Journal of Number Theory 132 (2012) 1731–1740

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Journal of Number Theory

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Proof of some conjectures of Z.-W. Sun on congruences forApéry polynomials

Victor J.W. Guo a,∗, Jiang Zeng b

a Department of Mathematics, East China Normal University, Shanghai 200062, People’s Republic of Chinab Université de Lyon, Université Lyon 1, Institut Camille Jordan, UMR 5208 du CNRS, 43, boulevard du 11 novembre 1918,F-69622 Villeurbanne Cedex, France

a r t i c l e i n f o a b s t r a c t

Article history:Received 23 January 2011Revised 11 February 2012Accepted 11 February 2012Available online 4 April 2012Communicated by D. Wan

MSC:11A0711B6505A1005A19

Keywords:Apéry polynomialsPfaff–Saalschütz identityLegendre symbolSchmidt numbersSchmidt polynomials

The Apéry polynomials are defined by An(x) = ∑nk=0

(nk

)2(n+kk

)2xk

for all nonnegative integers n. We confirm several conjectures ofZ.-W. Sun on the congruences for the sum

∑n−1k=0(−1)k(2k+1)Ak(x)

with x ∈ Z.© 2012 Elsevier Inc. All rights reserved.

1. Introduction

The Apéry polynomials [13] are defined by

An(x) =n∑

k=0

(n

k

)2(n + k

k

)2

xk (n ∈N).

* Corresponding author.E-mail addresses: jwguo@math.ecnu.edu.cn (V.J.W. Guo), zeng@math.univ-lyon1.fr (J. Zeng).URLs: http://math.ecnu.edu.cn/~jwguo (V.J.W. Guo), http://math.univ-lyon1.fr/~zeng (J. Zeng).

0022-314X/$ – see front matter © 2012 Elsevier Inc. All rights reserved.http://dx.doi.org/10.1016/j.jnt.2012.02.004

1732 V.J.W. Guo, J. Zeng / Journal of Number Theory 132 (2012) 1731–1740

Thus An := An(1) are the Apéry numbers [2]. Many people (see Chowla et al. [4], Gessel [5], andBeukers [3], for example) have studied congruences for Apéry numbers. Recently, among other things,Sun [13] proved that, for any integer x,

n−1∑k=0

(2k + 1)Ak(x) ≡ 0 (mod n), (1.1)

and proposed many remarkable conjectures. The main objective of this paper is to prove the followingresult, which was conjectured by Sun [13].

Theorem 1.1. Let x ∈ Z.

(i) If n ∈ Z+ , then

n−1∑k=0

(−1)k(2k + 1)Ak(x) ≡ 0 (mod n). (1.2)

(ii) If p is an odd prime and ( ·p ) is the Legendre symbol modulo p, then

1

p

p−1∑k=0

(−1)k(2k + 1)Ak(x) ≡(

1 − 4x

p

)(mod p). (1.3)

(iii) If p > 3 is a prime, then

1

p

p−1∑k=0

(−1)k(2k + 1)Ak ≡(

p

3

) (mod p2), (1.4)

1

p

p−1∑k=0

(−1)k(2k + 1)Ak(−2) ≡ 1 − 4

3

(2p−1 − 1

) (mod p2). (1.5)

We shall establish some preliminary results in Section 2 and prove Theorem 1.1 in Section 3.Then we give some generalizations of the congruences (1.1) and (1.2) in Section 4. Sun [13,14] alsoformulated conjectures for the values of

∑p−1k=0 Ak(x) modulo p2 with x = 1,−4,9. In particular, he

made the following conjecture.

Conjecture 1.2. (See Sun [13,14].) Let p be an odd prime. Then

p−1∑k=0

Ak ≡{

4x2 − 2p (mod p2), if p ≡ 1,3 (mod 8) and p = x2 + 2y2,

0 (mod p2), if p ≡ 5,7 (mod 8).

In an effort to prove the above conjecture, we found a single sum formula for∑p−1

k=0 Ak(x) mod-ulo p2, as given in the following theorem.

V.J.W. Guo, J. Zeng / Journal of Number Theory 132 (2012) 1731–1740 1733

Theorem 1.3. Let p > 3 be a prime and x ∈ Z. Then

p−1∑k=0

Ak(x) ≡p−1∑k=0

(2k)!4 p

(4k + 1)!k!4 xk ≡(p−1)/2∑

k=0

(p + 2k

4k + 1

)(2k

k

)2

xk (mod p2).

Clearly Theorem 1.3 implies the following result.

