Three families of q-supercongruences modulo the square and cube of a cyclotomic polynomial

In this paper, three parametric q-supercongruences for truncated very-well-poised basic hypergeometric series are proved, one of them modulo the square, the other two modulo the cube of a cyclotomic polynomial. The main ingredients of proof include a basic hypergeometric summation by George Gasper, the method of creative microscoping (a method recently introduced by the first author in collaboration with Wadim Zudilin), and the Chinese remainder theorem for coprime polynomials.


Introduction
More than one hundred years ago, in his second notebook, Ramanujan [19] mysteriously wrote 17 representations of 1/π (see [1, p. 352]), including for example, (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.
The cyclotomic polynomials n (q) are irreducible over the set of integers Z, and, for any positive integer n, we have s>1, s|n 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 3, r d − 4 (in particular, r may be negative) and gcd(d, r ) = 1. Let n be an integer such that n (d − r )/2 and n ≡ −r /2 (mod d). Then In this paper, we consider the r = d − 2 case of the left-hand side of (1.5) (in this case the condition r d − 4 for (1.5) is violated), and shall prove the following q-supercongruence. Theorem 1.1 Let d 3 be an odd integer and let n ≡ 1 (mod d) be a positive integer. Then (1.6) Further, the present authors [8] proved the following q-supercongruence: Let d and r be odd integers satisfying d 3, r d −4 (in particular, r may be negative) and gcd(d, r ) = 1. Let n be an integer such that n d − r and n ≡ −r (mod d). Then Note that in [8,Conjecture 3] it was even conjectured that the q-supercongruence (1.7) is true modulo n (q) 4 for d 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 3 be an odd integer and let n ≡ 2 (mod d) be a positive integer. Then
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 3, 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).
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 3 be an integer and let n ≡ 1 (mod d) be a positive integer. Then
Note that (1.10) is also true for d = 2. See [4,10] for a generalization of this case modulo [n] 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 ↔ a −1 ) into the terms of the series and prove that the congruences in Theorems 1.1 and 1.3 hold modulo 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 n (q)(1 − aq n )(a − q n ). By letting a = 1 the congruences are established modulo n (q) 3 . By showing that the congruence in (1.9) also holds modulo s (q) for any divisor s > 1 of n, we are able to lift the congruence modulo n (q) 3 to the claimed congruence modulo [n] 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 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 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 qshifted 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.

Preliminaries
We need the following result, which was first given by the present authors in [7, Lemma 2.1] (see also [9, Lemma 1]).
If gcd(d, n) = 1, then the above q-congruence also holds for a = 1.

A parametric generalization and proof of Theorem 1.1
We first give the following result.
Proof We shall prove the following parametric generalization of (3.1): 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 and similarly, 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 n (q), we arrive at the q-congruence (3.1).
We now give a parametric generalization of Theorem 1.1.

Theorem 3.2 Let d 3 be an odd integer and let n ≡ 1 (mod d) be a positive integer. Then, modulo n
Proof Since gcd(d, n) = 1, none of the numbers d, 2d, . . . , (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 , e m = q d−1 , n 1 = · · · = n m−1 = (2n − 2)/d and n m = (n − 1)/d and noticing N = n 1 + · · · + n m , we see that (3.4) is equal to the right-hand side of (3.3), where we have used the relation

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 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 n (q). Since [(d − 2)n] ≡ 0 (mod 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 n (q). Further, for any k satisfying (d − 2)(n − 1)/d < k n − 1, we have mod n (q)). This proves the q-congruence (3.3).

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 n (q) when a = 1. On the other hand, the polynomial (1 − aq n )(a − q n ) contains the factor n (q) 2 when a = 1. Hence, taking a = 1 in (3.3), we obtain Furthermore, applying Lemma 3.1 (d − 3)/2 times, modulo n (q) 2 , we have This, together with [(d − 2)n] ≡ 0 (mod n (q)), proves (1.6).

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 3 be an odd integer and let n ≡ 2 (mod d) be a positive integer. Then
Proof For a = q −n or a = q n , the left-hand side of (4.1) can be written as

by (2.4) and (2.5), the summation (4.2) is equal to
which is just the a = q −n or a = q n case of the right-hand side of (4.1). This proves the theorem.
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 n (q) for a = 1. Thus, letting a = 1 in (4.1), we conclude that (1.8) is true modulo n (q) 2 .

A parametric generalization and proof of Theorem 1.3
We need the following result.
The proof is similar to that of Lemma 3.1. Here we only give its parametric version of (5.1): We have the following parametric generalization of Theorem 1.3.

Theorem 5.2 Let d
3 be an integer and let n ≡ 1 (mod d) be a positive integer. Then, modulo n (q) ( Proof For a = q −n or a = q n , the left-hand side of (5.2) can be written as where M = (dn − n − r )/d or n − 1.
Here we point out that (6.1) is also true for d 3 and r = d − 1 (and thus n ≡ 1 (mod d)). This is because the proof of [7, Theorem 3.1] is also valid for r = d − 1, since 2r + 2n dn still holds for n d + 1 in this case [(6.1) clearly holds for n = 1].
Funding Open access funding provided by Austrian Science Fund (FWF).
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