1 Introduction

More than one hundred years ago, in his second notebook, Ramanujan [19] mysteriously wrote 17 representations of \(1/\pi \) (see [1, p. 352]), including for example,

$$\begin{aligned} \sum _{k=0}^{\infty }(6k+1)\frac{(\frac{1}{2})_k^3}{k!^3 4^k} =\frac{4}{\pi }, \end{aligned}$$
(1.1)

which he published in [19] later. Here, \((a)_n=a(a+1)\cdots (a+n-1)\) denotes the rising factorial. Van Hamme [21] numerically observed 13 amazing p-adic analogues of Ramanujan-type formulas, such as

$$\begin{aligned} \sum _{k=0}^{(p-1)/2}(6k+1)\frac{(\frac{1}{2})_k^3}{k!^3 4^k}&\equiv p(-1)^{(p-1)/2}\pmod {p^4}, \end{aligned}$$
(1.2)

where \(p>3\) is a prime. But Van Hamme also said, “We have no real explanation for our observations." He himself proved three of these formulas. Supercongruences like (1.2) are now usually called Ramanujan-type supercongruences (see [24]). The proof of (1.2) was first given by Long [16] using hypergeometric evaluation identities. We refer the reader to [18] for historical remarks on Van Hamme’s supercongruences.

Recently, many classical supercongruences have been extended to supercongruences on rational functions in q, which we call q-analogues of supercongruences, or simply q-supercongruences (see [4, 5, 7,8,9,10,11,12,13,14,15, 17, 22, 23]). In particular, the first author [4] gave a q-analogue of (1.2) as follows: for odd n,

$$\begin{aligned}&\sum _{k=0}^{(n-1)/2}q^{k^2}[6k+1]\frac{(q;q^2)_k^2 (q^2;q^4)_k }{(q^4;q^4)_k^3} \nonumber \\&\qquad \quad \equiv (-q)^{(1-n)/2}[n] +\frac{(n^2-1)(1-q)^2}{24}(-q)^{(1-n)/2}[n]^3 \pmod {[n]\Phi _n(q)^3}. \end{aligned}$$
(1.3)

Here and throughout the paper, we adopt the standard q-notation: \([n]=1+q+\cdots +q^{n-1}\) is the q-integer; \((a;q)_n=(1-a)(1-aq)\cdots (1-aq^{n-1})\) is the q-shifted factorial, with the condensed notation \((a_1,a_2,\ldots ,a_m;q)_n=(a_1;q)_n (a_2;q)_n\cdots (a_m;q)_n\); and \(\Phi _n(q)\) denotes the n-th cyclotomic polynomial in q, which may be given by

$$\begin{aligned} \Phi _n(q)=\prod _{\begin{array}{c} 1\leqslant k\leqslant n\\ \gcd (k,n)=1 \end{array}}(q-\zeta ^k), \end{aligned}$$

where \(\zeta \) is an n-th primitive root of unity.

The cyclotomic polynomials \(\Phi _n(q)\) are irreducible over the set of integers \({\mathbb {Z}}\), and, for any positive integer n, we have

$$\begin{aligned} \prod _{s>1,\;s\mid n}\Phi _s(q)=[n], \end{aligned}$$
(1.4)

which is a property we implicitly make use of in this paper.

Very recently, the present authors [9] proved the following result: Let d and r be odd integers satisfying \(d\geqslant 3\), \(r\leqslant d-4\) (in particular, r may be negative) and \(\gcd (d,r)=1\). Let n be an integer such that \(n\geqslant (d-r)/2\) and \(n\equiv -r/2\pmod {d}\). Then

$$\begin{aligned} \sum _{k=0}^{n-1}[2dk+r]\frac{(q^r;q^d)_k^d}{(q^d;q^d)_k^d}q^{d(d-r-2)k/2} \equiv 0 \pmod {\Phi _{n}(q)^3}. \end{aligned}$$
(1.5)

In this paper, we consider the \(r=d-2\) case of the left-hand side of (1.5) (in this case the condition \(r\leqslant d-4\) for (1.5) is violated), and shall prove the following q-supercongruence.

Theorem 1.1

Let \(d\geqslant 3\) be an odd integer and let \(n\equiv 1\pmod {d}\) be a positive integer. Then

$$\begin{aligned}&\sum _{k=0}^{n-1}[2dk+d-2]\frac{(q^{d-2};q^d)_k^d}{(q^d;q^d)_k^d} \nonumber \\&\qquad \quad \equiv \frac{[(d-2)n](q^{d-2},q^d;q^d)_{(d-2)(n-1)/d}}{(q,q^{d-1};q^d)_{(n-1)/d}(q^2,q^d;q^d)_{(2n-2)/d}^{(d-3)/2}} q^{-((d-1)^2+(d-3)^2n)(n-1)/(2d)} \pmod {\Phi _{n}(q)^3}. \end{aligned}$$
(1.6)

Further, the present authors [8] proved the following q-supercongruence: Let d and r be odd integers satisfying \(d\geqslant 3\), \(r\leqslant d-4\) (in particular, r may be negative) and \(\gcd (d,r)=1\). Let n be an integer such that \(n\geqslant d-r\) and \(n\equiv -r\pmod {d}\). Then

$$\begin{aligned} \sum _{k=0}^{n-1}[2dk+r]\frac{(q^r;q^d)_k^d}{(q^d;q^d)_k^d}q^{d(d-r-2)k/2} \equiv 0 \pmod {\Phi _{n}(q)^3}. \end{aligned}$$
(1.7)

Note that in [8, Conjecture 3] it was even conjectured that the q-supercongruence (1.7) is true modulo \(\Phi _n(q)^4\) for \(d\geqslant 5\), which still remains open.

In this paper, we consider the \(r=d-2\) case of the left-hand side of (1.7), and shall establish the following result.

Theorem 1.2

Let \(d\geqslant 3\) be an odd integer and let \(n\equiv 2\pmod {d}\) be a positive integer. Then

$$\begin{aligned}&\sum _{k=0}^{n-1}[2dk+d-2]\frac{(q^{d-2};q^d)_k^d}{(q^d;q^d)_k^d} \nonumber \\&\quad \equiv -\frac{(q^d;q^d)_{(dn-n-d+2)/d}(1+q^{n^2(d-1)/2}) }{(1-q)(q^d;q^d)_{(n-2)/d}^{d-1}} q^{-d{(dn-n+2)/d\atopwithdelims ()2}} \pmod {\Phi _{n}(q)^2}. \end{aligned}$$
(1.8)

Further, the present authors (see [6, Theorem 1] and [7, Theorem 4.2]) proved the following result: Let d, r, n be integers satisfying \(d\geqslant 3\), \(r\leqslant d-2\) (in particular, r may be negative), and \(n\geqslant d-r\), such that d and r are coprime, and \(n\equiv -r\pmod {d}\). Then

$$\begin{aligned} \sum _{k=0}^{M}[2dk+r] \frac{(q^r;q^d)_k^{2d}}{(q^d;q^d)_k^{2d}}q^{d(d-1-r)k} \equiv 0 \pmod {[n]\Phi _{n}(q)^3}, \end{aligned}$$
(1.9)

where \(M=(dn-n-r)/d\) or \(M=n-1\).

In this paper, we consider the \(r=d-1\) case of the left-hand side of (1.9), and shall build the following q-supercongruence.

Theorem 1.3

Let \(d\geqslant 3\) be an integer and let \(n\equiv 1\pmod {d}\) be a positive integer. Then

$$\begin{aligned}&\sum _{k=0}^{M}[2dk+d-1]\frac{(q^{d-1};q^d)_k^{2d}}{(q^d;q^d)_k^{2d}} \nonumber \\&\quad \equiv [(d-1)n]\frac{(q^{d-1},q^{d};q^d)_{(d-1)(n-1)/d}}{(q,q^d;q^d)_{(n-1)/d}^{d-1}}q^{-((d-2)n+d)(d-1)(n-1)/(2d)}\pmod {[n]\Phi _n(q)^2}, \end{aligned}$$
(1.10)

where \(M=(d-1)(n-1)/d\) or \(M=n-1\).

Note that (1.10) is also true for \(d=2\). See [4, 10] for a generalization of this case modulo \([n]\Phi _n(q)^3\).

In this paper we employ the method of creative microscoping, recently devised by the first author in collaboration with Zudilin [11], to prove Theorems 1.1–1.3. Here we appropriately introduce a parameter a (such that the series satisfies the symmetry \(a\leftrightarrow a^{-1}\)) into the terms of the series and prove that the congruences in Theorems 1.1 and 1.3 hold modulo \(\Phi _n(q)\), modulo \(1-aq^n\), and modulo \(a-q^n\). Thus, by the Chinese remainder theorem for coprime polynomials, the congruences hold modulo the product \(\Phi _n(q)(1-aq^n)(a-q^n)\). By letting \(a= 1\) the congruences are established modulo \(\Phi _n(q)^3\). By showing that the congruence in (1.9) also holds modulo \(\Phi _s(q)\) for any divisor \(s>1\) of n, we are able to lift the congruence modulo \(\Phi _n(q)^3\) to the claimed congruence modulo \([n]\Phi _n(q)^2\). This again is justified by appealing to the Chinese remainder theorem for coprime polynomials, in combination with (1.4). For the congruence in Theorem 1.2, we similarly give a parametric generalization modulo \((1-aq^n)(a-q^n)\).

