Some Q-Supercongruences for Truncated Basic Hypergeometric Series

Some q-supercongruences for truncated basic hypergeometric series

Victor J. W. Guo School of Mathematical Sciences Huaiyin Normal University Huaian, Jiangsu 223300 People’s Republic of China E-mail: [email protected] Jiang Zeng Universit´ede Lyon Universit´eLyon 1 Institut Camille Jordan, UMR 5208 du CNRS 43, boulevard du 11 novembre 1918 F-69622 Villeurbanne Cedex, France E-mail: [email protected]

Abstract For any odd prime p we obtain q-analogues of van Hamme’s and Rodriguez-Villegas’ supercongruences involving products of three binomial coeﬃcients such as

p−1 2 3 2k X 2k q 2 2 2 2 2 ≡ 0 (mod [p] ) for p ≡ 3 (mod 4), k 2 (−q ; q ) (−q; q) k=0 q k 2k p−1 2 3 2 3 3k X 2k (q; q )k(q ; q )kq 2 6 6 2 ≡ 0 (mod [p] ) for p ≡ 2 (mod 3), k 3 (q ; q ) k=0 q k

p−1 n−1 where [p] = 1+q+···+q and (a; q)n = (1−a)(1−aq) ··· (1−aq ). We also prove q-analogues of the Sun brothers’ generalizations of the

2010 Mathematics Subject Classiﬁcation: Primary 11B65; Secondary 05A10, 05A30. Key words and phrases: p-adic integer, congruences, supercongruences, basic hypergeometric series, q-binomial theorem, q-Chu-Vandermonde formula.

1 2 V. J. W. Guo and J. Zeng

above supercongruences. Our proofs are elementary in nature and use the theory of basic hypergeometric series and combinatorial q- binomial identities including a new q-Clausen-type summation formula.

1 Introduction

We shall follow the standard q-notations from [4]. The q-shifted factorial n−1 is deﬁned by (a; q)n = (1 − a)(1 − aq) ··· (1 − aq ) for n = 1, 2,..., and 1−qn (a; q)0 = 1, while the q-integer is denoted by [n] := 1−q . In a previous paper [6], we proposed several q-analogues of Rodriguez-Villegas and Morten- son type congruences for truncated hypergeometric series conjectured by Rodriguez-Villegas [11, 9] and proved the following q-analogue of one of their supercongruences:

p−1 2 2 2 X (q; q )k −1 1−p 2 (1.1) ≡ q 4 (mod [p] ). (q2; q2)2 p k=0 k · Here and in what follows p denotes an odd prime, and ( p ) is the Legendre symbol modulo p. Recently, by using the properties of generalized Legendre polynomials, Z.-H. Sun [13, Theorem 2.5] proved the following remarkable congruence:

p−1 X 2ka−1 − a 1 (1.2) ≡ 0 (mod p2), k k k 4k k=0 where a is a p-adic integer such that the least nonnegative residue of a modulo p is odd. It is interesting to note that (1.2) is a common generalization of several congruences due to van Hamme and Rodriguez-Villegas. On the other hand, van Hamme [17] proved the following congruence (1.3) p−1 2 3 ( X 2k 1 4x2 − 2p (mod p2), if p = x2 + y2 with x odd, ≡ k 64k 0 (mod p2), if p ≡ 3 (mod 4), k=0 of which a generalization was recently conjectured by H. Swisher [16, (H.3)]. The aim of this paper is to prove some q-supercongruences for certain truncated basic hypergeometric series generalizing the above results. n Recall that the q-binomial coeﬃcients k are then deﬁned by  n−k+1 (q ; q)k n n  , if 0 6 k 6 n, = = (q; q)k k k q 0, otherwise. truncated basic hypergeometric series 3

The ﬁrst aim of this paper is to give a uniﬁed q-analogue of (1.3) and Z.-W. Sun’s generalization [14, Theorem 1.1(i)] of (1.3).

p−1 Theorem 1.1. Let 0 6 s 6 2 . Then the following congruence holds modulo [p]2: (1.4)

p−1 2 2 X 2k 2k q2k k k + s (−q2; q2)2(−q; q)2 k=0 q2 q2 k 2k  2 2 2 2 2 p−1 p−2s−1 p+2s−1 p−1 2 (q ; q ) (q ; q )  s −s 2 2 2 p−1 (−1) q 2 , if s ≡ mod 2, p−2s−1 (q4; q4)2 2 ≡ 4 q4 p−1  2 0, otherwise.

When s = 0, the congruence (1.4) reduces to the following result.

Corollary 1.2. There holds

p−1 2 3 X 2k q2k (1.5) k (−q2; q2)2(−q; q)2 k=0 q2 k 2k  p−1 2  p−1 2 1 q 2 , if p ≡ 1 (mod 4), p−1 (−q2; q2)2 2 ≡ 4 q4 p−1 (mod [p] ).  2 0, otherwise,

To see that (1.5) is a q-analogue of (1.3) we need to recall a known result. Let p be a prime such that p ≡ 1 (mod 4) and p = x2 + y2 with x ≡ 1 (mod 4), then we have the so-called Beukers-Chowla-Dwork-Evans congruence [3, 10]:

p−1 2p−1 + 1 p 2 ≡ 2x − (mod p2), p−1 2 2x 4 which easily implies Sun’s congruence [12, Lemma 3.4])

2 p−1 1 (1.6) 2 ≡ 4x2 − 2p (mod p2). p−1 2p−1 4 Using (1.6), it is clear that (1.5) reduces to (1.3) when q → 1. Generalizing a result of Mortenson [7, 8], Z.-W. Sun [15, (1.4)] obtained the following congruence

p−1 2 2k2k+2s X −1 p − 1 (1.7) k k+s ≡ (mod p2), for 0 s . 42k+s p 6 6 2 k=0 4 V. J. W. Guo and J. Zeng

For any p-adic integer x, let hxip denote the least nonnegative residue of x modulo p. The second aim of this paper is to give a uniﬁed q-analogue of (1.2) and (1.7).