Corollary 1.4. If p > 3 is a prime, then Conjecture 1.2 is equivalent to the following congruence for binomialsums

(p−1)/2∑k=0

(p + 2k

4k + 1

)(2k

k

)2

≡{

4x2 − 2p (mod p2), if p ≡ 1,3 (mod 8) and p = x2 + 2y2,

0 (mod p2), if p ≡ 5,7 (mod 8).

2. Two preliminary results

Theorem 2.1. Let n ∈ Z+ . Then

1

n

n−1∑k=0

(−1)k(2k + 1)Ak(x) = (−1)n−1n−1∑m=0

(2m

m

)xm

m∑k=0

(m

k

)(m + k

k

)(n − 1

m + k

)(n + m + k

m + k

).

Proof. For �,m ∈ N, a special case of the Pfaff–Saalschütz identity (see [11, p. 44, Exercise 2.d] and[1]) reads

(�

m

)(� + m

m

)=

m∑k=0

(2m

m + k

)(� − m

k

)(� + m + k

k

). (2.1)

Multiplying both sides of (2.1) by(

�m

)(�+m

m

), we obtain another expression of the Apéry polynomials:

A�(x) =�∑

m=0

(2m

m

)xm

m∑k=0

(m

k

)(m + k

k

)(�

m + k

)(� + m + k

m + k

). (2.2)

Thus,

n−1∑�=0

(2� + 1)(−1)� A�(x)

=n−1∑m=0

(2m

m

)xm

m∑k=0

(m

k

)(m + k

k

) n−1∑�=m+k

(−1)�(2� + 1)

(�

m + k

)(� + m + k

m + k

).

The result then follows by applying the formula

n−1∑�=k

(−1)�(2� + 1)

(�

k

)(� + k

k

)= (−1)n−1n

(n − 1

k

)(n + k

k

), (2.3)

which can be easily verified by induction. �

1734 V.J.W. Guo, J. Zeng / Journal of Number Theory 132 (2012) 1731–1740

Theorem 2.2. Let p be an odd prime and x ∈ Z. Then

1

p

p−1∑k=0

(−1)k(2k + 1)Ak(x) ≡p−1∑k=0

(2k

k

)xk (

mod p2).

Proof. Suppose that 0 � k � m < p. If 0 � m + k � p − 1, then

(p − 1

m + k

)(p + m + k

m + k

)=

m+k∏i=1

p2 − i2

i2≡ (−1)m+k (

mod p2).

If m + k � p, then(p−1

m+k

)(p+m+km+k

) = 0. Therefore, by Theorem 2.1, we get

1

p

p−1∑�=0

(2� + 1)(−1)� A�(x) ≡p−1∑m=0

(2m

m

)xm

p−m−1∑k=0

(−1)m+k(

m

k

)(m + k

k

) (mod p2). (2.4)

Note that, for 0 � k � p − 1 and p � m + k < 2p, we have

(2m

m

)≡

(m + k

k

)≡ 0 (mod p),

which means that

(2m

m

)(m + k

k

)≡ 0

(mod p2). (2.5)

Hence, the congruence (2.4) may be written as

1

p

p−1∑�=0

(2� + 1)(−1)� A�(x) ≡p−1∑m=0

(2m

m

)xm

m∑k=0

(−1)m+k(

m

k

)(m + k

k

)

=p−1∑m=0

(2m

m

)xm (

mod p2),

where we have used the Chu–Vandermonde summation formula. �3. Proof of Theorem 1.1

It is clear that the congruence (1.2) is an immediate consequence of Theorem 2.1. To prove (1.3),we first give the following congruence that was implicitly given by Sun and Tauraso [16] (see alsoSun [12]).

Lemma 3.1. Let p be an odd prime and let a ∈ Z+ and x ∈ Z. Then

pa−1∑k=0

(2k

k

)xk ≡

(1 − 4x

p

)a

(mod p). (3.1)

V.J.W. Guo, J. Zeng / Journal of Number Theory 132 (2012) 1731–1740 1735

Proof. If p | x, then (3.1) clearly holds. If p � x, then there is a positive integer b such that bx ≡1 (mod p) and

pa−1∑k=0

(2k

k

)xk ≡

pa−1∑k=0

(2k

k

)b−k (mod p).

By [16, Theorem 1.1], we have

pa−1∑k=0

(2k

k

)b−k ≡

(b2 − 4b

p

)a

(mod p).

Since ( x2

p ) = 1, we obtain

(b2 − 4b

p

)=

(x2

p

)(b2 − 4b

p

)=

(1 − 4x

p

).

