Proof of two conjectures of Guo and Schlosser

Abstract

Let $[n]=(1-q^n)/(1-q)$ denote the q-integer and $\Phi _n(q)$ the nth cyclotomic polynomial in q. Recently, Guo and Schlosser provided two conjectures: For any odd integer $n>3$, modulo $[n]\Phi _n(q)(1-aq^n)(a-q^n)$,

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

and modulo $\Phi _n(q)^2(1-aq^n)(a-q^n)$,

$$\begin{aligned} \sum _{k=0}^{(n+1)/2}[4k+1]\frac{(aq^{-1};q^2)_k(q^{-1}/a;q^2)_k(q^{-1};q^2)_k(q;q^2)_k}{(aq^4;q^2)_k(q^4/a;q^2)_k(q^4;q^2)_k(q^2;q^2)_k}q^{6k} \equiv 0, \end{aligned}$$

where $(a;q)_k=(1-a)(1-aq)\cdots (1-aq^{k-1})$. In this paper, we confirm these two conjectures and further give their generalizations involving two free parameters. Our proof uses Guo and Zudilin’s ‘creative microscoping’ method and the Chinese remainder theorem for coprime polynomials.

1 Introduction

In the past few years, many q-congruences and q-supercongruences have been established by different authors. See, for example, [2, 4,5,6,7,8, 11,12,13,14,15,16,17,18,19,20,21,22,23,24,25]. In particular, Guo and Schlosser [9, Theorem 5.5] proved that, for any odd integer $n>3$,

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

(1.1)

where the q-shifted factorial is defined by

$$\begin{aligned} (a;q)_\infty =\prod _{j \geqslant 0}(1-aq^j) \quad \mathrm {and} \quad (a;q)_k = \frac{(a;q)_\infty }{(aq^k;q)_\infty }. \end{aligned}$$

For convenience, we adopt the notation

$$\begin{aligned} (a_1,a_2 ,\ldots ,a_r;q)_k=(a_1;q)_k(a_2;q)_k \cdots (a_r;q)_k, \quad k \in \mathbb {C} \cup \infty . \end{aligned}$$

The q-integer is defined by $[n]=[n]_q=(1-q^n)/(1-q)=1+q+\cdots +q^{n-1}$. Moreover, $\Phi _n(q)$ represents the nth cyclotomic polynomial in q, which may be defined as

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

with $\zeta$ being an nth primitive root of unity.

Guo and Schlosser [9, Conjecture 5.6] also proposed the following conjecture: for any odd integer $n>3$,

$$\begin{aligned} \sum _{k=0}^{(n+1)/2}[4k+1]\frac{(aq^{-1};q^2)_k(q^{-1}/a;q^2)_k(q;q^2)_k^2}{(aq^4;q^2)_k(q^4/a;q^2)_k(q^2;q^2)_k^2}q^{4k}\equiv 0 \pmod {[n]\Phi _n(q)(1-aq^n)(a-q^n)}. \end{aligned}$$

(1.2)

It is known from [9] that, letting $a \rightarrow 1$ in (1.2), we are led to

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

which is a refinement of the q-supercongruence (1.1).

Additionally, Guo and Schlosser [9, Conjecture 5.8] provided another similar conjecture: for any odd integer $n>3$,

$$\begin{aligned}&\sum _{k=0}^{(n+1)/2}[4k+1]\frac{(aq^{-1};q^2)_k(q^{-1}/a;q^2)_k(q^{-1},q;q^2)_k}{(aq^4;q^2)_k(q^4/a;q^2)_k(q^4,q^2;q^2)_k}q^{6k}\equiv 0 \pmod {\Phi _n(q)^2(1-aq^n)(a-q^n)}. \end{aligned}$$

(1.3)

Likewise, when $a \rightarrow 1$ in (1.3), we obtain

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

Inspired by Guo and Schlosser’s work, we shall prove these two conjectures in this paper. Our proof relies on the following theorem, which may be deemed as a generalization of the q-supercongruences (1.2) and (1.3).

Theorem 1

Let $n>3$ be an odd integer and $r\in \{0,2\}$. Then, modulo $\Phi _n(q)^2(1-aq^n)(a-q^n)$, we have

$$\begin{aligned}&\sum _{k=0}^{(n+1)/2}[4k+1]\frac{(aq^{-1},q^{-1}/a,q^{1-r},cq,dq,q;q^{2})_k}{(q^4/a,aq^4,q^{2+r},q^2/c,q^2/d,q^2;q^{2})_k}\bigg (\frac{q^{5+r}}{cd}\bigg )^k\nonumber \\&\qquad \equiv [n]Z_q(a,n)\frac{(aq^{-1},q^{-1}/a,q^{1-r},q/cd;q^2)_{(r+4)/2}}{(q,q^{r+2};q^2)_2(q^2/c,q^2/d;q^2)_{(r+4)/2}}\nonumber \\&\qquad \times \sum _{k=0}^{\frac{n-r-3}{2}}\frac{(aq^{r+3},q^{r+3}/a,q^5,q^{r+5}/cd;q^{2})_k}{(q^2,q^{6+r},q^{6+r}/c,q^{6+r}/d;q^{2})_k}q^{2k}, \end{aligned}$$

