Proof of some conjectural supercongruences via curious identities
aa r X i v : . [ m a t h . N T ] A ug PROOF OF SOME CONJECTURAL SUPERCONGRUENCESVIA CURIOUS IDENTITIES
CHEN WANG AND ZHI-WEI SUN
Abstract.
In this paper, we prove several supercongruences conjectured by Z.-W. Sun tenyears ago via certain strange hypergeometric identities. For example, for any prime p >
3, weshow that p − X k =0 (cid:0) k k +1 (cid:1)(cid:0) kk (cid:1) k ≡ p ) , and p − X k =0 (cid:0) kk (cid:1)(cid:0) kk (cid:1) k ≡ (cid:0) (2 p − / p − / (cid:1) (mod p ) if p ≡ ,p/ (cid:0) (2 p +2) / p +1) / (cid:1) (mod p ) if p ≡ . We also obtain some other results of such types. Introduction
For n, r ∈ N = { , , , . . . } and α , . . . , α r , β , . . . , β r , z ∈ C the truncated hypergeometricseries r +1 F r are defined by r +1 F r (cid:20) α α · · · α r β · · · β r (cid:12)(cid:12)(cid:12)(cid:12) z (cid:21) n := n X k =0 ( α ) k · · · ( α r ) k ( β ) k · · · ( β r ) k · z k k ! , where ( α ) k = α ( α + 1) · · · ( α + k −
1) is the Pochhammer symbol (rising factorial). Since( − α ) k / (1) k = ( − k (cid:0) αk (cid:1) , sometimes we may write the truncated hypergeometric series as sumsinvolving products of binomial coefficients. In the past decades, supercongruences involvingtruncated hypergeometric series have been widely studied (cf. for example, [5, 6, 9, 10, 11, 14,15, 24, 25, 26, 30]).Via the p -adic Gamma function and the Gross-Koblitz formula, E. Mortenson [14, 15] provedthat for any prime p > F (cid:20)
12 12 (cid:12)(cid:12)(cid:12)(cid:12) (cid:21) p − ≡ (cid:18) − p (cid:19) (mod p ) , F (cid:20)
13 23 (cid:12)(cid:12)(cid:12)(cid:12) (cid:21) p − ≡ (cid:18) − p (cid:19) (mod p ) , F (cid:20)
14 34 (cid:12)(cid:12)(cid:12)(cid:12) (cid:21) p − ≡ (cid:18) − p (cid:19) (mod p ) , F (cid:20)
16 56 (cid:12)(cid:12)(cid:12)(cid:12) (cid:21) p − ≡ (cid:18) − p (cid:19) (mod p ) , (1.1) Mathematics Subject Classification.
Primary 11B65, 33C20; Secondary 05A19, 11A07, 33E50.
Key words and phrases.
Truncated hypergeometric series, p -adic Gamma function, congruences, binomialcoefficients.The work is supported by the National Natural Science Foundation of China (grant no. 11971222). where ( · p ) denotes the Legendre symbol. Actually, these congruences were first conjectured in[18] motivated by hypergeometric families of Calabi-Yau manifolds. Z.-W. Sun [26] gave someextensions of (1.1). For any prime p >
3, Z.-W. Sun [25] showed further that ( p − / X k =0 (cid:0) kk (cid:1) k ≡ ( − ( p − / + p E p − (mod p ) , (1.2) X p/
3. The only known example we can recall is Wolstenholme’s congruence H p − ≡ p ) (cf. [31]) for any prime p >
3, where H n denotes the harmonic number P Let p > be a prime. Then p − X k =0 (cid:0) k k +1 (cid:1)(cid:0) kk (cid:1) k ≡ p ) . (1.5) Remark . Sun [29, Conjecture 17] conjectured further that for any prime p > p − X k =0 (cid:0) k k +1 (cid:1)(cid:0) kk (cid:1) k ≡ p B p − (cid:18) (cid:19) (mod p ) (1.6)and p p − X k =1 k k (2 k − (cid:0) k k (cid:1)(cid:0) kk (cid:1) ≡ (cid:16) p (cid:17) + 4 p (mod p ) , (1.7) ROOF OF SOME CONJECTURAL SUPERCONGRUENCES VIA CURIOUS IDENTITIES 3 where B p − ( x ) is the Bernoulli polynomial of degree p − 2. It is worth mentioning that (1.7)is related to Sun’s conjectural identity ∞ X k =1 k k (2 k − (cid:0) k k (cid:1)(cid:0) kk (cid:1) = 152 ∞ X k =1 ( k ) k (1.8)(cf. [28, (1.23)]) and Sun would like to offer $480 as the prize for the first proof of (1.8).Our second theorem concerns a variant of the second congruence in (1.1) and confirms aconjecture of Z.