Theorem 1.3. Let m and r two positive integers with p - m. Let s 6 r m−r r min{h− m ip, h− m ip} be a nonnegative integer. If h− m ip ≡ s + 1 (mod 2), then

p−1 2 m m r m m−r m mk X (q ; q )2k(q ; q )k(q ; q )kq 2 (1.8) m m m m 2m 2m 2 ≡ 0 (mod [p] ), (q ; q )k−s(q ; q )k+s(q ; q ) k=s k p−1 m m r m m−r m mk X (q ; q )2k(q ; q )k(q ; q )kq 2 (1.9) m m m m 2m 2m 2 ≡ 0 (mod [p] ). (q ; q )k−s(q ; q )k+s(q ; q ) k=s k

r If h− m ip ≡ s (mod 2), then the following congruence holds:

p−1 2 m m r m m−r m mk X (q ; q )2k(q ; q )k(q ; q )kq (1.10) m m m m 2m 2m 2 (q ; q )k−s(q ; q )k+s(q ; q ) k=s k (s+ p−1 )m m 2m m 2m 2 r ≡ q (q ; q ) h− ip−s (q ; q ) m−r m h− m ip−s 2 2 −mh− r i m −mh− m−r i m (q m p ; q ) (q m p ; q ) × s s (mod [p]). 2m 2m r 2m 2m (q ; q ) h− ip+s (q ; q ) m−r m h− m ip+s 2 2

r Letting s = 0, − m = a and q → 1 in (1.9), we obtain (1.2). On the other hand, it is not diﬃcult to see that (see [6]), for any prime p > 5, h− 1 i −3 h− 1 i −2 h− 1 i −1 (−1) 3 p = , (−1) 4 p = , (−1) 6 p = . p p p

Taking r = 1 and m = 3, 4, 6 in (1.8), we obtain

Corollary 1.4. Let > 5 be a prime and let s be a nonnegative integer. Then the following congruences hold modulo [p]2:

p−1 2 3 2 3 3k −3 X 2k (q; q )k(q ; q )kq p − 1 1 + ( p ) ≡ 0, if s and s ≡ mod 2, k + s (q6; q6)2 6 3 2 k=s q3 k p−1 2 4 3 4 4k −2 X 2k (q; q )k(q ; q )kq p − 1 1 + ( p ) ≡ 0, if s and s ≡ mod 2, k + s (q8; q8)2 6 4 2 k=s q4 k p−1 2 6 5 6 6k −1 X 2k (q; q )k(q ; q )kq p − 1 1 + ( p ) ≡ 0, if s and s ≡ mod 2. k + s (q12; q12)2 6 6 2 k=s q6 k truncated basic hypergeometric series 5

The proof of Theorem 1.3 is based on the following q-Clausen type summation formula, which seems new and interesting in its own right.

Theorem 1.5. Let n and s be nonnegative integers with s 6 n. Then

n −2n 2 k ! n −2n 2 k ! X (q ; q )k(x; q)kq X (q ; q )k(q/x; q)kq (1.11) (q; q)k−s(q; q)k+s (q; q)k−s(q; q)k+s k=s k=s n 2 2 2 −n2 n k 2 2 k2−2nk (−1) (q ; q )nq X (−1) (q ; q )n+k(x; q)k(q/x; q)kq = 2 2 2 2 2 2 . (q ; q )n−s(q ; q )n+s (q ; q )n−k(q; q)k−s(q; q)k+s(q; q)2k k=s

We also have the following q-analogue of (1.7), which reduces to (1.1) when s = 0.

p−1 Theorem 1.6. Let p be an odd prime and let 0 6 s 6 2 . Then

p−1 2 2 2 2 X (q; q )k(q; q )k+s −1 1−p 4 2 2 2 2 2 ≡ q (mod [p] ). (q ; q )k(q ; q )k+s p k=0

Finally, we shall prove the following result.

Theorem 1.7. Let p be an odd prime and let m, r be positive integers with m−r p - m and r < m. Then for any integer s with 0 6 s 6 h− m ip, there holds

(1.12)

p−s−1 r m m−r m r r X (q ; q )k(q ; q )k+s r −mh− m ip(h− m ip+1) h− m ip 2 m m m m ≡ (−1) q (mod [p]). (q ; q )k(q ; q )k+s k=0

In particular, if p ≡ ±1 (mod m), then

p−s−1 r m m−r m 2 X (q ; q )k(q ; q )k+s r r(m−r)(1−p ) h− m ip 2m (1.13) m m m m ≡ (−1) q (mod [p]). (q ; q )k(q ; q )k+s k=0

