SOME SUPERCONGRUENCES MODULO p2 ZHI-Hong Sun School of the Mathematical Sciences, Huaiyin Normal University, 1 Huaian, Jiangsu 223001, PR China 1 0 Email: zhihongsun@yahoo.com 2 Homepage: http://www.hytc.edu.cn/xsjl/szh n a J Abstract. Let p > 3 be a prime, and let m be an integer with p ∤ m. In the paper we 5 prove some supercongruences concerning ] p−1 2k 3k p−1 2k 4k p−1 3k 6k T X (cid:0)k(cid:1)(cid:0)k(cid:1), X (cid:0)k(cid:1)(cid:0)2k(cid:1), X (cid:0)k(cid:1)(cid:0)3k(cid:1), N 54k 128k 432k k=0 k=0 k=0 h. p−1 2k 2 3k p−1 2k 2 4k p−1 2k 3k 6k at X (cid:0)km(cid:1) k(cid:0)k(cid:1),X (cid:0)km(cid:1) k(cid:0)2k(cid:1), X (cid:0)k(cid:1)(cid:0)mkk(cid:1)(cid:0)3k(cid:1) (mod p2). m k=0 k=0 k=0 [ Thus we solve some conjectures of Zhi-Wei Sun and the author. 1 v MSC: Primary 11A07, Secondary 05A10, 05A19, 11E25 0 Keywords: Congruence; binomial coefficient; binary quadratic form 5 0 1 . 1 1. Introduction. 0 1 For positive integers a,b and n, if n = ax2 +by2 for some integers x and y, we briefly 1 say that n = ax2 + by2. Let p > 3 be a prime. In 2003, Rodriguez-Villegas[RV] posed : v some conjectures on supercongruences modulo p2. Three of his conjectures are equivalent i X to r a (1.1) p−1 2k 2 3k 4A2 2p (mod p2) if p = A2 +3B2 1 (mod 3), k k − ≡ 108k ≡ 0 (mod p2) if p 2 (mod 3), k=0 (cid:0) (cid:1) (cid:0) (cid:1) (cid:26) ≡ X (1.2) p−1 2k 2 4k 4c2 2p (mod p2) if p = c2 +2d2 1,3 (mod 8), k 2k − ≡ 256k ≡ 0 (mod p2) if p 5,7 (mod 8), k=0 (cid:0) (cid:1) (cid:0) (cid:1) (cid:26) ≡ X (1.3) p−1 2k 3k 6k (p)(4a2 2p) (mod p2) if p = a2 +b2 1 (mod 4) and 2 ∤ a, k k 3k 3 − ≡ 1728k ≡ 0 (mod p2) if p 3 (mod 4), k=0 (cid:0) (cid:1)(cid:0) (cid:1)(cid:0) (cid:1) (cid:26) ≡ X where( a )istheJacobisymbol. TheaboveconjectureshavebeensolvedbyMortenson[M] m and Zhi-Wei Sun[Su2]. The author is supported by the Natural Sciences Foundation of China (grant No. 10971078). 1 Let Z be the set of integers, and let [x] be the greatest integer function. For a prime p let Z be the set of rational numbers whose denominator is coprime to p. Recently the p author’s brother Zhi-Wei Sun posed many conjectures ([Su1]) involving p−1 2k 3k p−1 2k 4k p−1 3k 6k k k , k 2k , k 3k , 54k 128k 432k (cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) k=0 k=0 k=0 X X X p−1 2k 2 3k p−1 2k 2 4k p−1 2k 3k 6k k k , k 2k , k k 3k (mod p2), mk mk mk (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1)(cid:0) (cid:1) k=0 k=0 k=0 X X X where p > 3 is a prime and m Z with p ∤ m. For example, Zhi-Wei Sun conjectured ∈ ([Su1, Conjectures A8 and A9]) that for any prime p > 3, (1.4) p−1 2k 2 3k 0 (mod p2) if p 2 (mod 3), k k ≡ ( 192)k ≡ L2 2p (mod p2) if p 1 (mod 3) and so 4p = L2 +27M2, k=0 (cid:0)−(cid:1) (cid:0) (cid:1) (cid:26) − ≡ X (1.5) p−1 (6k)! 0 (mod p2) if ( p ) = 1, 19 − ( 96)3k(3k)!k!3 ≡ (−6)(x2 2p) (mod p2) if ( p ) = 1 and so 4p = x2 +19y2. ( k=0 − p − 19 X In [S3], the author proved (1.4) and (1.5) modulo p. Let p be an odd prime and let x be a variable. In the paper we establish the following general congruences: p−1 2 p−1 2k 3k 2k 3k 2 (x(1 27x))k xk (mod p2), k k − ≡ k k kX=0(cid:18) (cid:19) (cid:18) (cid:19) (cid:16)kX=0(cid:18) (cid:19)(cid:18) (cid:19) (cid:17) p−1 2 p−1 2k 4k 2k 4k 2 (x(1 64x))k xk (mod p2), k 2k − ≡ k 2k kX=0(cid:18) (cid:19) (cid:18) (cid:19) (cid:16)kX=0(cid:18) (cid:19)(cid:18) (cid:19) (cid:17) p−1 p−1 2k 3k 6k 3k 6k 2 (x(1 432x))k xk (mod p2). k k 3k − ≡ k 3k kX=0(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19) (cid:16)kX=0(cid:18) (cid:19)(cid:18) (cid:19) (cid:17) As an application, using the work in [S2,S3] we prove many congruences modulo p2. For example, (1.4) is true for p 2 (mod 3) and (1.5) is true when ( p ) = 1. ≡ 19 − (2k)2(3k) (2k)2(4k) 2. Congruences for p−1 k k and p−1 k 2k (mod p2). k=0 mk k=0 mk Lemma 2.1. Let m be a nonnegative integer. Then P P m 2 m 2k 3k k 2k 3k 2(m k) 3(m k) ( 27)m−k = − − . k k m k − k k m k m k k=0(cid:18) (cid:19) (cid:18) (cid:19)(cid:18) − (cid:19) k=0(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) − (cid:19)(cid:18) − (cid:19) X X We prove the lemma by using WZ method and Mathematica. Clearly the result is true for m = 0,1. Since both sides satisfy the same recurrence relation 81(m+1)(3m+2)(3m+4)S(m) 3(2m+3)(9m2+27m+22)S(m+1)+(m+2)3S(m+2) = 0, − 2 we see that the lemma is true. The proof certificate for the left hand side is 729k2(m+2)(m 2k)(m 2k +1) − − , − (m k +1)(m k +2) − − and the proof certificate for the right hand side is 9k2(3m 3k+1)(3m 3k +2)(9m2 9mk +30m 14k +24) − − − − . (m k +1)2(m k +2)2 − − Theorem 2.1. Let p be an odd prime and let x be a variable. Then p−1 2 p−1 2k 3k 2k 3k 2 (x(1 27x))k xk (mod p2). k k − ≡ k k Xk=0(cid:18) (cid:19) (cid:18) (cid:19) (cid:16)Xk=0(cid:18) (cid:19)(cid:18) (cid:19) (cid:17) Proof. It is clear that p−1 2 2k 3k (x(1 27x))k k k − k=0(cid:18) (cid:19) (cid:18) (cid:19) X p−1 2 k 2k 3k k = xk ( 27x)r k k r − k=0(cid:18) (cid:19) (cid:18) (cid:19) r=0(cid:18) (cid:19) X X 2(p−1) min{m,p−1} 2 2k 3k k = xm ( 27)m−k. k k m k − m=0 k=0 (cid:18) (cid:19) (cid:18) (cid:19)(cid:18) − (cid:19) X X Suppose p m 2p 