-W. Sun in [25, Conjecture 5.13] and [29, Conjecture 16(i)]. Theorem 1.2. Let p > be a prime. Then p − X k =0 (cid:0) kk (cid:1)(cid:0) kk (cid:1) k ≡ (cid:0) (2 p − / p − / (cid:1) (mod p ) if p ≡ ,p/ (cid:0) (2 p +2) / p +1) / (cid:1) (mod p ) if p ≡ . (1.9) Remark . By Sun [26, (1.20)], for any prime p > p − X k =0 (cid:0) kk (cid:1)(cid:0) kk (cid:1) k ≡ (cid:16) p (cid:17) p − X k =0 (cid:0) kk (cid:1)(cid:0) kk (cid:1) ( − k (mod p ) . Theorem 3.2 of Sun [27] with x = y = − z and a = 1 gives the following p -adic analogue ofthe Clausen identity (cf. [1, p. 116]): F (cid:20) α − α (cid:12)(cid:12)(cid:12)(cid:12) z (cid:21) p − ! ≡ F (cid:20) α − α (cid:12)(cid:12)(cid:12)(cid:12) z (1 − z ) (cid:21) p − (mod p ) (1.10)for any odd prime p and α, z ∈ Z p ; this was given by Z.-H. Sun in the cases α = 1 / , / , / α = 1 / z = 9 / / k (2 / k / (1) k = (cid:0) kk (cid:1)(cid:0) kk (cid:1) / k we obtain that p − X k =0 (cid:0) kk (cid:1)(cid:0) kk (cid:1) k ! ≡ p − X k =0 (cid:0) kk (cid:1) (cid:0) kk (cid:1) ( − k (mod p ) . (1.11)It is known (cf. [2, 3]) that for any prime p ≡ p = x + 27 y ( x, y ∈ Z ) wehave (cid:18) (2 p − / p − / (cid:19) ≡ (cid:16) x (cid:17) (cid:16) px − x (cid:17) (mod p ) . (1.12)Combining Theorem 1.2, (1.11) and (1.12) we immediately obtain the following result whichwas conjectured by Z.-W. Sun in [25, Conjecture 5.6] and [29, Conjecture 24(i)] and partiallyproved by Z.-H. Sun [21, Theorem 4.2]. Corollary 1.1. Let p > be a prime. Then p − X k =0 (cid:0) kk (cid:1) (cid:0) kk (cid:1) ( − k ≡ ( x − p (mod p ) if p ≡ p = x + 27 y ( x, y ∈ Z ) , p ) if p ≡ . CHEN WANG AND ZHI-WEI SUN The next result gives a companion of Theorem 1.2. Theorem 1.3. Let p > be a prime. Then p − X k =0 ( k + 1) (cid:0) kk (cid:1)(cid:0) kk (cid:1) k ≡ p/ (cid:0) (2 p − / p − / (cid:1) (mod p ) if p ≡ , − ( p + 1) (cid:0) (2 p +2) / p +1) / (cid:1) (mod p ) if p ≡ . (1.13)Combining Theorems 1.2 and 1.3 we confirm the following two conjectures of Sun [25,Conjecture 5.13]. Corollary 1.2. Let p > be a prime. Then p − X k =0 (cid:0) kk (cid:1)(cid:0) kk (cid:1) ( k + 1)24 k ≡ (cid:18) p − (cid:0) p (cid:1) ) / p − (cid:0) p (cid:1) ) / (cid:19) (mod p ) . (1.14) When p ≡ and p = x + 27 y with x, y ∈ Z and x ≡ , we have p − X k =0 k + 224 k (cid:18) kk (cid:19)(cid:18) kk (cid:19) ≡ x (mod p ) . (1.15) Remark . To obtain (1.14), we note the following congruence relation obtained by Sun [26] p − X k =0 (cid:0) kk (cid:1)(cid:0) kk (cid:1) ( k + 1) m k ≡ p + m − p − X k =0 k (cid:0) kk (cid:1)(cid:0) kk (cid:1) m k (mod p ) , where p > m is an integer with p ∤ m . To get (1.15) we only need to substitute(1.12) into (1.9) and (1.13).The proofs of the above three theorems depend on some new hypergeometric identitiesmotivated by the strange identities obtained by S. B. Ekhad [7] and the Pfaff transformation(cf. [1, p. 68]). We can also prove some other results similar to Theorems 1.1–1.3 in the sameway, however, we will not give the detailed proofs of them. We shall list these congruencesand sketch their proofs in the last section.Our next theorem gives a general result concerning hypergeometric series F . Theorem 1.4. Let p be an odd prime and x, t ∈ Z p and let a denote h− x i p . If p ∤ s = ( x + a ) /p ,then we have F (cid:20) x (cid:12)(cid:12)(cid:12)(cid:12) t (cid:21) p − ≡ (1 − t ) a s − s (1 − t ) p − st p − stp a X k =1 k (1 − t ) k ! (mod p ) . (1.16)Clearly, Theorem 1.4 has many applications. For example, we can prove the following result. Corollary 1.3. For any prime p > we have F (cid:20) (cid:12)(cid:12)(cid:12)(cid:12) (cid:21) p − − F (cid:20) (cid:12)(cid:12)(cid:12)(cid:12) (cid:21) p − ≡ p − p ) . We shall prove Theorems 1.1–1.4 in Sections 2–5, respectively. In the last section, we willlist some more congruences which can be proved similarly as Theorems 1.1 –1.3. ROOF OF SOME CONJECTURAL SUPERCONGRUENCES VIA CURIOUS IDENTITIES 5 Proof of Theorem 1.1 Let us recall the concept of the p -adic Gamma function introduced by Y. Morita [13] as a p -adic analogue of the classical Gamma function. For each integer n > 1, define the p -adicGamma function Γ p ( n ) := ( − n Y ≤ k For any prime p > and α, t ∈ Z p we have Γ p ( α + tp ) ≡ Γ p ( α ) (cid:0) tp (Γ ′ p (0) + H p − −h− α i p ) (cid:1) (mod p ) , (2.3) where H n = P nk =1 /k is the n th harmonic number.Proof. It is known (cf. [10, Theorem 14]) thatΓ p ( α + tp ) ≡ Γ p ( α )(1 + tp Γ ′ p ( α )) (mod p ) . By [30, Lemma 2.4] we haveΓ ′ p ( α )Γ p ( α ) ≡ Γ ′ p (0) + H p − −h− α i p (mod p ) . Combining the above we immediately get the desired result. (cid:3) The following hypergeometric identity plays a key role in the proof of Theorem 1.1. Lemma 2.2. For any nonnegative integer n and δ ∈ { , } , we have n X k =0 (4 n + 2 δ − k + 1)( − n ) k ( − δ − n ) k (2 n + δ + k + 1)(1) k ( − n − δ ) k (cid:18) (cid:19) k = (cid:18) (cid:19) n + δ Γ( )Γ(2 n + 1 + δ )Γ(2 n + δ + ) . (2.4) CHEN WANG AND ZHI-WEI SUN Proof. Denote the left-hand side of (2.4) by S ( n ). Via Zeilberger algorithm (cf. [16]) in Mathematica , we find that9( n + 1)(2 n + 2 δ + 1) S ( n ) − n + 2 δ + 1)(4 n + 2 δ + 3) S ( n + 1) = 0for any n ∈ N and δ ∈ { , } . Then the identities can be verified by induction on n . (cid:3) For a prime p and an integer a ≡ p ), we use q p ( a ) to denote the Fermat quotient( a p − − /p . Lemma 2.3 ([8]) . For any prime p > , we have the following congruences H ⌊ p/ ⌋ ≡ − q p (2) , H ⌊ p/ ⌋ ≡ − q p (3) , H ⌊ p/ ⌋ ≡ − q p (2) , H ⌊ p/ ⌋ ≡ − q p (2) − q p (3) modulo p .Proof of Theorem 1.1 . Note that (1 / k (3 / k / (1) k = (cid:0) k k (cid:1)(cid:0) kk (cid:1) / k . Thus we can rewrite (1.5)as follows: p − X k =0 k ( ) k ( ) k (2 k + 1)(1) k (cid:18) (cid:19) k ≡ p ) . Let f k ( x ) := ( x − k )( − x ) k ( − x ) k ( x + k + )(1 − x ) k (1) k (cid:18) (cid:19) k and f ( x ) := p − X k =0 k =( p − / f k ( x ) . Clearly, for any t ∈ Z p we have f ( tp ) ≡ f (0) + tpf ′ (0) (mod p ) , where f ′ ( x ) stands for the derivative of f ( x ). Thus we can easily obtain that f (0) ≡ 12 (3 f ( p ) − f (3 p )) (mod p ) . (2.5)On the other hand, it is easy to see that f (0) = − p − X k =0 k ( ) k ( ) k (2 k + 1)(1) k (cid:18) (cid:19) k + ǫ, (2.6)where ǫ := ( p − ) ( p − / ( ) ( p − / p (1) p − / (cid:18) (cid:19) ( p − / . ROOF OF SOME CONJECTURAL SUPERCONGRUENCES VIA CURIOUS IDENTITIES 7 Combining (2.5) and (2.6) we arrive at p − X k =0 k ( ) k ( ) k (2 k + 1)(1) k (cid:18) (cid:19) k ≡ ǫ − f ( p ) + 12 f (3 p ) (mod p ) . (2.7)We first consider ǫ modulo p . Clearly,Γ( − + p )Γ( + p )Γ( )Γ( ) = p Γ p ( − + p )Γ p ( + p )4Γ p ( )Γ p ( ) . Thus by Lemma 2.1 we have ǫ = ( p − − + p )Γ( + p )Γ(1) p Γ( )Γ( )Γ( + p ) (cid:18) (cid:19) ( p − / = ( p − p Γ p ( − + p )Γ p ( + p )Γ p (1) p Γ p ( )Γ p ( )Γ p ( + p ) (cid:18) (cid:19) ( p − / ≡ ( − p − p ( ) (cid:0) pH ⌊ p/ ⌋ − pH ( p − / (cid:1) (cid:18) (cid:19) ( p − / (mod p ) . Note that (cid:18) (cid:19) ( p − / = 3 ( p − / (cid:18) (cid:19) p − ≡ ( p − / (1 + p q p (2) − p q p (3)) (mod p ) . Then by (2.1) and Lemma 2.3 we immediately obtain ǫ ≡ ( − ( p − / (1 + p − p q p (3)) (mod p ) . (2.8)Now we evaluate f ( p ) modulo