2+\ell } } \\&\quad =\frac{(1-q^{2n})\,(q^{3-6\ell };q^4)_{(n-1)/2+\ell }}{(1-q^{2-4\ell })\,(q^{5-2\ell };q^4)_{(n-1)/2+\ell }}\,q^{(2\ell -1)((n-1)/2+\ell )}. \end{aligned}$$

This proves that the q-congruence (2.2) holds modulo $1-aq^{2n}$ or $a-q^{2n}$.

On the other hand, by Lemma 2.2, for $0\leqslant k\leqslant (n-1)/2+\ell$, modulo $\Phi _n(q)$ we have

$$\begin{aligned}&\frac{(aq^{1-2\ell };q^2)_{(n-1)/2+\ell -k}}{(q^2/a;q^2)_{(n-1)/2+\ell -k}}\\&\quad =\frac{(aq^{1-2\ell };q^2)_{\ell }(aq;q^2)_{(n-1)/2-k}}{(q^{n+1-2k}/a;q^2)_{\ell }(q^2/a;q^2)_{(n-1)/2-k}} \\&\quad \equiv (-a)^{(n-1)/2-2k}\frac{(aq^{1-2\ell };q^2)_{\ell }(aq;q^2)_k}{(q^{n+1-2k}/a;q^2)_{\ell }(q^2/a;q^2)_k}\,q^{(n-1)^2/4+k} \\&\quad =(-a)^{(n-1)/2-2k}\frac{(aq^{1-2\ell };q^2)_k (aq^{2k-2\ell +1};q^2)_\ell }{(q^{n+1-2k}/a;q^2)_{\ell }(q^2/a;q^2)_k}\,q^{(n-1)^2/4+k} \\&\quad \equiv (-a)^{(n-1)/2+\ell -2k}\frac{(aq^{1-2\ell };q^2)_k }{(q^2/a;q^2)_k}\,q^{(n-1)^2/4+k+(2k-\ell )\ell }, \end{aligned}$$

where we used $q^n\equiv 1\;({\text {mod}}\Phi _n(q))$ in the last step. Using the above q-congruence we can easily check that, for odd $n>1$ and $0\leqslant k\leqslant (n-1)/2+\ell$, sum of the kth and $((n-1)/2+\ell -k)$th summands on the left-hand side of (2.2) is congruent to 0 modulo $\Phi _n(-q)$ (or modulo $\Phi _n(q^2)$ if $n\equiv 3-2\ell \;({\text {mod}}4)$). It follows that

$$\begin{aligned}&\sum _{k=0}^{(n-1)/2+\ell } \frac{(1+q^{4k-2\ell +1})\,(aq^{2-4\ell };q^4)_k (q^{2-4\ell }/a;q^4)_k (q^{2-4\ell };q^4)_k}{(1+q^{1-2\ell })\,(aq^4;q^4)_k (q^4/a;q^4)_k (q^4;q^4)_k}\,q^{(6\ell +1)k} \\&\quad \equiv 0 {\left\{ \begin{array}{ll} \;({\text {mod}}\Phi _n(-q)) &{}\quad \text { if } n+2\ell \equiv 1\;({\text {mod}}4), \\ \;({\text {mod}}\Phi _n(q^2)) &{}\quad \text { if } n+2\ell \equiv 3\;({\text {mod}}4). \end{array}\right. } \end{aligned}$$

Clearly, the right-hand side of (2.1) is congruent to 0 modulo $\Phi _n(-q)$ if $n+2\ell \equiv 1\;({\text {mod}}4)$ and modulo $\Phi _n(q^2)$ if $n+2\ell \equiv 3\;({\text {mod}}4)$. Therefore, the q-congruence (2.2) holds modulo $\Phi _n(-q)$ if $n+2\ell \equiv 1\;({\text {mod}}4)$ and modulo $\Phi _n(q^2)$ if $n+2\ell \equiv 3\;({\text {mod}}4)$. Since the polynomials $1-aq^{2n}$, $a-q^{2n}$ and $\Phi _n(-q)$ (or $\Phi _n(q^2)$) are pairwise coprime, we complete the proof of (2.2). $\square$

Proof of Theorem 2.1

We assume that $n>1$, since the $n=1$ case (making $\ell =0$ only possible) is trivial. The limits of the denominators on both sides of (2.2) as $a\rightarrow 1$ are relatively prime to $\Phi _n(q^2)$, since k is in the range $0\leqslant k\leqslant (n-1)/2+\ell$. On the other hand, the limit of $(1-aq^{2n})(a-q^{2n})$ as $a\rightarrow 1$ contains the factor $\Phi _n(q^2)^2$. $\square$

Proof of Theorem 1.4

Take $b=c=\ell =1$ in Eq. (2.3). $\square$

Discussion

The method of creative microscoping used in our proofs indicates the origin of q-congruences from infiniteq-hypergeometric identities; for example, the q-congruence (1.7) corresponds to the identity

$$\begin{aligned} \sum _{k=0}^\infty \frac{(1+q^{4k+1})\,(q^2;q^4)_k^3}{(1+q)\,(q^4;q^4)_k^3}\,q^{k} =\frac{(q^2;q^4)_\infty ^2(q^3;q^4)_\infty ^2}{(1+q)\,(q;q^4)_\infty ^2(q^4;q^4)_\infty ^2}, \end{aligned}$$

