aa r X i v : . [ m a t h . N T ] J a n Congruences for truncated hypergeometric series F Department of Mathematics, East China Normal University, Shanghai 200241, PR China [email protected]
Abstract.
Rodriguez-Villegas conjectured four supercongruences associated to certain el-liptic curves, which were first confirmed by Mortenson by using the Gross-Koblitz formula.In this paper, we aim to prove four supercongruences between two truncated hypergeo-metric series F . This generalizes these four Rodriguez-Villegas supercongruences. Keywords : Supercongruences; Hypergeometric series; Fermat quotient
MR Subject Classifications : Primary 11A07; Secondary 33C05
In 2003, Rodriguez-Villegas [13] studied hypergeometric families of Calabi-Yau manifolds.He observed numerically some remarkable supercongruences between the values of thetruncated hypergeometric series and expressions derived from the number of F p -points ofthe associated Calabi-Yau manifolds. A number of supercongruences for hypergeometricCalabi-Yau manifolds have been conjectured by Rodriguez-Villegas. For manifolds ofdimension d = 1, he conjectured four interesting supercongruences associated to certainelliptic curves. These four supercongruences were first confirmed by Mortenson [9, 10] byusing the Gross-Koblitz formula.We first define the truncated hypergeometric series. For complex numbers a i , b j and z , with none of the b j being negative integers or zero, the truncated hypergeometric seriesare given by r F s (cid:20) a , a , · · · , a r b , b , · · · , b s ; z (cid:21) n = n − X k =0 ( a ) k ( a ) k · · · ( a r ) k ( b ) k ( b ) k · · · ( b s ) k · z k k ! , where ( a ) = 1 and ( a ) k = a ( a + 1) · · · ( a + k −
1) for k ≥ Theorem 1.1 (Rodriguez-Villegas-Mortenson) Let p ≥ be a prime. Then F (cid:20) , (cid:21) p ≡ (cid:18) − p (cid:19) (mod p ) , F (cid:20) , (cid:21) p ≡ (cid:18) − p (cid:19) (mod p ) , F (cid:20) , (cid:21) p ≡ (cid:18) − p (cid:19) (mod p ) , F (cid:20) , (cid:21) p ≡ (cid:18) − p (cid:19) (mod p ) , where (cid:16) · p (cid:17) denotes the Legendre symbol. p were obtained in [15,16]. For some interesting q -analogues of Theorem1.1, one refers to [6, 7]. By studying the generalized Legendre polynomials, Z.-H. Sun [14]extended Theorem 1.1 as follows. Theorem 1.2 (Z.-H. Sun) Let p ≥ be a prime. For any p -adic integer x , we have F (cid:20) x, − x (cid:21) p ≡ ( − h− x i p (mod p ) , (1.1) where h a i p denotes the least non-negative integer r with a ≡ r (mod p ) . Observe that ( − h− / i p = (cid:18) − p (cid:19) , ( − h− / i p = (cid:18) − p (cid:19) , ( − h− / i p = (cid:18) − p (cid:19) , ( − h− / i p = (cid:18) − p (cid:19) . Then Theorem 1.2 reduces to Theorem 1.1 when x = , , , .Ap´ery introduced the numbers A n = P nk =0 (cid:0) nk (cid:1) (cid:0) n + kk (cid:1) in his ingenious proof [1] of theirrationality of ζ (3). These numbers are known as Ap´ery numbers. Since the appearanceof these numbers, some interesting arithmetic properties have been gradually discovered.For example, Gessel [5] proved that for any prime p ≥ A np ≡ A n (mod p ) , which confirmed a conjecture by Chowla et al. [4].In this paper, we aim to prove a similar type of supercongruences for the truncatedhypergeometric series F , which generalizes these four supercongruences in Theorem 1.1as follows. Theorem 1.3