Throughout the paper we will often use the fact that for any prime p, the q-integer [p] is always an irreducible polynomial in Q[q]. Namely, Q[q]/[p] is a ﬁeld. Therefore, rational functions a(q)/b(q) are well deﬁned modulo [p] r or [p] (r > 1) on condition that b(q) is relatively prime to [p]. The rest of this paper is organized as follows. In Sections 2–5 we prove Theorem 1.1, Theorem 1.5, Theorem 1.3, and Theorems 1.6 and 1.7, respectively. We conclude the paper with some open problems. 6 V. J. W. Guo and J. Zeng

2 Proof of Theorem 1.1

We ﬁrst establish two lemmas.

p−1 Lemma 2.1. Let 0 6 k 6 2 . Then

p−1 kp−k2 2 + k k 2k q 2 (2.1) ≡ (−1) 2 (mod [p] ). 2k q2 k q2 (−q; q)2k Proof. Since

(1 − qp−2j+1)(1 − qp+2j−1) + (1 − q2j−1)2qp−2j+1 = (1 − qp)2, we have

(1 − qp−2j+1)(1 − qp+2j−1) ≡ −(1 − q2j−1)2qp−2j+1 (mod [p]2).

It follows that p−1 Qk (1 − qp−2j+1)(1 − qp+2j−1) 2 + k j=1 = 2 2 2k q2 (q ; q )2k Qk (1 − q2j−1)2qp−2j+1 k j=1 ≡ (−1) 2 2 (q ; q )2k kp−k2 k 2k q 2 = (−1) 2 (mod [p] ), k q2 (−q; q)2k as desired.

Lemma 2.2. Let n and s be nonnegative integers with s 6 n. Then

n k (n−k) X n + k 2k 2k (−1) q 2 (2.2) 2k k k + s (−q; q)2 k=0 k  2 n2−s2 n (q; q)n−s(q; q)n+s  s 2 (−1) q n−s 2 2 2 , if n ≡ s (mod 2), = 2 q2 (q ; q )n 0, otherwise.

Proof. We may rewrite the left-hand side of (2.2) as

n −n n+1 1 1 k+(n) X (q ; q)k(q ; q)k(q 2 ; q)k(−q 2 ; q)kq 2 (q; q)k(q; q)k−s(q; q)k+s(−q; q)k k=s n −n n+1 2 s+( ) s−n n+s+1 s+ 1 s+ 1 (q ; q)s(q ; q)s(q; q )2sq 2 q , q , q 2 , −q 2 = 2 2 4φ3 s+1 s+1 2s+1 ; q, q . (q ; q )s(q; q)2s q , −q , q truncated basic hypergeometric series 7

The result then follows from Andrews’ terminating q-analogue of Watson’s formula [4, (II.17)]:  0, if n is odd, q−n, a2qn+1, b, −b  n 2 2 2 2 (2.3) 4φ3 2 ; q, q = b (q, a q /b ; q ) aq, −aq, b n/2 , if n is even  2 2 2 2 (a q , b q; q )n/2

s s+ 1 with the substitution of n, a and b by n − s, q and q 2 , respectively.

Proof of Theorem 1.1. By the congruence (2.1), we have

p−1 2 2 X 2k 2k q2k k k + s (−q2; q2)2(−q; q)2 k=0 q2 q2 k 2k p−1 2 p−1 k X + k 2k 2k (−1) 2 ≡ 2 qk +2k−pk (mod [p]2). 2k k k + s (−q2; q2)2 k=0 q2 q2 q2 k p−1 2 The proof then follows from (2.2) with n = 2 and q → q .

3 Proof of Theorem 1.5

We ﬁrst establish four lemmas to make the proof easier. Lemma 3.1. Let n be a nonnegative integer. Then n n−k+1 n ( 2 ) X n−k n (aq ; q)kq (ax; q)n(q; q)n (3.1) (−1) −k = −n , k (a; q)k(1 − xq ) (a; q)n(xq ; q)n+1 k=0 n (n+1) X n k (q; q)n−k q 2 n−k (2) (3.2) (−1) q = n . k (x; q)n−k+1 1 − xq k=0 Proof. For (3.1), by the partial fraction decomposition we have n (ax; q)n(q; q)n X ak (3.3) −n = −k (a; q)n(xq ; q)n+1 1 − xq k=0 with −k n (1 − xq )(ax; q)n(q; q)n n−k+1 n (aq ; q)k n−k ( 2 ) ak = lim −n = (−1) q . x→qk (a; q)n(xq ; q)n+1 k (a; q)k By the Gauss or q-binomial inversion (see, for example, [1, p. 77, Exercise 2.47]), the identity (3.2) is equivalent to n (q; q)n X n k+1 1 k ( 2 ) = (−1) q k , (x; q)n+1 k 1 − xq k=0 q n which corresponds to the a = 0 case of (3.3) with x → xq . 8 V. J. W. Guo and J. Zeng

Lemma 3.2. Let n be a positive integer. Then (3.4)

