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Supercongruences motivated by e
Journal of Number Theory 147 (2015) 326–341
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Journal of Number Theory www.elsevier.com/locate/jnt
Supercongruences motivated by e ✩ Zhi-Wei Sun Department of Mathematics, Nanjing University, Nanjing 210093, People’s Republic of China
a r t i c l e
i n f o
Article history: Received 10 April 2014 Received in revised form 21 July 2014 Accepted 21 July 2014 Available online 6 September 2014 Communicated by David Goss MSC: primary 11B65 secondary 05A10, 11A07, 11B68 Keywords: Supercongruences Binomial coefficients Bernoulli numbers
a b s t r a c t In this paper we establish some new supercongruences motivated by the well-known fact limn→∞ (1 + 1/n)n = e. Let p > 3 be a prime. We prove that p−1 k=0
−1/(p + 1)p+1 ≡0 k
mod p5
and p−1 k=0
1/(p − 1)p−1 2 ≡ p4 Bp−3 3 k
mod p5
where B0 , B1 , B2 , . . . are Bernoulli numbers. We also show that for any a ∈ Z with p a we have p−1 k=1 p−1 k=1
a k 1 1+ ≡ −1 k k
(mod p)
1 1 a k ≡1+ 1 + k2 k 2a
and
(mod p).
© 2014 Elsevier Inc. All rights reserved.
✩ Supported by the National Natural Science Foundation of China (grant 11171140) and the PAPD of Jiangsu Higher Education Institutions. E-mail address: [email protected]. URL: http://math.nju.edu.cn/~zwsun.
http://dx.doi.org/10.1016/j.jnt.2014.07.013 0022-314X/© 2014 Elsevier Inc. All rights reserved.
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
327
1. Introduction A p-adic congruence (with p prime) is called a supercongruence if it happens to hold modulo higher powers of p. Here is a classical example due to Wolstenholme (cf. [W] or [HT]): p−1 1 ≡0 k
mod p2
2p − 1 p−1
and
k=1
≡1
mod p3
for every prime p > 3. In 1900 Glaisher [G1,G2] showed further that
2p − 1 p−1
2 ≡ 1 − p3 Bp−3 3
mod p4
for any prime p > 3, where B0 , B1 , B2 , . . . are Bernoulli numbers. (See [IR, pp. 228–241] for an introduction to Bernoulli numbers.) The reader may consult [L,Su1,Su3,T] for some other known supercongruences. In this paper we establish some new supercongruences modulo prime powers motivated by the well-known formula lim
n→∞
1 1+ n
n = e.
Now we state our main results. Theorem 1.1. Let p > 3 be a prime. Then p+1 p−1 −1/(p + 1) k
k=0
≡0
mod p5 .
(1.1)
Moreover, if p > 5 then p+1 p−1 −1/(p + 1) k=0
k
≡
p5 Bp−3 18
mod p6 .
(1.2)
Theorem 1.2. Let p > 3 be a prime and let m be an integer not divisible by p. Then we have p−1 k=0
km
(−1)
m (m − 1)(7m − 5) 4 p/m − 1 ≡ p Bp−3 36m2 k
mod p5 .
(1.3)
In particular, p−1 p−1 1/(p − 1) k=0
k
≡
2 4 p Bp−3 3
mod p5 .
(1.4)
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
328
Remark 1.1. Note that if p is a prime and m is an integer with p m then
p/m − 1 −1 ≡ = (−1)k ≡ 0 (mod p) for all k = 0, 1, . . . , p − 1. k k
(1.1)–(1.4) are interesting since supercongruences modulo p5 are very rare. We conjecture that there are no composite numbers p > 1 satisfying (1.1) or the congruence p−1 p−1 1/(p − 1) k=0
k
≡0
mod p4 .
(1.5)
Theorem 1.3. Let p > 3 be a prime and let m be an integer not divisible by p. (i) If p > 5, then m p−1 (−1)km p/m − 1 k2
k=1
11 p k p−1
≡
k
mod p3 .
(1.6)
k=1
Also, for any n = 1, . . . , (p − 3)/2 we have m p−1 (−1)km p/m − 1 k2n
k=1
k
≡−
p Bp−1−2n 2n + 1
mod p2 .
(1.7)
mod p3 .
(1.8)
(ii) For every n = 1, . . . , (p − 3)/2, we have m p−1 (−1)km p/m − 1 k2n−1 k 2 p n 1−m (2n + 1) Bp−1−2n ≡ 1+ 2m 2n + 1
k=1
Remark 1.2. If n is a positive integer and p > 2n +1 is a prime, then (1.8) with m = p ±1 yields the congruences p−1 k=1
1
mod p3
p+1 −1/(p + 1) p2 n Bp−1−2n ≡ k 2n + 1
k2n−1
p−1 2p2 n2 1/(p − 1) Bp−1−2n ≡− 2n + 1 k
(1.9)
and p−1 k=1
1 k2n−1
mod p3 .
(1.10)
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
329
Theorem 1.4. Let p be an odd prime and let a ∈ Z with p a. Then k p−1 a 1 1+ ≡ −1 k k
(mod p).
(1.11)
k=1
If p > 3, then k p−1 a 1 1 1 + ≡1+ k2 k 2a
(mod p).
(1.12)
k=1
Remark 1.3. (1.11) with a = 1 yields the congruence p−1 (k + 1)k
kk+1
k=1
≡ −1
(mod p).
(1.13)
We will show Theorems 1.1–1.2 in the next section. Theorems 1.3 and 1.4 will be proved in Sections 3 and 4 respectively. To conclude this section, we pose two related conjectures for further research. Conjecture 1.1. Let m > 2 and q > 0 be integers with m even or q odd. Then, for any prime p > mq we have the supercongruence p−1
km
(−1)
k=0
p/m − q k
m ≡0
mod p3 .
(1.14)
Remark 1.4. Clearly (1.14) with q = 1 follows from (1.3). For a prime p and a p-adic number x, as usual we let νp (x) denote the p-adic valuation (i.e., p-adic order) of x. Conjecture 1.2. Let p be a prime and let n be a positive integer. Then
νp
n−1
−1/(p + 1)p+1 k=0
k
cp
where ⎧ ⎨1 cp = 3 ⎩ 5
if p = 2, if p = 3, if p 5.
νp (n) + 1 , 2
(1.15)
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
330
If p > 3, then we have νp
n−1
1/(p − 1)p−1
νp (n) + 1 . 4 2
k
k=0
(1.16)
2. Proofs of Theorems 1.1 and 1.2 For m = 1, 2, 3, . . . and n = 0, 1, 2, . . . , we define
Hn(m) :=
0
1 km (1)
and call it a harmonic number of order m. Those Hn = Hn (n = 0, 1, 2, . . .) are usually called harmonic numbers. Lemma 2.1. Let p > 3 be a prime. Then Hp−1 ≡ −
p2 Bp−3 3
mod p3
(2)
Hp−1 ≡
and
2 pBp−3 3
mod p2 .
(2.1)
mod p2 ,
(2.2)
Also, p−1
(2)
Hk
≡ p2 Bp−3
p−1
mod p3 ,
k=1
(3)
Hk
k=1
2 ≡ − pBp−3 3
and p−1
(4)
Hk
(3)
≡ Hp−1 ≡ 0 (mod p).
(2.3)
k=1
Proof. The two congruences in (2.1) are known results due to Glaisher [G2], see also Theorem 5.1 and Corollary 5.1 of [S]. For m = 2, 3, 4 we have p−1
(m) Hk
k=1
p−1 p−1 k 1 = = jm j=1 j=1 k=1
p−1 k=j jm
1
=
p−1 p−j j=1
jm
(m)
Note also that (p−1)/2 (3)
Hp−1 =
k=1
1 1 + 3 k (p − k)3
(m−1)
= pHp−1 − Hp−1
≡ 0 (mod p).
Combining these with (2.1), we immediately get (2.2) and (2.3). 2
.
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331
Lemma 2.2. Let p > 3 be a prime. Set
Σ1 =
1 1 , + ij 2 i2 j
p−1
Σ2 =
k=1 1i
Σ3 =
1 1 + ij 3 i3 j
p−1
p−1
k=1 1i
and
Σ4 =
k=1 1i
p−1
1 i2 j 2
k=1 1i
,
(2)
Hk . ij
Then we have Σ1 ≡ pBp−3
mod p2 ,
Σ2 ≡ Bp−3
(mod p)
(2.4)
and Σ3 ≡ Σ4 ≡ −Bp−3
(mod p).
(2.5)
Proof. With the help of Lemma 2.1,
Σ1 =
1i
=p
1 1 + 2 2 ij i j
1i
p−1
1 1 + 2 2 ij i j
1
k=j
1i
(2) (3) = p Hp−1 Hp−1 − Hp−1 −
1 1 + 2 ij i
1 p−1−i − ij i=1 i2 p−1
1i
≡0−
−
1 2 (2) (2) Hp−1 − Hp−1 − (p − 1)Hp−1 + Hp−1 ≡ pBp−3 2
Recall the congruence
Σ2 =
p−1 k=1
1i
mod p2 .
Hk /k2 ≡ Bp−3 (mod p) (cf. [ST, (5.4)]). Note that p−1 1 1=p i2 j 2 k=j
1i
1 − 2 i j2
p−1 p−1 Hp−1 − Hi Hk ≡ ≡ Bp−3 ≡− 2 i k2 i=1
1i
1 i2 j
(mod p)
k=1
and Σ3 =
1i
1 1 + 3 3 ij i j
p−1 k=j
1=
1i
1 1 + 3 (p − j) ij 3 i j
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
332
≡−
1i
≡−
p−1 Hk
k=1
1 1 + 3 ij 2 i (3)
p−1 p−1 Hj − 1/j p − 1 − i =− − j2 i3 j=1 i=1
(2)
+ 2Hp−1 + Hp−1 ≡ −Bp−3
k2
(mod p).
In light of (2.2),
Σ4 =
1i
p−1
j−1 1 (2) (2) Hk − Hs ij s=1 k=1
≡−
1i
=−
1 ij
1ts
(2)
Hj−1 Hj
+
j=1
p−1 j=1
1 =− t2
1i
j 1 j−t ij t=1 t2
1 Hj Hj − mod p2 . j j
For every k = 1, . . . , p − 1, we have Hp−k = Hp−1 −
0
1 ≡ Hk−1 p−j
(mod p)
and (2)
(2)
Hp−k = Hp−1 −
0
1 (2) ≡ −Hk−1 (p − j)2
(mod p).
Thus p−1
(2)
Hk−1 Hk
≡
k=1
p−1
(2)
Hp−k Hk
=
k=1
≡−
p−1
p−1
(2)
Hk Hp−k
k=1 (2)
Hk Hk−1 = −
k=1
p−1
k=1
1 (2) Hk Hk − 2 k
(mod p)
and hence Σ4 ≡
p−1
(2)
H k Hk − 2
k=1
p−1 Hk k=1
k2
+
p−1 H2 k
k=1
k
(mod p).
(2.6)
Since p−1 H2 k
k=1
k
=
p−1 2 Hp−k k=1
p−k
≡−
2 p−1 p−1 p−1 1 1 Hk2 Hk (3) Hk − +2 =− − Hp−1 k k k k2
k=1
k=1
k=1
(mod p)
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
333
(3)
and Hp−1 ≡ 0 (mod p), we have p−1 H2 k
k=1
k
≡
p−1 Hk k=1
k2
≡ Bp−3
(mod p).
(2.7)
p−1 (2) Combining this with (2.6) and the congruence ≡ 0 (mod p) (cf. [Su2, k=1 Hk Hk (2.8)]), we obtain Σ4 ≡ −Bp−3 (mod p). This concludes the proof. 2 Proof of Theorem 1.1. (1.1) in the case p = 5 can be verified directly; in fact, 5+1 5−1 −1/(5 + 1) k
k=0
≡ 55
mod 56 .
Below we assume p > 5. As (1.2) implies (1.1), we only need to prove (1.2). ∞ n n For any p-adic integer x, we can write (1 + px)m as the p-adic series n=0 m n p x . Thus, for each k = 1, . . . , p − 1, we have
p+1 p+1 −1/(p + 1) p/(p + 1) − 1 = k k p+1 k p 1− = (p + 1)j j=1 k
p p+1 p2 p+1 p3 1− + ≡ − j 2 (p + 1)2 j 2 (p + 1)3 j 3 3 j=1 p+1 p4 + (p + 1)4 j 4 4 k p p3 p4 (p − 1) p5 (p − 1)(p − 2) = 1− + − + j 2(p + 1)j 2 6(p + 1)2 j 3 24(p + 1)3 j 4 j=1 ≡
k j=1
≡
k j=1
p p3 (p2 − p + 1) p4 2p5 1− + − (p − 1)(1 − 2p) + j 2j 2 6j 3 24j 4 1−
p p5 − p4 + p3 p4 − 3p5 p5 + + + 2 3 j 2j 6j 12j 4
and hence
p+1 k p −1/(p + 1) 1− − j k j=1 ≡
p5 − p4 + p3 (2) p4 − 3p5 (3) p5 (4) Hk + Hk + Hk 2 6 12
mod p6
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
334
p(p4 − p3 ) + 2 +
p5 2
1 p5 1 − + ij 2 i2 j 6
1i
1i1
1 i1 i2
k j=1
1 1 1 − 2− 2 j2 i1 i2
1 1 + 3 ij 3 i j
1i
mod p6 .