Suppose p ≥ is a prime and n is a positive integer. For x ∈ { , , , } ,we have F (cid:20) x, − x (cid:21) np ≡ ( − h− x i p · F (cid:20) x, − x (cid:21) n (mod p ) . (1.2)Theorem 1.3 reduces to Theorem 1.1 when n = 1. Replacing n by p r − in (1.2) andthen using induction, we immediately get the following results: Corollary 1.4
Suppose p ≥ is a prime and r is a positive integer. For x ∈ { , , , } ,we have F (cid:20) x, − x (cid:21) p r ≡ ( − h− x i p · r (mod p ) . The organization of this paper is as follows. In the next section, we first recall someproperties of the Fermat quotients and some combinatorial identities involving harmonicnumbers, and then prove two congruences. In Section 3, we will give a new proof ofTheorem 1.2 by using combinatorial identities. The proof of Theorem 1.3 will be given inthe Section 4. We make some concluding remarks in the last section.2
Some lemmas
The Fermat quotient of an integer a with respect to an odd prime p is given by q p ( a ) = a p − − p , which plays an important role in the study of cyclotomic fields. Lemma 2.1 (Eisenstein) Suppose p is an odd prime and r is a positive integer. Fornon-zero p -adic integers a and b , we have q p ( ab ) ≡ q p ( a ) + q p ( b ) (mod p ) ,q p ( a r ) ≡ rq p ( a ) (mod p ) . Lemma 2.2 (Lehmer [8]) Let H n = P nk =1 1 k be the n -th harmonic number. For anyprime p ≥ , we have H ⌊ p/ ⌋ ≡ − q p (2) (mod p ) , H ⌊ p/ ⌋ ≡ − q p (3) (mod p ) ,H ⌊ p/ ⌋ ≡ − q p (2) (mod p ) , H ⌊ p/ ⌋ ≡ − q p (2) − q p (3) (mod p ) , where ⌊ x ⌋ denotes the greatest integer less than or equal to a real number x . Lemma 2.3 If n is a positive integer, then n X k =0 ( − k (cid:18) nk (cid:19)(cid:18) n + kk (cid:19) = ( − n , (2.1) n X k =1 ( − k (cid:18) nk (cid:19)(cid:18) n + kk (cid:19) H k = 2( − n H n , (2.2) n X k =1 ( − k (cid:18) nk (cid:19)(cid:18) n + kk (cid:19) k X i =1 n + i = ( − n H n . (2.3) Proof.
Prodinger [12] has given a proof of (2.1)-(2.2) by partial fraction decomposition andcreative telescoping, see also [11]. Using the same method, Prodinger [12] also obtainedthe following identity: n X k =1 ( − k (cid:18) nk (cid:19)(cid:18) n + kk (cid:19) H n + k = 2( − n H n . (2.4)It follows that n X k =1 ( − k (cid:18) nk (cid:19)(cid:18) n + kk (cid:19) k X i =1 n + i = n X k =1 ( − k (cid:18) nk (cid:19)(cid:18) n + kk (cid:19) ( H n + k − H n )= ( − n H n . (by (2.1) and (2.4))3his concludes the proof of (2.3). (cid:3) Lemma 2.4
Let p ≥ be a prime, r and ≤ k ≤ p − be non-negative integers. For x ∈ { , , , } , we have ( x ) k + rp (1 − x ) k + rp (1) k + rp ≡ ( x ) r (1 − x ) r (1) r · ( x ) k (1 − x ) k (1) k × rpH ⌊ px ⌋ − rpH k + rp k − X i =0 (cid:18) x + i + 11 − x + i (cid:19)! (mod p ) . (2.5) Proof.
