Proof of two supercongruences conjectured by Z.-W. Sun
aa r X i v : . [ m a t h . N T ] O c t Proof of two supercongruences conjectured by Z.-W. Sun
Guo-Shuai Mao Department of Mathematics, Nanjing University of Information Science and Technology, Nanjing210044, People’s Republic of China [email protected]
Chen-Wei Wen Department of Mathematics, Nanjing University, Nanjing 210093, People’s Republic of China [email protected]
Abstract.
In this paper, we prove two supercongruences conjectured by Z.-W. Sun viathe Wilf-Zeilberger method. One of them is, for any prime p > p − X n =0 n + 1256 n (cid:18) nn (cid:19) ≡ p ( − ( p − / − p E p − (mod p ) . In fact, this supercongruence is a generalization of a supercongruence of van Hamme.
Keywords : Supercongruence; Binomial coeficients; Wilf-Zeilberger method; Euler num-bers.
AMS Subject Classifications:
1. Introduction
In the past decade, many researchers studied supercongruences via the Wilf-Zeilberger(WZ) method. For instance, W. Zudilin [15] proved several Ramanujan-type supercon-gruences by the WZ method. One of them, conjectured by van Hamme, says that for anyodd prime p , ( p − / X k =0 (4 k + 1)( − k (cid:0) (cid:1) k k ! ! ≡ ( − ( p − / p (mod p ) , (1.1)where ( a ) n = a ( a + 1) . . . ( a + n − n ∈ { , , . . . } ) with ( a ) = 1 is the raising factorialfor a ∈ C .Chen, Xie and He [4] confirmed a supercongruence conjetured by Z.-W. Sun [11], whichsays that for any prime p > p − X k =0 k + 1( − k (cid:18) kk (cid:19) ≡ p ( − ( p − / + p E p − (mod p ) , E n are the Euler numbers defined by E = 1 , and E n = − ⌊ n/ ⌋ X k =1 (cid:18) n k (cid:19) E n − k for n ∈ { , , . . . } . For n ∈ N , define H n := X
Theorem 1.1.
Let p > be a prime. Then p − X n =0 n + 1256 n (cid:18) nn (cid:19) ≡ p ( − ( p − / − p E p − (mod p ) . (1.5)2ia the WZ method, Zudilin [15] proved that for any odd prime p , p − X n =0 (cid:0) (cid:1) n (cid:0) (cid:1) n n ! n + 32 n ≡ p ( − ( p − / (mod p ) . (1.6)Z.-W. Sun [12] used the WZ pair which was used by Zudilin [15] to prove the followingsupercongruence, for any prime p > ( p − / X k =0 k + 3( − ) k (cid:18) kk, k, k, k (cid:19) ≡ p ( − ( p − (2 p − + 2 − (2 p − − ) (mod p ) . Our second result is the following congruence which generalizes Sun’s result.
Theorem 1.2.
Let p be an odd prime. Then p − X n =0 n + 3( − ) n (cid:18) nn, n, n, n (cid:19) ≡ p ( − ( p − / + 3 p E p − (mod p ) . Remark 1.3.
In [12], Sun said that he ever proved the congruence in Theorem 1.2, buthe lost the draft containing the complicated details. Recently, he told us to prove thiscongruence.Our main tool in this paper is the WZ method. We shall prove Theorem 1.1 in thenext Section, and the last Section is devoted to prove Theorem 1.2.
2. Proof of Theorem 1.1
We will use the following WZ pair which appears in [5] to prove Theorem 1.1. Fornonnegative integers n, k , define F ( n, k ) = (6 n − k + 1)2 n − k (cid:0) nn (cid:1)(cid:0) n +2 kn + k (cid:1)(cid:0) n − kn − k (cid:1)(cid:0) n + kn (cid:1)(cid:0) kk (cid:1) and G ( n, k ) = n (cid:0) nn (cid:1)(cid:0) n +2 kn + k (cid:1)(cid:0) n − kn − k (cid:1)(cid:0) n + kn (cid:1) n − k − (2 n + 2 k − (cid:0) kk (cid:1) . Clearly F ( n, k ) = G ( n, k ) = 0 if n < k . It is easy to check that F ( n, k − − F ( n, k ) = G ( n + 1 , k ) − G ( n, k ) (2.1)for all nonnegative integer n and k > n from 0 to p − p − X n =0 F ( n, k − − p − X n =0 F ( n, k ) = G ( p, k ) − G (0 , k ) = G ( p, k ) . k from 1 to p −
1, we obtain p − X n =0 F ( n,
0) = F ( p − , p −
1) + p − X k =1 G ( p, k ) . (2.2) Lemma 2.1.
Let p > be a prime. Then F ( p − , p − ≡ − p − p + 18 p q p (2) (mod p ) , where q p (2) stands for the Fermat quotient (2 p − − /p .Proof. By the definition of F ( n, k ), we have F ( p − , p −
1) = 4 p − p − (cid:18) p − p − (cid:19)(cid:18) p − p − (cid:19) = p (cid:0) p − p − (cid:1)(cid:0) p − p − (cid:1) p − = p (cid:0) p − p − (cid:1)(cid:0) p − p − (cid:1) (4 p − p − . It is known that Jocobsthal’s congruence is as follows: For primes p >
3, integers a, b andintegers r, s ≥ (cid:18) ap r bp s (cid:19) / (cid:18) ap r − bp s − (cid:19) ≡ p r + s + min { r , s } ) . (2.3)Thus, (cid:18) p − p − (cid:19) = 12 (cid:18) p p (cid:19) ≡ (cid:18) (cid:19) ≡ p ) . This, with (1.2) and 2 p − = 1 + pq p (2) yields that F ( p − , p − ≡ p (4 p − p − ≡ − p − p + 18 p q p (2) (mod p ) . Therefore the proof of Lemma 2.1 is complete. ✷ By the definition of G ( n, k ) we have G ( p, k ) = p (cid:0) pp (cid:1)(cid:0) p +2 kp + k (cid:1)(cid:0) p − kp − k (cid:1)(cid:0) p + kp (cid:1) p − − k (2 p + 2 k − (cid:0) kk (cid:1) = p (cid:0) pp (cid:1)(cid:0) p +2 kp (cid:1)(cid:0) p − kp − k (cid:1)(cid:0) p +2 kk (cid:1) p − − k (2 p + 2 k − (cid:0) kk (cid:1) , (2.4)where we used the binomial transformation (cid:18) nk (cid:19)(cid:18) kj (cid:19) = (cid:18) nj (cid:19)(cid:18) n − jk − j (cid:19) . Lemma 2.2. ( [10, (3.1)]) ( p − / X k =1 k k (2 k − (cid:0) kk (cid:1) ≡ E p − (mod p ) . emark 2.3. The congruence in Lemma 2.2 is often used when we use the WZ methodto prove some supercongruences. For instance, see [4, 6].
Lemma 2.4.
For any primes p > , we have ( p − / X k =1 G ( p, k ) ≡ − p E p − (mod p ) . Proof.
It is easy to see that (cid:0) p − kp − k (cid:1) ≡ p ) for each 1 ≤ k ≤ ( p − /
2. Then by(2.4), (2.3) and Lucas congruence, we have G ( p, k ) ≡ p (cid:0) p − kp − k (cid:1) (2 p + 2 k − p − − k (mod p ) . In view of [11, Lemma 2.1], k (cid:18) kk (cid:19)(cid:18) p − k ) p − k (cid:19) ≡ ( − ⌊ k/p ⌋− p (mod p ) for all k = 1 , . . . , p − . Then for each 1 ≤ k ≤ ( p − / (cid:18) p − kp − k (cid:19) ≡ − pk (cid:0) kk (cid:1) (mod p ) . Hence G ( p, k ) ≡ − p p − − k k (2 k − (cid:0) kk (cid:1) ≡ − p k k (2 k − (cid:0) kk (cid:1) (mod p ) . Therefore we immediately obtain the desired result with Lemma 2.2. ✷ Lemma 2.5.
Let p > be a prime. Then G ( p, ( p + 1) / ≡ ( − ( p − / p (cid:0) − pq p (2) + 6 p q p (2) (cid:1) (mod p ) . Proof.
In view of (2.4) and (2.3), we have G ( p, ( p + 1) /
2) = p (cid:0) pp (cid:1)(cid:0) p +1 p (cid:1)(cid:0) p − p − / (cid:1)(cid:0) p +1( p +1) / (cid:1) p − (2 p + p ) (cid:0) p +1( p +1) / (cid:1) = p (cid:0) pp (cid:1)(cid:0) p +12 p +1 (cid:1)(cid:0) p − p − / (cid:1)(cid:0) p +1( p +1) / (cid:1) p − (2 p + p ) (cid:0) p +1( p +1) / (cid:1) = (3 p + 1)( p + 1) (cid:0) p +1( p +1) / (cid:1) (2 p + 1)2 p − = (3 p + 1) (cid:0) p (3 p +1) / (cid:1) p − = p (cid:0) p − p − / (cid:1) p − . It is easy to see that (cid:18) p − p − / (cid:19) ≡ ( − ( p − / (cid:16) − pH ( p − / + 2 p H p − / − p H (2)( p − / (cid:17) (mod p ) .
5n view of [9], we have H ( p − / ≡ − q p (2) + pq p (2) (mod p ) and H (2)( p − / ≡ p ) . Thus, (cid:18) p − p − / (cid:19) ≡ ( − ( p − / (cid:0) pq p (2) + 6 p q p (2) (cid:1) (mod p ) . Therefore G ( p, ( p + 1) / ≡ ( − ( p − / p (cid:0) − pq p (2) + 6 p q p (2) (cid:1) (mod p )since 1 / p − ≡ − pq p (2) + 28 p q p (2) (mod p ). ✷ Lemma 2.6. ( [10, (1.1) and (1.7)])
Let p > be a prime. Then ( p − / X k =0 (cid:0) kk (cid:1) (2 k + 1)4 k ≡ − ( − ( p − / q p (2) (mod p ) , ( p − / X k =0 (cid:0) kk (cid:1) (2 k + 1) k ≡ − ( − ( p − / q p (2) p ) . Lemma 2.7. ( [6]) n X k =1 ( − k (cid:0) nk (cid:1) H k (2 k + 1) = − n (2 n + 1) (cid:0) nn (cid:1) n X k =1 (cid:0) kk (cid:1) k k , n X k =1 ( − k (cid:0) nk (cid:1) H k (2 k + 1) = − n (2 n + 1) (cid:0) nn (cid:1) H n n X k =1 (cid:0) kk (cid:1) k k ! . Lemma 2.8. ( [8])
For any prime p > , we have (cid:18) p − p − / (cid:19) ≡ ( − ( p − / p − (mod p ) . Lemma 2.9.
For any prime p > , we have p − X k =( p +3) / G ( p, k ) ≡ p (1 + 4 p − pq p (2)) + ( − ( p − / p q p (2)(1 − pq p (2)) (mod p ) . Proof.
Again by (2.4), we have p − X k = p +32 G ( p, k ) = 2 p p − p − X k = p +32 k (cid:0) p +2 kp (cid:1)(cid:0) p +2 kk (cid:1)(cid:0) p − kp − k (cid:1) (2 p + 2 k − (cid:0) kk (cid:1) = 2 p p − p − X k =1 (cid:0) p − kp (cid:1)(cid:0) p − kp − k (cid:1)(cid:0) kk (cid:1) k (4 p − k − (cid:0) p − kp − k (cid:1) .