p . Obviously, f ( p − / ( p ) = 0 since (cid:16) − j p k(cid:17) ( p − / = 0 . Taking n = ⌊ p/ ⌋ and δ = (1 − ( − p )) / f ( p ) = f ( p ) + f ( p − / ( p ) = (cid:18) (cid:19) ( p − / Γ( )Γ( p +12 )Γ( p ) = − ( p − / p − Γ p ( )Γ p ( p +12 )Γ p ( p ) ≡ − ( p − / Γ p (cid:18) (cid:19) (cid:16) p H ( p − / (cid:17) (1 − p q p (2)) ≡ ( − ( p − / (1 − p q p (2)) (mod p ) . (2.9)Finally, we consider f (3 p ) modulo p . Taking n = ⌊ p/ ⌋ and δ = (1 + ( − p )) / f (3 p ) + f ( p − / (3 p ) = (cid:18) (cid:19) (3 p − / Γ( )Γ( p +12 )Γ( p ) . (2.10) CHEN WANG AND ZHI-WEI SUN Via a similar discussion as in (2.9) we deduce that (cid:18) (cid:19) (3 p − / Γ( )Γ( p +12 )Γ( p ) = (cid:18) (cid:19) (3 p − / p Γ p ( )Γ p ( p +12 ) p Γ p ( p )Γ p (1) ≡ 32 ( − ( p − / (1 + p q p (3) − p q p (2)) (mod p ) . (2.11)Observe that f ( p − / (3 p ) = (5 p + 1)Γ( − p )Γ( − − p )Γ(1)Γ(1 − p )4 p Γ( − p )Γ( − p )Γ( + p )Γ( − p ) (cid:18) (cid:19) ( p − / = − p (5 p + 1)Γ p ( − p )Γ p ( − − p )Γ p (1)Γ p (1 − p )4 p Γ p ( − p )Γ p ( − p )Γ p ( + p )Γ p ( − p ) (cid:18) (cid:19) ( p − / , where we note that Γ( − p )Γ( − − p )Γ( − p )Γ( − p ) = − p Γ p ( − p )Γ p ( − − p )2Γ p ( − p )Γ p ( − p ) . Furthermore, in view of (2.1), Lemmas 2.1 and 2.3, we arrive at f ( p − / (3 p ) ≡ ( − ( p − / p − p q p (2) − p q p (3)) (mod p ) . (2.12)Now combining (2.7) and (2.8)–(2.12) we finally obtain p − X k =0 k ( ) k ( ) k (2 k + 1)(1) k (cid:18) (cid:19) k ≡ p ) . This completes the proof. (cid:3) Proof of Theorem 1.2 Lemma 3.1. For any nonnegative integer n we have F (cid:20) − n − n − n (cid:12)(cid:12)(cid:12)(cid:12) (cid:21) n = ( ) n n ( ) n (3.1) and F (cid:20) − n − − n − n − (cid:12)(cid:12)(cid:12)(cid:12) (cid:21) n = ( ) n n ( ) n . (3.2) Proof. Denote the left-hand sides of (3.1) and (3.2) by F ( n ) and G ( n ) respectively. By applyingZeilberger algorithm in Mathematica , we find that − n + 1) F ( n ) + 4(3 n + 1) F ( n + 1) = 0and − (6 n + 5) G ( n ) + 4(3 n + 2) G ( n + 1) = 0 . Then Lemma 3.1 follows by induction on n . (cid:3) ROOF OF SOME CONJECTURAL SUPERCONGRUENCES VIA CURIOUS IDENTITIES 9 The next lemma is a p -adic analogue of the classical Gauss multiplication formula. Lemma 3.2. [17, p. 371] Let p be an odd prime. Then for any x ∈ Z p and m ∈ Z + we have Y ≤ j 2) (mod p ) , (3.7)where in the last step we note that Γ p (1 / p (2 / 3) = ( − p − / = − H p − − k ≡ H k (mod p ) for any k ∈ { , , . . . , p − } . We now evaluate σ . Similarly, by (3.6),Lemmas 2.1 and 2.3 we have σ =2 (2 − p ) / Γ( + p )Γ( )Γ( )Γ( p ) = Γ( + p ) Γ( + p )Γ( )Γ( + p )Γ(1)Γ( + p )Γ( p )Γ( )Γ( )Γ( + p ) ≡ Γ p ( + p )Γ p ( + p )Γ p ( )Γ p ( + p )Γ p ( + p )Γ p ( p )Γ p ( )Γ p ( ) ≡ Γ p ( ) Γ p ( ) Γ p ( )Γ p ( )Γ p ( ) (cid:16) − p H ( p − / + pH ( p − / (cid:17) ≡ Γ p ( ) Γ p ( ) Γ p ( )Γ p ( )Γ p ( ) (1 − p q p (2) − p q p (3)) (mod p ) . In view of Lemma 3.2, we arrive atΓ p ( ) Γ p ( ) Γ p ( )Γ p ( )Γ p ( ) = Γ p ( )Γ p ( ) Γ p ( )Γ p ( ) = Γ p ( ) Γ p ( ) = − Γ p (cid:18) (cid:19) . So we have σ ≡ − Γ p (cid:18) (cid:19) (1 − p q p (2) − p q p (3)) (mod p ) . (3.8)Substituting (3.7) and (3.8) into (3.5) we obtain p − X k =0 ( ) k ( ) k (1) k (cid:18) (cid:19) k ≡ − Γ p (cid:18) (cid:19) (mod p ) . Thus it suffices to show that (cid:18) (2 p − / p − / (cid:19) ≡ − Γ p (cid:18) (cid:19) (mod p ) . ROOF OF SOME CONJECTURAL SUPERCONGRUENCES VIA CURIOUS IDENTITIES 11 In fact, it is routine to verify that (cid:18) (2 p − / p − / (cid:19) = − Γ p ( + p )Γ p ( + p ) ≡ − Γ p (cid:18) (cid:19) (cid:18) p (cid:0) H (2 p − / − H ( p − / (cid:1)(cid:19) ≡ − Γ p (cid:18) (cid:19) (mod p ) . This proves the case p ≡ p ≡ F (cid:20) − p − p − p (cid:12)(cid:12)(cid:12)(cid:12) (cid:21) p − ≡ p − X k =0 ( ) k ( ) k (1) k (cid:18) (cid:19) k − p k − X j =0 j + 1 − p k − X j =0 j + 2 + 2 pH k ! (mod p )and F (cid:20) − p − p − p (cid:12)(cid:12)(cid:12)(cid:12) (cid:21) p − ≡ p − X k =0 ( ) k ( ) k (1) k (cid:18) (cid:19) k − p k − X j =0 j + 1 − p k − X j =0 j + 2 + pH k ! (mod p ) . Combining the above two congruences and letting n = (2 p − / n = ( p − / p − X k =0 ( ) k ( ) k (1) k (cid:18) (cid:19) k ≡ σ − σ (mod p ) , (3.9)where σ := 2 (5 − p ) / ( ) ( p − / ( ) ( p − / and σ := 2 (1 − p ) / ( ) (2 p − / ( ) (2 p − / . We first consider σ . Note that( ) (2 p − / (1) (2 p − / = (cid:0) (4 p − / p − / (cid:1) (2 p − / = 4 (1 − p ) / Γ( p +13 )Γ( p +23 ) . Therefore, in view of (2.2) we have σ =4 × p − × (1 − p ) / Γ( + p )Γ( )Γ( )Γ( p ) = 4 × p − Γ( + p )Γ( + p )Γ( + p )Γ( )Γ( )Γ( + p )Γ( )Γ( p ) ≡ p Γ p ( ) Γ p ( ) p ( )Γ p ( )Γ p ( ) (mod p ) , where we note that Γ(1 / p/ / Γ(1 / 2) contains the factor p/ 2, Γ(2 / p/ / Γ(1 / p/ /p and Γ(1 / p/ / Γ(5 / 6) contains the factor p/ p ≡ p (1 / p (2 / 3) = ( − p +1) / = 1, we obtainΓ p ( ) Γ p ( ) Γ p ( )Γ p ( )Γ p ( ) = Γ p ( ) Γ p ( ) = Γ p (cid:18) (cid:19) . Thus we get σ ≡ p p (cid:18) (cid:19) (mod p ) . (3.10)We now consider σ . Clearly, σ =2 − p (cid:18) (4 p − / p − / (cid:19) (1) (2 p − / ( ) (2 p − / = 2 − p Γ( + p )Γ( )Γ( + p )Γ( p )=2 − p p Γ p ( + p )Γ p ( )3Γ p ( + p )Γ p ( p ) ≡ p Γ p ( ) p ( ) = p p (cid:18) (cid:19) (mod p ) . (3.11)Substituting (3.10) and (3.11) into (3.9) we have p − X k =0 ( ) k ( ) k (1) k (cid:18) (cid:19) k ≡ p p (cid:18) (cid:19) (mod p ) . (3.12)On the other hand, p (cid:0) (2 p +2) / p +1) / (cid:1) = − p Γ p ( + p ) Γ p ( + p ) ≡ p p (cid:18) (cid:19) (mod p ) . This together with (3.12) proves the case p ≡ (cid:3) Proof of Theorem 1.3 Lemma 4.1. Let n be a nonnegative integer. Then n X k =0 (3 n + k + 2) ( − n ) k ( − n ) k (1) k ( − n ) k (cid:18) (cid:19) k = 3 × / − n Γ( )Γ( + n )Γ( )Γ( + n ) (4.1) and n X k =0 (3 n + k + 3) ( − n ) k ( − − n ) k (1) k ( − n − k (cid:18) (cid:19) k = 3 × − n Γ( )Γ( + n )Γ( )Γ( + n ) . (4.2) Proof. Denote the left-hand sides of (4.1) and (4.2) by J ( n ) and K ( n ) respectively. ViaZeilberger algorithm, we find that(6 n + 7) J ( n ) − n + 1) J ( n + 1) = 0 ROOF OF SOME CONJECTURAL SUPERCONGRUENCES VIA CURIOUS IDENTITIES 13 and 3(2 n + 3) K ( n ) − n + 2) K ( n + 1) = 0 . Then the identities can be checked by induction on n . (cid:3) The following lemma is the well-known Gauss multiplication formula. Lemma 4.2. [17, p. 371] For any z ∈ C and m ∈ { , , . . . } , we have Y ≤ j On the other hand, it is routine to check that (cid:18) (2 p + 2) / p + 1) / (cid:19) ≡ − p ) Γ p ( + p )Γ p ( + p ) ≡ − p )Γ p ( ) (mod p ) . Comparing the above two congruences we immediately obtain the desired result.The proof of Theorem 1.3 is now complete. (cid:3) Proofs of Theorem 1.4 and Corollary 1.3 We need the following identity. Lemma 5.1. For any n ∈ N and t ∈ C we have n X k =0 ( − n ) k (1) k t k H k = (1 − t ) n H n − n X k =1 (1 − t ) n − k k . (5.1) Proof. Denote the left-hand side of (5.1) by S n . Using the summation package Sigma (cf. [19])in Mathematica we find that( n + 1)( t − S n + (2 n + 3)( t − S n +1 + ( n + 2) S n +2 = − t. Then we can easily prove the identity by induction on n . (cid:3) Proof of Theorem 1.4 . By the binomial theorem we have p − X k =0 ( − a ) k (1) k t k = (1 − t ) a and p − X k =0 ( − a − p ) k (1) k t