2 and 0 k p 1. If k > p, then p 2k and so p2 2k 2. If ≤ ≤ − ≤ ≤ − 2 | k | k k < p, then m k p k > k and so k = 0. Thus, from the above and Lemma 2.1 2 − ≥ − m−k (cid:0) (cid:1) (cid:0) (cid:1) we deduce (cid:0) (cid:1) p−1 2 2k 3k (x(1 27x))k k k − k=0(cid:18) (cid:19) (cid:18) (cid:19) X p−1 m 2 2k 3k k xm ( 27)m−k ≡ k k m k − m=0 k=0(cid:18) (cid:19) (cid:18) (cid:19)(cid:18) − (cid:19) X X p−1 m 2k 3k 2(m k) 3(m k) = xm − − k k m k m k m=0 k=0(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) − (cid:19)(cid:18) − (cid:19) X X p−1 p−1 2k 3k 2(m k) 3(m k) = xk − − xm−k k k m k m k k=0(cid:18) (cid:19)(cid:18) (cid:19) m=k(cid:18) − (cid:19)(cid:18) − (cid:19) X X p−1 p−1−k 2k 3k 2r 3r = xk xr k k r r k=0(cid:18) (cid:19)(cid:18) (cid:19) r=0 (cid:18) (cid:19)(cid:18) (cid:19) X X p−1 p−1 p−1 2k 3k 2r 3r 2r 3r = xk xr xr k k r r − r r Xk=0(cid:18) (cid:19)(cid:18) (cid:19) (cid:16)Xr=0(cid:18) (cid:19)(cid:18) (cid:19) r=Xp−k(cid:18) (cid:19)(cid:18) (cid:19) (cid:17) p−1 p−1 p−1 2k 3k 2 2k 3k 2r 3r = xk xk xr (mod p2). k k − k k r r (cid:16)kX=0(cid:18) (cid:19)(cid:18) (cid:19) (cid:17) kX=0(cid:18) (cid:19)(cid:18) (cid:19) r=Xp−k(cid:18) (cid:19)(cid:18) (cid:19) 3 If 2p k p 1, then 2k 3k = (3k)! 0 (mod p2). If 0 k p and p k r p 1, 3 ≤ ≤ − k k k!3 ≡ ≤ ≤ 3 − ≤ ≤ − then 2p r p 1 and so 2r 3r = (3r)! 0 (mod p2). If p < k < 2p and p k r 3 ≤ ≤ − (cid:0) (cid:1)(cid:0) r(cid:1) r r!3 ≡ 3 3 − ≤ ≤ p 1, then r p k > p, 2k 3k = (3k)! 0 (mod p) and 2r 3r = (3r)! 0 (mod p). − ≥ − 3 (cid:0)k (cid:1)(cid:0)k (cid:1) k!3 ≡ r r r!3 ≡ Hence, for 0 k p 1 and p k r p 1 we have p2 2k 3k 2r 3r and so ≤ ≤ − (cid:0) (cid:1)(cid:0)−(cid:1) ≤ ≤ − (cid:0)| k(cid:1)(cid:0) (cid:1)k r r (cid:0) (cid:1)(cid:0) (cid:1)(cid:0) (cid:1)(cid:0) (cid:1) p−1 p−1 2k 3k 2r 3r xk xr 0 (mod p2). k k r r ≡ k=0(cid:18) (cid:19)(cid:18) (cid:19) r=p−k(cid:18) (cid:19)(cid:18) (cid:19) X X Therefore the result follows. Corollary 2.1. Let p > 3 be a prime and m Z with m 0 (mod p). Then p ∈ 6≡ p−1 2k 2 3k p−1 2k 3k 1 1 108/m k 2 k k − − (mod p2). mk ≡ k k 54 kX=0 (cid:0) (cid:1) (cid:0) (cid:1) (cid:16)kX=0(cid:18) (cid:19)(cid:18) (cid:19)(cid:16) p (cid:17) (cid:17) 1−√1−108/m Proof. Taking x = in Theorem 2.1 we deduce the result. 54 Corollary 2.2. Let p > 3 be a prime and m Z with m 0 (mod p). Then p ∈ 6≡ [p/3] 2k 2 3k p−1 2k 2 3k k k 0 (mod p) implies k k 0 (mod p2). mk ≡ mk ≡ (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) k=0 k=0 X X (2k)2(3k) Proof. Clearly 2k 3k = (3k)! 