Note that ( x ) n (1 − x ) n (1) n = S x ( n ) a nx , where S x ( n ) = (cid:0) nn (cid:1) , if x = , (cid:0) nn (cid:1)(cid:0) nn (cid:1) , if x = , (cid:0) nn (cid:1)(cid:0) n n (cid:1) , if x = , (cid:0) nn (cid:1)(cid:0) n n (cid:1) , if x = ,and a / = 16 , a / = 27 , a / = 64 and a / = 432. We see that (2.5) is equivalent to S x ( k + rp ) ≡ a r ( p − x S x ( r ) S x ( k ) × rpH ⌊ px ⌋ − rpH k + rp k − X i =0 (cid:18) x + i + 11 − x + i (cid:19)! (mod p ) . (2.6)By Lemma 2.1 and 2.2, we have a r ( p − x ≡ pq p ( a rx ) ≡ rpq p ( a x ) ≡ − rpH ⌊ px ⌋ (mod p ) . So in order to prove (2.6), it suffices to show that S x ( k + rp ) ≡ S x ( r ) S x ( k ) × − rpH k + rp k − X i =0 (cid:18) x + i + 11 − x + i (cid:19)! (mod p ) . (2.7)4oting that k − X j =0 (cid:18) j + 1 / j + 1 / (cid:19) = 4 H k − H k , k − X j =0 (cid:18) j + 1 / j + 2 / (cid:19) = 3 H k − H k , k − X j =0 (cid:18) j + 1 / j + 3 / (cid:19) = 4 H k − H k , k − X j =0 (cid:18) j + 1 / j + 5 / (cid:19) = 6 H k − H k − H k + H k , (2.7) becomes the following congruences modulus p (cid:18) rp + 2 krp + k (cid:19) ≡ (cid:18) rr (cid:19) (cid:18) kk (cid:19) (1 + rp (4 H k − H k )) , (2.8) (cid:18) rp + 2 krp + k (cid:19)(cid:18) rp + 3 krp + k (cid:19) ≡ (cid:18) rr (cid:19)(cid:18) rr (cid:19)(cid:18) kk (cid:19)(cid:18) kk (cid:19) (1 + rp (3 H k − H k )) , (2.9) (cid:18) rp + 2 krp + k (cid:19)(cid:18) rp + 4 k rp + 2 k (cid:19) ≡ (cid:18) rr (cid:19)(cid:18) r r (cid:19)(cid:18) kk (cid:19)(cid:18) k k (cid:19) (1 + rp (4 H k − H k − H k )) , (2.10) (cid:18) rp + 3 krp + k (cid:19)(cid:18) rp + 6 k rp + 3 k (cid:19) ≡ (cid:18) rr (cid:19)(cid:18) r r (cid:19)(cid:18) kk (cid:19)(cid:18) k k (cid:19) (1 + rp (6 H k − H k − H k − H k )) . (2.11)Next we only prove (2.8). The proof of (2.9)-(2.11) runs analogously.Note that (cid:18) rp + 2 krp + k (cid:19) = (cid:18) rprp (cid:19) k Y i =1 (2 rp + i ) / k Y i =1 ( rp + i ) ≡ (cid:18) rr (cid:19) k Y i =1 (2 rp + i ) / k Y i =1 ( rp + i ) (mod p ) , (2.12)where we have utilized the Babbage’s theorem [2] (cid:18) apbp (cid:19) ≡ (cid:18) ab (cid:19) (mod p ) . Now we consider the following rational function: f ( x ) = k Y i =1 (2 rx + i ) / k Y i =1 ( rx + i ) . (2.13)5aking the logarithmic derivative on both sides of (2.13) gives f ′ ( x ) f ( x ) = k X i =1 r rx + i − k X i =1 rrx + i . (2.14)From (2.13) and (2.14), we have f (0) = (cid:18) kk (cid:19) and f ′ (0) = r (cid:18) kk (cid:19) (2 H k − H k ) . Now we get the first two terms of the Taylor expansion for f ( x ): f ( x ) = (cid:18) kk (cid:19) + rx (cid:18) kk (cid:19) (2 H k − H k ) + O (cid:0) x (cid:1) . (2.15)Combining (2.12) and (2.15), we immediately get (cid:18) rp + 2 krp + k (cid:19) ≡ (cid:18) rr (cid:19)(cid:18) kk (cid:19) (1 + rp (2 H k − H k )) (mod p ) . It follows that (cid:18) rp + 2 krp + k (cid:19) ≡ (cid:18) rr (cid:19) (cid:18) kk (cid:19) (1 + rp (4 H k − H k )) (mod p ) . This concludes the proof of (2.8). (cid:3)
Lemma 2.5
Suppose p ≥ is a prime and ≤ k ≤ p − is an integer. For x ∈{ , , , } , we have ( x ) k (1 − x ) k (1) k ≡ ( − k (cid:18) ⌊ px ⌋ k (cid:19)(cid:18) ⌊ px ⌋ + kk (cid:19) (mod p ) . (2.16) Proof.