(1.4)

where

$$\begin{aligned} Z_q(a,n)&=(-1)^{\frac{r+2}{2}}q^{\frac{-2n+r^2+10r+26}{4}}\Bigg \{\frac{n(1-aq^2)(1-q^2/a)(1-a)a^{(n-1)/2}}{(aq,q/a;q^2)_{(r+2)/2}(1-a^n)}-\frac{(1-q^2)^2}{(q;q^2)_{(r+2)/2}^2}\Bigg \}\\&\quad \times \frac{(1-aq^n)(a-q^n)}{(1-a)^2}+q^{\frac{(r-1)(n+1)}{2}+r+8}\frac{(q^{n-2};q^4)_{2}}{(q^{n-r-1};q^2)_{r+2}}. \end{aligned}$$

Further, for the $r=0$ and $a \rightarrow 1$ case of Theorem 1, we have the following stronger conclusion.

Theorem 2

Let $n>3$ be an odd integer. Then

$$\begin{aligned}&\sum _{k=0}^{M}[4k+1]\frac{(q^{-1};q^2)_k^2(q;q^2)_k^2(cq,dq;q^{2})_k}{(q^{4};q^2)_k^2(q^2;q^2)_k^2(q^2/c,q^2/d;q^{2})_k}\bigg (\frac{q^{5}}{cd}\bigg )^k\nonumber \\&\quad \equiv S_q(n)\frac{(q^{-1};q^2)_2^2(q/cd;q^2)_{2}}{(q^{2},q^2/c,q^2/d;q^2)_{2}}\nonumber \\&\quad \times \sum _{k=0}^{\frac{n-3}{2}}\frac{(q^{3};q^2)_k^2(q^{5},q^5/cd;q^{2})_k}{(q^2,q^6,q^6/c,q^6/d;q^{2})_k}q^{2k} \pmod {[n]\Phi _n(q)^3}, \end{aligned}$$

(1.5)

where $M \in \{(n+1)/2,n-2\}$ and

$$\begin{aligned} S_q(n)=&\, q^{\frac{13-n}{2}}\frac{n^2(1-q^2)^2-(1+24q+22q^2+24q^3+q^4)}{24}[n]^3\\&+q^{\frac{15-n}{2}}\frac{[n-2][n+2]}{[n-1][n+1]}[n]. \end{aligned}$$

Putting $c \rightarrow 1$ and $d \rightarrow \infty$ in Theorem 2, we obtain the following result.

Corollary 3

Let $n>3$ be an odd integer. Then, modulo $[n]\Phi _n(q)^3$,

$$\begin{aligned}&\sum _{k=0}^{M}(-1)^k[4k+1]\frac{(q^{-1};q^2)_k^2(q;q^2)_k^3}{(q^{4};q^2)_k^2(q^2;q^2)_k^3}q^{5k+k^2}\nonumber \\&\quad \equiv S_q(n)\frac{(q^{-1};q^2)_2^2}{(q^{2};q^2)_{2}^2} \sum _{k=0}^{\frac{n-3}{2}}\frac{(q^{3};q^2)_k^2(q^{5};q^{2})_k}{(q^2;q^2)_k(q^6;q^{2})_k^2}q^{2k}. \end{aligned}$$

(1.6)

When $M=(p^l+1)/2$ and $q \rightarrow 1$ in (1.6), we are led to the following: for odd prime $p>3$,

$$\begin{aligned} \sum _{k=0}^{(p^l+1)/2}(-1)^k(4k+1)\frac{(-\frac{1}{2})_k^2(\frac{1}{2})_k^3}{k!^3(k+1)!^2}\equiv \frac{p^l}{4}\sum _{k=0}^{(p^l-3)/2}\frac{(\frac{3}{2})_k^2(\frac{5}{2})_k}{k!(k+2)!^2} \pmod {p^{l+3}}, \end{aligned}$$

where l is a positive integer, and the notation will be used frequently in this section.

Moreover, the case $c \rightarrow 1$, $d \rightarrow 1$ of Theorem 2 yields the following q-supercongruence.

Corollary 4

Let $n>3$ be an odd integer. Then, modulo $[n]\Phi _n(q)^3$,

$$\begin{aligned}&\sum _{k=0}^{M}[4k+1]\frac{(q^{-1};q^2)_k^2(q;q^2)_k^4}{(q^{4};q^2)_k^2(q^2;q^2)_k^4}q^{5k}\nonumber \\&\quad \equiv S_q(n)\frac{(q^{-1};q^2)_2^2(q;q^2)_2}{(q^{2};q^2)_{2}^3}\sum _{k=0}^{\frac{n-3}{2}}\frac{(q^{3};q^2)_k^2(q^{5};q^{2})_k^2}{(q^2;q^2)_k(q^6;q^{2})_k^3}q^{2k}. \end{aligned}$$

(1.7)

Putting $M=(p^l+1)/2$ and $q \rightarrow 1$ in (1.7), we gain the following: for odd prime $p>3$,

$$\begin{aligned} \sum _{k=0}^{(p^l+1)/2}(4k+1)\frac{(-\frac{1}{2})_k^2(\frac{1}{2})_k^4}{k!^4(k+1)!^2}\equiv \frac{3p^l}{16}\sum _{k=0}^{(p^l-3)/2}\frac{(\frac{3}{2})_k^2(\frac{5}{2})_k^2}{k!(k+2)!^3} \pmod {p^{l+3}}. \end{aligned}$$

In addition, letting $r=2$ and $a \rightarrow 1$ in Theorem 1, and applying the L’Hospital rule, we arrive at the following conclusion.