Thus, in view of Lemmas 2.1 and 2.2, we obtain p+1 p−1 −1/(p + 1)
p−1 k k k=1 k=1 p5 2 p 5 − p4 + p3 2 p4 − 3p5 p Bp−3 + − pBp−3 + ×0 ≡ 2 6 3 12 −
p−1
(−1)k
p5 − p4 p5 p5 pBp−3 − (−Bp−3 ) + −Bp−3 − (−Bp−3 ) 2 6 2 p5 Bp−3 mod p6 ≡ 18 p−1 and hence (1.2) follows since k=0 (−1)k p−1 = (1 − 1)p−1 = 0. k The proof of Theorem 1.1 is now complete. 2 +
Proof of Theorem 1.2. For each k ∈ {1, . . . , p − 1}, obviously (−1)km
m
p/m − 1 k
=
k
1−
j=1
p jm
m
is congruent to k
1−
j=1
=
3 4 pm m(m − 1) p2 m m p p + · 2 2− + jm 2 j m 3 j 3 m3 4 j 4 m4
k
p m − 1 p2 (m − 1)(m − 2) p3 (m − 1)(m − 2)(m − 3) p4 + · 2 − · + · 4 j 2m j 6m2 j3 24m3 j
1−
j=1
modulo p5 , and hence (−1)km ≡
m
p/m − 1 k
−
k j=1
1−
p j
m − 1 2 (2) (m − 1)(m − 2) 3 (3) (m − 1)(m − 2)(m − 3) 4 (4) p Hk − p Hk + p Hk 2m 6m2 24m3 1 (m − 1)2 4 m−1 3 1 1 p + − + p 2 2 2 2 2m ij i j 4m i j2 1i
1i
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
+
+
1 1 + ij 3 i3 j
(m − 1)(m − 2) 4 p 6m2 m−1 4 p 2m
335
1i
1i1
1 i1 i2
k j=1 j=i1 ,i2
1 j2
mod p5 .
Therefore, applying (2.2) and (2.3) we get p−1
(−1)km
k=1
m
p/m − 1 k
−
p−1 k=1
(−1)k
p−1 k
(m − 1)(m − 2) 3 2 m−1 3 m − 1 2 2 p p Bp−3 − p Σ1 p − pBp−3 − ≡ 2 2m 6m 3 2m +
(m − 1)2 4 (m − 1)(m − 2) 4 m−1 4 p (Σ4 − Σ3 ) p Σ2 + p Σ3 + 4m2 6m2 2m
mod p5 ,
where Σ1 , Σ2 , Σ3 , Σ4 are defined in Lemma 2.2. Combining this with Lemma 2.2, we finally obtain p−1
(−1)km
k=0
m
p/m − 1 k
− (1 − 1)p−1
m−1 4 (m − 1)(m − 2) 4 p Bp−3 + p Bp−3 2m 9m2 m−1 3 (m − 1)2 4 (m − 1)(m − 2) 4 p (pBp−3 ) + − p Bp−3 + p (−Bp−3 ) 2m 4m2 6m2 (m − 1)(m − 2) 4 (m − 1)2 p Bp−3 − mod p5 , = 2 2 4m 18m ≡
which gives (1.3). Putting m = p − 1 in (1.3) we immediately get (1.4). This concludes the proof. 2 3. Proof of Theorem 1.3 Lemma 3.1. Let m and n be positive integers with m 2n, and let p > 2n + 1 be a prime. Then p−1 k=1
(m)
Hk
k2n+1−m
≡
(−1)m−1 2n + 1 Bp−1−2n 2n + 1 m
(mod p).
(3.1)
When m < 2n we have p−1 (m) Hk pBp−1−2n m n − m 2n + 1 n + (−1) ≡ k2n−m 2n + 1 m+1 m
k=1
mod p2 .
(3.2)
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
336
p−1 Proof. Since k=1 ks ≡ 0 (mod p) for any integer s ≡ 0 (mod p − 1) (see, e.g., [IR, p−1 p. 235]), we have k=1 1/k 2n+1 ≡ 0 (mod p). Hence p−1 k=1
(m)
Hk
k2n+1−m
≡
p−1 k=1
≡
p−1 k=1
1 k2n+1
+
p−1 k=1
1 k2n+1−m
j p−1−m
0
p−1−m p − m 1 Bi kp−m−i k2n+1−m (p − m) i=0 i
by [IR, p. 230]
p−1−m p−1 p−m 1 Bi kp−1−2n−i i m i=0 k=1 1 1 p−m p−m ≡ Bi = Bp−1−2n m m 2n + 1 − m i
≡−
0ip−1−m p−1|2n+i
≡
1 (−1)m−1 2n + 1 −m Bp−1−2n = Bp−1−2n m 2n + 1 − m 2n + 1 m
(mod p).
This proves (3.1). Now assume that m < 2n. As m, 2n − m ∈ {1, . . . , p − 2}, we have p−1 p−1 1 1 ≡ ≡ 0 (mod p). m 2n−m j k j=1 k=1
It is known that p−1 ps 1 Bp−1−s ≡ ks s+1
mod p2 for each s = 1, . . . , p − 2.
k=1
(See, e.g., [G2] or [S, Corollary 5.1].) Thus p−1 p−1 p−1 p−1 2n 1 1 1 1 pBp−1−2n + ≡ ≡ m 2n−m 2n 2n j k k k 2n +1 j=1 k=1
k=1
k=1
On the other hand, p−1 p−1 p−1 1 1 1 + − m 2n−m j k k2n j=1 k=1
=
1kjp−1
=
k=1
1jkp−1
1 j m k2n−m
=
1
1jkp−1
1jkp−1
1 (p − j)m (p − k)2n−m
(p + j)m (p + k)2n−m − j 2 )m (p2 − k2 )2n−m
(p2
j m k2n−m
mod p2 .
(3.3)
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
≡
337
(j m + pmj m−1 )(k2n−m + p(2n − m)k2n−m−1 ) j 2m k2(2n−m) 1jkp−1
≡
1jkp−1
1 j m k2n−m
pm p(2n − m) + m+1 2n−m + m 2n−m+1 j k j k
mod p2 .
Therefore, with the help of (3.1) we have H 2n k pBp−1−2n − 2 2n + 1 k2n−m p−1
(m)
k=1
≡ pm
p−1 (m+1) H k=1
k k2n−m
+ p(2n − m)
p−1 k=1
(m)
Hk
k2n−m+1
(−1)m 2n + 1 (−1)m−1 2n + 1 Bp−1−2n + p(2n − m) Bp−1−2n 2n + 1 m + 1 2n + 1 m 2(m − n) 2n + 1 pBp−1−2n mod p2 = (−1)m m+1 2n + 1 m ≡ pm
and hence (3.2) holds. 2 p−1 2 Remark 3.1. By [ST, (5.4)], k=1 Hk /k ≡ Bp−3 (mod p) for any prime p > 3. By p−1 3 [M, (5)], k=1 Hk /k ≡ −pBp−5 /10 (mod p2 ) for any prime p > 5. Obviously these two results are particular cases of Lemma 3.1. Lemma 3.2. Let p > 5 be a prime. Then p−1 1 − pHk
k2
k=1
≡
Hp−1 p
mod p3 .
(3.4)
Proof. In view of Theorem 5.1(a) and Remark 5.1 of [S], (2) Hp−1 B2p−4 Bp−3 Hp−1 ≡p −2 ≡− 2 2p − 4 p−3 p
mod p3
(3)
and Hp−1 ≡ 0 (mod p2 ). Also, p−1 Hk−1 k=1
k2
=
1j
by [T, Theorem 2.3]. So we have
Hp−1 1 ≡ −3 2 2 jk p
mod p2
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
338
p−1 1 − pHk
(2)
k2
k=1
(3)
= Hp−1 − pHp−1 − p
p−1 Hk−1
k2
k=1
≡
Hp−1 p
mod p3 .