(x; q)n + (a/x; q)n = (x; q)n(a/x; q)n n−1 n k X (x; q)k(a/x; q)k(1 − q ) X k j j (2) k+j + n−k (−1) q (aq ; q)n−k, (q; q)k(1 − q ) j k=0 j=0 (3.5) (x; q)n + (a/x; q)n = (x; q)n(a/x; q)n + (a; q)n n−1 n−k (j)+kj j X X n − k − 1 k + j − 1 q 2 a + (x; q) (a/x; q) (1 − qn) (−1)j . k k j − 1 j − 1 1 − qj k=1 j=1 −m Proof. We ﬁrst prove (3.4). Taking x = q (0 6 m 6 n − 1), we have (3.6) n−1 n k X (x; q)k(a/x; q)k(1 − q ) X k j j (2) k+j n−k (−1) q (aq ; q)n−k (q; q)k(1 − q ) j k=0 j=0 n−1 −m m n k X (q ; q)k(aq ; q)k(1 − q ) X k j j (2) k+j = n−k (−1) q (aq ; q)n−k (q; q)k(1 − q ) j k=0 j=0 m m n k j X m k (aq ; q)k(1 − q ) X k j (aq ; q)n k (2)−mk j (2) = (−1) q n−k (−1) q j k (1 − q ) j (aq ; q)k k=0 j=0 m m m n X m j X m − j k (aq ; q)k(1 − q ) j (2) j k (2)−mk = (−1) q (aq ; q)n (−1) q j n−k . j k − j (aq ; q)k(1 − q ) j=0 k=j It follows from (3.1) that

m m n X m − j k (aq ; q)k(1 − q ) k (2)−mk (−1) q j n−k k − j (aq ; q)k(1 − q ) k=j m m m+j n (aq ; q)j X m − j m−k+1 m+1 (aq ; q)k−j(1 − q ) k ( 2 )−( 2 ) = j (−1) q 2j n−k (aq ; q)j k − j (aq ; q)k−j(1 − q ) k=j m m n+j n −(m+1) (−1) (aq ; q)j(aq ; q)m−j(q; q)m−j(1 − q )q 2 = j 2j n−m (aq ; q)j(aq ; q)m−j(q ; q)m−j+1 m n+j n −(m+1) (−1) (a; q)j(aq ; q)m−j(q; q)m−j(1 − q )q 2 = n−m . (a; q)m(q ; q)m−j+1 Therefore, the right-hand side of (3.6) can be simpliﬁed as (3.7) m+1 m n −( 2 ) (a; q)m+n(1 − q )q X m−j m j (q; q)m−j m (−1) q(2) = (aq ; q) , (a; q) j (qn−m; q) n m j=0 m−j+1 truncated basic hypergeometric series 9

−m where the equality follows from (3.2). Noticing that (q ; q)n = 0 for 0 6 −m m 6 n − 1, we have proved that both sides of (3.4) are equal for x = q m (0 6 m 6 n − 1), and by symmetry, for x = aq (0 6 m 6 n − 1) too. Furthermore, both sides of (3.4) are of the form x−nP (x) with P (x) being a n polynomial in x of degree 2n with the leading coeﬃcient (−1)nq(2). Hence, they must be identical. This proves (3.4). By the q-binomial theorem (see, for example, [2, Theorem 3.3]), for k > 1, we have k X j k j k+j (3.8) (−1) q(2)(aq ; q) j n−k j=0 k n−k X j k j X i n − k i +(k+j)i i = (−1) q(2) (−1) q(2) a j i j=0 i=0 n−k k X i n − k i +ik i X j k j +ij = (−1) q(2) a (−1) q(2) i j i=0 j=0 n−k X i n − k i i +ik i = (−1) (q ; q) q(2) a . i k i=1

Moreover, for k = 0, the left-hand side of (3.8) is clearly equal to (a; q)n. Noticing that i n − k (q ; q)k n − k − 1 k + i − 1 1 n−k = i , i (q; q)k(1 − q ) i − 1 i − 1 1 − q we complete the proof of (3.5).

Let n and h be positive integers and let m and s be nonnegative integers such that s 6 m and h 6 n − m (so n > m). Let

−n −n f(x; j, k) := (x; q)j(x; q)k(q ; q)j(q ; q)k j−m−h+1 k−m−h+1 2j+k 2k+j (q ; q)h−1(q ; q)h−1(q − q ) × m−s−1 2 , (−1) (q; q)n(q; q)h−1(q; q)j−s(q; q)j+s(q; q)k−s(q; q)k+s and, for integers a, b > s, let a b X X La,b(x) := f(x; j, k). j=s k=s

m2+3m−s2+s 2 Lemma 3.3. Let A = 2 − m(n + h) − h + h. Then s+1 n−s−h A (x; q)s(x; q)m+h(q /x; q)n−s−hx q (3.9) Lm,n(x) = . (q; q)m−s(q; q)m+s(q; q)n−s(q; q)n+s(q; q)n−m−h 10 V. J. W. Guo and J. Zeng

Proof. Without loss of generality, we assume that q is a complex number with |q| < 1. We first note that f(x; j, k) = −f(x; k, j), and so Lm,m(x) = 0. Since both sides of (3.9) are polynomials in x of degree m+n with the same leading coefficient, it suffices to show that these two polynomials have m+n common roots, counted with multiplicity. We proceed by dividing the roots of the right-hand side of (3.9) into four cases.

2 • If s > 1, then it is easily seen that (x; q)s divides Lm,n(x), which −r means that the numbers q (0 6 r 6 s − 1) are roots of Lm,n(x) with multiplicity 2.

−r • For r with s 6 r 6 m, we have (q ; q)k = 0 if k > m, and so −r −r Lm,n(q ) = Lm,m(q ) = 0.