Our paper is arranged as follows: in Sect. 2, we list some tools we require in our proof of Theorems 1.1–1.3. These include a lemma on a simple q-congruence modulo a cyclotomic polynomial \(\Phi _n(q)\), and a very-well-poised Karlsson–Minton type summation by Gasper of which we require some special cases. In Sect. 3, we first prove a parametric generalization of Theorem 1.1 involving the insertion of different powers of the parameter a in the respective q-shifted factorials. Afterwards we show how Theorem 1.1 can be deduced from this parametric generalization. We prove Theorems 1.2 and 1.3 similarly in Sects. 4 and 5, respectively.

2 Preliminaries

We need the following result, which was first given by the present authors in [7, Lemma 2.1] (see also [9, Lemma 1]).

Lemma 2.1

Let d, m and n be positive integers with \(m\leqslant n-1\). Let r be an integer satisfying \(dm\equiv -r\pmod {n}\). Then, for \(0\leqslant k\leqslant m\) and any indeterminate a, we have

$$\begin{aligned} \frac{(aq^r;q^d)_{m-k}}{(q^d/a;q^d)_{m-k}} \equiv (-a)^{m-2k}\frac{(aq^r;q^d)_k}{(q^d/a;q^d)_k} q^{m(dm-d+2r)/2+(d-r)k} \pmod {\Phi _n(q)}. \end{aligned}$$

If \(\gcd (d,n)=1\), then the above q-congruence also holds for \(a=1\).

We will also use a very-well-poised Karlsson–Minton type summation of Gasper [2, Eq. (5.13)] (see also [3, Ex. 2.33 (i)]):

$$\begin{aligned}&\sum _{k=0}^\infty \frac{(a,q\sqrt{a},-q\sqrt{a},b,a/b,d,e_1,aq^{n_1+1}/e_1,\dots , e_m,aq^{n_m+1}/e_m;q)_k}{(q,\sqrt{a},-\sqrt{a},aq/b,bq,aq/d,aq/e_1,e_1q^{-n_1}, \dots ,aq/e_m,e_mq^{-n_m};q)_k}\left( \frac{q^{1-N}}{d}\right) ^k\nonumber \\&\qquad \quad =\frac{(q,aq,aq/bd,bq/d;q)_\infty }{(bq,aq/b,aq/d,q/d;q)_\infty } \prod _{j=1}^m\frac{(aq/be_j,bq/e_j;q)_{n_j}}{(aq/e_j,q/e_j;q)_{n_j}}, \end{aligned}$$
(2.1)

where \(n_1,\dots ,n_m\) are non-negative integers, \(N=n_1+\cdots +n_m\), and the convergence condition \(|q^{1-N}/d|<1\) is necessary if the series does not terminate. The terminating \(d=q^{-N}\) case of (2.1) reduces to

$$\begin{aligned}&\sum _{k=0}^N\frac{(a,q\sqrt{a},-q\sqrt{a},b,a/b,e_1,aq^{n_1+1}/e_1,\dots , e_m,aq^{n_m+1}/e_m,q^{-N};q)_k}{(q,\sqrt{a},-\sqrt{a},aq/b,bq,aq/e_1,e_1q^{-n_1}, \dots ,aq/e_m,e_mq^{-n_m},aq^{N+1};q)_k}q^{k} \nonumber \\&\qquad \quad =\frac{(q,aq;q)_N}{(bq,aq/b;q)_N}\prod _{j=1}^m \frac{(aq/be_j,bq/e_j;q)_{n_j}}{(aq/e_j,q/e_j;q)_{n_j}}. \end{aligned}$$
(2.2)

Note that an elliptic extension of (2.2) was provided by Rosengren and the second author [20, Eq. (1.7)].

In particular, taking \(b\rightarrow \infty \) we get the following summation formula:

$$\begin{aligned}&\sum _{k=0}^N\frac{(a,q\sqrt{a},-q\sqrt{a},e_1,aq^{n_1+1}/e_1,\dots , e_m,aq^{n_m+1}/e_m,q^{-N};q)_k}{(q,\sqrt{a},-\sqrt{a},aq/e_1,e_1q^{-n_1}, \dots ,aq/e_m,e_mq^{-n_m},aq^{N+1};q)_k} \nonumber \\&\qquad \quad =\frac{(q,aq;q)_N q^{{n_1\atopwithdelims ()2}+\cdots +{n_m\atopwithdelims ()2}-{N\atopwithdelims ()2}}}{\prod _{j=1}^m(aq/e_j,q/e_j;q)_{n_j}e_j^{n_j}}, \end{aligned}$$
(2.3)

where \(N=n_1+\cdots +n_m\).

Other special cases of (2.2) that we require are the following (see [3, (1.9.9), (1.9.12)]):

$$\begin{aligned} \sum _{k=0}^N\frac{(q^{-N},b_1q^{n_1},\dots ,b_m q^{n_m};q)_k}{(q,b_1,\ldots ,b_m;q)_k}q^k =(-1)^N\frac{(q;q)_N \,b_1^{n_1}\cdots b_m^{n_m}}{(b_1;q)_{n_1}\cdots (b_m;q)_{n_m}}q^{{n_1\atopwithdelims ()2}+\cdots +{n_m\atopwithdelims ()2}},\nonumber \\ \quad \end{aligned}$$
(2.4)

where \(N=n_1+\cdots +n_m\), and

$$\begin{aligned} \sum _{k=0}^N\frac{(q^{-N},b_1q^{n_1},\dots ,b_m q^{n_m};q)_k}{(q,b_1,\ldots ,b_m;q)_k} =(-1)^N\frac{(q;q)_N \,q^{-{N+1\atopwithdelims ()2}}}{(b_1;q)_{n_1}\cdots (b_m;q)_{n_m}}, \end{aligned}$$
(2.5)

where \(N\geqslant n_1+\cdots +n_m\).

3 A parametric generalization and proof of Theorem 1.1

We first give the following result.

Lemma 3.1

Let \(d\geqslant 3\) be an odd integer and let \(n\equiv 1\pmod {d}\) be a positive integer. Then, for any integer A, modulo \(\Phi _n(q)^2\),

$$\begin{aligned} (q^{2+An},q^{d+An};q^d)_{(2n-2)/d}\equiv q^{2An(n-1)/d} (q^2,q^d;q^d)_{(2n-2)/d} \pmod {\Phi _n(q)^2}.\qquad \end{aligned}$$
(3.1)

Proof

We shall prove the following parametric generalization of (3.1):

$$\begin{aligned} (aq^2,aq^d;q^d)_{(2n-2)/d}\equiv a^{2(n-1)/d}(q^2,q^d;q^d)_{(2n-2)/d} \pmod {(1-a)(1-aq^{2n})}. \qquad \end{aligned}$$
(3.2)

It is clear that both sides of (3.2) are equal for \(a=1\). This means that the congruence (3.2) holds modulo \(1-a\). Further, for \(a=q^{-2n}\) the left-hand side of (3.2) is equal to \((q^{2-2n},q^{d-2n};q^d)_{(2n-2)/d}\). But

$$\begin{aligned} (q^{2-2n};q^d)_{(2n-2)/d}&=(1-q^{2-2n})(1-q^{2-2n+d})\cdots (1-q^{-d}) \\&=q^{-(d+2n-2)(n-1)/d}(1-q^{2n-2})(1-q^{2n-2-d})\cdots (1-q^d) \\&=q^{-(d+2n-2)(n-1)/d}(q^d;q^d)_{(2n-2)/d}, \end{aligned}$$

and similarly,

$$\begin{aligned} (q^{d-2n};q^d)_{(2n-2)/d} =q^{-(2+2n-d)(n-1)/d}(q^2;q^d)_{(2n-2)/d}. \end{aligned}$$

This proves that for \(a=q^{-2n}\) both sides of (3.2) are also equal, and so the congruence (3.2) is true modulo \(1-aq^{2n}\). Since \(1-a\) and \(1-aq^{2n}\) are relatively prime polynomials, we obtain (3.2).

Let \(a=q^{An}\) in (3.2). Noticing that both \(1-q^{An}\) and \(1-q^{(A+2)n}\) contain the factor \(\Phi _n(q)\), we arrive at the q-congruence (3.1). \(\square \)

We now give a parametric generalization of Theorem 1.1.

Theorem 3.2

Let \(d\geqslant 3\) be an odd integer and let \(n\equiv 1\pmod {d}\) be a positive integer. Then, modulo \(\Phi _n(q)(1-aq^n)(a-q^n)\),

$$\begin{aligned}&\sum _{k=0}^{n-1}[2dk+d-2] \frac{(a^{d-2}q^{d-2}, a^{d-4}q^{d-2},\ldots , aq^{d-2};q^d)_k}{(a^{d-2}q^d, a^{d-4}q^d,\ldots ,aq^d;q^d)_k}\nonumber \\&\qquad \times \frac{(a^{2-d}q^{d-2}, a^{4-d}q^{d-2},\ldots , a^{-1}q^{d-2};q^d)_k (q^{d-2};q^d)_k}{(a^{2-d}q^d, a^{4-d}q^d,\ldots , a^{-1}q^d;q^d)_k (q^d;q^d)_k } \nonumber \\&\quad \equiv \frac{[(d-2)n](q^{d-2},q^d;q^d)_{(d-2)(n-1)/d} \,q^{-(d-1)^2(n-1)/(2d)}}{(q,q^{d-1};q^d)_{(n-1)/d}\prod _{j=1}^{(d-3)/2} (q^{d+(d-2j-2)n},q^{2+(d-2j-2)n};q^d)_{(2n-2)/d}}. \end{aligned}$$
(3.3)