Theorem 5

Let $n>3$ be an odd integer. Then

$$\begin{aligned}&\sum _{k=0}^{(n+1)/2}[4k+1]\frac{(q^{-1};q^2)_k^3(q,cq,dq;q^{2})_k}{(q^{4};q^2)_k^3(q^2,q^2/c,q^2/d;q^{2})_k}\bigg (\frac{q^{7}}{cd}\bigg )^k\nonumber \\&\quad \equiv T_q(n)\frac{(q^{-1};q^2)_3^3(q/cd;q^2)_{3}}{(q,q^{4};q^2)_2(q^2/c,q^2/d;q^2)_{3}}\nonumber \\&\quad \times \sum _{k=0}^{\frac{n-5}{2}}\frac{(q^{5};q^2)_k^3(q^7/cd;q^{2})_k}{(q^2,q^8,q^8/c,q^8/d;q^{2})_k}q^{2k}\pmod {\Phi _n(q)^4}, \end{aligned}$$

(1.8)

where

$$\begin{aligned} T_q(n)&=q^{\frac{25-n}{2}}\frac{(n^2-1)(1+q)^2(1-q^3)^2-24q(1+3q+7q^2+9q^3+7q^4+3q^5+q^6)}{24(1-q^3)^2(1+q+q^2)^2}[n]^3\\&\quad +q^{\frac{n+21}{2}}\frac{(q^{n-2};q^4)_{2}}{(q^{n-3};q^2)_4}[n]. \end{aligned}$$

Taking $c \rightarrow q^{-2}$ and $d \rightarrow \infty$ in Theorem 5, we obtain the following result.

Corollary 6

Let $n>3$ be an odd integer. Then

$$\begin{aligned}&\sum _{k=0}^{(n+1)/2}(-1)^k[4k+1]\frac{(q^{-1};q^2)_k^4(q;q^2)_k}{(q^{4};q^2)_k^4(q^2;q^2)_k}q^{9k+k^2}\nonumber \\&\qquad \equiv T_q(n)\frac{(q^{-1};q^2)_3^3}{(q,q^4;q^2)_{2}(q^4;q^2)_3}\nonumber \\&\qquad \times \sum _{k=0}^{\frac{n-5}{2}}\frac{(q^{5};q^{2})_k^3}{(q^2,q^8,q^{10};q^{2})_k}q^{2k}\pmod {\Phi _n(q)^4}. \end{aligned}$$

(1.9)

Letting $n=p^l$ be an odd prime power greater than 3 and $q \rightarrow 1$ in (1.9), we get

$$\begin{aligned} \sum _{k=0}^{(p^l+1)/2}(-1)^k(4k+1)\frac{(-\frac{1}{2})_k^4(\frac{1}{2})_k}{k!(k+1)!^4} \equiv&\,\frac{p^l}{8}\sum _{k=0}^{(p^l-5)/2}\frac{(\frac{5}{2})_k^3}{k!(k+3)!(k+4)!} \pmod {p^{4}}. \end{aligned}$$

Likewise, putting $c \rightarrow q^{-2}$ and $d \rightarrow q^{-2}$ in Theorem 5, we get the following q-supercongruence.

Corollary 7

Let $n>3$ be an odd integer. Then

$$\begin{aligned}&\sum _{k=0}^{(n+1)/2}[4k+1]\frac{(q^{-1};q^2)_k^5(q;q^2)_k}{(q^{4};q^2)_k^5(q^2;q^2)_k}q^{11k}\nonumber \\&\quad \equiv T_q(n)\frac{(q^{-1};q^2)_3^3(q^{5};q^2)_3}{(q,q^4;q^2)_{2}(q^4;q^2)_3^2}\nonumber \\&\quad \times \sum _{k=0}^{\frac{n-5}{2}}\frac{(q^{5};q^{2})_k^3(q^{11};q^2)_k}{(q^2,q^8;q^2)_k(q^{10};q^{2})_k^2}q^{2k}\pmod {\Phi _n(q)^4}. \end{aligned}$$

(1.10)

When $n=p^r$ is an odd prime power greater than 3 and $q \rightarrow 1$ in (1.10), we get

$$\begin{aligned} \sum _{k=0}^{(p^l+1)/2}(4k+1)\frac{(-\frac{1}{2})_k^5(\frac{1}{2})_k}{k!(k+1)!^5} \equiv&\frac{315p^l}{64}\sum _{k=0}^{(p^l-5)/2}\frac{(\frac{5}{2})_k^3(\frac{11}{2})_k}{k!(k+3)!(k+4)!^2} \pmod {p^{4}}. \end{aligned}$$

The rest of this paper is arranged as follows. Firstly, we present some lemmas which will be needed in the proof of Theorem 1 in Sect. 2. Then, we prove Theorem 1 and Guo and Schlosser’s conjectures (1.2) and (1.3) in Sect. 3. Our proof makes use of the ‘creative microscoping’ method which was recently introduced by Guo and Zudilin [10], and the Chinese remainder theorem for coprime polynomials. Finally, in Sect. 4, we give a proof of Theorem 2.