0 (mod p) for p < k < p. Suppose [p/3] k k k k k!3 ≡ 3 k=0 mk ≡ 0 (mod p). Then (cid:0) (cid:1)(cid:0) (cid:1) P p−1 2k 2 3k [p/3] 2k 2 3k k k k k 0 (mod p). mk ≡ mk ≡ (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) k=0 k=0 X X Using Corollary 2.1 we see that p−1 2k 3k 1 1 108/m k − − 0 (mod p) k k 54 ≡ kX=0(cid:18) (cid:19)(cid:18) (cid:19)(cid:16) p (cid:17) and so the result follows from Corollary 2.1. Theorem 2.2. Let p 1 (mod 3) be a prime and so p = A2+3B2 with A 1 (mod 3). ≡ ≡ Then p−1 2k 3k p k k 2A (mod p2). 54k ≡ − 2A (cid:0) (cid:1)(cid:0) (cid:1) k=0 X Proof. Clearly 2k 3k = (3k)! 0 (mod p) for p < k < p. Using [S1, Theorem 2.5] k k k!3 ≡ 3 we have p−1(cid:0)2k(cid:1)(cid:0)3k(cid:1) p−1 [p/3] (3k)! (3k)! k k = 2A (mod p). 54k 54k k!3 ≡ 54k k!3 ≡ k=0 (cid:0) (cid:1)(cid:0) (cid:1) k=0 · k=0 · X X X 4 (2k)(3k) Set p−1 k k = 2A+qp. Then k=0 54k P p−1 2k 3k 2 k k = (2A+qp)2 4A2 +4Aqp (mod p2). 54k ≡ (cid:0) (cid:1)(cid:0) (cid:1) (cid:16)kX=0 (cid:17) Taking m = 108 in Corollary 2.1 we get p−1 2k 2 3k p−1 2k 3k 2 k k k k (mod p2). 108k ≡ 54k (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) Xk=0 (cid:16)Xk=0 (cid:17) Thus, by (1.1) and the above we have p−1 2k 2 3k p−1 2k 3k 2 4A2 2p k k k k 4A2 +4Aqp (mod p2) − ≡ 108k ≡ 54k ≡ (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) kX=0 (cid:16)kX=0 (cid:17) and hence q 1 (mod p). So the theorem is proved. ≡ −2A Remark 2.1 Theorem 2.2 was conjectured by the author in [S1]. When p is a prime of (2k)(3k) the form 6k +5, it was conjectured in [S1] that p−1 k k 0 (mod p2). This was k=0 54k ≡ recently confirmed by Zhi-Wei Sun[Su3]. P Theorem 2.3. Let p 5 (mod 6) be a prime. Then ≡ p−1 2k 2 3k k k 0 (mod p2). ( 192)k ≡ k=0 (cid:0)−(cid:1) (cid:0) (cid:1) X Proof. Since 2k 3k = (3k)! 0 (mod p) for p < k < p, from [S3, Theorem 4.3] we k k k!3 ≡ 3 know that (cid:0) (cid:1)(cid:0) (cid:1) p−1 2k 3k p−1 [p/3] (3k)! (3k)! k k = 0 (mod p). ( 216)k ( 216)k k!3 ≡ ( 216)k k!3 ≡ k=0 (cid:0)−(cid:1)(cid:0) (cid:1) k=0 − · k=0 − · X X X Thus, taking x = 1 in Theorem 2.1 we obtain the result. −216 Lemma 2.2. Let m be a nonnegative integer. Then m 2 m 2k 4k k 2k 4k 2(m k) 4(m k) ( 64)m−k = − − . k 2k m k − k 2k m k 2(m k) k=0(cid:18) (cid:19) (cid:18) (cid:19)(cid:18) − (cid:19) k=0(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) − (cid:19)(cid:18) − (cid:19) X X We prove the lemma by using WZ method and Mathematica. Clearly the result is true for m = 0,1. Since both sides satisfy the same recurrence relation 1024(m+1)(2m+1)(2m+3)S(m) 8(2m+3)(8m2 +24m+19)S(m+1) − +(m+2)3S(m+2) = 0, we see that Lemma 2.2 is