Proof

Since \(\gcd (d,n)=1\), none of the numbers \(d,2d,\ldots , (n-1)d\) are divisible by n. Therefore, the denominators of the left-hand side of (3.3) do not have the factor \(1-aq^n\) nor \(1-a^{-1}q^n\). Thus, for \(a=q^{-n}\) or \(a=q^n\), the left-hand side of (3.3) can be written as

$$\begin{aligned}&\sum _{k=0}^{(d-2)(n-1)/d}[2dk+d-2] \frac{(q^{d-2-(d-2)n}, q^{d-2-(d-4)n},\ldots ,q^{d-2-n};q^d)_k }{(q^{d-(d-2)n}, q^{d-(d-4)n},\ldots , q^{d-n};q^d)_k } \nonumber \\&\quad \qquad \times \frac{(q^{(d-2)n+d-2}, q^{(d-4)n+d-2},\ldots ,q^{n+d-2};q^d)_k (q^{d-2};q^d)_k}{(q^{(d-2)n+d}, q^{(d-4)n+d},\ldots , q^{n+d};q^d)_k (q^d;q^d)_k }, \end{aligned}$$
(3.4)

where we have used \((q^{d-2-(d-2)n};q^d)_k=0\) for \(k>(d-2)(n-1)/d\). Specializing the parameters in (2.3) by \(N=(d-2)(n-1)/d\), \(q\mapsto q^d\), \(a=q^{d-2}\), \(m=(d-1)/2\), \(e_j=q^{d-2-(d-2j-2)n}\) (\(1\leqslant j\leqslant m-1\)), \(e_m=q^{d-1}\), \(n_1=\cdots =n_{m-1}=(2n-2)/d\) and \(n_m=(n-1)/d\) and noticing \(N=n_1+\cdots +n_m\), we see that (3.4) is equal to the right-hand side of (3.3), where we have used the relation

$$\begin{aligned}{}[d-2](q^{2d-2};q^d)_{(d-2)(n-1)/d}=[(d-2)n](q^{d-2};q^d)_{(d-2)(n-1)/d}. \end{aligned}$$

This proves that (3.3) holds modulo \((1-aq^n)(a-q^n)\).

For \(M=(d-2)(n-1)/d\), by Lemma 2.1, we can easily verify that

$$\begin{aligned}&[2d(M-k)+d-2]\frac{(a^{d-2}q^{d-2}, a^{d-4}q^{d-2},\ldots , aq^{d-2};q^d)_{M-k}}{(a^{d-2}q^d, a^{d-4}q^d,\ldots ,aq^d;q^d)_{M-k}}\\&\qquad \times \frac{(a^{2-d}q^{d-2}, a^{4-d}q^{d-2},\ldots ,a^{-1}q^{d-2};q^d)_{M-k} (q^{d-2};q^d)_{M-k}}{(a^{2-d}q^d, a^{4-d}q^d,\ldots , a^{-1}q^d;q^d)_{M-k} (q^d;q^d)_{M-k} } \\&\quad \equiv -[2dk+d-2]\frac{(a^{d-2}q^{d-2}, a^{d-4}q^{d-2},\ldots , aq^{d-2};q^d)_k}{(a^{d-2}q^d, a^{d-4}q^d,\ldots ,aq^d;q^d)_k}\\&\qquad \times \frac{(a^{2-d}q^{d-2}, a^{4-d}q^{d-2},\ldots , a^{-1}q^{d-2};q^d)_k (q^{d-2};q^d)_k}{(a^{2-d}q^d, a^{4-d}q^d,\ldots , a^{-1}q^d;q^d)_k (q^d;q^d)_k } \pmod {\Phi _n(q)}. \end{aligned}$$

It is now evident that the sum of the k-th and \((M-k)\)-th summands on the left-hand side of (3.3) vanishes modulo \(\Phi _n(q)\). Since \([(d-2)n]\equiv 0\pmod {\Phi _n(q)}\), we conclude that the partial sum of the left-hand side of (3.3) for k up to \((d-2)(n-1)/d\) is congruent to 0 modulo \(\Phi _n(q)\). Further, for any k satisfying \((d-2)(n-1)/d<k\leqslant n-1\), we have \((q^{d-2};q^d)_k/(q^d;q^d)_k\equiv 0\pmod {\Phi _n(q)}\). This proves the q-congruence (3.3). \(\square \)

Proof of Theorem 1.1

In view of \(\gcd (n,d)=1\), the factors involving a in the denominators of the left-hand side of (3.3) are all relatively prime to \(\Phi _n(q)\) when \(a=1\). On the other hand, the polynomial \((1-aq^n)(a-q^n)\) contains the factor \(\Phi _n(q)^2\) when \(a=1\). Hence, taking \(a=1\) in (3.3), we obtain

$$\begin{aligned}&\sum _{k=0}^{n-1}[2dk+d-2]\frac{(q^{d-2};q^d)_k^d}{(q^d;q^d)_k^d} \nonumber \\&\quad \equiv \frac{[(d-2)n](q^{d-2},q^d;q^d)_{(d-2)(n-1)/d}\, q^{-(d-1)^2(n-1)/(2d)}}{(q,q^{d-1};q^d)_{(n-1)/d}\prod _{j=1}^{(d-3)/2} (q^{d+(d-2j-2)n},q^{2+(d-2j-2)n};q^d)_{(2n-2)/d}} \pmod {\Phi _{n}(q)^3}. \end{aligned}$$