2 Some Lemmas

We shall make use of Watson’s $_8\phi _7$ transformation [1, Appendix (III.18)], which can be expressed as

$$\begin{aligned}&_8\phi _7 \left[ \begin{array}{l} a,\,\,\,\,\, qa^{\frac{1}{2}},\,\,\,\,\,\, \,-qa^{\frac{1}{2}},\,\,\,\,\,\, \,\,\,b, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,d,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,e\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,q^{-N} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,a^{\frac{1}{2}},\,\,\,\,\,\,\,\, -a^{\frac{1}{2}},\,\,\,\,\,\,\,\,a q/b,\,\,\,\,\,\,\,\,a q/c,\,\,\,\,\,\,\,\,a q/d, \,\,\,\,\,\,\,\,aq/e, \,\,\,\,\,\,\,\,aq^{N+1}\end{array}; q,\,\frac{a^2q^{N+2}}{bcde} \right] \nonumber \\&\quad =\frac{(aq,aq/de;q)_{N}}{(aq/d,aq/e;q)_N}{_4\phi _3} \left[ \begin{array}{l} aq/bc,\,\, \,\,\,\, d,\,\,\,\,\,\,\, \,\,\,\,\,e,\,\,\,\,\,\,\,\,\,\,\,\, q^{-N} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, aq/b,\,\,\, aq/c,\,\,\,deq^{-N}/a\, \end{array}; q,\,q \right] , \end{aligned}$$

(2.1)

where the basic hypergeometric series $_{s+1}\phi _s$ is defined as

$$\begin{aligned} _{s+1}\phi _{s}\left[ \begin{array}{c} a_1,a_2,\ldots ,a_{s+1}\\ b_1,b_2,\ldots ,b_{s} \end{array};q,\, z \right] =\sum _{k=0}^{\infty }\frac{(a_1,a_2,\ldots , a_{s+1};q)_k z^k}{(q,b_1,\ldots ,b_{s};q)_k} \end{aligned}$$

with $0<|z|<1$.

In this section, we shall give three lemmas, which will play an important role in our proof of Theorem 1.

Lemma 1

Let $n>3$ be an odd integer and $r\in \{0,2\}$. Then

$$\begin{aligned} \sum _{k=0}^{M}[4k+1]\frac{(aq^{-1},q^{-1}/a,q^{1-r}/b,cq,dq,q;q^{2})_k}{(q^4/a,aq^4,bq^{2+r},q^2/c,q^2/d,q^2;q^{2})_k}\bigg (\frac{bq^{5+r}}{cd}\bigg )^k \equiv 0 \pmod {\Phi _n(q)}, \end{aligned}$$

(2.2)

where $M \in \{(n+1)/2,n-2\}$.

Proof

Let $c_q(k)$ be the kth term on left-hand side of (2.2), i.e.,

$$\begin{aligned} c_q(k)=[4k+1]\frac{(aq^{-1},q^{-1}/a,q^{1-r}/b,cq,dq,q;q^{2})_k}{(q^4/a,aq^4,bq^{2+r},q^2/c,q^2/d,q^2;q^{2})_k}\bigg (\frac{bq^{5+r}}{cd}\bigg )^k. \end{aligned}$$

Using the following q-congruence due to Guo and Schlosser [9, Lemma 3.1]

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

we have

$$\begin{aligned} c_q(k) \equiv -c_q((n-1)/2-k) \pmod {\Phi _n(q)}. \end{aligned}$$

This proves that

$$\begin{aligned} \begin{aligned}&\sum _{k=0}^{(n-1)/2}[4k+1]\frac{(aq^{-1},q^{-1}/a,q^{1-r}/b,cq,dq,q;q^{2})_k}{(q^4/a,aq^4,bq^{2+r},q^2/c,q^2/d,q^2;q^{d})_k}\bigg (\frac{bq^{5+r}}{cd}\bigg )^k \equiv 0\quad \pmod {\Phi _n(q)}. \end{aligned} \end{aligned}$$

(2.3)

Since the numerator of $c_q(k)$ contains the factor $(q;q^2)_k$, we see that $c_q(k)$ is congruent to 0 modulo $\Phi _n(q)$ for $(n+1)/2\le k\le n-2$. Thus, the proof of Lemma 1 is completed. $\square$

Lemma 2

Let $n>3$ be an odd integer and $r\in \{0,2\}$. Then

$$\begin{aligned} \begin{aligned}&\sum _{k=0}^{M}[4k+1]\frac{(aq^{-1},q^{-1}/a,q^{1-r}/b,cq,dq,q;q^{2})_k}{(q^4/a,aq^4,bq^{2+r},q^2/c,q^2/d,q^2;q^{2})_k}\bigg (\frac{bq^{5+r}}{cd}\bigg )^k\\&\quad \equiv [n]b^{\frac{n+1}{2}}q^{\frac{(r-1)(n+1)}{2}}\frac{(1-q^{n-2})(1-q^{n+2})(q^{-2-r}/b;q^2)_{(n+1)/2}}{(1-q^{-3})(1-q^{-1})(bq^{2+r};q^2)_{(n+1)/2}}\\&\qquad \times \sum _{k=0}^{(n+1)/2}\frac{(aq^{-1},q^{-1}/a,q^{1-r}/b,q/cd;q^{2})_k}{(q^2,q^2/c,q^2/d,q^{-2-r}/b;q^{2})_k}q^{2k} \pmod {\Phi _n(q)(1-aq^n)(a-q^n)}, \end{aligned} \end{aligned}$$

(2.4)

where $M \in \{(n+1)/2,n-2\}$.