k = (1 − t ) a + p . On the other hand, p − X k =0 ( − a ) k (1) k t k = p − X k =0 ( x − sp ) k (1) k t k ≡ p − X k =0 ( x ) k (1) k t k − sp k − X j =0 x + j ! (mod p ) . Thus we have sp p − X k =0 ( x ) k (1) k t k k − X j =0 x + j ≡ p − X k =0 ( x ) k (1) k t k − (1 − t ) a (mod p ) . (5.2)Moreover, p − X k =0 ( − a − p ) k (1) k t k = p − X k =0 ( − a − p ) k (1) k t k + p − X k =0 ( − a − p ) p + k (1) p + k t p + k = Σ + Σ , (5.3) where Σ := p − X k =0 ( − a − p ) k (1) k t k and Σ := ( − a − p ) p (1) p t p p − X k =0 ( − a ) k (1 + p ) k t k . Clearly,Σ = p − X k =0 ( x − ( s + 1) p ) k (1) k t k ≡ p − X k =0 ( x ) k (1) k t k − ( s + 1) p k − X j =0 x + j ! (mod p ) . (5.4)Also, by Lemma 5.1 we haveΣ ≡ − t p a Y k =1 (cid:16) pk (cid:17) p − X k =0 ( − a ) k (1) k t k (1 − pH k ) ≡ − t p (1 + pH a ) (1 − t ) a − p (1 − t ) a H a + p a X k =1 (1 − t ) a − k k ! ≡ − t p (1 − t ) a − tp a X k =1 (1 − t ) a − k k (mod p ) . (5.5)Combining (5.2)–(5.5) we complete the proof. (cid:3) Proof of Corollary 1.3 . We only prove the case p ≡ F (cid:20) (cid:12)(cid:12)(cid:12)(cid:12) (cid:21) p − ≡ ( − ( p − / (cid:18) − 23 (1 + p q p (2)) − p (cid:0) H ( p − / − H ( p − / (cid:1)(cid:19) ≡ p q p (2) (mod p )and F (cid:20) (cid:12)(cid:12)(cid:12)(cid:12) (cid:21) p − ≡ ( − (2 p − / (cid:18) − 43 (1 + p q p (2)) − p (cid:0) H ( p − / − H ( p − / (cid:1)(cid:19) ≡ − p q p (2) (mod p ) . Then the desired result follows at once. (cid:3) More results similar to Theorems 1.1–1.3 In this section, we list some congruences that can also be proved by some strange identities.However, we only give the outlines of their proofs since the proofs are quite similar to the onesof Theorems 1.1–1.3.In [25, Conjecture 5.14(i)], Sun posed the following conjecture as a variant of the thirdcongruence in (1.1). ROOF OF SOME CONJECTURAL SUPERCONGRUENCES VIA CURIOUS IDENTITIES 17 Conjecture 6.1. Let p > be a prime. If p ≡ and p = x +3 y with x ≡ ,then we have p − X k =0 (cid:0) kk (cid:1)(cid:0) k k (cid:1) k ≡ x − p x (mod p ) (6.1) and p − X k =0 k + 148 k (cid:18) kk (cid:19)(cid:18) k k (cid:19) ≡ x (mod p ) . (6.2) If p ≡ , then we have p − X k =0 (cid:0) kk (cid:1)(cid:0) k k (cid:1) k ≡ p (cid:0) ( p +1) / p +1) / (cid:1) (mod p ) . (6.3)(6.1) has been proved by G.-S. Mao and H. Pan [11] and its proof depends on the resultsconcerning Legendre polynomials obtained by M. J. Coster and L. Van Hamme [4]. In fact,we can confirm Conjecture 6.1 completely by using some strange hypergeometric identities. Theorem 6.1. For any prime p > we have p − X k =0 (cid:0) kk (cid:1)(cid:0) k k (cid:1) k ≡ (cid:0) ( p − / p − / (cid:1) (cid:0) p q p (2) − p q p (3) (cid:1) (mod p ) if p ≡ , p/ (cid:0) (cid:0) ( p +1) / p +1) / (cid:1)(cid:1) (mod p ) if p ≡ , (6.4) and p − X k =0 (2 k + 1) (cid:0) kk (cid:1)(cid:0) k k (cid:1) k ≡ p/ (cid:0) ( p − / p − / (cid:1) (mod p ) if p ≡ , (cid:0) ( p +1) / p +1) / (cid:1) (cid:0) − − p − p q p (2) + p q p (3) (cid:1) (mod p ) if p ≡ . (6.5) Remark . It is known [2, p. 283] that for any prime p = x + 3 y ≡ x ≡ (cid:18) ( p − / p − / (cid:19) ≡ (cid:16) x − p x (cid:17) (cid:18) − p q p (2) + 3 p q p (3) (cid:19) (mod p ) . This together with (6.4) gives (6.1) and (6.2).Applying (1.10) with α = and z = and noting that (1 / k (3 / k = (cid:0) k k (cid:1)(cid:0) kk (cid:1) / k wearrive at p − X k =0 (cid:0) k k (cid:1)(cid:0) kk (cid:1) k ! ≡ p − X k =0 (cid:0) kk (cid:1) (cid:0) kk (cid:1) ( − k (mod p ) . (6.6) Therefore, combining (6.6) with Theorem 6.1 we can easily obtain the following result whichwas conjectured and partially proved by Z.