• For r with m + 1 6 r 6 m + h − 1, we have r − m − h + 1 6 0, and so k−m−h+1 (q ; q)h−1 = 0 for m + 1 6 k 6 r, while for r < k 6 n we have −r −r −r (q ; q)k = 0. Hence, we again get Lm,n(q ) = Lm,m(q ) = 0.

r • For r with s + 1 6 r 6 n − h, it is clear that Lm,n(q ) = 0 follows from the identity

n −n r k−m−h+1 k−j 2j+k X (q ; q)k(q ; q)k(q ; q)h−1(1 − q )q (3.10) = 0, (q; q)k−s(q; q)k+s k=s which will be proved as follows: The left-hand side of (3.10) can be written as (3.11)

−n r (q ; q)s(q ; q)s n −n+s r+s k−m−h+1 k−j 2j+k X (q ; q)k−s(q ; q)k−s(q ; q)h−1(1 − q )q × (q; q)k−s(q; q)k+s k=s n−s −n r X k n − s −(n−s)k+ k = (q ; q) (q ; q) (−1) q (2)R , s s k k k=0 where r+s k+s−m−h+1 k+s−j 2j+k+s (q ; q)k(q ; q)h−1(1 − q )q Rk = . (q; q)k+2s Since (qr+s; q) (qk+2s+1; q) k = r−s−1 , (q; q)k+2s (q; q)r+s−1 truncated basic hypergeometric series 11

k we see that Rk is a polynomial in q of degree r−s−1+h−1+2 6 n−s with constant term 0. By the q-binomial theorem (see [2, Theorem 3.3]),

n X k n k+1 k (−1) q( 2 )x = (xq; q) , k n k=0 q

we have n ( X k n k+1 −ik 0, for 1 6 i 6 n, (3.12) (−1) q( 2 ) = k (q; q) , for i = 0. k=0 q n

It follows that the right-hand side of (3.11) is equal to 0. Namely, the identity (3.10) holds.

Thus, we have found out all the m + n roots of Lm,n(x), which are clearly the same as those of the right-hand side of (3.9). This completes the proof.

Remark. We can use the identity (3.12) to give a short proof of Jackson’s terminating q-analogue of Dixon’s identity, see [5].

Lemma 3.4. Let n and h be positive integers and let m and s be nonnegative integers with h 6 n − m and s 6 m. Then (3.13) m n −2n 2 −2n 2 k−j j+k+jh X X (q ; q )j(q ; q )k(1 − q )q k − m − 1 m + h − j − 1 (q; q)j−s(q; q)j+s(q; q)k−s(q; q)k+s h − 1 h − 1 j=sk=m+h n−m−h 2 2 2 m2−n2−2mn+mh (−1) (q ; q )n(−q; q)2n−hq = 2 2 2 2 2 2 . (q; q)m−s(q; q)m+s(q ; q )n−s(q ; q )n+s(q; q)h−1(q , q )n−m−h

k−m−1 Proof. By the deﬁnition of q-binomial coeﬃcients, there holds h−1 = 0 for m+1 6 k < m+h. Hence, the left-hand side of (3.13) remains unchanged Pn Pn if we replace k=m+h by k=m+1. Furthermore, we have

k − m − 1m + h − j − 1 h − 1 h − 1 j−m−h+1 k−m−h+1 (m−j)(h−1)−(h) (q ; q)h−1(q ; q)h−1q 2 = h−1 2 . (−1) (q; q)h−1

−n The proof then follows from the identity (3.9) with x = −q . 12 V. J. W. Guo and J. Zeng

Proof of Theorem 1.5. The left-hand side of (1.11) may be expanded as

(3.14) n X (q−2n; q2)2q2k k (x; q) (q/x; q) (q; q)2 (q; q)2 k k k=s k−s k+s −2n 2 −2n 2 j+k X (q ; q )j(q ; q )kq (x; q)j(q/x; q)k + (x; q)k(q/x; q)j + . (q; q)j−s(q; q)j+s(q; q)k−s(q; q)k+s s6j

For 0 6 j < k, from (3.5) we deduce that

(3.15) (x; q)j(q/x; q)k + (x; q)k(q/x; q)j j j+1 = (x; q)j(q/x; q)j (xq ; q)k−j + (q /x; q)k−j 2j+1 = (x; q)k(q/x; q)k + (x; q)j(q/x; q)j(q ; q)k−j k−j−1 X k−j + (x; q)j+i(q/x; q)j+i(1 − q ) i=1 k−j−i (h+1)+(i+2j)h X k − j − i − 1 i + h − 1 q 2 × (−1)h , h − 1 h − 1 1 − qh h=1

= (x; q)k(q/x; q)k + (x; q)j(q/x; q)j k−j−1 X k−j + (x; q)j+i(q/x; q)j+i(1 − q ) i=0 k−j−i (h+1)+(i+2j)h X k − j − i − 1 i + h − 1 q 2 × (−1)h , h − 1 h − 1 1 − qh h=1 where in the last step we have used the q-binomial theorem:

k−j 2j+1 X h k − j h+1 +2jh (q ; q) = 1 + (−1) q( 2 ) . k−j h h=1

Pn By (3.15), we may write (3.14) as m=s am(x; q)m(q/x; q)m, where

n −2n 2 −2n 2 j+m X (q ; q )j(q ; q )mq (3.16) a = m (q; q) (q; q) (q; q) (q; q) j=s j−s j+s m−s m+s m n −2n 2 −2n 2 k−j j+k X X (q ; q )j(q ; q )k(1 − q )q + (q; q)j−s(q; q)j+s(q; q)k−s(q; q)k+s j=s k=m+1 k−m (h+1)+(m+j)h X k − m − 1 m + h − j − 1 q 2 × (−1)h . h − 1 h − 1 1 − qh h=1 truncated basic hypergeometric series 13

It is easy to see that

n −2n 2 j −2n 2 s n −2n+2s 2 j−s X (q ; q )jq (q ; q )sq X (q ; q )jq = (q; q) (q; q) (q; q) (q; q) (q2s+1; q) j=s j−s j+s 2s j=s j−s j−s −2n 2 s −n+s −n+s (q ; q )sq q , −q , = 2φ1 2s+1 ; q, q (q; q)2s q −2n 2 n+s+1 s−(n−s)2 n−s (q ; q )s(−q ; q)n−sq = (−1) 2s+1 (q; q)2s(q ; q)n−s by the q-Chu-Vandermonde summation formula [4, Appendix (II.6)]. Hence,

n −2n 2 −2n 2 j+m X (q ; q )j(q ; q )mq (3.17) (q; q) (q; q) (q; q) (q; q) j=s j−s j+s m−s m+s (q−2n; q2) (q−2n; q2) (−qn+s+1; q) qm+s−(n−s)2 = (−1)n−s s m n−s (q; q)m−s(q; q)m+s(q; q)n+s n−m 2 2 2 m2−n2−2mn (−1) (q ; q )n(−q; q)2nq = 2 2 2 2 2 2 . (q; q)m−s(q; q)m+s(q ; q )n−s(q ; q )n+s(q , q )n−m Substituting (3.17) and (3.13) into (3.16), we obtain (3.18) 2 2 n−m h+1 n−m 2 2 2 m −n −2mn ( 2 )+2mh (−1) (q ; q )nq X (−q; q)2n−hq am = 2 2 2 2 h 2 2 . (q; q)m−s(q; q)m+s(q ; q )n−s(q ; q )n+s (−1) (q; q)h(q , q )n−m−h h=0 Replacing h by n − m − h, we have (3.19) n−m h (h+1)+2mh X (−1) (−q; q)2n−hq 2 2 2 (q; q)h(q , q )n−m−h h=0 n−m m−n n+m+1 (−q; q)n+m n−m n−m+1 +2m(n−m) X (q ; q)h(−q ; q)h −2hm = (−1) q( 2 ) q (q; q)n−m (−q; q)h(q; q)h h=0 −(n−m) m+n+1 (−q; q)n+m n−m (n−m+1)+2m(n−m) q , −q −2m = (−1) q 2 2φ1 ; q, q (q; q)n−m −q −n−m (−q; q)n+m(q ; q)n−m n−m n−m+1 +2m(n−m) = (−1) q( 2 ) (q; q)n−m(−q; q)n−m 2 2 (q ; q )m+n = 2 2 , (q ; q )n−m(q; q)2m where we have used the q-Chu-Vandermonde summation formula [4, Ap- pendix (II.7)]. It follows from (3.18) and (3.19) that am is just the coeﬃ- cient of (x; q)m(q/x; q)m in the right-hand side of (1.11). This completes the proof. 14 V. J. W. Guo and J. Zeng

4 Proof of Theorem 1.3

We ﬁrst give a congruence modulo [p].

Lemma 4.1. Let m and r be two positive integers with p - m. Let s 6 r m−r min{h− m ip, h− m ip} be a nonnegative integer. Then the following congruence holds modulo [p]:

(4.1)

p−1 2 m 2m r m mk X (q ; q )k(q ; q )kq m m m m (q ; q )k−s(q ; q )k+s k=s r  (h− m ip+s)m r m 2m −mh− ip m q 2 (q ; q ) h− r i −s (q m ; q )  m p s  2 r  , if h− ip ≡ s mod 2, ≡ 2m 2m r m (q ; q ) h− m ip+s  2 0 otherwise.

r m−r p−1 Proof. It is easy to see that h− m ip+h− m ip = p−1, and so s 6 2 . Since p m 2m p+1 is an odd prime, we see that (q ; q )k ≡ 0 (mod [p]) for 2 6 k 6 p−s−1, which means that

p−1 2 m 2m r m mk p−s−1 m 2m r m mk X (q ; q )k(q ; q )kq X (q ; q )k(q ; q )kq m m m m ≡ m m m m (mod [p]). (q ; q )k−s(q ; q )k+s (q ; q )k−s(q ; q )k+s k=s k=s

r mh− m ip+r r Let a = p . Then m|r − ap and r − ap = −mh− m ip 6 0. It is clear r m r−ap m r−ap m r that (q ; q )k ≡ (q ; q )k (mod [p]) and (q ; q )k = 0 for k > h− m ip. m−r r Moreover, we have p − s − 1 > p − h− m ip − 1 = h− m ip > s, and therefore

p−s−1 m 2m r−ap m mk X (q ; q )k(q ; q )kq m m (q ; q )k−s(q; q)k+s k=s h− r i m p m 2m r−ap m mk X (q ; q )k(q ; q )kq ≡ m m m m (q ; q )k−s(q ; q )k+s k=s r m 2m −mh− ip m ms (q ; q )s(q m ; q )sq = m m (q ; q )2s −m(h− r i −s) (s+ 1 )m (s+ 1 )m q m p , q 2 , −q 2 × φ ; qm, qm (mod [p]). 3 2 0, q(2s+1)m