6t is easy to check that (cid:18) p − kp (cid:19)(cid:18) p − kp − k (cid:19) = (4 p − k ) . . . (2 p − k + 1) p !( p − k )!= 6 p (3 p + p − k ) . . . (3 p + 1)(3 p − . . . (2 p + 1)(2 p − . . . (2 p − ( k − p − p − k )! ≡ p ( p − k )!(1 + 3 pH p − k )( p − pH p − )( − k ( k − − pH k − )( p − p − k )! ≡ ( − k p (1 + 3 pH p − k − pH k − ) k (cid:0) p − kk (cid:1) (mod p ) . Hence p − X k =( p +3) / G ( p, k ) ≡ − p p − p − / X k =1 (1 + 3 pH p − k − pH k − ) (cid:0) kk (cid:1) ( − k (4 p − k − k (cid:0) p − kp − k (cid:1)(cid:0) p − kk (cid:1) (mod p ) . By simple computation, we have (cid:0) p − kp − k (cid:1)(cid:0) p − kk (cid:1) p = (2 p − k ) . . . ( p + 1)( p − . . . ( p − k + 1)( p − k )! k ! ≡ − (1 + pH p − k − pH k − ) (cid:0) kk (cid:1) (2 k ) (cid:0) p − kk (cid:1) (mod p )and (cid:18) p − kk (cid:19) = ( p − k ) . . . ( p − k + 1) k ! = ( − k (2 k − − p ( H k − − H k − )) k !( k − ≡ ( − k (cid:18) kk (cid:19) (1 − pH k − + pH k − ) (mod p ) . Thus p − X k =( p +3) / G ( p, k ) ≡ p p − p − / X k =1 (1 + 3 pH p − k − pH k − − pH k − + pH k − ) (cid:0) kk (cid:1) k (4 p − k − ≡ p p − p − / X k =1 (1 + 2 pH k − pH k ) (cid:0) kk (cid:1) k (4 p − k −
1) (mod p ) , where we used H p − − k ≡ H k (mod p ) for all k ∈ { , , . . . , p − } .Continuing to calculate the above congruence, we have the following congruence mod-ulo p p − X k = p +32 G ( p, k ) ≡ − p p − p − X k =1 (cid:0) kk (cid:1) (2 k + 1)4 k + 4 p p − X k =1 (cid:0) kk (cid:1) (2 k + 1) k + p p − X k =1 (cid:0) kk (cid:1) (2 H k − H k )(2 k + 1)4 k .
7n view of Lemma 2.7, we have p − X k =1 ( − k (cid:0) p − k (cid:1) (2 H k − H k )(2 k + 1) ≡ p − X k =1 ( − k (cid:0) p − k (cid:1) (2 H k − H k )(2 k + 1) − ( − p − p (2 H p − − H p − ) ≡ − p − p (cid:0) p − p − / (cid:1) H ( p − / + ( − p − p H p − (mod p ) . Hence by Lemma 2.8, 2 p − = 1 + pq p (2) and H ( p − / ≡ − q p (2) (mod p ) p − X k =1 ( − k (cid:0) p − k (cid:1) (2 H k − H k )(2 k + 1) ≡ − − ( p − / q p (2) (mod p ) . (2.5)Therefore we immediately get the desired result with Lemma 2.6. ✷ Proof of Theorem 1.1 . Combining (2.2) with Lemmas 2.1, 2.4, 2.5 and 2.9, we immediatelyobtain that p − X k =0 k + 1256 k (cid:18) kk (cid:19) ≡ ( − ( p − / p − p E p − (mod p ) . Therefore the proof of Theorem 1.1 is finished. ✷
3. Proof of Theorem 1.2
To prove Theorem 1.2, we should use the following WZ pair which appears in [12](see also [15]). For nonnegative integers n, k , define F ( n, k ) = ( − n + k (20 n − k + 3)4 n − k (cid:0) nn (cid:1)(cid:0) n +2 k n + k (cid:1)(cid:0) n − kn (cid:1)(cid:0) n + k k (cid:1)(cid:0) kk (cid:1) and G ( n, k ) = ( − n + k n (cid:0) n − n − (cid:1)(cid:0) n +2 k − n + k − (cid:1)(cid:0) n − k − n − (cid:1)(cid:0) n + k − k (cid:1) n − k − (cid:0) kk (cid:1) . Clearly F ( n, k ) = G ( n, k ) = 0 if n < k . It is easy to check that F ( n, k − − F ( n, k ) = G ( n + 1 , k ) − G ( n, k ) (3.1)for all nonnegative integer n and k > n from 0 to p − k from 1 to p −
1, we have p − X n =0 F ( n,
0) = F ( p − , p −
1) + p − X k =1 G ( p, k ) . (3.2)8 emma 3.1. Let p > be a prime. Then F ( p − , p − ≡ p ( − − p + 8 pq p (2)) (mod p ) . Proof.
By the definition of F ( n, k ), we have F ( p − , p −
1) = 18 p − p − (cid:18) p − p − (cid:19)(cid:18) p − p − (cid:19) = 3 p (cid:0) p − p − (cid:1)(cid:0) p − p − (cid:1) p − = p (cid:0) p − p − (cid:1)(cid:0) p − p − (cid:1) · p − = p (cid:0) p − p − (cid:1)(cid:0) p − p − (cid:1) p − = 3 p (cid:0) p − p − (cid:1)(cid:0) p − p − (cid:1) (6 p − p − = p (cid:0) p p (cid:1)(cid:0) p p (cid:1) (6 p − p − . In view of (2.3), we have (cid:18) p p (cid:19)(cid:18) p p (cid:19) ≡ (cid:18) (cid:19)(cid:18) (cid:19) ≡
60 (mod p ) . This, with 2 p − = 1 + pq p (2) yields that F ( p − , p − ≡ p (6 p − p − ≡ p ( − − p + 8 pq p (2)) (mod p ) . Therefore the proof of Lemma 3.1 is complete. ✷ By the definition of G ( n, k ) we have G ( p, k ) = ( − k +1 p (cid:0) p − p − (cid:1)(cid:0) p +2 k − p + k − (cid:1)(cid:0) p − k − p − (cid:1)(cid:0) p + k − k (cid:1) p − − k (cid:0) kk (cid:1) = ( − k +1 p (cid:0) p − p − (cid:1)(cid:0) p +2 k − k (cid:1)(cid:0) p − k − p − (cid:1)(cid:0) p − p − k − (cid:1) p − − k (cid:0) kk (cid:1) = ( − k +1 p (cid:0) p − p − (cid:1)(cid:0) p +2 k − k (cid:1)(cid:0) p − p − (cid:1)(cid:0) p − p − k (cid:1) p − − k (cid:0) kk (cid:1) , where we used the binomial transformation (cid:18) nk (cid:19)(cid:18) kj (cid:19) = (cid:18) nj (cid:19)(cid:18) n − jk − j (cid:19) . Note that (cid:0) nk (cid:1) = ( − k (cid:0) − n + k − k (cid:1) , then we have (cid:18) p + 2 k − k (cid:19) = (cid:18) − p + 12 k (cid:19) = 4 p (4 p − k (2 k − (cid:18) − p − k − (cid:19) . So by (2.3) we have G ( p, k ) ≡ ( − k +1 p (cid:0) − p − k − (cid:1)(cid:0) p − p − k (cid:1) p − − k k (2 k − (cid:0) kk (cid:1) (mod p ) . (3.3)9 emma 3.2. For any primes p > , we have ( p − / X k =1 G ( p, k ) ≡ − p E p − (mod p ) . Proof.
It is easy to see that (cid:18) − p − k − (cid:19)(cid:18) p − p − k (cid:19) ≡ ( − p − k (mod p ) . Then by (3.3), we have G ( p, k ) ≡ p k k (2 k − (cid:0) kk (cid:1) (mod p ) . Therefore we immediately obtain the desired result with Lemma 2.2. ✷ Lemma 3.3.
Let p > be a prime. Then G ( p, ( p + 1) / ≡ ( − ( p − / p (cid:0) − pq p (2) + 15 p q p (2) (cid:1) (mod p ) . Proof.
In view of (3.3), we have the following congruence modulo p G ( p, ( p + 1) / ≡ ( − ( p − / p (cid:0) − p − p − (cid:1)(cid:0) p − p − / (cid:1) p − p +12 p (cid:0) p +1( p +1) / (cid:1) = ( − ( p − / p (cid:0) − p − p − (cid:1)(cid:0) p − p − / (cid:1) p − (cid:0) p − p − / (cid:1) . It is easy to see that (cid:18) p − p − / (cid:19) ≡ ( − ( p − / (cid:16) − pH ( p − / + 3 p H p − / − p H (2)( p − / (cid:17) (mod p ) . In view of [9], we have H ( p − / ≡ − q p (2) + pq p (2) (mod p ) and H (2)( p − / ≡ p ) . Thus, (cid:18) p − p − / (cid:19) ≡ ( − ( p − / (cid:0) pq p (2) + 15 p q p (2) (cid:1) (mod p ) . Therefore by Lemma 2.8, (2.3) and 1 / p − ≡ − pq p (2) + 66 p q p (2) (mod p ), wehave G ( p, ( p + 1) / ≡ ( − ( p − / p (cid:0) − pq p (2) + 15 p q p (2) (cid:1) (mod p ) . Now the proof of Lemma 3.3 is completed. ✷ emma 3.4. For any prime p > , we have p − X k =( p +3) / G ( p, k ) ≡ p (1 + 6 p − pq p (2)) + ( − ( p − / p ( q p (2) − pq p (2) ) (mod p ) . Proof.