Proof

When $a=q^{-n}$ or $q^n$, the left-hand side of (2.4) equals

$$\begin{aligned}&_8\phi _7 \left[ \begin{array}{l} q,\,\,\,\,\,\,\, q^{\frac{5}{2}},\,\,\,\,\,\,\,\, \,-q^{\frac{5}{2}},\,\,\,\,\,\, \,\,\,cq, \,\,\,\,\,\,\,\,\,\,\,\,\,dq,\,\,\,\,\,\,\,\,\,\,\,\,\,q^{1-r}/b,\,\,\,\,\,\,q^{-1-n},\,\,\,\,\,\,\,\,\,q^{-1+n} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,q^{\frac{1}{2}},\,\,\,\,\,\,\,\, -q^{\frac{1}{2}},\,\,\,\,\,\,\,\,q^2/c,\,\,\,\,\,\,\,\,q^2/d,\,\,\,\,\,\,\,\,\,\,bq^{2+r}, \,\,\,\,\,\,\,\,\,\,q^{4+n}, \,\,\,\,\,\,\,\,\,\,\,\,q^{4-n}\end{array}; q^2,\,\frac{bq^{5+r}}{cd} \right] . \end{aligned}$$

Applying Watson’s $_8\phi _7$ transformation (2.1), we obtain

$$\begin{aligned}&\sum _{k=0}^{M}[4k+1]\frac{(q^{-1+n},q^{-1-n},q^{1-r}/b,cq,dq,q;q^{2})_k}{(q^{4-n},aq^{4+n},bq^{2+r},q^2/c,q^2/d,q^2;q^{2})_k}\bigg (\frac{bq^{5+r}}{cd}\bigg )^k\\&\quad =\frac{(q^3,bq^{3+r-n};q^2)_{(n+1)/2}}{(bq^{2+r},q^{4-n};q^2)_{(n+1)/2}}\sum _{k=0}^{(n+1)/2}\frac{(q^{-1+n},q^{-1-n},q^{1-r}/b,q/cd;q^{2})_k}{(q^2,q^2/c,q^2/d,q^{-2-r}/b;q^{2})_k}q^{2k} \\&\quad =[n]b^{\frac{n+1}{2}}\frac{(1-q^{n-2})(1-q^{n+2})(q^{-2-r}/b;q^2)_{(n+1)/2}}{(1-q^{-3})(1-q^{-1})(bq^{2+r};q^2)_{(n+1)/2}}q^{\frac{(r-1)(n+1)}{2}}\\ \nonumber&\qquad \times \sum _{k=0}^{(n+1)/2}\frac{(q^{-1+n},q^{-1-n},q^{1-r}/b,q/cd;q^{2})_k}{(q^2,q^2/c,q^2/d,q^{-2-r}/b;q^{2})_k}q^{2k}, \end{aligned}$$

which means that the q-supercongruence (2.4) holds modulo $1-aq^n$ and $a-q^n$. The proof then follows from Lemma 1 and the fact that $\Phi _n(q)$, $1-aq^n$ and $a-q^n$ are pairwise relatively prime polynomials. $\square$

Lemma 3

Let $n>3$ be an odd integer and $r\in \{0,2\}$. Then

$$\begin{aligned} \begin{aligned}&\sum _{k=0}^{M}[4k+1]\frac{(aq^{-1},q^{-1}/a,q^{1-r}/b,cq,dq,q;q^{2})_k}{(q^4/a,aq^4,bq^{2+r},q^2/c,q^2/d,q^2;q^{2})_k}\bigg (\frac{bq^{5+r}}{cd}\bigg )^k\\&\quad \equiv [n]\frac{(bq^2;q^2)_{r/2}( b;q^2)_{(r+4)/2}(q;q^2)_{(n-1)/2}^2}{(q;q^2)_{2}(q^4/a;q^2)_{(n+r-1)/2}(aq^4;q^2)_{(n+r-1)/2}}\\&\quad \times \sum _{k=0}^{(n+r-1)/2}\frac{(aq^{-1},q^{-1}/a,q^{1-r}/b,q/cd;q^{2})_k}{(q^2,q^2/c,q^2/d,q^{-2-r}/b;q^{2})_k}q^{2k} \pmod {b-q^n}, \end{aligned} \end{aligned}$$

where $M \in \{(n+1)/2,n-2\}$.