The proof then follows from Andrews’ identity (2.3). truncated basic hypergeometric series 15

p−1 Proof of Theorems 1.3. By Lemma 2.1, for 0 6 k 6 2 , we have m m (q ; q )2k 2k 1 2m 2m 2 = m m (q ; q )k k q2m (−q ; q )2k p−1 k mk2−mkp 2 + k m m ≡ (−1) q (−q ; q )2k 2k q2m k mk2−mkp 2m 2m (−1) q (q ; q ) p−1 2 +k 2 = 2m 2m m m (mod [p] ), (q ; q ) p−1 (q ; q )2k 2 −k and so (4.2) p−1 2 m m r m m−r m mk X (q ; q )2k(q ; q )k(q ; q )kq m m m m 2m 2m 2 (q ; q )k−s(q ; q )k+s(q ; q ) k=s k p−1 k m m r m m−r m mk2−mk(p−1) 2 (−1) (q ; q ) p−1 (q ; q )k(q ; q )kq X 2 +k 2 ≡ 2m 2m m m m m m m (mod [p] ). (q ; q ) p−1 (q ; q )k−s(q ; q )k+s(q ; q )2k k=s 2 −k

m r p−1 Letting q → q , x = q and n = 2 in Theorem 1.5, we see that the right-hand side of (4.2) can be written as (4.3) 2 p−1 2m 2m 2m 2m (p−1) (−1) 2 (q ; q ) p−1 (q ; q ) p−1 q 4 2 −s 2 +s 2m 2m 2 (q ; q ) p−1 2  p−1  p−1  2 m(1−p) 2m r m mk 2 m(1−p) 2m m−r m mk X (q ; q )k(q ; q )kq X (q ; q )k(q ; q )kq × m m m m  m m m m  . (q ; q )k−s(q ; q )k+s (q ; q )k−s(q ; q )k+s k=s k=s

r If h− m ip ≡ s + 1 (mod 2), then by the congruence (4.1), we have

p−1 p−1 2 m(1−p) 2m r m mk 2 m 2m r m mk X (q ; q )k(q ; q )kq X (q ; q )k(q ; q )kq m m m m ≡ m m m m ≡ 0 mod [p], (q ; q )k−s(q ; q )k+s (q ; q )k−s(q ; q )k+s k=s k=s

m−r and also h− m ip ≡ s + 1 (mod 2) which means that

p−1 2 m(1−p) 2m m−r m mk X (q ; q )k(q ; q )kq m m m m ≡ 0 (mod [p]). (q ; q )k−s(q ; q )k+s k=s

2m 2m Noticing the fact (q ; q ) p−1 6≡ 0 (mod [p]), we conclude that the right- 2 hand side of (4.2) is congruent to 0 modulo [p]2. This proves (1.8). To prove (1.9), just observe that (see the proof of Lemma 4.1)

r m m−r m (q ; q )k ≡ (q ; q )k ≡ 0 (mod [p]) 16 V. J. W. Guo and J. Zeng

r m−r for max h− m ip, h− m ip < k 6 p − 1, and

m m r m m−r m (q ; q )2k (q ; q )k(q ; q )k ≡ m m m m ≡ 0 (mod [p]) (q ; q )k−s(q ; q )k+s

p−1 r m−r for 2 < k 6 max h− m ip, h− m ip . Finally, the proof of (1.10) follows from factorizing (4.2) into (4.3), apply- ing the ﬁrst case of the congruence (4.1), and then using the aforementioned r m−r relation h− m ip + h− m ip = p − 1.

5 Proof of Theorems 1.6 and 1.7

The following lemma is probably known. For the reader’s convenience, we include a proof.

Lemma 5.1. Let m, n and s be nonnegative integers with s 6 n. Then n X k n m + k k −nk n − n+1 (5.1) (−1) q(2) = (−1) q ( 2 ), k n k=0 n k2−k−2nk X k n + k 2k + 2s q n −n(n+1) (5.2) (−1) 2k+1 = (−1) q . 2k 2 k + s 2 (−q ; q)2s k=0 q q

Proof. It is not diﬃcult to see that the two identities (5.1) and (5.2) are equivalent, respectively, to

n X k n m + n − k k+1 (5.3) (−1) q( 2 ) = 1, k n k=0 n 2n−2k+1 2 X n 2n − k (q ; q )s k+1 k 2( 2 ) (5.4) (−1) 2n−2k+2 2 q = 1. k 2 n 2 (q ; q )s k=0 q q

m+n−k −k Since n can be written as a polynomial in q of degree n with constant term 1/(q; q)n. Identity (5.3) then follows from (3.12). On the other hand, since 0 6 s 6 n, we see that

2n−2k+1 2 2n−2k+2s+2 2 2n−2k+1 2 2n − k (q ; q )s (q ; q )n−s(q ; q )s 2n−2k+2 2 = 2 2 n q2 (q ; q )s (q ; q )n

−2k 1 is a polynomial in q of degree n with constant term 2 2 . Therefore, (q ;q )n the identity (5.4) follows from (3.12) with q → q2. This completes the proof. truncated basic hypergeometric series 17