(3.1)

which is just a particular instance of (2.3). Note that the limiting cases as $q\rightarrow -1$ and $q\rightarrow 1$ of (3.1) give the formulas (1.1) and

$$\begin{aligned} \sum _{k=0}^\infty \frac{(\frac{1}{2})_k^3}{k!^3} =\frac{\Gamma (1/4)^4}{4\pi ^3} =\frac{8L(f,1)}{\pi } =\frac{16L(f,2)}{\pi ^2} \end{aligned}$$

(3.2)

where

$$\begin{aligned} f(\tau )=q\prod _{j=1}^\infty (1-q^{4j})^6=\sum _{n=1}^\infty a(n)q^n, \quad \text {with}\; q=\exp (2\pi i\tau ), \end{aligned}$$

is the CM modular form from the introduction and L(f, s) denotes its L-function. This means that the q-identity (3.1) presents a common q-extension of evaluations (1.1) and (3.2)—the fact that makes it less surprising that the q-congruence (1.7) simultaneously extends (1.2) and (1.3).

The intermediate use of parametricq-hypergeometric identities in our proof of Theorem 2.1 based on the q-Dixon sum suggests that different q-congruences underlying (3.1) are possible. This is indeed the case when we analyze the formula (3.1) as the $a=1$ specialization of

$$\begin{aligned}&\sum _{k=0}^\infty \frac{(1+q^{4k+1})\,(aq;q^2)_k(q/a;q^2)_k(-q;q^2)_k^2(q^2;q^4)_k}{(1+q)\,(q^2;q^2)_k^2(-aq^2;q^2)_k(-q^2/a;q^2)_k(q^4;q^4)_k}\,q^k \nonumber \\&\quad =\frac{(-q;q^2)_\infty ^2(aq^3;q^4)_\infty ^2(q^3/a;q^4)_\infty ^2}{(1+q)\,(-aq^2;q^2)_\infty (-q^2/a;q^2)_\infty (q^2;q^2)_\infty ^2} \end{aligned}$$

(3.3)

which originates from a q-analogue of Watson’s $_3F_2$ sum [3, Appendix (II.16)]. When we choose $a=q^n$ (or $a=q^{-n}$) in (3.3), for $n>1$ odd, we get the sum terminating after $(n-1)/2$ terms on the left-hand side of (3.3), while the right-hand side vanishes if n is of the form $4m+3$ and it becomes equal to

$$\begin{aligned} \frac{(-q;q^2)_\infty ^2(q^{4m+4};q^4)_\infty ^2(q^{2-4m};q^4)_\infty ^2}{(1+q)\,(-q^{4m+3};q^2)_\infty (-q^{1-4m};q^2)_\infty (q^2,q^4;q^4)_\infty ^2} =[4m+1]\,\frac{(q^2;q^4)_m^2}{(q^4;q^4)_m^2} \end{aligned}$$

if $n=4m+1$. This means that modulo $(a-q^n)(1-aq^n)$ we have

$$\begin{aligned}&\sum _{k=0}^N\frac{(1+q^{4k+1})\,(aq;q^2)_k(q/a;q^2)_k(-q;q^2)_k^2(q^2;q^4)_k}{(1+q)\,(q^2;q^2)_k^2(-aq^2;q^2)_k(-q^2/a;q^2)_k(q^4;q^4)_k}\,q^k \\&\quad \equiv {\left\{ \begin{array}{ll} [4m+1]\,\dfrac{(q^2;q^4)_m^2}{(q^4;q^4)_m^2} &{}\quad \text {if}\; n=4m+1, \\ 0 &{}\quad \text {if}\; n\equiv 3\;({\text {mod}}4), \end{array}\right. } \end{aligned}$$

for any $N\ge (n-1)/2$. The limiting $a\rightarrow 1$ case of the congruences can be shown to be

$$\begin{aligned} \sum _{k=0}^{(n-1)/2}\frac{(1+q^{4k+1})\,(q^2;q^4)_k^3}{(1+q)\,(q^4;q^4)_k^3}\,q^k \equiv {\left\{ \begin{array}{ll} [4m+1]\,\dfrac{(q^2;q^4)_m^2}{(q^4;q^4)_m^2} &{}\quad \text {if}\; n=4m+1, \\ 0 &{}\quad \text {if}\; n\equiv 3\;({\text {mod}}4), \end{array}\right. } \end{aligned}$$

(3.4)

modulo $\Phi _n(q)^2\Phi _n(-q)$. This is quite similar in spirit to (1.5), though still far from constructing q-analogues for the coefficients a(p) in (1.6) of the modular form $f(\tau )$. The latter means that a hunt for q-rational functions, which equal the left-hand side of (1.5) or (3.4) modulo $\Phi _n(q)^2$ and specialize to a(n) as $q\rightarrow 1$ (at least for n prime), is still on its way. Such q-rational functions are also expected to be self-reciprocal, that is, invariant under the involution $q\mapsto 1/q$, as all the left- and right-hand sides in (1.5), (1.7), (3.4) and also (2.1) are.