Again by (3.3), we have p − X k = p +32 G ( p, k ) ≡ − p p − p − X k = p +32 ( − k (cid:0) − p − k − (cid:1)(cid:0) p − p − k (cid:1) k (2 k − (cid:0) kk (cid:1) = − p p − p − / X k =1 ( − p − k (cid:0) − p − p − k − (cid:1)(cid:0) p − k (cid:1) ( p − k )(2 p − k − (cid:0) p − kp − k (cid:1) = 6 p p − p − / X k =1 (cid:0) − p − p − k − (cid:1)(cid:0) p − k (cid:1) ( − k ( p − k )(2 p − k − (cid:0) p − kp − k (cid:1) (mod p ) . It is easy to check that (cid:18) p − k (cid:19) = k Y j =1 p − jj = ( − k k Y j =1 (cid:18) − pj (cid:19) ≡ ( − k (1 − pH p − k ) (mod p )and (cid:18) − p − p − k − (cid:19) = p − k − Y j =1 − p − jj = p − k − Y j =1 (cid:18) pj (cid:19) = 5 p − Y j =1 (cid:18) pj (cid:19) p − k − Y j = p +1 (cid:18) pj (cid:19) ≡ pH p − )(1 + 4 pH p − k − ) ≡ pH p − k − ) (mod p ) . Hence p − X k =( p +3) / G ( p, k ) ≡ p p − p − / X k =1 (1 − pH k + 4 pH p − k − )4 k ( p − k )(2 p − k − (cid:0) p − kp − k (cid:1) (mod p ) . By simple computation, we have (cid:0) p − kp − k (cid:1) p = (2 p − k ) . . . ( p + 1)( p − . . . ( p − k + 1)( p − k )! ≡ ( − k − (1 + pH p − k − pH k − ) k (cid:0) p − kk (cid:1) (mod p )11nd (cid:18) p − kk (cid:19) = ( p − k ) . . . ( p − k + 1) k ! = ( − k (2 k − − p ( H k − − H k − )) k !( k − ≡ ( − k (cid:18) kk (cid:19) (1 − pH k − + pH k − ) (mod p ) . Then modulo p we have p − X k =( p +3) / G ( p, k ) ≡ − p p − p − / X k =1 (1 − pH k + 4 pH k +1 − pH k − + 2 pH k − ) k (cid:0) kk (cid:1) k ( p − k )(2 p − k − , where we used H p − − k ≡ H k (mod p ) for all k ∈ { , , . . . , p − } .It is easy to see that ( p − / X k =1 k (cid:0) kk (cid:1) k ( p − k )(2 p − k − ≡ ( p − / X k =1 (cid:0) kk (cid:1) k (2 k + 1) + ( p − / X k =1 p (cid:0) kk (cid:1) k (2 k + 1) + ( p − / X k =1 p (cid:0) kk (cid:1) k k (2 k + 1) (mod p ) . So modulo p we have p − X k = p +32 G ( p, k ) ≡ − p p − p − X k =1 (cid:0) kk (cid:1) k (2 k + 1) + p − X k =1 p (cid:0) kk (cid:1) (2 H k − H k )4 k (2 k + 1) + p − X k =1 p (cid:0) kk (cid:1) k (2 k + 1) . Then we immediately get the desired result with Lemma 2.6 and (2.5). ✷ Proof of Theorem 1.2 . Combining (3.2) with Lemmas 3.1, 3.2, 3.3 and 3.4, we immediatelyget that for any prime p > p − X n =0 n + 3( − ) n (cid:18) nn, n, n, n (cid:19) ≡ ( − ( p − / p + 3 p E p − (mod p ) .p = 3 is easy to check. Therefore the proof of Theorem 1.2 is complete. ✷ Acknowledgments.
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