Proof of Theorem 1.6. It is easy to see that

2 (q; q )k 2k 1 2 2 = . (q ; q )k k q2 (−q; q)2k Hence, by Lemmas 2.1 and 5.1, we have

p−1 p−1 2 2 2 2 X (q; q )k(q; q )k+s X 2k 2k + 2s 1 2 2 2 2 = (q ; q )k(q ; q )k+s k 2 k + s 2 (−q; q)2k(−q; q)2k+2s k=0 k=0 q q p−1 2 p−1 k2−kp X + k 2k + 2s (−q; q)2kq ≡ (−1)k 2 2k 2 k + s 2 (−q; q)2k+2s k=0 q q 2 p−1 1−p 2 = (−1) 2 q 4 (mod [p] ), as desired. r mh− m ip+r Proof of Theorem 1.7. Again, let a = p . Then a is a positive integer, m|ap − r, and so

r m k mj+r−m (q ; q )k Y 1 − q (5.5) = (qm; qm) 1 − qmj k j=1 k Y (1 − qap−mj−r+m)qmj+r−m ≡ (−1)k 1 − qmj j=1 ap−r k mk(k−1) +kr = (−1) m q 2 k qm r k h− ip mk(k−1) −mkh− r i ≡ (−1) m q 2 m p (mod [p]), k qm

m−r m k+s ap+mj−r ap−r (q ; q )k+s Y 1 − q + k + s (5.6) ≡ = m (qm; qm) 1 − qmj k + s k+s j=1 qm h− r i + k + s = m p (mod [p]). k + s qm By the congruences (5.5) and (5.6), we have

p−s−1 r m m−r m X (q ; q )k(q ; q )k+s m m m m (q ; q )k(q ; q )k+s k=0 p−s−1 r r X k h− ip h− ip + k + s mk(k−1) −mkh− r i ≡ (−1) m m q 2 m p k k + s k=0 qm qm −mh− r i (h− r i +1) h− r i m p m p = (−1) m p q 2 (mod [p]), 18 V. J. W. Guo and J. Zeng

m−r r where in the last step we have used p − s − 1 > p − h− m ip − 1 = h− m ip and the identity (5.1). This proves (1.12). r(m−r)(1−p2) To prove (1.13), just notice that if p ≡ ±1 (mod m), then 2m is an integer and

−mh− r i (h− r i + 1) r(m − r)(1 − p2) m p m p ≡ (mod p). 2 2m

6 Concluding remarks and open problems

It seems that the congruence (1.9) can be further generalized as follows.

Conjecture 6.1. Let m and r be two positive integers with p - m. Let r s 6 p − 1 be a nonnegative integer. If h− m ip ≡ s + 1 (mod 2), then

p−1 m m r m m−r m mk X (q ; q )2k(q ; q )k(q ; q )kq 2 (6.1) m m m m 2m 2m 2 ≡ 0 (mod [p] ). (q ; q )k−s(q ; q )k+s(q ; q ) k=s k

r m−r Note that, if s > max{h− m ip, h− m ip}, then the congruence (6.1) is obviously true, since in this case each summand in the left-hand side is congruent to 0 modulo [p]2. It is easy to see that, when r = 1, m = 3, 4, 6 and q → 1, Conjecture 6.1 reduces to a result of Z.-W. Sun [14, Theorem 1.3(i)]. We conjecture that Theorem 1.7 can be further strengthened.

Conjecture 6.2. Let m and r be positive integers with p ≡ ±1 (mod m) m−r and r < m. Then for any integer s with 0 6 s 6 h− m ip, there holds

p−s−1 r m m−r m 2 X (q ; q )k(q ; q )k+s r r(m−r)(1−p ) h− m ip 2m 2 m m m m ≡ (−1) q (mod [p] ). (q ; q )k(q ; q )k+s k=0 Like [6, Conjecture 7.1], Conjecture 6.2 seems to have a further generalization as follows:

Conjecture 6.3. Let m and |r| be positive integers with p - m and m - r. Then there exists a unique integer fp,m,r such that, for any s with 0 6 s 6 m−r h− m ip, there holds

p−s−1 r m m−r m X (q ; q )k(q ; q )k+s r h− m ip fp,m,r 2 m m m m ≡ (−1) q (mod [p] ). (q ; q )k(q ; q )k+s k=0 truncated basic hypergeometric series 19

Furthermore, the numbers fp,m,r satisfy the symmetry fp,m,r = fp,m,m−r and the recurrence relation: ( −fp,m,r, if r ≡ 0 (mod p), fp,m,m+r = fp,m,r − r, otherwise.

Here we give some values of fp,m,r:

f3,2,1 = −2, f3,2,3 = −3, f3,2,5 = 3, f3,2,7 = −2, f3,2,9 = −9,

f3,2,11 = 9, f3,2,13 = −2, f5,3,1 = −8, f5,3,2 = −8, f5,3,4 = −9,

f5,3,5 = −10, f5,3,7 = −13, f5,3,8 = 10, f5,3,10 = −20, f5,3,11 = 2,

f5,3,13 = 20, f5,3,14 = −9, f5,3,16 = 7, f5,3,17 = −23, f5,3,19 = −9,

f5,8,1 = −23, f7,9,1 = −54, f7,9,2 = −21, f7,9,4 = −37, f7,9,5 = −37,

f7,9,7 = −21, f7,9,8 = −54, f7,9,10 = −55, f7,9,11 = −23 f7,9,13 = −41,

f7,9,14 = −42, f7,9,16 = −22, f7,9,17 = −33.

Finally, supercongruences (or q-supercongruences) have now been around for quite a long time, and it would be desirable to have some more con- ceptual proofs of these phenomena, such as, combinatorial interpretations, connections to elliptic curves or representations of p-adic groups.

Acknowledgement

This work was partially supported by the National Natural Science Foun- dation of China (grant no. 11371144).

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