-H. Sun (cf. [20, Conjecture 2.2] and [22, Theorem5.1]). Corollary 6.1. Let p > be a prime. Then p − X k =0 (cid:0) kk (cid:1) (cid:0) k k (cid:1) ( − k ≡ ( x − p (mod p ) if p ≡ p = x + 3 y ( x, y ∈ Z ) , p ) if p ≡ . To show Theorem 6.1, we need the following identities which can be easily checked byZeilberger algorithm. Lemma 6.1. Let n ne a nonnegative integer. Then we have the following identities. n X k =0 ( − n ) k ( − n ) k (1) k ( − n ) k (cid:18) (cid:19) k = (cid:18) (cid:19) n ( ) n ( ) n , (6.7) n X k =0 ( − n ) k ( − − n ) k (1) k ( − n − k (cid:18) (cid:19) k = (cid:18) (cid:19) n +1 ( ) n +1 ( ) n +1 , (6.8) n X k =0 (2 n + k + 1)( − n ) k ( − n ) k (1) k ( − n ) k (cid:18) (cid:19) k = 3 n +1 ( ) n n ( ) n , (6.9) n X k =0 (2 n + k + 2)( − n ) k ( − − n ) k (1) k ( − n − k (cid:18) (cid:19) k = 9 n +1 ( ) n +2 n +1 ( ) n +1 . (6.10) Proof of Theorem 6.1 . We should divide the proof into four cases that p ≡ , , , 11 (mod 12).We only prove (6.4) for p ≡ p − X k =0 (cid:0) kk (cid:1)(cid:0) k k (cid:1) k ≡ (cid:18) (cid:19) ( p − / ( ) ( p − / ( ) ( p − / − (cid:18) (cid:19) (3 p − / ( ) (3 p − / ( ) (3 p − / (mod p ) . From [12, Lemma 4.1] we know that for any positive integer n and integer a not divisible by p we have a ( p − / ≡ (cid:18) ap (cid:19) n − X k =0 (cid:18) k (cid:19) ( p q p ( a )) k (mod p n ) . (6.11)Thus we have32 (cid:18) (cid:19) ( p − / ( ) ( p − / ( ) ( p − / ≡ (cid:18) p q p (3) − p q p (2) (cid:19) Γ p ( + p )Γ p ( )Γ p ( )Γ p ( p ) ≡ − (cid:16) − p q p (2) − p q p (3) (cid:17) Γ p (cid:18) (cid:19) (mod p ) ROOF OF SOME CONJECTURAL SUPERCONGRUENCES VIA CURIOUS IDENTITIES 19 and 12 (cid:18) (cid:19) (3 p − / ( ) (3 p − / ( ) (3 p − / ≡ (cid:18) p q p (3) − p q p (2) (cid:19) p Γ p ( + p )Γ p ( ) p Γ p ( )Γ p ( p ) ≡ − (cid:18) − p q p (2) − p q p (3) (cid:19) Γ p (cid:18) (cid:19) (mod p ) . Thus we have p − X k =0 (cid:0) kk (cid:1)(cid:0) k k (cid:1) k ≡ − Γ p (cid:18) (cid:19) (mod p ) . On the other hand, it is routine to check that (cid:18) ( p − / p − / (cid:19) (cid:18) p q p (2) − p q p (3) (cid:19) ≡ − Γ p (cid:18) (cid:19) (mod p ) . This completes the proof. (cid:3) In [25, Conjecture 5.14(iii)] Sun made the following conjecture. Conjecture 6.2. For any prime p > , if p ≡ and p = x + 4 y ( x, y ∈ Z ) with x ≡ , then p − X k =0 (cid:0) kk (cid:1)(cid:0) k k (cid:1) k ≡ (cid:18) p (cid:19) (cid:16) x − p x (cid:17) (mod p ) (6.12) and p − X k =0 (1 − k ) (cid:0) kk (cid:1)(cid:0) k k (cid:1) k ≡ (cid:18) p (cid:19) x (mod p ); (6.13) if p ≡ , then p − X k =0 (cid:0) kk (cid:1)(cid:0) k k (cid:1) k ≡ (cid:18) p (cid:19) p (cid:0) ( p +1) / p +1) / (cid:1) (mod p ) . (6.14)We shall prove these congruences by establish the following result. Theorem 6.2. For any prime p > we have p − X k =0 (cid:0) kk (cid:1)(cid:0) k k (cid:1) k ≡ ( p ) (cid:0) ( p − / p − / (cid:1) (cid:0) − p q p (2) (cid:1) (mod p ) if p ≡ , p ( p ) / (cid:0) (cid:0) ( p +1) / p +1) / (cid:1)(cid:1) (mod p ) if p ≡ , (6.15) and p − X k =0 (2 k − (cid:0) kk (cid:1)(cid:0) k k (cid:1) k ≡ − p ( p ) / (cid:0) ( p − / p − / (cid:1) (mod p ) if p ≡ , ( p ) (cid:0) ( p +1) / p +1) / (cid:1) (cid:0) + p − p q p (2) (cid:1) (mod p ) if p ≡ . (6.16) Remark . From [2, p. 281] we know for any prime p > 3, if p ≡ p = x + 4 y with x ≡ (cid:18) ( p − / p − / (cid:19) ≡ (cid:18) x − x (cid:19) (cid:16) p q p (2) (cid:17) (mod p ) . Combining this with Theorem 6.2 we obtain (6.12) and (6.13).Applying (1.10) with α = and z = we arrive at p − X k =0 (cid:0) k k (cid:1)(cid:0) kk (cid:1) k ! ≡ p − X k =0 (cid:0) kk (cid:1) (cid:0) kk (cid:1) k (mod p ) . (6.17)Therefore, combining (6.17) with Theorem 6.2 we can easily obtain the following result whichwas conjectured and partially proved by Z.