A New Family of q-Supercongruences Modulo the Fourth Power of a Cyclotomic Polynomial

We establish a new family of q-supercongruences modulo the fourth power of a cyclotomic polynomial, and give several related results. Our main ingredients are q-microscoping and the Chinese remainder theorem for polynomials.

The first author and Zudilin [13,Theorem 4.2] also gave a two-parameter generalization of (1.3) as follows: for odd n, where m = (n − 1)/2 or (n − 1). Recently, based on the above q-congruence, by applying the Chinese remainder theorem for coprime polynomials, the first author [6, Theorem 1.1] succeeded in giving a full parametric generalization of (1. where m = (n−1)/2 or (n−1). Moreover, the present authors [8, Theorem 1.1, Equation (2a)] showed that, for odd n > 1, The main purpose of this paper is to establish a new family of qsupercongruences modulo the fourth power of a cyclotomic polynomial, which may somewhat be deemed a generalization of (1.3) and (1.7) modulo [n]Φ n (q) 3 . Theorem 1.1. Let d, n, r be integers satisfying d 2, r d − 2 (in particular, r may be negative), and n d − r, such that d and r are coprime, and n ≡ −r (mod d).
The proof is given in Sect. 3.
It is easy to see that the q-factorial (q 2r ; q d ) (dn−n−r)/d in (1.8) contains the factor 1 − q (d−2)n for d 3.
for d 3. Our proof of Theorem 1.1 implies that the above q-congruence is further true modulo [n]Φ n (q). Or the reader may check that the denominator of the reduced form of the fraction (q 2r ; We should point out that the present authors [10] have given a different generalization of (1.3) and (1.7) modulo Φ n (q) 4 as follows: for d 3 and n, r satisfying the same condition as in Theorem 1.1, we have For d 3 and r = ±1, Theorem 1.1 can be stated as follows: where M = (dn − n − 1)/d or n − 1.

Corollary 1.3. Let d and n be positive integers with d 3 and n ≡ 1 (mod d). Then
We shall also prove the following q-congruence, which was originally conjectured by the first author and Zudilin [ Note that the a = 1 case of (1.10) modulo Φ n (q) 2 has already been proved by the present authors [11,Theorem 2.4].

Some Lemmas
We first give the following result which is a generalization of [11, Proof. In view of q dm+r ≡ q n ≡ 1 (mod Φ n (q)), we have which together with (2.1) establishes the assertion.
Proof. It is clear that Lemma 2.2 is true for n = 1 or r = 0. We now assume that n > 1 and r = 0. By Lemma 2.1 (which is clearly also true for m = 0) This proves that the q-congruence (2.2) holds modulo Φ n (q). Moreover, since dm ≡ −r (mod n), the expression (q r ; q d ) k contains a factor of the form 1 − q αn for m < k n − 1, and is therefore congruent to 0 modulo Φ n (q). At the same time the expression (q d ; q d ) k is relatively prime to Φ n (q) for m < k n − 1. Therefore, each summand in (2.3) with k in the range m < k n − 1 is congruent to 0 modulo Φ n (q). This together with (2.2) modulo Φ n (q) establishes the q-congruence (2.3) modulo Φ n (q).
We are now able to prove (2.2) and (2.3) modulo [n]. Let ζ = 1 be an nth root of unity, not necessarily primitive. Namely, ζ is a primitive root of unity of degree s with s | n and s > 1. Let c q (k) denote the kth term on the left-hand side of (2.3), i.e., The q-congruences (2.2) and (2.3) modulo Φ n (q) with n → s imply that This means that the sums n−1 k=0 c q (k) and m k=0 c q (k) are both divisible by the cyclotomic polynomial Φ s (q). Since this is true for any divisor s > 1 of n, we deduce that they are divisible by Proof. By the condition in the lemma, we have r = 0. Recall that Jackson's 6 φ 5 summation formula can be written as Namely, when a = q dn−n or a = q n−dn the two sides of (2.6) are equal.
Proof. Letting q → q d and taking a = q r , b = aq r , c = q r /a and N = (dn − n − r)/d in ( Namely, when b = q dn−n both sides of (2.9) are equal. This proves the desired q-congruence.

Proof of Theorems 1.1 and 1.4
With the help of Lemmas 2.3 and 2.4, we can prove the main theorems in this paper now. We need to establish the following parametric generalization of Theorem 1.1. where M = (dn − n − r)/d or n − 1.
Proof. It is clear that the polynomials [n](1−aq dn−n )(a−q dn−n ) and b−q dn−n are coprime. By the Chinese remainder theorem for coprime polynomials, we can determine the remainder of the left-hand side of (2.6) modulo [n](1 − aq dn−n )(a − q dn−n )(b − q dn−n ) from (2.6) and (2.9). To this end, we require the following q-congruences: