Some supercongruences modulo $p^2$

0.14 MB·

by  Zhi-Hong Sun

SOME SUPERCONGRUENCES MODULO p2
ZHI-Hong Sun
School of the Mathematical Sciences, Huaiyin Normal University,
1 Huaian, Jiangsu 223001, PR China
1
0 Email: [email protected]
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)