Combining the above three identities yields (3.1). �Proof of (1.3). Theorem 2.2 implies that

p−1∑k=0

(−1)k(2k + 1)Ak(x) ≡ pp−1∑k=0

(2k

k

)xk (

mod p3). (3.2)

The proof then follows from the a = 1 case of Lemma 3.1. �Proof of (1.4). Letting x = 1 in (3.2), we have

p−1∑k=0

(−1)k(2k + 1)Ak ≡ pp−1∑k=0

(2k

k

) (mod p3).

The proof then follows form the a = 1 case of the congruence [17, (1.9)]

pa−1∑k=0

(2k

k

)≡

(pa

3

) (mod p2).

Note that the condition p > 3 is not necessary in this case. �Proof of (1.5). Letting x = −2 in (3.2), we have

p−1∑k=0

(−1)k(2k + 1)Ak(−2) ≡ pp−1∑k=0

(2k

k

)(−2)k (

mod p3). (3.3)

Similarly to (3.1), the first congruence in [12, Theorem 1.1] implies that

1736 V.J.W. Guo, J. Zeng / Journal of Number Theory 132 (2012) 1731–1740

pa−1∑k=0

(2k

k

)xk ≡

(1 − 4x

p

)a

+(

1 − 4x

p

)a−1

up−( 1−4xp )

(mod p2), (3.4)

where p � x and the sequence {un}n�0 is defined as

u0 = 0, u1 = 1, and un+1 = (b − 2)un − un−1, n ∈ Z+, where bx ≡ 1(mod p2).

If b2 − 4b �≡ 0 (mod p2), then

un = (b − 2 + √b2 − 4b)n − (b − 2 − √

b2 − 4b)n

2n√

b2 − 4b.

Now suppose that p > 3. For a = 1 and x = −2, the congruence (3.4) reduces to

p−1∑k=0

(2k

k

)(−2)k ≡ 1 + up−1 = 1 − 4p−1 − 1

3 · 2p−2≡ 1 − 4

3

(2p−1 − 1

) (mod p2). (3.5)

Combining (3.3) and (3.5), we complete the proof. �Remark. The following stronger version of (3.5):

p−1∑k=0

(2k

k

)(−2)k ≡ 1 − 4

3

(2p−1 − 1

) (mod p3) for any prime p > 3,

was independently obtained by Mattarei and Tauraso [7] and Z.-W. Sun [arXiv:0911.5665v52, Remarkafter Conjecture A69].

4. Generalizations of the congruences (1.1) and (1.2)

Following Schmidt [10] we call the numbers S(r)n = ∑n

k=0

(nk

)r(n+kk

)rthe Schmidt numbers, and

define the Schmidt polynomials as follows:

S(r)n (x) =

n∑k=0

(n

k

)r(n + k

k

)r

xk.

As generalizations of (1.1) and (1.2), we have the following congruences for Schmidt polynomials.

Theorem 4.1. Let n ∈ Z+ , r � 2 and x ∈ Z. Then

n−1∑k=0

(2k + 1)S(r)k (x) ≡ 0 (mod n), (4.1)

n−1∑k=0

(−1)k(2k + 1)S(r)k (x) ≡ 0 (mod n). (4.2)

V.J.W. Guo, J. Zeng / Journal of Number Theory 132 (2012) 1731–1740 1737

Remark. When r = 1, the polynomial Dn(x) := S(1)n (x) is called the Delannoy polynomial. Sun [15,

(1.15)] proved that

1

n

n−1∑k=0

(2k + 1)Dk(x) =n−1∑k=0

(n

k + 1

)(n + k

k

)xk ∈ Z[x] for n = 1,2,3, . . . .

To prove Theorem 4.1, we need the following lemma.

Lemma 4.2. For �,m ∈N and r � 2, there exist integers a(r)m,k (m � k � rm) such that

(�

m

)r(� + m

m

)r

=rm∑

k=m

a(r)m,k

(�

k

)(� + k

k

). (4.3)

Proof. We proceed by mathematical induction on r. For r = 2, the identity (2.2) gives

a(2)

m,m+k =(

2m

m

)(m

k

)(m + k

k

), 0 � k � m.

Suppose that (4.3) holds for r. A special case of the Pfaff–Saalschütz identity [1, (1.4)] reads

(� + m

m + n

)(� + n

n

)=

n∑k=0

(� − m

k

)(m

k + m − n

)(� + m + k

�

),

which can be rewritten as

(�

n

)(� + n

n

)=

n∑k=0

(m + n)!k!(m + k)!n!