Furthermore, applying Lemma 3.1\((d-3)/2\) times, modulo \(\Phi _n(q)^2\), we have

$$\begin{aligned} \prod _{j=1}^{(d-3)/2}(q^{2+(d-2j-2)n},q^{d+(d-2j-2)n};q^d)_{(2n-2)/d} \equiv (q^2,q^d;q^d)_{(2n-2)/d}^{(d-3)/2}\,q^{(d-3)^2n(n-1)/(2d)}. \end{aligned}$$

This, together with \([(d-2)n]\equiv 0\pmod {\Phi _n(q)}\), proves (1.6). \(\square \)

4 A parametric generalization and proof of Theorem 1.2

As before, we first give a parametric generalization of Theorem 1.2.

Theorem 4.1

Let \(d\geqslant 3\) be an odd integer and let \(n\equiv 2\pmod {d}\) be a positive integer. Then, modulo \((1-aq^n)(a-q^n)\),

$$\begin{aligned}&\sum _{k=0}^{n-1}[2dk+d-2]\frac{(a^{d-1}q^{d-2}, a^{d-3}q^{d-2},\ldots , a^2q^{d-2};q^d)_k}{(a^{d-2}q^d, a^{d-4}q^d,\ldots ,aq^d;q^d)_k}\nonumber \\&\qquad \times \frac{(a^{1-d}q^{d-2}, a^{3-d}q^{d-2},\ldots , a^{-2}q^{d-2};q^d)_k (q^{d-2};q^d)_k}{(a^{2-d}q^d, a^{4-d}q^d,\ldots , a^{-1}q^d;q^d)_k (q^d;q^d)_k } \nonumber \\&\quad \equiv -\frac{(q^d;q^d)_{(dn-n-d+2)/d}(1+q^{n^2(d-1)/2}) q^{-d{(dn-n+2)/d\atopwithdelims ()2}} }{(1-q)(a^{d-2}q^d,a^{d-4}q^d,\ldots , a^{4-d}q^d,a^{2-d}q^d;q^d)_{(n-2)/d}} . \end{aligned}$$
(4.1)

Proof

For \(a=q^{-n}\) or \(a=q^n\), the left-hand side of (4.1) can be written as

$$\begin{aligned}&\sum _{k=0}^{(dn-n-d+2)/d}[2dk+d-2]\frac{(q^{d-2-(d-1)n}, q^{d-2-(d-3)n},\ldots ,q^{d-2-2n};q^d)_k }{(q^{d-(d-2)n}, q^{d-(d-4)n},\ldots , q^{d-n};q^d)_k } \nonumber \\&\quad \qquad \times \frac{(q^{(d-1)n+d-2}, q^{(d-3)n+d-2},\ldots , q^{2n+d-2};q^d)_k(q^{d-2};q^d)_k}{(q^{(d-2)n+d}, q^{(d-4)n+d},\ldots , q^{n+d};q^d)_k (q^d;q^d)_k }, \end{aligned}$$
(4.2)

where we have used \((q^{d-2-(d-1)n};q^d)_k=0\) for \(k>(dn-n-d+2)/d\). Since

$$\begin{aligned}{}[2dk+d-2]&=\frac{1-q^{2dk+d-2}}{1-q} \\&=\frac{1-q^{(d-2)/2}}{1-q}\frac{(q^{d+(d-2)/2};q^d)_k}{(q^{(d-2)/2};q^d)_k}q^{dk+(d-2)/2}\\&\quad +\frac{1-q^{(d-2)/2}}{1-q}\frac{(q^{d+(d-2)/2};q^d)_k}{(q^{(d-2)/2};q^d)_k}, \end{aligned}$$

by (2.4) and (2.5), the summation (4.2) is equal to

$$\begin{aligned}&\frac{q^{(d-2)/2}(-1)^{(dn-n-d+2)/d}(q^d;q^d)_{(dn-n-d+2)/d} \,q^{(d-2)/2+(n-2)(d-1)}}{(1-q)(q^{d-(d-2)n}, q^{d-(d-4)n},\ldots , q^{d+(d-4)n},q^{d+(d-2)n};q^d)_{(n-2)/d} } q^{d(d-1){(n-2)/d\atopwithdelims ()2}} \\&\quad \qquad +\frac{(-1)^{(dn-n-d+2)/d}(q^d;q^d)_{(dn-n-d+2)/d} \,q^{-d{(dn-n+2)/d\atopwithdelims ()2}}}{(1-q)(q^{d-(d-2)n}, q^{d-(d-4)n}, \ldots , q^{d+(d-4)n},q^{d+(d-2)n};q^d)_{(n-2)/d} }, \end{aligned}$$

which is just the \(a=q^{-n}\) or \(a=q^n\) case of the right-hand side of (4.1). This proves the theorem. \(\square \)

Proof of Theorem 1.2

Since \(\gcd (n,d)=1\), the factors related to a in the denominators of the left-hand side of (4.1) are relatively prime to \(\Phi _n(q)\) for \(a=1\). Thus, letting \(a=1\) in (4.1), we conclude that (1.8) is true modulo \(\Phi _n(q)^2\). \(\square \)

5 A parametric generalization and proof of Theorem 1.3

We need the following result.

