The $q$-WZ pairs and divisibility properties of certain polynomials

1.   Introduction

It was conjectured by Van Hamme ^[24] that, for any odd prime $p$,

The congruence (1.1) was first proved by Mortenson ^[19] using a technical evaluation of a quotient of Gamma functions, and later reproved by Zudilin ^[28] via the WZ (Wilf-Zeilberger) method. Using the same WZ-pair as Zudilin, Z.-W. Sun ^[23] proves the following generalization of (1.1): for any positive integer $n$,

Moreover, Z.-W. Sun [22,Conjecture 5.1(ⅰ)] proposed the following conjecture: for $n\geqslant 1$,

The above three congruences were later proved by He ^[13,14] using the WZ method again. Recently, still via the WZ method, Sun [21,Theorem 1.1] gave further generalizations of (1.3)–(1.5), such as: for $n\geqslant 1$ and $0\leqslant j\leqslant n$,

It is easy to see that the $j = 0$ case of (1.6) reduces to (1.5). The reader is referred to ^[1] for a collection of some other interesting applications of the WZ method in recent years.

On the other hand, by finding a $q$-analogue of the WZ pair in Zudilin's proof of (1.1), Guo ^[3] gave the following $q$-analogue of (1.2):

Here and in what follows, the $q$-shifted factorials are defined by

the $q$-integers are defined as $[n] = [n]_q = (1-q^n)/(1-q)$, and the $q$-binomial coefficients are given by

In his subsequent work ^[4,5,6], Guo also gave similar $q$-analogues of (1.3)–(1.5). For more recent progress on $q$-congruences, see ^{[7,8,9,10,11,15,16,17,18,20,25,26,29]}.

In this paper, we shall give $q$-analogues of Sun's generalizations of (1.3)–(1.5), including a $q$-analogue of (1.6). We shall also give a further generalization of (1.7). Our main results can be stated as follows.

Theorem 1.1. Let $n$ be a positive integer and $0\leqslant j\leqslant n$. Then modulo $(1+q^n)^2[2n+1]{2n\brack n}$,

It is easy to see that, when $q = 1$ the congruence (1.10) reduces to

where $(a)_k = a(a+1)\cdots(a+k-1)$ for $k\geqslant 1$ and $(a)_0 = 1$. This congruence is exactly an equivalent form of (1.6). Similarly, the congruences (1.8) and (1.9) in the $q = 1$ case reduce to the other two results in [21,Theorem 1.1].

Theorem 1.2. Let $n$ be a positive integer and $0\leqslant j\leqslant n$. Then

It is clear that the $q = 1$ case of (1.11) gives

which is a generalization of (1.2), and was neglected by Sun ^[21].

The rest of the paper is organized as follows. We first establish a general divisibility results based on the $q$-WZ machinery in the next section. We shall prove Theorems 1.1 and 1.2 in Sections 3 and 4, respectively. Finally, we shall propose some open problems in Section 5.

2.   The $q$-WZ pair and an auxillary result

Let $A = A(n, k)$ be a double-indexed sequence with values in a suitable ground-field containing the rational number field and $q$. Recall that the sequence $A$ is called $q$-hypergeometric in both parameters if both quotients

are rational functions in $q^n$ and $q^k$ over certain field for all $n$ and $k$ whenever the quotients are well-defined. We say that two $q$-hypergeometric functions $F(n, k)$ and $G(n, k)$ form a $q$-WZ pair if they satisfy the following relation:

Wilf and Zeilberger ^[27] showed that in this case there exists a rational function $C(n, k)$ in $q^n$ and $q^k$ such that $F(n, k) = C(n, k)G(n, k)$. The function $C(n, k)$ is usually called the certificate of the pair $(F, G)$.

We have the following $q$-version of [21,Theorem 2.1].

Theorem 2.1. Let $F(n, k)$ and $G(n, k)$ be a $q$-WZ pair.Assume that $F(n, k) = G(n, k) = 0$ for $n < k$.Let $A_N(q)$ be a polynomial in $q$ with integer coefficients such that $A_N(q)G(N+1, k)$ is also a polynomial in $q$ with integer coefficients for all $k\geqslant 1$.If $P_N(q)$ is a polynomial in $q$ with integer coefficients satisfying

(ⅰ) $A_N(q)G(N+1, k) \equiv0 \mod{P_N(q)}$, for all $k\geqslant 1$;

(ⅱ) $A_N(q)F(N, N) \equiv0 \mod{P_N(q)}$.