This concludes the proof. 2 Proof of Theorem 1.3. Let k ∈ {1, . . . , p − 1}. Then km
(−1)
m
p/m − 1 k
=
k j=1
≡
k
p 1− jm 1−
j=1
m
p m(m − 1) p2 + · 2 2 j 2 j m
≡ 1 − pHk +
m − 1 2 (2) p Hk + p 2 2m
1i
1 ij
mod p3 .
Thus, for every r = 1, . . . , p − 3 we have m p−1 (−1)km p/m − 1 kr
k=1
≡
k
p−1 1 − pHk
+
kr
k=1
p−1 p−1 (2) (2) m − 1 2 Hk p2 Hk2 − Hk p + 2m kr 2 kr k=1
mod p3 .
(3.5)
k=1
(i) Now suppose that n ∈ {1, . . . , (p − 3)/2}. Then (3.5) with r = 2n yields m p−1 (−1)km p/m − 1 k=1
k2n
k
p−1 1 − pHk
≡
k2n
k=1
mod p2 .
(3.6)
By (3.3) and Lemma 3.1, p−1 1 − pHk k=1
k2n
≡
pBp−1−2n 2n − 1 pBp−1−2n = − 2n + 1 2n + 1
mod p2 .
So (1.7) follows from (3.6). If n < (p − 3)/2, then 2
p−1 (2) H k=1
k k2n
≡
p−1 (2) H k=1
k k2n
(2)
+
Hp−k
(p − k)2n
≡
p−1 1/k 2 k=1
k2n
≡ 0 (mod p),
and hence by (3.5) with r = 2n we obtain m p−1 (−1)km p/m − 1 k=1
k2n
k
≡
p−1 1 − pHk k=1
k2n
+
p−1 p2 Hk2 2 k2n k=1
mod p3 .
(3.7)
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339
When p > 5, (3.7) in the case n = 1, together with (3.4) and the subtle congruence p−1 H2 k
k=1
k2
≡0
(mod p)
of Sun [Su2, (1.5)], yields (1.6). (ii) Fix n ∈ {1, . . . , (p − 3)/2}. Putting r = 2n − 1 in (3.5) we get m p−1 (−1)km p/m − 1 k=1
≡
k2n−1
k
p−1 1 − pHk k=1
k2n−1
+
p−1 p−1 (2) (2) m − 1 2 Hk p2 Hk2 − Hk p + 2m k2n−1 2 k2n−1 k=1
mod p3 .
(3.8)
k=1
In view of a known result (cf. [G2] or [S, Theorem 5.1(a)]), p−1 k=1
1 k2n−1
≡
n − 2n2 2 p Bp−1−2n 2n + 1
mod p3 .
By Lemma 3.1, p−1 1 + 3n − 2n2 Hk pBp−1−2n ≡ k2n−1 2(2n + 1)
mod p2
k=1
and p−1 (2) Hk ≡ −nBp−1−2n k2n−1
(mod p).
k=1
Note also that p−1 p−1 p−1 2 Hp−k Hk2 (Hk − 1/k)2 = ≡ − k2n−1 (p − k)2n−1 k2n−1
k=1
k=1
(mod p)
k=1
and hence p−1 p−1 p−1 p−1 1 1 Hk2 Hk Hk ≡ − ≡ ≡ Bp−1−2n k2n−1 k2n 2 k2n+1 k2n
k=1
k=1
k=1
(mod p)
k=1
with the help of (3.1) in the case m = 1. Combining all these we obtain (1.8) from (3.8). The proof of Theorem 1.3 is now complete. 2
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
340
4. Proof of Theorem 1.4 Lemma 4.1. Let p be an odd prime. Then, for any positive integers d and r with d +r < p, we have k
p−1 k=r
r
kr+d
≡0
(mod p).
(4.1)
Proof. Observe that p−1 k k=r
r
k−r−d =
p−1−r
p−1−r r+s −r − 1 (−1)s (r + s)−r−d = (r + s)−r−d s s s=0
s=0
≡
p−1−r
p−1−r−d p−1−r (−1)s s − 1 − (p − 1 − r) s
s=0
=
p−1−r
p−1−r (−1)p−1−r−k (−1 − k)p−1−r−d k
k=0
= (−1)d
p−1−r
p−1−r (−1)k (k + 1)p−1−r−d k
k=0
(mod p).
It is known that for any positive integer n we have n n k=0
k
(−1)k km = 0
for all m = 0, 1, . . . , n − 1.
(See, e.g., [vLW, pp. 125–126].) Therefore, (4.1) follows from the above. 2 Proof of Theorem 1.4. Let d ∈ {1, 2}. With the help of Lemma 4.1, we have k p−1 p−1 a 1 (k + a)k 1 + = kd k kk+d k=1 k=1
p−1 k 1 k k−r r k k + = a k kk+d r r=1 k=1
p−1 k p−1 p−1 1 r r + a = kd r=1 kr+d k=1
≡
p−1 k=1
Thus,
k=r
1 + kd
p−1 r=p−d
a
r
p−1 k=r
k r
kr+d
(mod p).
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
k (p−1)/2 p−1 1 a 1 ap−1 1 1+ + + ≡ ≡ −1 k k k p−k (p − 1)p
k=1
341
(mod p)
k=1
in view of Fermat’s little theorem. When p > 3, we have and hence
p−1 k=1
1/k 2 ≡ 0 (mod p) by [W],
p−1 k p−2 p−1 p−1 a ap−1 1 1 p−2 p−2 1+ ≡ + + ap−2 + k2 k k2 (p − 1)p+1 (p − 2)p (p − 1)p k=1 k=1 1 1 1 +1 =1+ (mod p) ≡0+1+ a −2 2a as desired. This concludes the proof. 2 Acknowledgment The author would like to thank the referee for helpful comments. References [G1] [G2] [HT] [IR] [L] [M] [S] [Su1] [Su2] [Su3]
[ST] [T] [vLW] [W]
J.W.L. Glaisher, Congruences relating to the sums of products of the first n numbers and to other sums of products, Q. J. Math. 31 (1900) 1–35. J.W.L. Glaisher, On the residues of the sums of products of the first p − 1 numbers, and their powers, to modulus p2 or p3 , Q. J. Math. 31 (1900) 321–353. C. Helou, G. Terjanian, On Wolstenholme’s theorem and its converse, J. Number Theory 128 (2008) 475–499. K. Ireland, M. Rosen, A Classical Introduction to Modern Number Theory, second ed., Grad. Texts in Math., vol. 84, Springer, New York, 1990. L. Long, Hypergeometric evaluation identities and supercongruences, Pacific J. Math. 249 (2011) 405–418. R. Meštrović, Proof of a congruence for harmonic numbers conjectured by Z.-W. Sun, Int. J. Number Theory 8 (2012) 1081–1085. Z.-H. Sun, Congruences concerning Bernoulli numbers and Bernoulli polynomials, Discrete Appl. Math. 105 (2000) 193–223. Z.-W. Sun, On sums of Apéry polynomials and related congruences, J. Number Theory 132 (2012) 2673–2699. Z.-W. Sun, Arithmetic theory of harmonic numbers, Proc. Amer. Math. Soc. 140 (2012) 415–428. Z.-W. Sun, Conjectures and results on x2 mod p2 with 4p = x2 + dy 2 , in: Y. Ouyang, C. Xing, F. Xu, P. Zhang (Eds.), Number Theory and Related Area, in: Adv. Lectures Math., vol. 27, Higher Education Press and Internat. Press, Beijing, Boston, 2013, pp. 149–197. Z.-W. Sun, R. Tauraso, New congruences for central binomial coefficients, Adv. in Appl. Math. 45 (2010) 125–148. R. Tauraso, More congruences for central binomial coefficients, J. Number Theory 130 (2010) 2639–2649. J.H. van Lint, R.M. Wilson, A Course in Combinatorics, second ed., Cambridge Univ. Press, Cambridge, 2001. J. Wolstenholme, On certain properties of prime numbers, Quart. J. Appl. Math. 5 (1862) 35–39.
Contents lists available at ScienceDirect
Journal of Number Theory www.elsevier.com/locate/jnt
Supercongruences motivated by e ✩ Zhi-Wei Sun Department of Mathematics, Nanjing University, Nanjing 210093, People’s Republic of China
a r t i c l e
i n f o
Article history: Received 10 April 2014 Received in revised form 21 July 2014 Accepted 21 July 2014 Available online 6 September 2014 Communicated by David Goss MSC: primary 11B65 secondary 05A10, 11A07, 11B68 Keywords: Supercongruences Binomial coefficients Bernoulli numbers
a b s t r a c t In this paper we establish some new supercongruences motivated by the well-known fact limn→∞ (1 + 1/n)n = e. Let p > 3 be a prime. We prove that p−1 k=0
−1/(p + 1)p+1 ≡0 k
mod p5
and p−1 k=0
1/(p − 1)p−1 2 ≡ p4 Bp−3 3 k
mod p5
where B0 , B1 , B2 , . . . are Bernoulli numbers. We also show that for any a ∈ Z with p a we have p−1 k=1 p−1 k=1
a k 1 1+ ≡ −1 k k
(mod p)
1 1 a k ≡1+ 1 + k2 k 2a
and
(mod p).
© 2014 Elsevier Inc. All rights reserved.
✩ Supported by the National Natural Science Foundation of China (grant 11171140) and the PAPD of Jiangsu Higher Education Institutions. E-mail address: [email protected]. URL: http://math.nju.edu.cn/~zwsun.
http://dx.doi.org/10.1016/j.jnt.2014.07.013 0022-314X/© 2014 Elsevier Inc. All rights reserved.
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
327
1. Introduction A p-adic congruence (with p prime) is called a supercongruence if it happens to hold modulo higher powers of p. Here is a classical example due to Wolstenholme (cf. [W] or [HT]): p−1 1 ≡0 k
mod p2
2p − 1 p−1
and
k=1
≡1
mod p3
for every prime p > 3. In 1900 Glaisher [G1,G2] showed further that
2p − 1 p−1
2 ≡ 1 − p3 Bp−3 3
mod p4
for any prime p > 3, where B0 , B1 , B2 , . . . are Bernoulli numbers. (See [IR, pp. 228–241] for an introduction to Bernoulli numbers.) The reader may consult [L,Su1,Su3,T] for some other known supercongruences. In this paper we establish some new supercongruences modulo prime powers motivated by the well-known formula lim
n→∞
1 1+ n
n = e.
Now we state our main results. Theorem 1.1. Let p > 3 be a prime. Then p+1 p−1 −1/(p + 1) k
k=0
≡0
mod p5 .
(1.1)
Moreover, if p > 5 then p+1 p−1 −1/(p + 1) k=0
k
≡
p5 Bp−3 18
mod p6 .
(1.2)
Theorem 1.2. Let p > 3 be a prime and let m be an integer not divisible by p. Then we have p−1 k=0
km
(−1)
m (m − 1)(7m − 5) 4 p/m − 1 ≡ p Bp−3 36m2 k
mod p5 .
(1.3)
In particular, p−1 p−1 1/(p − 1) k=0
k
≡
2 4 p Bp−3 3
mod p5 .
(1.4)
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
328
Remark 1.1. Note that if p is a prime and m is an integer with p m then
p/m − 1 −1 ≡ = (−1)k ≡ 0 (mod p) for all k = 0, 1, . . . , p − 1. k k
(1.1)–(1.4) are interesting since supercongruences modulo p5 are very rare. We conjecture that there are no composite numbers p > 1 satisfying (1.1) or the congruence p−1 p−1 1/(p − 1) k=0
k
≡0
mod p4 .
(1.5)
Theorem 1.3. Let p > 3 be a prime and let m be an integer not divisible by p. (i) If p > 5, then m p−1 (−1)km p/m − 1 k2
k=1
11 p k p−1
≡
k
mod p3 .
(1.6)
k=1
Also, for any n = 1, . . . , (p − 3)/2 we have m p−1 (−1)km p/m − 1 k2n
k=1
k
≡−
p Bp−1−2n 2n + 1
mod p2 .
(1.7)
mod p3 .
(1.8)
(ii) For every n = 1, . . . , (p − 3)/2, we have m p−1 (−1)km p/m − 1 k2n−1 k 2 p n 1−m (2n + 1) Bp−1−2n ≡ 1+ 2m 2n + 1
k=1
Remark 1.2. If n is a positive integer and p > 2n +1 is a prime, then (1.8) with m = p ±1 yields the congruences p−1 k=1
1
mod p3
p+1 −1/(p + 1) p2 n Bp−1−2n ≡ k 2n + 1
k2n−1
p−1 2p2 n2 1/(p − 1) Bp−1−2n ≡− 2n + 1 k
(1.9)
and p−1 k=1
1 k2n−1
mod p3 .
(1.10)
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
329
Theorem 1.4. Let p be an odd prime and let a ∈ Z with p a. Then k p−1 a 1 1+ ≡ −1 k k
(mod p).
(1.11)
k=1
If p > 3, then k p−1 a 1 1 1 + ≡1+ k2 k 2a
(mod p).
(1.12)
k=1
Remark 1.3. (1.11) with a = 1 yields the congruence p−1 (k + 1)k
kk+1
k=1
≡ −1
(mod p).
(1.13)
We will show Theorems 1.1–1.2 in the next section. Theorems 1.3 and 1.4 will be proved in Sections 3 and 4 respectively. To conclude this section, we pose two related conjectures for further research. Conjecture 1.1. Let m > 2 and q > 0 be integers with m even or q odd. Then, for any prime p > mq we have the supercongruence p−1
km
(−1)
k=0
p/m − q k
m ≡0
mod p3 .
(1.14)
Remark 1.4. Clearly (1.14) with q = 1 follows from (1.3). For a prime p and a p-adic number x, as usual we let νp (x) denote the p-adic valuation (i.e., p-adic order) of x. Conjecture 1.2. Let p be a prime and let n be a positive integer. Then
νp
n−1
−1/(p + 1)p+1 k=0
k
cp
where ⎧ ⎨1 cp = 3 ⎩ 5
if p = 2, if p = 3, if p 5.
νp (n) + 1 , 2
(1.15)
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
330
If p > 3, then we have νp
n−1
1/(p − 1)p−1
νp (n) + 1 . 4 2
k
k=0
(1.16)
2. Proofs of Theorems 1.1 and 1.2 For m = 1, 2, 3, . . . and n = 0, 1, 2, . . . , we define
Hn(m) :=
0
1 km (1)
and call it a harmonic number of order m. Those Hn = Hn (n = 0, 1, 2, . . .) are usually called harmonic numbers. Lemma 2.1. Let p > 3 be a prime. Then Hp−1 ≡ −
p2 Bp−3 3
mod p3
(2)
Hp−1 ≡
and
2 pBp−3 3
mod p2 .
(2.1)
mod p2 ,
(2.2)
Also, p−1
(2)
Hk
≡ p2 Bp−3
p−1
mod p3 ,
k=1
(3)
Hk
k=1
2 ≡ − pBp−3 3
and p−1
(4)
Hk
(3)
≡ Hp−1 ≡ 0 (mod p).
(2.3)
k=1
Proof. The two congruences in (2.1) are known results due to Glaisher [G2], see also Theorem 5.1 and Corollary 5.1 of [S]. For m = 2, 3, 4 we have p−1
(m) Hk
k=1
p−1 p−1 k 1 = = jm j=1 j=1 k=1
p−1 k=j jm
1
=
p−1 p−j j=1
jm
(m)
Note also that (p−1)/2 (3)
Hp−1 =
k=1
1 1 + 3 k (p − k)3
(m−1)
= pHp−1 − Hp−1
≡ 0 (mod p).
Combining these with (2.1), we immediately get (2.2) and (2.3). 2
.
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
331
Lemma 2.2. Let p > 3 be a prime. Set
Σ1 =
1 1 , + ij 2 i2 j
p−1
Σ2 =
k=1 1i
Σ3 =
1 1 + ij 3 i3 j
p−1
p−1
k=1 1i
and
Σ4 =
k=1 1i
p−1
1 i2 j 2
k=1 1i
,
(2)
Hk . ij
Then we have Σ1 ≡ pBp−3
mod p2 ,
Σ2 ≡ Bp−3
(mod p)
(2.4)
and Σ3 ≡ Σ4 ≡ −Bp−3
(mod p).
(2.5)
Proof. With the help of Lemma 2.1,
Σ1 =
1i
=p
1 1 + 2 2 ij i j
1i
p−1
1 1 + 2 2 ij i j
1
k=j
1i
(2) (3) = p Hp−1 Hp−1 − Hp−1 −
1 1 + 2 ij i
1 p−1−i − ij i=1 i2 p−1
1i
≡0−
−
1 2 (2) (2) Hp−1 − Hp−1 − (p − 1)Hp−1 + Hp−1 ≡ pBp−3 2
Recall the congruence
Σ2 =
p−1 k=1
1i
mod p2 .
Hk /k2 ≡ Bp−3 (mod p) (cf. [ST, (5.4)]). Note that p−1 1 1=p i2 j 2 k=j
1i
1 − 2 i j2
p−1 p−1 Hp−1 − Hi Hk ≡ ≡ Bp−3 ≡− 2 i k2 i=1
1i
1 i2 j
(mod p)
k=1
and Σ3 =
1i
1 1 + 3 3 ij i j
p−1 k=j
1=
1i
1 1 + 3 (p − j) ij 3 i j
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
332
≡−
1i
≡−
p−1 Hk
k=1
1 1 + 3 ij 2 i (3)
p−1 p−1 Hj − 1/j p − 1 − i =− − j2 i3 j=1 i=1
(2)
+ 2Hp−1 + Hp−1 ≡ −Bp−3
k2
(mod p).
In light of (2.2),
Σ4 =
1i
p−1
j−1 1 (2) (2) Hk − Hs ij s=1 k=1
≡−
1i
=−
1 ij
1ts
(2)
Hj−1 Hj
+
j=1
p−1 j=1
1 =− t2
1i
j 1 j−t ij t=1 t2
1 Hj Hj − mod p2 . j j
For every k = 1, . . . , p − 1, we have Hp−k = Hp−1 −
0
1 ≡ Hk−1 p−j
(mod p)
and (2)
(2)
Hp−k = Hp−1 −
0
1 (2) ≡ −Hk−1 (p − j)2
(mod p).
Thus p−1
(2)
Hk−1 Hk
≡
k=1
p−1
(2)
Hp−k Hk
=
k=1
≡−
p−1
p−1
(2)
Hk Hp−k
k=1 (2)
Hk Hk−1 = −
k=1
p−1
k=1
1 (2) Hk Hk − 2 k
(mod p)
and hence Σ4 ≡
p−1
(2)
H k Hk − 2
k=1
p−1 Hk k=1
k2
+
p−1 H2 k
k=1
k
(mod p).
(2.6)
Since p−1 H2 k
k=1
k
=
p−1 2 Hp−k k=1
p−k
≡−
2 p−1 p−1 p−1 1 1 Hk2 Hk (3) Hk − +2 =− − Hp−1 k k k k2
k=1
k=1
k=1
(mod p)
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
333
(3)
and Hp−1 ≡ 0 (mod p), we have p−1 H2 k
k=1
k
≡
p−1 Hk k=1
k2
≡ Bp−3
(mod p).
(2.7)
p−1 (2) Combining this with (2.6) and the congruence ≡ 0 (mod p) (cf. [Su2, k=1 Hk Hk (2.8)]), we obtain Σ4 ≡ −Bp−3 (mod p). This concludes the proof. 2 Proof of Theorem 1.1. (1.1) in the case p = 5 can be verified directly; in fact, 5+1 5−1 −1/(5 + 1) k
k=0
≡ 55
mod 56 .
Below we assume p > 5. As (1.2) implies (1.1), we only need to prove (1.2). ∞ n n For any p-adic integer x, we can write (1 + px)m as the p-adic series n=0 m n p x . Thus, for each k = 1, . . . , p − 1, we have
p+1 p+1 −1/(p + 1) p/(p + 1) − 1 = k k p+1 k p 1− = (p + 1)j j=1 k
p p+1 p2 p+1 p3 1− + ≡ − j 2 (p + 1)2 j 2 (p + 1)3 j 3 3 j=1 p+1 p4 + (p + 1)4 j 4 4 k p p3 p4 (p − 1) p5 (p − 1)(p − 2) = 1− + − + j 2(p + 1)j 2 6(p + 1)2 j 3 24(p + 1)3 j 4 j=1 ≡
k j=1
≡
k j=1
p p3 (p2 − p + 1) p4 2p5 1− + − (p − 1)(1 − 2p) + j 2j 2 6j 3 24j 4 1−
p p5 − p4 + p3 p4 − 3p5 p5 + + + 2 3 j 2j 6j 12j 4
and hence
p+1 k p −1/(p + 1) 1− − j k j=1 ≡
p5 − p4 + p3 (2) p4 − 3p5 (3) p5 (4) Hk + Hk + Hk 2 6 12
mod p6
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
334
p(p4 − p3 ) + 2 +
p5 2
1 p5 1 − + ij 2 i2 j 6
1i
1i1
1 i1 i2
k j=1
1 1 1 − 2− 2 j2 i1 i2
1 1 + 3 ij 3 i j
1i
mod p6 .