Proof

Letting $b=q^n$ in the left-hand side of the above relation, we have

$$\begin{aligned}&\sum _{k=0}^{M}[4k+1]\frac{(aq^{-1},q^{-1}/a,q^{1-r-n},cq,dq,q;q^{2})_k}{(q^4/a,aq^4,q^{2+r+n},q^2/c,q^2/d,q^2;q^{2})_k}\bigg (\frac{q^{5+r+n}}{cd}\bigg )^k\\&\quad =\frac{[n](q^{2+n};q^2)_{r/2}( q^n;q^2)_{(r+4)/2}(q;q^2)_{(n-1)/2}^2}{(q;q^2)_{2}(q^4/a;q^2)_{(n+r-1)/2}(aq^4;q^2)_{(n+r-1)/2}}\\&\qquad \times \sum _{k=0}^{\frac{n+r-1}{2}}\frac{(aq^{-1},q^{-1}/a,q^{1-r-n},q/cd;q^{2})_k}{(q^2,q^2/c,q^2/d,q^{-2-r-n};q^{2})_k}q^{2k}, \end{aligned}$$

which follows from the substitutions $a= q, q \mapsto q^2, b= cq, c = dq, d = aq^{-1}, e = q^{-1}/a$, and $N = (n+r-1)/2$ in Watson’s $_8\phi _7$ transformation (2.1). Namely, Lemma 3 is true. $\square$

3 Proof of Theorem 1

Firstly, we need the following two q-congruences:

$$\begin{aligned}&\frac{(b-q^n)(ab-1-a^2+aq^n)}{(a-b)(1-ab)} \equiv 1 \pmod {(1-aq^n)(a-q^n)}, \\&\frac{(1-aq^n)(a-q^n)}{(a-b)(1-ab)} \equiv 1 \pmod {(b-q^n)}, \end{aligned}$$

which can be found in Guo [3]. Employing the Chinese remainder theorem for coprime polynomials and combining Lemmas 2 and 3, we conclude that, modulo $\Phi _n(q)(1-aq^n)(a-q^n)(b-q^n)$,

$$\begin{aligned}&\sum _{k=0}^{(n+1)/2}[4k+1]\frac{(aq^{-1},q^{-1}/a,q^{1-r}/b,cq,dq,q;q^{2})_k}{(q^4/a,aq^4,bq^{2+r},q^2/c,q^2/d,q^2;q^{2})_k}\bigg (\frac{bq^{5+r}}{cd}\bigg )^k\nonumber \\&\qquad \equiv [n]W_q(a,b,n)\sum _{k=0}^{\frac{n+1}{2}}\frac{(aq^{-1},q^{-1}/a,q^{1-r}/b,q/cd;q^{2})_k}{(q^2,q^2/c,q^2/d,q^{-2-r}/b;q^{2})_k}q^{2k}, \end{aligned}$$

(3.1)

where

$$\begin{aligned} W_q(a,b,n)&=b^{\frac{n+1}{2}}\frac{(1-q^{n-2})(1-q^{n+2})(q^{-2-r}/b;q^2)_{(n+1)/2}}{(1-q^{-3})(1-q^{-1})(bq^{2+r};q^2)_{(n+1)/2}}q^{\frac{(r-1)(n+1)}{2}}\\&\quad \times \frac{(b-q^n)(ab-1-a^2+aq^n)}{(a-b)(1-ab)}\\&\quad +\frac{(bq^2;q^2)_{r/2}( b;q^2)_{(r+4)/2}(q;q^2)_{(n-1)/2}^2}{(q;q^2)_{2}(q^4/a;q^2)_{(n+r-1)/2}(aq^4;q^2)_{(n+r-1)/2}} \frac{(1-aq^n)(a-q^n)}{(a-b)(1-ab)}. \end{aligned}$$

Obviously, the q-supercongruence (3.1) can be expressed as modulo $\Phi _n(q)(1-aq^n)(a-q^n)(b-q^n)$,

$$\begin{aligned}&\sum _{k=0}^{(n+1)/2}[4k+1]\frac{(aq^{-1},q^{-1}/a,q^{1-r}/b,cq,dq,q;q^{2})_k}{(q^4/a,aq^4,bq^{2+r},q^2/c,q^2/d,q^2;q^{2})_k}\bigg (\frac{bq^{5+r}}{cd}\bigg )^k\nonumber \\&\quad \equiv [n] W_q(a,b,n) \sum _{k=0}^{(r+2)/2}\frac{(aq^{-1},q^{-1}/a,q^{1-r}/b,q/cd;q^{2})_k}{(q^2,q^2/c,q^2/d,q^{-2-r}/b;q^{2})_k}q^{2k}\nonumber \\&\qquad +[n] W_q(a,b,n)\sum _{k=0}^{(n-r-3)/2}\frac{(aq^{-1},q^{-1}/a,q^{1-r}/b,q/cd;q^{2})_{k+(r+4)/2}}{(q^2,q^2/c,q^2/d,q^{-2-r}/b;q^{2})_{k+(r+4)/2}}q^{2k+r+4} . \end{aligned}$$

(3.2)