-H. Sun (cf. [20, Conjecture 2.1] and [22, Theorem5.1]). Corollary 6.2. Let p > be a prime. Then p − X k =0 (cid:0) kk (cid:1) (cid:0) k k (cid:1) k ≡ ( x − p (mod p ) if p ≡ p = x + 4 y with x, y ∈ Z , p ) if p ≡ . To prove Theorem 6.2 we need the following preliminary result which can also be showedby Zeilberger algorithm. Lemma 6.2. Let n be a nonnegative integer. Then n X k =0 ( − n ) k ( − n ) k (1) k ( − n ) k (cid:18) (cid:19) k = Γ( )Γ(3 n )3 n − Γ(2 n + )Γ( n ) , (6.18) n X k =0 ( − n ) k ( − − n ) k (1) k ( − n − k (cid:18) (cid:19) k = 3 n +1 Γ( n + )Γ( n + )2Γ( )Γ(2 n + ) , (6.19) n X k =0 (10 n − k + 3)( − n ) k ( − n ) k (1) k ( − n ) k (cid:18) (cid:19) k = 3 n +2 Γ( n + )Γ( n + )Γ( )Γ(2 n + ) , (6.20) n X k =0 (10 n − k + 8)( − n ) k ( − − n ) k (1) k ( − n − k (cid:18) (cid:19) k = 2Γ( )Γ(3 n + 3)9 n Γ(2 n + )Γ( n + 1) . (6.21) Proof of Theorem 6.2 . We only prove (6.15) for p ≡ p ≡ ROOF OF SOME CONJECTURAL SUPERCONGRUENCES VIA CURIOUS IDENTITIES 21 By Lemma 6.2 we can easily check that p − X k =0 (cid:0) kk (cid:1)(cid:0) k k (cid:1) k ≡ Γ( )Γ( p − )2 × ( p − / Γ( p )Γ( p − ) − (3 p +1) / Γ( p +112 )Γ( p +512 )4Γ( )Γ( p ) (mod p ) . Clearly, by Lemma 2.3 and (6.11) we haveΓ( )Γ( p − )2 × ( p − / Γ( p )Γ( p − ) = Γ p ( )Γ p ( p − )2 × ( p − / Γ p ( p )Γ p ( p − ) ≡ (cid:18) p (cid:19) (cid:18) − p q p (3) (cid:19) (cid:18) p H ( p +3) / − p H ( p − / (cid:19) Γ p ( )Γ p ( − )Γ p ( − ) ≡ ( − ( p +3) / (cid:18) p (cid:19) (cid:18) − p q p (3) − p q p (2) (cid:19) Γ p (cid:18) (cid:19) Γ p (cid:18) (cid:19) (mod p ) . Moreover, by Lemma 4.2 we have3 (3 p +1) / Γ( p +112 )Γ( p +512 )Γ( p +34 )4Γ( )Γ( p )Γ( p +34 ) = 3 (2 − p ) / Γ( )Γ( + p )2Γ( p )Γ( + p ) = 3 (2 − p ) / × p Γ p ( )Γ p ( + p )2 × p Γ p ( p )Γ p ( + p ) ≡ ( − ( p +3) / (cid:18) p (cid:19) (cid:18) − p q p (3) (cid:19) (cid:18) p H ( p − / (cid:19) Γ p (cid:18) (cid:19) Γ p (cid:18) (cid:19) ≡ ( − ( p +3) / (cid:18) p (cid:19) (cid:18) − p q p (3) − p q p (2) (cid:19) Γ p (cid:18) (cid:19) Γ p (cid:18) (cid:19) (mod p ) . Thus we obtain p − X k =0 (cid:0) kk (cid:1)(cid:0) k k (cid:1) k ≡ (cid:18) p (cid:19) Γ p (cid:18) (cid:19) Γ p (cid:18) (cid:19) (mod p ) , since (cid:18) p (cid:19) (cid:18) p (cid:19) = (cid:18) p (cid:19) and (cid:18) p (cid:19) = ( − ( p − / − ( p − / = ( − ( p − / . On the other hand, one may easily check that (cid:18) ( p − / p − / (cid:19) (cid:16) − p q p (2) (cid:17) ≡ Γ p (cid:18) (cid:19) Γ p (cid:18) (cid:19) (mod p ) . This completes the proof. (cid:3) References [1] G. E. Andrews, R. Askey and R. Roy, Special Functions , Encyclopedia of Mathematics and its Applications71, Cambridge University Press, Cambridge.[2] B. C. Berndt, R. J. Evans and K. S. Williams, Gauss and Jacobi Sums , John Wiley & Sons, 1998.[3] M. J. Coster, Generalisation of a congruence of Gauss , J. Number Theory (1988), no.3, 300–310.[4] M. J. Coster and L. Van Hamme, Supercongruences of Atkin and Swinnerton-Dyer type for Legendrepolynomials , J. Number Theory (1991), 265–286 [5] V. J. W. 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(Chen Wang) Department of Mathematics, Nanjing University, Nanjing 210093, People’sRepublic of China E-mail address : [email protected] (Zhi-Wei Sun) Department of Mathematics, Nanjing University, Nanjing 210093, People’sRepublic of China E-mail address ::