Thus, in view of Lemmas 2.1 and 2.2, we obtain p+1 p−1 −1/(p + 1)
p−1 k k k=1 k=1 p5 2 p 5 − p4 + p3 2 p4 − 3p5 p Bp−3 + − pBp−3 + ×0 ≡ 2 6 3 12 −
p−1
(−1)k
p5 − p4 p5 p5 pBp−3 − (−Bp−3 ) + −Bp−3 − (−Bp−3 ) 2 6 2 p5 Bp−3 mod p6 ≡ 18 p−1 and hence (1.2) follows since k=0 (−1)k p−1 = (1 − 1)p−1 = 0. k The proof of Theorem 1.1 is now complete. 2 +
Proof of Theorem 1.2. For each k ∈ {1, . . . , p − 1}, obviously (−1)km
m
p/m − 1 k
=
k
1−
j=1
p jm
m
is congruent to k
1−
j=1
=
3 4 pm m(m − 1) p2 m m p p + · 2 2− + jm 2 j m 3 j 3 m3 4 j 4 m4
k
p m − 1 p2 (m − 1)(m − 2) p3 (m − 1)(m − 2)(m − 3) p4 + · 2 − · + · 4 j 2m j 6m2 j3 24m3 j
1−
j=1
modulo p5 , and hence (−1)km ≡
m
p/m − 1 k
−
k j=1
1−
p j
m − 1 2 (2) (m − 1)(m − 2) 3 (3) (m − 1)(m − 2)(m − 3) 4 (4) p Hk − p Hk + p Hk 2m 6m2 24m3 1 (m − 1)2 4 m−1 3 1 1 p + − + p 2 2 2 2 2m ij i j 4m i j2 1i
1i
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
+
+
1 1 + ij 3 i3 j
(m − 1)(m − 2) 4 p 6m2 m−1 4 p 2m
335
1i
1i1
1 i1 i2
k j=1 j=i1 ,i2
1 j2
mod p5 .
Therefore, applying (2.2) and (2.3) we get p−1
(−1)km
k=1
m
p/m − 1 k
−
p−1 k=1
(−1)k
p−1 k
(m − 1)(m − 2) 3 2 m−1 3 m − 1 2 2 p p Bp−3 − p Σ1 p − pBp−3 − ≡ 2 2m 6m 3 2m +
(m − 1)2 4 (m − 1)(m − 2) 4 m−1 4 p (Σ4 − Σ3 ) p Σ2 + p Σ3 + 4m2 6m2 2m
mod p5 ,
where Σ1 , Σ2 , Σ3 , Σ4 are defined in Lemma 2.2. Combining this with Lemma 2.2, we finally obtain p−1
(−1)km
k=0
m
p/m − 1 k
− (1 − 1)p−1
m−1 4 (m − 1)(m − 2) 4 p Bp−3 + p Bp−3 2m 9m2 m−1 3 (m − 1)2 4 (m − 1)(m − 2) 4 p (pBp−3 ) + − p Bp−3 + p (−Bp−3 ) 2m 4m2 6m2 (m − 1)(m − 2) 4 (m − 1)2 p Bp−3 − mod p5 , = 2 2 4m 18m ≡
which gives (1.3). Putting m = p − 1 in (1.3) we immediately get (1.4). This concludes the proof. 2 3. Proof of Theorem 1.3 Lemma 3.1. Let m and n be positive integers with m 2n, and let p > 2n + 1 be a prime. Then p−1 k=1
(m)
Hk
k2n+1−m
≡
(−1)m−1 2n + 1 Bp−1−2n 2n + 1 m
(mod p).
(3.1)
When m < 2n we have p−1 (m) Hk pBp−1−2n m n − m 2n + 1 n + (−1) ≡ k2n−m 2n + 1 m+1 m
k=1
mod p2 .
(3.2)
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
336
p−1 Proof. Since k=1 ks ≡ 0 (mod p) for any integer s ≡ 0 (mod p − 1) (see, e.g., [IR, p−1 p. 235]), we have k=1 1/k 2n+1 ≡ 0 (mod p). Hence p−1 k=1
(m)
Hk
k2n+1−m
≡
p−1 k=1
≡
p−1 k=1
1 k2n+1
+
p−1 k=1
1 k2n+1−m
j p−1−m
0
p−1−m p − m 1 Bi kp−m−i k2n+1−m (p − m) i=0 i
by [IR, p. 230]
p−1−m p−1 p−m 1 Bi kp−1−2n−i i m i=0 k=1 1 1 p−m p−m ≡ Bi = Bp−1−2n m m 2n + 1 − m i
≡−
0ip−1−m p−1|2n+i
≡
1 (−1)m−1 2n + 1 −m Bp−1−2n = Bp−1−2n m 2n + 1 − m 2n + 1 m
(mod p).
This proves (3.1). Now assume that m < 2n. As m, 2n − m ∈ {1, . . . , p − 2}, we have p−1 p−1 1 1 ≡ ≡ 0 (mod p). m 2n−m j k j=1 k=1
It is known that p−1 ps 1 Bp−1−s ≡ ks s+1
mod p2 for each s = 1, . . . , p − 2.
k=1
(See, e.g., [G2] or [S, Corollary 5.1].) Thus p−1 p−1 p−1 p−1 2n 1 1 1 1 pBp−1−2n + ≡ ≡ m 2n−m 2n 2n j k k k 2n +1 j=1 k=1
k=1
k=1
On the other hand, p−1 p−1 p−1 1 1 1 + − m 2n−m j k k2n j=1 k=1
=
1kjp−1
=
k=1
1jkp−1
1 j m k2n−m
=
1
1jkp−1
1jkp−1
1 (p − j)m (p − k)2n−m
(p + j)m (p + k)2n−m − j 2 )m (p2 − k2 )2n−m
(p2
j m k2n−m
mod p2 .
(3.3)
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
≡
337
(j m + pmj m−1 )(k2n−m + p(2n − m)k2n−m−1 ) j 2m k2(2n−m) 1jkp−1
≡
1jkp−1
1 j m k2n−m
pm p(2n − m) + m+1 2n−m + m 2n−m+1 j k j k
mod p2 .
Therefore, with the help of (3.1) we have H 2n k pBp−1−2n − 2 2n + 1 k2n−m p−1
(m)
k=1
≡ pm
p−1 (m+1) H k=1
k k2n−m
+ p(2n − m)
p−1 k=1
(m)
Hk
k2n−m+1
(−1)m 2n + 1 (−1)m−1 2n + 1 Bp−1−2n + p(2n − m) Bp−1−2n 2n + 1 m + 1 2n + 1 m 2(m − n) 2n + 1 pBp−1−2n mod p2 = (−1)m m+1 2n + 1 m ≡ pm
and hence (3.2) holds. 2 p−1 2 Remark 3.1. By [ST, (5.4)], k=1 Hk /k ≡ Bp−3 (mod p) for any prime p > 3. By p−1 3 [M, (5)], k=1 Hk /k ≡ −pBp−5 /10 (mod p2 ) for any prime p > 5. Obviously these two results are particular cases of Lemma 3.1. Lemma 3.2. Let p > 5 be a prime. Then p−1 1 − pHk
k2
k=1
≡
Hp−1 p
mod p3 .