Lemma 5.1

Let \(d\geqslant 3\) be an integer and let \(n\equiv 1\pmod {d}\) be a positive integer. Then, for any integer A, modulo \(\Phi _n(q)^2\),

$$\begin{aligned} (q^{1+An},q^{d+An};q^d)_{(n-1)/d}\equiv q^{An(n-1)/d} (q,q^d;q^d)_{(n-1)/d} \pmod {\Phi _n(q)^2}. \end{aligned}$$
(5.1)

Proof

The proof is similar to that of Lemma 3.1. Here we only give its parametric version of (5.1):

$$\begin{aligned} (aq,aq^d;q^d)_{(n-1)/d}\equiv a^{(n-1)/d}(q,q^d;q^d)_{(n-1)/d} \pmod {(1-a)(1-aq^n)}. \end{aligned}$$

\(\square \)

We have the following parametric generalization of Theorem 1.3.

Theorem 5.2

Let \(d\geqslant 3\) be an integer and let \(n\equiv 1\pmod {d}\) be a positive integer. Then, modulo \(\Phi _n(q)(1-aq^n)(a-q^n)\),

$$\begin{aligned}&\sum _{k=0}^{M}[2dk+d-1]\frac{(a^{d-1}q^{d-1}, a^{d-2}q^{d-1},\ldots , aq^{d-1},q^{d-1};q^d)_k}{(a^{d-1}q^d, a^{d-2}q^d,\ldots ,aq^d,q^d;q^d)_k}\nonumber \\&\qquad \times \frac{(a^{1-d}q^{d-1}, a^{2-d}q^{d-1},\ldots , a^{-1}q^{d-1},q^{d-1};q^d)_k }{(a^{1-d}q^d, a^{2-d}q^d,\ldots , a^{-1}q^d,q^d;q^d)_k } \nonumber \\&\quad \equiv [(d-1)n]\frac{(q^{d-1},q^d;q^d)_{(d-1)(n-1)/d}\,q^{-(d-1)(n-1)/2}}{\prod _{j=1}^{d-1}(q^{d+(d-j-1)n},q^{1+(d-j-1)n};q^d)_{(n-1)/d}}, \end{aligned}$$
(5.2)

where \(M=(d-1)(n-1)/d\) or \(M=n-1\).

Proof

For \(a=q^{-n}\) or \(a=q^n\), the left-hand side of (5.2) can be written as

$$\begin{aligned}&\sum _{k=0}^{(d-1)(n-1)/d}[2dk+d-1]\frac{(q^{d-1-(d-1)n}, q^{d-1-(d-2)n}, \ldots ,q^{d-1-n},q^{d-1};q^d)_k }{(q^{d-(d-1)n}, q^{d-(d-2)n},\ldots , q^{d-n},q^d;q^d)_k } \nonumber \\&\quad \qquad \times \frac{(q^{(d-1)n+d-1}, q^{(d-2)n+d-1},\ldots ,q^{n+d-1},q^{d-1};q^d)_k }{(q^{(d-1)n+d}, q^{(d-2)n+d},\ldots , q^{n+d}, q^d;q^d)_k }, \end{aligned}$$
(5.3)

where we have utilized \((q^{d-1-(d-1)n};q^d)_k=0\) for \(k>(d-1)(n-1)/d\). Performing the parameter substitutions \(N=(d-1)(n-1)/d\), \(q\mapsto q^d\), \(a=q^{d-1}\), \(m=d-1\), \(e_j=q^{d-1-(d-j-1)n}\) (\(1\leqslant j\leqslant m\)), \(n_1=\cdots =n_{m}=(n-1)/d\) in (2.3), we see that (5.3) is equal to the right-hand side of (5.2), where we have used the relation

$$\begin{aligned}{}[d-1](q^{2d-1};q^d)_{(d-1)(n-1)/d}=[(d-1)n](q^{d-1};q^d)_{(d-1)(n-1)/d}. \end{aligned}$$

This proves that (5.2) is true modulo \((1-aq^n)(a-q^n)\).