It suffices to show that for 0 ≤ k ≤ p − ⌊ px ⌋ + 1) k ( −⌊ px ⌋ ) k ≡ ( x ) k (1 − x ) k (mod p ) . (2.17)For any prime p ≥
5, there exists ε ∈ { , − } such that p ≡ ε (mod 2 , , , x = . The proof of other three cases runs similarly.If p ≡ ⌊ p/ ⌋ = p − , and hence (cid:18) p + 23 (cid:19) k (cid:18) − p + 13 (cid:19) k ≡ (cid:18) (cid:19) k (cid:18) (cid:19) k (mod p ) . If p ≡ − ⌊ p/ ⌋ = p − , and so (cid:18) p + 13 (cid:19) k (cid:18) − p + 23 (cid:19) k ≡ (cid:18) (cid:19) k (cid:18) (cid:19) k (mod p ) . This implies that (2.17) is true for x = . (cid:3) A new proof of Theorem 1.2
Letting x → − x in Theorem 1.2, (1.1) is equivalent to F (cid:20) − x, x (cid:21) p ≡ ( − h x i p (mod p ) . (3.1)It is easy to see that ( − x ) k (1 + x ) k (1) k = ( − k (cid:18) xk (cid:19)(cid:18) x + kk (cid:19) . Let δ denote the number δ = ( x − h x i p ) /p . It is clear that δ is a p -adic integer and x = h x i p + δp . Note that (cid:18) xk (cid:19)(cid:18) x + kk (cid:19) = (cid:18) h x i p + δpk (cid:19)(cid:18) h x i p + δp + kk (cid:19) = k Y i =1 ( h x i p + δp + 1 − i ) k Y i =1 ( h x i p + δp + i ) k Y i =1 i − ! ≡ (cid:18) h x i p k (cid:19)(cid:18) h x i p + kk (cid:19) δp k X i =1 h x i p + i + k X i =1 h x i p + 1 − i !! (mod p ) . It follows thatLHS (3.1) ≡ p − X k =0 ( − k (cid:18) h x i p k (cid:19)(cid:18) h x i p + kk (cid:19) × δp k X i =1 h x i p + i + k X i =1 h x i p + 1 − i !! (mod p ) . (3.2)Let b = p − h x i p . It is clear that h x i p ≡ − b (mod p ) and 0 ≤ b − ≤ p −
1. By (2.3),we have p − X k =0 ( − k (cid:18) h x i p k (cid:19)(cid:18) h x i p + kk (cid:19) k X i =1 h x i p + 1 − i ≡ − p − X k =0 (cid:18) − bk (cid:19)(cid:18) − b + kk (cid:19) k X i =1 b − i (mod p )= − p − X k =0 (cid:18) b − k (cid:19)(cid:18) b − kk (cid:19) k X i =1 b − i = ( − b H b − ≡ ( − h x i p +1 H h x i p (mod p ) , (3.3)7here we have used the fact that (cid:0) − bk (cid:1)(cid:0) − b + kk (cid:1) = (cid:0) b − k (cid:1)(cid:0) b − kk (cid:1) in the second step and H p − k − ≡ H k (mod p ) for 0 ≤ k ≤ p − p − X k =0 ( − k (cid:18) h x i p k (cid:19)(cid:18) h x i p + kk (cid:19) = ( − h x i p , (3.4)and p − X k =0 ( − k (cid:18) h x i p k (cid:19)(cid:18) h x i p + kk (cid:19) k X i =1 h x i p + i = ( − h x i p H h x i p . (3.5)Substituting (3.3)-(3.5) into (3.2), we complete the proof of (3.1). Assume x ∈ { , , , } . We first prove that for any non-negative integer r ( r +1) p − X k = rp ( x ) k (1 − x ) k (1) k ≡ ( x ) r (1 − x ) r (1) r · F (cid:20) x, − x (cid:21) p (mod p ) . (4.1)Letting k → k + rp on the left-hand side of (4.1) and then applying Lemma 2.4, we obtain ( r +1) p − X k = rp ( x ) k (1 − x ) k (1) k ≡ ( x ) r (1 − x ) r (1) r p − X k =0 ( x ) k (1 − x ) k (1) k × rpH ⌊ px ⌋ − rpH k + rp k − X i =0 (cid:18) x + i + 11 − x + i (cid:19)! (mod p ) . (4.2)Using the following identity [19, Theorem 1]:( x ) k (1 − x ) k (1) k k − X i =0 (cid:18) x + i + 11 − x + i (cid:19) = k − X i =0 ( x ) i (1 − x ) i (1) i · k − i ,