(m

n − k

)(� − m

k

)(� + m + k

k

). (4.4)

Now, multiplying both sides of (4.3) by(

�m

)(�+m

m

)and applying (4.4), we have

(�

m

)r+1(� + m

m

)r+1

=rm∑

k=m

a(r)m,k

(�

k

)(� + k

k

)(�

m

)(� + m

m

)

=rm∑

k=m

k∑i=0

a(r)m,k

(m + k)!i!(m + i)!k!

(m

k − i

)(� − m

i

)(� + m + i

i

)(�

m

)(� + m

m

)

=rm∑

k=m

k∑i=0

a(r)m,k

(m + k

k

)(m

k − i

)(m + i

i

)(�

m + i

)(� + m + i

m + i

),

which implies that

a(r+1)m,m+i =

rm∑k=m

(m + k

k

)(m

k − i

)(m + i

i

)a(r)

m,k ∈ Z, 0 � i � rm.

The proof of the inductive step is then completed. �

1738 V.J.W. Guo, J. Zeng / Journal of Number Theory 132 (2012) 1731–1740

Proof of Theorem 4.1. Applying (4.3) and the easily checked formula (see [12, Lemma 2.1])

n−1∑�=k

(2� + 1)

(�

k

)(� + k

k

)= n

(n

k + 1

)(n + k

k

),

we have

n−1∑�=0

(2� + 1)S(r)� (x) =

n−1∑�=0

(2� + 1)

�∑m=0

(�

m

)r(� + m

m

)r

xm

=n−1∑m=0

xmn−1∑�=m

rm∑k=m

a(r)m,k(2� + 1)

(�

k

)(� + k

k

)

= nn−1∑m=0

xmrm∑

k=m

a(r)m,k

(n

k + 1

)(n + k

k

).

Similarly, applying (4.3) and (2.3), we have

n−1∑�=0

(−1)�(2� + 1)S(r)� (x) = (−1)n−1n

n−1∑m=0

xmrm∑

k=m

a(r)m,k

(n − 1

k

)(n + k

k

).

Therefore the congruences (4.1) and (4.2) hold. �Furthermore, similarly to the proof of [6, Theorem 1.2], we can prove the following generalization

of Theorem 4.1.

Theorem 4.3. Let n ∈ Z+ , a ∈ N, r � 2 and x ∈ Z. Then

n−1∑k=0

εk(2k + 1)ka(k + 1)a S(r)k (x) ≡ 0 (mod n),

n−1∑k=0

εk(2k + 1)2a+1 S(r)k (x) ≡ 0 (mod n),

where ε = ±1.

We conclude this section with the following conjecture, which is due to Sun [12] in the case r = 2.

Conjecture 4.4. Let ε ∈ {±1}, m,n ∈ Z+ , r � 2, and x ∈ Z. Then

n−1∑k=0

(2k + 1)εk S(r)k (x)m ≡ 0 (mod n).

V.J.W. Guo, J. Zeng / Journal of Number Theory 132 (2012) 1731–1740 1739

5. Proof of Theorem 1.3

From the Lagrange interpolation formula for the polynomial∏m

k=1(x − k) at the values −i (0 �i � m) of x we immediately derive that

m∑k=0

(m

k

)(m + k

k

)(−1)m−k

x + k= 1

x

m∏k=1

x − k

x + k. (5.1)

A proof of (5.1) by using the creative telescoping method was given in [8, Lemma 3.1].

Remark. The identity (5.1) plays an important role in Mortenson’s proof [8] of Rodriguez-Villegas’sconjectures [9] on supercongruences for hypergeometric Calabi–Yau manifolds of dimension d � 3.Letting x = m + 1

2 in (5.1), we obtain

m∑k=0

(m

k

)(m + k

k

)(−1)m−k

2m + 2k + 1= (2m)!3

(4m + 1)!m!2 . (5.2)

Let p > 3 be a prime. Similarly to the proof of Theorem 2.1, applying (2.2) and the trivial identity

p−1∑�=m

(�

m + k

)(� + m + k

m + k

)=

(2m + 2k

m + k

)(p + m + k

2m + 2k + 1

),

we have

p−1∑�=0

A�(x) =p−1∑m=0

(2m

m

)xm

m∑k=0

(m

k

)(m + k

k

)(2m + 2k

m + k

)(p + m + k

2m + 2k + 1

). (5.3)

Note that, if 0 � m + k < p, then

(2m + 2k

m + k

)(p + m + k

2m + 2k + 1

)= (2m + 2k)!p ∏m+k

i=1 (p2 − i2)