For \(M=(d-1)(n-1)/d\), by Lemma 2.1, we can easily verify that

$$\begin{aligned}&[2d(M-k)+d-1]\frac{(a^{d-1}q^{d-1}, a^{d-2}q^{d-1},\ldots , aq^{d-1},q^{d-1};q^d)_{M-k}}{(a^{d-1}q^d, a^{d-2}q^d,\ldots ,aq^d,q^d;q^d)_{M-k}}\\&\qquad \times \frac{(a^{1-d}q^{d-1}, a^{2-d}q^{d-1},\ldots , a^{-1}q^{d-1},q^{d-1};q^d)_{M-k} }{(a^{1-d}q^d, a^{2-d}q^d,\ldots , a^{-1}q^d,q^d;q^d)_{M-k} } \\&\quad \equiv -[2dk+d-1]\frac{(a^{d-1}q^{d-1}, a^{d-2}q^{d-1},\ldots , aq^{d-1},q^{d-1};q^d)_k}{(a^{d-1}q^d, a^{d-2}q^d,\ldots ,aq^d,q^d;q^d)_k}\\&\qquad \times \frac{(a^{1-d}q^{d-1}, a^{2-d}q^{d-1},\ldots , a^{-1}q^{d-1},q^{d-1};q^d)_k }{(a^{1-d}q^d, a^{2-d}q^d,\ldots , a^{-1}q^d,q^d;q^d)_k } \pmod {\Phi _n(q)}. \end{aligned}$$

Since \([(d-2)n]\equiv 0\pmod {\Phi _n(q)}\), we conclude that (5.2) holds modulo \(\Phi _n(q)\). Further, for any k in the range \((d-1)(n-1)/d<k\leqslant n-1\), we have \((q^{d-1};q^d)_k/(q^d;q^d)_k\equiv 0\pmod {\Phi _n(q)}\). This means that the q-congruence (5.2) also holds modulo \(\Phi _n(q)\) for \(M=n-1\). \(\square \)

Proof of Theorem 1.3

Since \(\gcd (n,d)=1\), the factors related to a in the denominators of the left-hand side of (5.2) are coprime with \(\Phi _n(q)\) for \(a=1\). Thus, letting \(a=1\) in (5.2), we conclude that

$$\begin{aligned}&\sum _{k=0}^{M}[2dk+d-1]\frac{(q^{d-1};q^d)_k^{2d}}{(q^d;q^d)_k^{2d}} \nonumber \\&\quad \qquad \equiv [(d-1)n]\frac{(q^{d-1},q^d;q^d)_{(d-1)(n-1)/d}\, q^{-(d-1)(n-1)/2}}{\prod _{j=1}^{d-1} (q^{d+(d-j-1)n},q^{1+(d-j-1)n};q^d)_{(n-1)/d}}\pmod {\Phi _n(q)^3}, \end{aligned}$$
(5.4)

Moreover, by repeatedly using Lemma 5.1\(d-1\) times, we obtain

$$\begin{aligned}&\prod _{j=1}^{d-1}(q^{d+(d-j-1)n},q^{1+(d-j-1)n};q^d)_{(n-1)/d} \\&\quad \equiv (q,q^d;q^d)_{(n-1)/d}^{d-1}\, q^{(d-1)(d-2)n(n-1)/(2d)} \pmod {\Phi _n(q)^2}. \end{aligned}$$

Substituting this into (5.4), we see that the q-congruence (1.10) is true modulo \(\Phi _n(q)^3\). Finally, the validity of (1.10) modulo [n] follows from [7, Lemma 4.1]. \(\square \)

6 Concluding remarks

The present authors [7, Theorem 1.1] gave the following result: Let d, n, r be integers satisfying \(d\geqslant 2\), \(r\leqslant d-2\) (in particular, r may be negative), and \(n\geqslant d-r\), such that d and r are coprime, and \(n\equiv -r\pmod {d}\). Then

$$\begin{aligned}&\sum _{k=0}^{M}[2dk+r]\frac{(q^r;q^d)_k^4}{(q^d;q^d)_k^4}q^{(d-2r)k} \nonumber \\&\quad \equiv {\left\{ \begin{array}{ll} 0 \pmod {[n]\Phi _n(q)^3} &{}\text {if }d=2,\\ q^{r(n+r-dn)/d}\dfrac{(q^{2r};q^d)_{(dn-n-r)/d}}{(q^d;q^d)_{(dn-n-r)/d}}[dn-n] \pmod {[n]\Phi _n(q)^3} &{}\text {if }d\geqslant 3, \end{array}\right. } \end{aligned}$$
(6.1)

where \(M=(dn-n-r)/d\) or \(n-1\).

Here we point out that (6.1) is also true for \(d\geqslant 3\) and \(r=d-1\) (and thus \(n\equiv 1\pmod {d}\)). This is because the proof of [7, Theorem 3.1] is also valid for \(r=d-1\), since \(2r+2n\leqslant dn\) still holds for \(n\geqslant d+1\) in this case [(6.1) clearly holds for \(n=1\)].