8e have p − X k =0 ( x ) k (1 − x ) k (1) k k − X i =0 (cid:18) x + i + 11 − x + i (cid:19) = p − X k =0 k − X i =0 ( x ) i (1 − x ) i (1) i · k − i = p − X i =0 ( x ) i (1 − x ) i (1) i p − X k = i +1 k − i ≡ p − X i =0 ( x ) i (1 − x ) i (1) i H i (mod p ) , (4.3)since H p − − i ≡ H i (mod p ) and ( x ) i (1 − x ) i ≡ p ) for i = p − ( r +1) p − X k = rp ( x ) k (1 − x ) k (1) k ≡ ( x ) r (1 − x ) r (1) r p − X k =0 ( x ) k (1 − x ) k (1) k + rp p − X k =0 ( x ) k (1 − x ) k (1) k (2 H ⌊ px ⌋ − H k ) ! (mod p ) . (4.4)By (2.16), we have p − X k =0 ( x ) k (1 − x ) k (1) k (2 H ⌊ px ⌋ − H k ) ≡ p − X k =0 ( − k (cid:18) ⌊ px ⌋ k (cid:19)(cid:18) ⌊ px ⌋ + kk (cid:19) (2 H ⌊ px ⌋ − H k ) (mod p )= 0 . (by (2.1) and (2.2)) (4.5)Substituting (4.5) into (4.4), we complete the proof of (4.1).Taking the sum over r from 0 to n − F (cid:20) x, − x (cid:21) np ≡ F (cid:20) x, − x (cid:21) p · F (cid:20) x, − x (cid:21) n (mod p ) . (4.6)The proof of Theorem 1.3 then follows from (1.1) and (4.6). Numerical calculation suggests that supercongruence (1.2) can not be extended to any p -adic integer x in the direction of Theorem 1.2.9ecently, Z.-W. Sun [17, Conjecture 5.4] made four challenging conjectures whichextend Theorem 1.3 and some results proved by Z.-H. Sun [15] and Z.-W. Sun [16]. Conjecture 5.1 (Z.-W. Sun) Let p ≥ be a prime and n be a positive integer. Then n n (cid:0) nn (cid:1) np − X k =0 (cid:0) kk (cid:1) k − (cid:18) − p (cid:19) n − X k =0 (cid:0) kk (cid:1) k ! ≡ − p E p − (mod p ) , n n (cid:0) nn (cid:1)(cid:0) nn (cid:1) np − X k =0 (cid:0) kk (cid:1)(cid:0) kk (cid:1) k − (cid:18) − p (cid:19) n − X k =0 (cid:0) kk (cid:1)(cid:0) kk (cid:1) k ! ≡ − p B p − (cid:18) (cid:19) (mod p ) , n n (cid:0) nn (cid:1)(cid:0) n n (cid:1) np − X k =0 (cid:0) kk (cid:1)(cid:0) k k (cid:1) k − (cid:18) − p (cid:19) n − X k =0 (cid:0) kk (cid:1)(cid:0) k k (cid:1) k ! ≡ − p E p − (cid:18) (cid:19) (mod p ) , n n (cid:0) nn (cid:1)(cid:0) n n (cid:1) np − X k =0 (cid:0) kk (cid:1)(cid:0) k k (cid:1) k − (cid:18) − p (cid:19) n − X k =0 (cid:0) kk (cid:1)(cid:0) k k (cid:1) k ! ≡ − p E p − (mod p ) , where E m is the m -th Euler number, E m ( x ) and B m ( x ) denote the Euler polynomial andthe Bernoulli polynomial of degree m , respectively. Noting that (cid:0) kk (cid:1) k = (cid:0) (cid:1) k (1) k , (cid:0) kk (cid:1)(cid:0) kk (cid:1) k = (cid:0) (cid:1) k (cid:0) (cid:1) k (1) k , (cid:0) kk (cid:1)(cid:0) k k (cid:1) k = (cid:0) (cid:1) k (cid:0) (cid:1) k (1) k , (cid:0) kk (cid:1)(cid:0) k k (cid:1) k = (cid:0) (cid:1) k (cid:0) (cid:1) k (1) k , we can directly deduce Theorem 1.3 from Conjecture 5.1. Unfortunately, the methodin this paper is not applicable for proving Conjecture 5.1. We hope that the interestedreader can work on this problem. Acknowledgments.
The author would like to thank Professor V.J.W. Guo for valuablecomments which improved the presentation of the paper.
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