Letting $b \rightarrow 1$ in (3.2) and using the relation

$$\begin{aligned} (1-q^n)(1+a^2-a-aq^n)=(1-a)^2+(1-aq^n)(a-q^n), \end{aligned}$$

we attain modulo $\Phi _n(q)(1-aq^n)(a-q^n)(b-q^n)$,

$$\begin{aligned}&\sum _{k=0}^{(n+1)/2}[4k+1]\frac{(aq^{-1},q^{-1}/a,q^{1-r},cq,dq,q;q^{2})_k}{(q^4/a,aq^4,q^{2+r},q^2/c,q^2/d,q^2;q^{2})_k}\bigg (\frac{q^{5+r}}{cd}\bigg )^k\nonumber \\&\quad \equiv [n] Y_q(a,n)\frac{(aq^{-1},q^{-1},q^{1-r},q/cd;q^2)_{(r+4)/2}}{(q,q^{r+2};q^2)_2(q^2/c,q^2/d;q^2)_{(r+4)/2}}q^{r+4}\nonumber \\&\qquad \times \sum _{k=0}^{(n-r-3)/2}\frac{(aq^{r+3},q^{r+3}/a,q^{1-r},q^{r+5}/cd;q^{5})_{k}}{(q^2,q^{6+r}/c,q^{6+r}/d,q^{6+r};q^{2})_{k}}q^{2k}, \end{aligned}$$

(3.3)

where

$$\begin{aligned} Y_q(a,n)&=\Bigg \{\frac{(q^{n-2};q^4)_2}{(q^{n-r-1};q^2)_{r+2}}q^{\frac{(r-1)(n+1)}{2}+4}+(-1)^{(r+2)/2}\frac{(q;q^2)_{(n-1)/2}^2}{(q^4/a,aq^4;q^2)_{(n+r-1)/2}} \Bigg \}\\&\quad \times \frac{(1-aq^n)(a-q^n)}{(1-a)^2}+\frac{(q^{n-2};q^4)_2}{(q^{n-r-1};q^2)_{r+2}}q^{\frac{(r-1)(n+1)}{2}+4}. \end{aligned}$$

Noting $q^n \equiv 1 \pmod {\Phi _n(q)}$ and recalling

$$\begin{aligned}&(aq^2,q^2/a;q^2)_{(n-1)/2}\equiv (-1)^{(n-1)/2}\frac{(1-a^n)q^{-(n-1)^2/4}}{(1-a)a^{(n-1)/2}} \pmod {\Phi _n(q)},\\&(aq,q/a;q^2)_{(n-1)/2}\equiv (-1)^{(n-1)/2}\frac{(1-a^n)q^{(1-n^2)/4}}{(1-a)a^{(n-1)/2}} \pmod {\Phi _n(q)}, \end{aligned}$$

which were first observed by Guo [3, Lemma 2.1], we are led to the following q-congruence: modulo $\Phi _n(q)$,

$$\begin{aligned}&\frac{(q^{n-2};q^4)_2}{(q^{n-r-1};q^2)_{r+2}}q^{\frac{(r-1)(n+1)}{2}+4}+(-1)^{(r+2)/2}\frac{(q;q^2)_{(n-1)/2}^2}{(q^4/a,aq^4;q^2)_{(n+r-1)/2}} \nonumber \\&\quad \equiv (-1)^{\frac{r+2}{2}}q^{\frac{-2n+r^2+6r+10}{4}}\Bigg \{\frac{n(1-aq^2)(1-q^2/a)(1-a)a^{(n-1)/2}}{(aq,q/a;q^2)_{(r+2)/2}(1-a^n)}-\frac{(1-q^2)^2}{(q;q^2)_{(r+2)/2}^2}\Bigg \}. \end{aligned}$$

(3.4)

The proof then follows from (3.3) and (3.4).

Obviously, Theorem 1 also holds true when the summation in the left-hand side of (1.4) is from 0 to $n-2$.

We now prove (1.2) and (1.3) which were conjectured by Guo and Schlosser.

Proof of (1.2)

When $cd=q$ and $r=0$ in Theorem 1, we have

$$\begin{aligned} \sum _{k=0}^{(n+1)/2}[4k+1]\frac{(aq^{-1};q^2)_k(q^{-1}/a;q^2)_k(q;q^2)_k^2}{(aq^4;q^2)_k(q^4/a;q^2)_k(q^2;q^2)_k^2}q^{4k} \equiv 0 \pmod {\Phi _n(q)^2(1-aq^n)(a-q^n)}. \end{aligned}$$

It remains to show that

$$\begin{aligned} \sum _{k=0}^{(n+1)/2}[4k+1]\frac{(aq^{-1};q^2)_k(q^{-1}/a;q^2)_k(q;q^2)_k^2}{(aq^4;q^2)_k(q^4/a;q^2)_k(q^2;q^2)_k^2}q^{4k} \equiv 0 \pmod {[n]}. \end{aligned}$$

(3.5)

For $n>3$, let $\zeta \ne 1$ be an nth root of unity, not necessarily primitive. Then $\zeta$ must be a primitive mth root of unity with m|n. Let $\alpha _q(k)$ denote the kth term on the left-hand side of (3.5):

$$\begin{aligned} \alpha _q(k)=[4k+1]\frac{(aq^{-1};q^2)_k(q^{-1}/a;q^2)_k(q;q^2)_k^2}{(aq^4;q^2)_k(q^4/a;q^2)_k(q^2;q^2)_k^2}q^{4k}. \end{aligned}$$