true. The proof certificate for the left hand side is 4096k2(m+2)(m 2k)(m 2k +1) − − , − (m k +1)(m k +2) − − and the proof certificate for the right hand side is 16k2(4m 4k +1)(4m 4k+3)(16m2 16mk +55m 26k+46) − − − − . (m k +1)2(m k +2)2 − − 5 Theorem 2.4. Let p be an odd prime and let x be a variable. Then p−1 2 p−1 2k 4k 2k 4k 2 (x(1 64x))k xk (mod p2). k 2k − ≡ k 2k Xk=0(cid:18) (cid:19) (cid:18) (cid:19) (cid:16)Xk=0(cid:18) (cid:19)(cid:18) (cid:19) (cid:17) Proof. It is clear that p−1 2 2k 4k (x(1 64x))k k 2k − k=0(cid:18) (cid:19) (cid:18) (cid:19) X p−1 2 k 2k 4k k = xk ( 64x)r k 2k r − k=0(cid:18) (cid:19) (cid:18) (cid:19) r=0(cid:18) (cid:19) X X 2(p−1) min{m,p−1} 2 2k 4k k = xm ( 64)m−k. k 2k m k − m=0 k=0 (cid:18) (cid:19) (cid:18) (cid:19)(cid:18) − (cid:19) X X Suppose p m 2p 2 and 0 k p 1. If k > p, then p 2k and so p2 2k 2. If ≤ ≤ − ≤ ≤ − 2 | k | k k < p, then m k p k > k and so k = 0. Thus, from the above and Lemma 2.2 2 − ≥ − m−k (cid:0) (cid:1) (cid:0) (cid:1) we deduce (cid:0) (cid:1) p−1 2 2k 4k (x(1 64x))k k 2k − k=0(cid:18) (cid:19) (cid:18) (cid:19) X p−1 m 2 2k 4k k xm ( 64)m−k ≡ k 2k m k − m=0 k=0(cid:18) (cid:19) (cid:18) (cid:19)(cid:18) − (cid:19) X X p−1 m 2k 4k 2(m k) 4(m k) = xm − − k 2k m k 2(m k) m=0 k=0(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) − (cid:19)(cid:18) − (cid:19) X X p−1 p−1 2k 4k 2(m k) 4(m k) = xk − − xm−k k 2k m k 2(m k) k=0(cid:18) (cid:19)(cid:18) (cid:19) m=k(cid:18) − (cid:19)(cid:18) − (cid:19) X X p−1 p−1−k 2k 4k 2r 4r = xk xr k 2k r 2r k=0(cid:18) (cid:19)(cid:18) (cid:19) r=0 (cid:18) (cid:19)(cid:18) (cid:19) X X p−1 p−1 p−1 2k 4k 2r 4r 2r 4r = xk xr xr k 2k r 2r − r 2r Xk=0(cid:18) (cid:19)(cid:18) (cid:19) (cid:16)Xr=0(cid:18) (cid:19)(cid:18) (cid:19) r=Xp−k(cid:18) (cid:19)(cid:18) (cid:19) (cid:17) p−1 p−1 p−1 2k 4k 2 2k 4k 2r 4r = xk xk xr (mod p2). k 2k − k 2k r 2r (cid:16)kX=0(cid:18) (cid:19)(cid:18) (cid:19) (cid:17) kX=0(cid:18) (cid:19)(cid:18) (cid:19) r=Xp−k(cid:18) (cid:19)(cid:18) (cid:19) Now suppose 0 k p 1 and p k r p 1. If k 3p, then p2 ∤ (2k)!, ≤ ≤ − − ≤ ≤ − ≥ 4 p3 (4k)! and so 2k 4k = (4k)! 0 (mod p2). If k < p, then r p k 3p and so | k 2k (2k)!k!2 ≡ 4 ≥ − ≥ 4 2r 4r = (4r)! 0 (mod p2). If p < k < p, then r p k > p, p ∤ (2k)!, p (4k)!, r 2r (2r)!r!2(cid:0)≡(cid:1)(cid:0) (cid:1) 4 2 ≥ − 2 | 6 (cid:0) (cid:1)(cid:0) (cid:1) p 2r and 2k 4k = (4k)! 0 (mod p). If p < k < 3p, then r p k > p, p 2k | r k 2k (2k)!k!2 ≡ 2 4 ≥ − 4 | k and 2r 4r = (4r)! 