Then, for all $m\geqslant 0$,

Proof. Our proof is similar to that of [21,Theorem 2.1]. For the sake of completeness, we provide it here. We proceed by induction on $m$.

Summing (2.1) over $k$ from $1$ to $N$, we obtain

Multiplying both sides of the above identity by $A_N(q)$ and then summing it over $n$ from $0$ to $N$, we get

where we have used $F(n, N) = 0$ for $n < N$ and $G(0, k) = 0$ for $k\geqslant 1$. From (2.3) and the conditions (ⅰ) and (ⅱ), we immediately deduce that

Namely, the congruence (2.2) holds for $m = 0$.

We now assume that (2.2) is true for some $m = k$ with $k\geqslant 0$. Similarly as before, multiplying both sides of (2.1) by $A_N(q)$, shifting $k\rightarrow k+1$, and then summing it over $n$ from $k$ to $N$, we have

Thus, by the condition (ⅰ) and the induction hypothesis, we get

This completes the inductive step, and therefore (2.2) is true for all $m\geqslant 0$.

3.   Proof of Theorem 1.1

Proof of (1.8). For positive integers $n$, define

The following functions $F$ and $G$ introduced in ^[6]:

satisfy the relation (2.1). Namely, they form a $q$-WZ pair.

Since $[N+1]{2N+2\brack N+1}/(1+q^{N+1}) = [2N+1]{2N\brack N}$ and ${2N\brack N}\equiv 0\pmod{(1+q^N)}$, we have

for $k = 1$ or $k\geqslant 2$. It is clear that $F(N, N) = 0$. The proof of (1.8) then follows from Theorem 2.1.

Proof of (1.9). We again use a $q$-WZ pair to prove (1.9). The $q$-WZ pair has already been given in ^[4]:

By [4,Lemma 3.2], for $1\leqslant k\leqslant N$, we have

because $(-q^2;q^2)_N/(-q^2;q^2)_{N-k+1}$ is clearly a polynomial in $q$ with integer coefficients.

It is easy to see that

By [5,Lemma 3.1], we have

and so

Therefore, by Theorem 2.1, the congruence (1.9) holds.

Proof of (1.10). The following functions, introduced in ^[5],

form a $q$-WZ pair. By [5,Lemma 3.2], for $1\leqslant k\leqslant N$, we have

It is easy to see that

By (3.1), we get

Therefore, by Theorem 2.1, we obtain

Namely, the congruence (1.10) holds.

4.   Proof of Theorem 1.2

The proof is similar to that of Theorem 1.1. This time we need the following $q$-WZ pair in ^[3]:

By [3,Lemma 4.2], for $1\leqslant k\leqslant N$, we have

Moreover, since

we immediately get

The proof of (1.11) then follows readily from Theorem 2.1.

5.   Some consequences and open problems

It is easy to see that

Letting $j = 1$ in Theorems 1.1 and 1.2, we obtain the following results.

Corollary 5.1. Let $n$ be a positive integer. Then, modulo $(1+q^n)(1+q^{2n})[2n+1]{2n\brack n}$,

Corollary 5.2. Let $n$ be a positive integer. Then, modulo $(1+q^n)^2[2n+1]{2n\brack n}$,

It is worth mentioning that Guo and the author [12,Theorem 1.4] proved that, for $n\geqslant 1$,

However, we are unable to prove the following generalization of (5.1):

Motivated by (1.11) and the above conjectural generalization of (5.1), we propose the following generalization of [20,Theorem 5.1].

Conjecture 5.3. Let $n$ be a positive integer and $r\geqslant 2$. Then, for $0\leqslant j\leqslant n$, modulo $(1+q^n)^{2r-2}[2n+1]{2n\brack n}$,

and modulo $(1+q^n)^{2r-1}[2n+1]{2n\brack n}$,

6.   Conclusions

In Sections 3 and 4, we give proofs of some divisibility properties of certain polynomials by using the $q$-WZ pairs. Note that the $q$-WZ pairs are difficult to find, but once a $q$-WZ pair is given it may play a key role in the proof of a congruence. The Section 5 provides a conjectural generalization of [20,Theorem 5.1].

Acknowledgments

The author was partially supported by the Natural Science Foundation of Inner Mongolia, China (grant 2020BS01012), Research Program of Science and Technology at Universities of Inner Mongolia Autonomous Region (grant NJZY22600).

Conflict of interest

The author declares that there is no conflict of interest in this paper.