(m + k)!2(2m + 2k + 1)!≡ (−1)m+k p

2m + 2k + 1

(mod p2),

while if m + k � p, then(2m+2k

m+k

)( p+m+k2m+2k+1

) = 0.From (5.3), we deduce that

p−1∑�=0

A�(x) ≡p−1∑m=0

(2m

m

)xm

p−m−1∑k=0

(m

k

)(m + k

k

)(−1)m+k p

2m + 2k + 1

(mod p2). (5.4)

Since p > 3, we have 2m + 2k + 1 �≡ 0 (mod p2) for any 0 � k � m � p − 1. Thus, by (2.5) and (5.2),the congruence (5.4) is equivalent to

1740 V.J.W. Guo, J. Zeng / Journal of Number Theory 132 (2012) 1731–1740

p−1∑�=0

A�(x) ≡p−1∑m=0

(2m

m

)xm

m∑k=0

(m

k

)(m + k

k

)(−1)m+k p

2m + 2k + 1

=p−1∑m=0

(2m)!4 p

(4m + 1)!m!4 xm (mod p2). (5.5)

On the other hand, it is not difficult to see that

(2m)!4 p

(4m + 1)!m!4 ≡ 0(mod p2) for p/2 < m < p,

and

p(2m)!2(4m + 1)! ≡ p

∏2mi=1(p2 − i2)

(4m + 1)! =(

p + 2m

4m + 1

) (mod p2) for p > 3 and 0 �m < p/2.

This proves that

p−1∑m=0

(2m)!4 p

(4m + 1)!m!4 xm ≡(p−1)/2∑

m=0

(2m)!4 p

(4m + 1)!m!4 xm

≡(p−1)/2∑

m=0

(p + 2m

4m + 1

)(2m

m

)2

xm (mod p2). (5.6)

Combining (5.5) and (5.6), we complete the proof.

Acknowledgments

This work was partially supported by the Fundamental Research Funds for the Central Universi-ties, Shanghai Rising-Star Program (#09QA1401700), Shanghai Leading Academic Discipline Project(#B407), and the National Science Foundation of China (#10801054).

References

[1] G.E. Andrews, Identities in combinatorics, I: On sorting two ordered sets, Discrete Math. 11 (1975) 97–106.[2] R. Apéry, Irrationalité de ζ(2) et ζ(3), Astérisque 61 (1979) 11–13.[3] F. Beukers, Another congruence for the Apéry numbers, J. Number Theory 25 (1987) 201–210.[4] S. Chowla, J. Cowles, M. Cowles, Congruence properties of Apéry numbers, J. Number Theory 12 (1980) 188–190.[5] I. Gessel, Some congruences for Apéry numbers, J. Number Theory 14 (1982) 362–368.[6] V.J.W. Guo, J. Zeng, New congruences for sums involving Apery numbers or central Delannoy numbers, preprint,

arXiv:1008.2894.[7] S. Mattarei, R. Tauraso, Congruences for central binomial sums and finite polylogarithms, preprint, arXiv:1012.1308.[8] E. Mortenson, Supercongruences between truncated 2 F1 by geometric functions and their Gaussian analogs, Trans. Amer.

Math. Soc. 355 (2003) 987–1007.[9] F. Rodriguez-Villegas, Hypergeometric families of Calabi–Yau manifolds, in: Calabi–Yau Varieties and Mirror Symmetry,

Toronto, ON, 2001, in: Fields Inst. Commun., vol. 38, Amer. Math. Soc., Providence, RI, 2003, pp. 223–231.[10] A.L. Schmidt, Generalized q-Legendre polynomials, J. Comput. Appl. Math. 49 (1993) 243–249.[11] R.P. Stanley, Enumerative Combinatorics, vol. I, Cambridge University Press, Cambridge, 1997.[12] Z.-W. Sun, Binomial coefficients, Catalan numbers and Lucas quotients, Sci. China Math. 53 (2010) 2473–2488.[13] Z.-W. Sun, On sums of Apéry polynomials and related congruences, preprint, arXiv:1101.1946.[14] Z.-W. Sun, Conjectures and results on x2 mod p2 with 4p = x2 + dy2, in: Y. Ouyang, C. Xing, F. Xu, P. Zhang (Eds.), Number

Theory and the Related Topics, Higher Education Press & International Press, Beijing and Boston, in press, online version isavailable from http://arxiv.org/abs/1103.4325.

[15] Z.-W. Sun, Congruences involving generalized trinomial coefficients, preprint, arXiv:1008.3887.[16] Z.-W. Sun, R. Tauraso, New congruences for central binomial coefficients, Adv. in Appl. Math. 45 (2010) 125–148.[17] Z.-W. Sun, R. Tauraso, On some new congruences for binomial coefficients, Int. J. Number Theory 7 (2011) 645–662.