0 (mod p). Hence we always have 2k 4k 2r 4r 0 (mod p2) (cid:0) r(cid:1) 2r (cid:0) (cid:1)(2(cid:0)r)!r(cid:1)!2 ≡ k 2k r 2r ≡ (cid:0) (cid:1) and so (cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1)(cid:0) (cid:1)(cid:0) (cid:1) p−1 p−1 2k 4k 2r 4r xk xr 0 (mod p2). k 2k r 2r ≡ k=0(cid:18) (cid:19)(cid:18) (cid:19) r=p−k(cid:18) (cid:19)(cid:18) (cid:19) X X Now combining all the above we obtain the result. Corollary 2.3. Let p > 3 be a prime and m Z with m 0 (mod p). Then p ∈ 6≡ p−1 2k 2 4k p−1 2k 4k 1 1 256/m k 2 k 2k − − (mod p2). mk ≡ k 2k 128 kX=0 (cid:0) (cid:1) (cid:0) (cid:1) (cid:16)kX=0(cid:18) (cid:19)(cid:18) (cid:19)(cid:16) p (cid:17) (cid:17) 1−√1−256/m Proof. Taking x = in Theorem 2.4 we deduce the result. 128 Corollary 2.4. Let p > 3 be a prime and m Z with m 0 (mod p). Then p ∈ 6≡ [p/4] 2k 2 4k p−1 2k 2 4k k 2k 0 (mod p) implies k 2k 0 (mod p2). mk ≡ mk ≡ (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) k=0 k=0 X X Proof. For p < k < p we see that 2k 2 4k = (4k)! 0 (mod p). Suppose 4 k 2k k!4 ≡ (2k)2(4k) [p/4] k 2k 0 (mod p). Then (cid:0) (cid:1) (cid:0) (cid:1) k=0 mk ≡ P p−1 2k 2 4k [p/4] 2k 2 4k k 2k k 2k 0 (mod p). mk ≡ mk ≡ (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) k=0 k=0 X X Using Corollary 2.3 we see that p−1 2k 4k 1 1 256/m k − − 0 (mod p) k 2k 128 ≡ kX=0(cid:18) (cid:19)(cid:18) (cid:19)(cid:16) p (cid:17) and so the result follows from Corollary 2.3. Theorem 2.5. Let p 1,3 (mod 8) be a prime and p = c2 + 2d2 with c,d Z and ≡ ∈ c 1 (mod 4). Then ≡ p−1 2k 4k k 2k ( 1)[p8]+p−21 2c p (mod p2). 128k ≡ − − 2c (cid:0) (cid:1)(cid:0) (cid:1) Xk=0 (cid:16) (cid:17) Proof. From the proof of Theorem 2.4 we know that p 2k 4k for p < k < p. By | k 2k 4 [S2, Theorem 2.1] we have (cid:0) (cid:1)(cid:0) (cid:1) p−1 2k 4k [p/4] 2k 4k k 2k k 2k ( 1)[p8]+p−212c (mod p). 128k ≡ 128k ≡ − (cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) k=0 k=0 X X 7 Set p−1 (2kk)(42kk) = ( 1)[p8]+p−212c+qp. Then k=0 128k − P p−1 2k 4k k 2k 2 = (( 1)[p8]+p−212c+qp)2 4c2 +( 1)[p8]+p−214cqp (mod p2). 128k − ≡ − (cid:0) (cid:1)(cid:0) (cid:1) (cid:16)kX=0 (cid:17) Taking m = 256 in Corollary 2.3 we get p−1 2k 2 4k p−1 2k 4k 2 k 2k k 2k (mod p2). 256k ≡ 128k (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) Xk=0 (cid:16)Xk=0 (cid:17) Thus, by (1.2) and the above, p−1 2k 2 4k p−1 2k 4k 4c2 2p k 2k k 2k 2 4c2 +( 1)[p8]+p−214cqp (mod p2) − ≡ 256k ≡ 128k ≡ − (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) kX=0 (cid:16)kX=0 (cid:17) and hence q ( 1)[p8]+p−21 1 (mod p). So the theorem is proved. ≡ − − 2c Remark 2.2 Theorem 2.5 is a conjecture of Zhi-Wei Sun ([Su1, Conjecture A49]). In (2k)(4k) [Su3], Zhi-Wei Sun showed that p−1 k 2k 0 (mod p2) for primes p 5,7 (mod 8). k=0 128k ≡ ≡ Theorem 2.6 ([S1, ConjecturPe 2.1]). Let p > 3 be a prime of the form 4k+3. Then p−1 2k 2 4k k 2k 0 (mod p2). 