(3.4)
Proof. In view of Theorem 5.1(a) and Remark 5.1 of [S], (2) Hp−1 B2p−4 Bp−3 Hp−1 ≡p −2 ≡− 2 2p − 4 p−3 p
mod p3
(3)
and Hp−1 ≡ 0 (mod p2 ). Also, p−1 Hk−1 k=1
k2
=
1j
by [T, Theorem 2.3]. So we have
Hp−1 1 ≡ −3 2 2 jk p
mod p2
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
338
p−1 1 − pHk
(2)
k2
k=1
(3)
= Hp−1 − pHp−1 − p
p−1 Hk−1
k2
k=1
≡
Hp−1 p
mod p3 .
This concludes the proof. 2 Proof of Theorem 1.3. Let k ∈ {1, . . . , p − 1}. Then km
(−1)
m
p/m − 1 k
=
k j=1
≡
k
p 1− jm 1−
j=1
m
p m(m − 1) p2 + · 2 2 j 2 j m
≡ 1 − pHk +
m − 1 2 (2) p Hk + p 2 2m
1i
1 ij
mod p3 .
Thus, for every r = 1, . . . , p − 3 we have m p−1 (−1)km p/m − 1 kr
k=1
≡
k
p−1 1 − pHk
+
kr
k=1
p−1 p−1 (2) (2) m − 1 2 Hk p2 Hk2 − Hk p + 2m kr 2 kr k=1
mod p3 .
(3.5)
k=1
(i) Now suppose that n ∈ {1, . . . , (p − 3)/2}. Then (3.5) with r = 2n yields m p−1 (−1)km p/m − 1 k=1
k2n
k
p−1 1 − pHk
≡
k2n
k=1
mod p2 .
(3.6)
By (3.3) and Lemma 3.1, p−1 1 − pHk k=1
k2n
≡
pBp−1−2n 2n − 1 pBp−1−2n = − 2n + 1 2n + 1
mod p2 .
So (1.7) follows from (3.6). If n < (p − 3)/2, then 2
p−1 (2) H k=1
k k2n
≡
p−1 (2) H k=1
k k2n
(2)
+
Hp−k
(p − k)2n
≡
p−1 1/k 2 k=1
k2n
≡ 0 (mod p),
and hence by (3.5) with r = 2n we obtain m p−1 (−1)km p/m − 1 k=1
k2n
k
≡
p−1 1 − pHk k=1
k2n
+
p−1 p2 Hk2 2 k2n k=1
mod p3 .
(3.7)
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
339
When p > 5, (3.7) in the case n = 1, together with (3.4) and the subtle congruence p−1 H2 k
k=1
k2
≡0
(mod p)
of Sun [Su2, (1.5)], yields (1.6). (ii) Fix n ∈ {1, . . . , (p − 3)/2}. Putting r = 2n − 1 in (3.5) we get m p−1 (−1)km p/m − 1 k=1
≡
k2n−1
k
p−1 1 − pHk k=1
k2n−1
+
p−1 p−1 (2) (2) m − 1 2 Hk p2 Hk2 − Hk p + 2m k2n−1 2 k2n−1 k=1
mod p3 .
(3.8)
k=1
In view of a known result (cf. [G2] or [S, Theorem 5.1(a)]), p−1 k=1
1 k2n−1
≡
n − 2n2 2 p Bp−1−2n 2n + 1
mod p3 .
By Lemma 3.1, p−1 1 + 3n − 2n2 Hk pBp−1−2n ≡ k2n−1 2(2n + 1)
mod p2
k=1
and p−1 (2) Hk ≡ −nBp−1−2n k2n−1
(mod p).
k=1
Note also that p−1 p−1 p−1 2 Hp−k Hk2 (Hk − 1/k)2 = ≡ − k2n−1 (p − k)2n−1 k2n−1
k=1
k=1
(mod p)
k=1
and hence p−1 p−1 p−1 p−1 1 1 Hk2 Hk Hk ≡ − ≡ ≡ Bp−1−2n k2n−1 k2n 2 k2n+1 k2n
k=1
k=1
k=1
(mod p)
k=1
with the help of (3.1) in the case m = 1. Combining all these we obtain (1.8) from (3.8). The proof of Theorem 1.3 is now complete. 2
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
340
4. Proof of Theorem 1.4 Lemma 4.1. Let p be an odd prime. Then, for any positive integers d and r with d +r < p, we have k
p−1 k=r
r
kr+d
≡0
(mod p).
(4.1)
Proof. Observe that p−1 k k=r
r
k−r−d =
p−1−r
p−1−r r+s −r − 1 (−1)s (r + s)−r−d = (r + s)−r−d s s s=0
s=0
≡
p−1−r
p−1−r−d p−1−r (−1)s s − 1 − (p − 1 − r) s
s=0
=
p−1−r
p−1−r (−1)p−1−r−k (−1 − k)p−1−r−d k
k=0
= (−1)d
p−1−r
p−1−r (−1)k (k + 1)p−1−r−d k
k=0
(mod p).
It is known that for any positive integer n we have n n k=0
k
(−1)k km = 0
for all m = 0, 1, . . . , n − 1.
(See, e.g., [vLW, pp. 125–126].) Therefore, (4.1) follows from the above. 2 Proof of Theorem 1.4. Let d ∈ {1, 2}. With the help of Lemma 4.1, we have k p−1 p−1 a 1 (k + a)k 1 + = kd k kk+d k=1 k=1
p−1 k 1 k k−r r k k + = a k kk+d r r=1 k=1
p−1 k p−1 p−1 1 r r + a = kd r=1 kr+d k=1
≡
p−1 k=1
Thus,
k=r
1 + kd
p−1 r=p−d
a
r
p−1 k=r
k r
kr+d
(mod p).
Z.-W. Sun / Journal of Number Theory 147 (2015) 326–341
k (p−1)/2 p−1 1 a 1 ap−1 1 1+ + + ≡ ≡ −1 k k k p−k (p − 1)p
k=1
341
(mod p)
k=1
in view of Fermat’s little theorem. When p > 3, we have and hence
p−1 k=1
1/k 2 ≡ 0 (mod p) by [W],
p−1 k p−2 p−1 p−1 a ap−1 1 1 p−2 p−2 1+ ≡ + + ap−2 + k2 k k2 (p − 1)p+1 (p − 2)p (p − 1)p k=1 k=1 1 1 1 +1 =1+ (mod p) ≡0+1+ a −2 2a as desired. This concludes the proof. 2 Acknowledgment The author would like to thank the referee for helpful comments. References [G1] [G2] [HT] [IR] [L] [M] [S] [Su1] [Su2] [Su3]
[ST] [T] [vLW] [W]
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