Putting $n=m$, $cd=q$, $r=0$ and $b \rightarrow 1$ in Lemma 1, and since $\alpha _\zeta (k)=0$ for $(m+1)/2<k \le m-1$, we are led to

$$\begin{aligned} \sum _{k=0}^{(m+1)/2}\alpha _\zeta (k)=\sum _{k=0}^{m-1}\alpha _\zeta (k)=0. \end{aligned}$$

Since

$$\begin{aligned} \frac{\alpha _\zeta (lm+k)}{\alpha _\zeta (lm)}=\lim _{q\rightarrow \zeta }\frac{\alpha _q(lm+k)}{\alpha _q(lm)}=\alpha _\zeta (k), \end{aligned}$$

we immediately obtain

$$\begin{aligned} \sum _{k=0}^{(n+1)/2}\alpha _\zeta (k)=\sum _{l=0}^{(n/m-3)/2}\alpha _\zeta (lm)\sum _{k=0}^{m-1}\alpha _\zeta (k)+\sum _{k=0}^{(m+1)/2}\alpha _\zeta ((n-m)/2+k)=0, \end{aligned}$$

which shows that the cyclotomic polynomial $\Phi _m(q)$ divides the sum $\sum _{k=0}^{(n+1)/2}\alpha _q(k)$. In view of

$$\begin{aligned} \prod _{m|n,m>1}\Phi _m(q)=[n], \end{aligned}$$

the proof of (3.5) is completed and therefore (1.2) is true. $\square$

Proof of (1.3)

Likewise, letting $cd=q$ and $r=2$ in Theorem 1, we get (1.3) immediately. $\square$

4 Proof of Theorem 2

Through the L’Hospital rule, we have

$$\begin{aligned}&\lim \limits _{a \rightarrow 1} \Bigg \{\frac{n(1-aq^2)(1-q^2/a)(1-a)a^{(n-1)/2}}{(1-aq)(1-q/a)(1-a^n)}-(1+q)^2\Bigg \} \times \frac{(1-aq^n)(a-q^n)}{(1-a)^2}\\&\quad =-\frac{n^2(1-q^2)^2-(1+24q+22q^2+24q^3+q^4)}{24}[n]^2. \end{aligned}$$

Hence, when $r=0$ and $a \rightarrow 1$ in Theorem 1, we know that (1.5) holds modulo $\Phi _n(q)^4$. It remains to show that

$$\begin{aligned} \sum _{k=0}^{M}[4k+1]\frac{(q^{-1};q^2)_k^2(q;q^2)_k^2(cq,dq;q^{2})_k}{(q^4;q^2)_k^2(q^2;q^2)_k^2(q^2/c,q^2/d;q^2)_k}\bigg (\frac{q^{5}}{cd}\bigg )^k \equiv 0 \pmod {[n]}, \end{aligned}$$

(4.1)

where $M \in \{(n+1)/2,n-2\}$. Just like the proof of (3.5), let $\zeta \ne 1$ be a primitive mth root of unity with m|n and let $\beta _q(k)$ be the kth term on the left-hand side of (4.1), i.e.,

$$\begin{aligned} \beta _q(k)=[4k+1]\frac{(q^{-1};q^2)_k^2(q;q^2)_k^2(cq,dq;q^{2})_k}{(q^4;q^2)_k^2(q^2;q^2)_k^2(q^2/c,q^2/d;q^2)_k}\bigg (\frac{q^{5}}{cd}\bigg )^k. \end{aligned}$$

Fixing $n=m$, $a \rightarrow 1$, $b\rightarrow 1$, and $x=0$ in Lemma 1, and noticing that $\beta _\zeta (k)=0$ for $(m+1)/2<k \le m-1$, we have

$$\begin{aligned} \sum _{k=0}^{(m+1)/2}\beta _\zeta (k)=\sum _{k=0}^{m-1}\beta _\zeta (k)=0. \end{aligned}$$

Similarly as before, we get

$$\begin{aligned}&\sum _{k=0}^{n-2}\beta _\zeta (k)=\sum _{l=0}^{n/m-2}\sum _{k=0}^{m-1}\beta _\zeta (lm+k)+\sum _{k=0}^{m-2}\beta _\zeta ((n-m)+k)=0,\\&\sum _{k=0}^{(n+1)/2}\beta _\zeta (k)=\sum _{l=0}^{(n/m-3)/2}\beta _\zeta (lm)\sum _{k=0}^{m-1}\beta _\zeta (k)+\sum _{k=0}^{(m+1)/2}\beta _\zeta ((n-m)/2+k)=0, \end{aligned}$$

which means that the cyclotomic polynomial $\Phi _m(q)$ divides the sums $\sum _{k=0}^{(n+1)/2}\beta _q(k)$ and $\sum _{k=0}^{n-2}\beta _q(k)$. Since

$$\begin{aligned} \prod _{m|n,m>1}\Phi _m(q)=[n], \end{aligned}$$

we immediately obtain (4.1). The q-supercongruence (1.5) then follows from the fact that [n] is coprime with the denominator of the right-hand side of (1.5).