648k ≡ (cid:0) (cid:1) (cid:0) (cid:1) k=0 X Proof. Since p 2k 4k for p > k > p, from [S2, Theorem 2.4] we know that | k 2k 4 (cid:0) (cid:1)(cid:0) (cid:1) p−1 2k 4k [p/4] 2k 4k k 2k k 2k 0 (mod p). 72k ≡ 72k ≡ (cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) k=0 k=0 X X Thus, taking x = 1 in Theorem 2.4 and applying the above we obtain the result. 72 Theorem 2.7 ([S1, Conjecture 2.2]). Let p be a prime of the form 6k +5. Then p−1 2k 2 4k k 2k 0 (mod p2). ( 144)k ≡ k=0 (cid:0)−(cid:1) (cid:0) (cid:1) X Proof. Since p 2k 4k for p > k > p, from [S2, Theorem 2.5] we know that | k 2k 4 (cid:0) (cid:1)(cid:0) (cid:1) p−1 2k 4k [p/4] 2k 4k k 2k k 2k 0 (mod p). 48k ≡ 48k ≡ (cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) k=0 k=0 X X Thus, taking x = 1 in Theorem 2.4 and applying the above we obtain the result. 48 8 Theorem2.8 ([S1, Conjecture2.3]). Let p > 3 be a prime such that p 3,5,6(mod7). ≡ Then p−1 2k 2 4k k 2k 0 (mod p2). ( 3969)k ≡ k=0 (cid:0)−(cid:1) (cid:0) (cid:1) X Proof. Since p 2k 4k for p > k > p, from [S2, Theorem 2.6] we know that | k 2k 4 (cid:0) (cid:1)(cid:0) (cid:1) p−1 2k 4k [p/4] 2k 4k k 2k k 2k 0 (mod p). 63k ≡ 63k ≡ (cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) k=0 k=0 X X Thus, taking x = 1 in Theorem 2.4 and applying the above we obtain the result. 63 (2k)(3k)(6k) 3. Congruences for p−1 k k 3k (mod p2). k=0 mk Lemma 3.1. Let m bePa nonnegative integer. Then m m 2k 3k 6k k 3k 6k 3(m k) 6(m k) ( 432)m−k = − − . k k 3k m k − k 3k m k 3(m k) k=0(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) − (cid:19) k=0(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) − (cid:19)(cid:18) − (cid:19) X X We prove the lemma by using WZ method and Mathematica. Clearly the result is true for m = 0,1. Since both sides satisfy the same recurrence relation 20736(m+1)(3m+1)(3m+5)S(m) 24(2m+3)(18m2 +54m+41)S(m+1) − +(m+2)3S(m+2) = 0, we see that Lemma 3.1 is true. The proof certificate for the left hand side is 186624k2(m+2)(m 2k)(m 2k +1) − − , − (m k +1)(m k +2) − − and the proof certificate for the right hand side is 144k2(6m 6k +1)(6m 6k +5)(36m2 36mk +129m 62k+114) − − − − . (m k +1)2(m k +2)2 − − For given prime p and integer n, if pα n but pα+1 ∤ n, we say that pα n. | k Lemma 3.2. Let p be an odd prime, k,r 0,1,... ,p 1 and k +r p. Then ∈ { − } ≥ 3k 6k 3r 6r 0 (mod p2). k 3k r 3r ≡ (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19) Proof. If k > 5p, then p5 (6k)!, p (2k)!, p2 (3k)! and so 3k 6k = (6k)! 6 | k k k 3k k!(2k)!(3k)! ≡ 0 (mod p2). If 2p k < 5p, then 2p 3k < 3p, 4p 6k < 5p, p4 (6k)!, p2 (3k)!, 3 ≤ 6 ≤ ≤ (cid:0) (cid:1)(cid:0)k (cid:1) k p (2k)! and so 3k 6k = (6k)! 0 (mod p). If p < k < 2p, then p < 3k < 2p, k k 3k k!(2k)!(3k)! ≡ 2 3 9 (cid:0) (cid:1)(cid:0) (cid:1) 3p < 6k < 4p, p3 (6k)!, p (2k)!, p (3k)! and so 3k 6k = (6k)! 0 (mod p). | k k k 3k k!(2k)!(3k)! ≡ If p k < p, then 2k < p, p 3k < 2p, 6k 2p, p2 (6k)!, p ∤ (2k)!, p (3k)! 3 ≤ 2 ≤ ≥ (cid:0) (cid:1)(cid:0) (cid:1)| k and so 3k 6k = (6k)! 0 (mod p). If p < k < p, then 3k < p, 6k > p and so k 3k k!(2k)!(3k)! ≡ 6 3 3k 6k = (6k)! 0 (mod p). (cid:0) (cid:1)(cid:0) (cid:1) k 3k k!(2k)!(3k)! ≡ (cid:0) (cid:1)(cid:0) (cid:1) From the above we see that p 3k 6k for k > p. Therefore, if k > p and r > p, | k 3k 6 6 6 then 3k 6k 3r 6r 0 (mod p2). If r < p, then k p r > 5p and so p2 3k 6k k 3k r 3r ≡ (cid:0) (cid:1)(cid:0) (cid:1) 6 ≥ − 6 | k 3k by the above. If k < p, then r p k > 5p and so p2 3r 6r by the above. (cid:0) (cid:1)(cid:0) (cid:1)(cid:0) (cid:1)(cid:0) (cid:1)6 ≥ − 6 | r 3r (cid:0) (cid:1)(cid:0) (cid:1) Now putting all the above together we prove the lemm(cid:0)a.(cid:1)(cid:0) (cid:1) Theorem 3.1. Let p be an odd prime and let x be a variable. Then p−1 p−1 2k 3k 6k 3k 6k 2 (x(1 432x))k xk (mod p2). k k 3k − ≡ k 3k kX=0(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19) (cid:16)kX=0(cid:18) (cid:19)(cid:18) (cid:19) (cid:17) Proof. It is clear that p−1 2k 3k 6k (x(1 432x))k k k 3k − k=0(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19) X p−1 k 2k 3k 6k k = xk ( 432x)r k k 3k r − k=0(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19) r=0(cid:18) (cid:19) X X 2(p−1) min{m,p−1} 2k 3k 6k k = xm ( 432)m−k. k k 3k m k − m=0 k=0 (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) − (cid:19) X X Suppose p m 2p 2 and 0 k p 1. If k 2p, then 2p 3k < 3p, 6k 4p, ≤ ≤ − ≤ ≤ − ≥ 3 ≤ ≥ p3 ∤ (3k)!, p4 (6k)! and so 2k 3k 6k = (6k)! 0 (mod p2). If p < k < 2p, then | k k 3k (3k)!k!3 ≡ 2 3 3k < 2p, 6k > 3p, p2 ∤ (3k)! and p3 (6k)! and so 2k 3k 6k = (6k)! 0 (mod p2). If (cid:0) (cid:1)(cid:0) |(cid:1)(cid:0) (cid:1) k k 3k (3k)!k!3 ≡ k < p, then m k p k > k and so k = 0. Thus, from the above and Lemma 3.1 2 − ≥ − m−k (cid:0) (cid:1)(cid:0) (cid:1)(cid:0) (cid:1) 10 (cid:0) (cid:1)

Some supercongruences modulo $p^2$

Similar Some supercongruences modulo $p^2$