Jul 22, 2014 - Theorem 1. ... In particular, if n = pr, Theorem 2 becomes Theorem 1. 2. ..... Wolstenholme's theorem, and p5 variations of Lucas' theorem, J.
arXiv:1407.5789v1 [math.NT] 22 Jul 2014
A CURIOUS CONGRUENCE INVOLVING ALTERNATING HARMONIC SUMS LIUQUAN WANG Abstract. Let p be a prime and Pp the set of positive integers which are prime to p. We establish the following interesting congruence X pr−1 (−1)i ≡ Bp−3 (mod pr ). ijk 2 r i+j+k=p i,j,k∈Pp
1. Introduction and Main Results Let Bn be the n-th Bernoulli number, which is defined by ∞ X Bn n x = x . x e − 1 n=0 n
In [4], Zhao found a curious congruence X 1 (1) ≡ −2Bp−3 (mod p), ijk i+j+k=p i,j,k>0
where p ≥ 3 is a prime. Let Pn denotes the set of positive integers which are prime to n. Recently, Wang and Cai [2] gave a generalization to this congruence, they have shown X 1 ≡ −2pr−1 Bp−3 (mod pr ). ijk r i+j+k=p i,j,k∈Pp
For more variant and generalizations of (1), we refer the reader to see [3] and [5]. In this paper, we consider the alternating sums and obtain the following analogous result. Theorem 1. Let p ≥ 3 be a prime and r a positive integer, then i
(2)
X
i+j+k=pr i,j,k∈Pp
pr−1 (−1) ≡ Bp−3 (mod pr ). ijk 2
Furthermore, we have Date: June 2, 2014. 2010 Mathematics Subject Classification. Primary 11A07, 11A41. Key words and phrases. congruences, Bernoulli numbers, harmonic sums. 1
2
LIUQUAN WANG
Theorem 2. Let n be a positive integer and p ≥ 3 a prime, if pr |n, r ≥ 1 is an integer, then X (−1)i n ≡ Bp−3 (mod pr ). ijk 2p i+j+k=n i,j,k∈Pp
In particular, if n = pr , Theorem 2 becomes Theorem 1. 2. Preliminaries Lemma 1 (cf. Theorem 5.2 in [1]). For any prime p ≥ 3 we have (p−1)/2
X
x=1
1 ≡ −2Bp−3 (mod p). x3
The following Lemma is contained in [2], for the sake of completeness, we provide the proof here. Lemma 2. Let p ≥ 3 be a prime and r a positive integer, for 1 ≤ x ≤ p − 1, define X 1 , S(x, pr ) = i i≡x ( mod p) 1≤i≤pr −1
we have S(x, pr ) ≡ pr−1 x−1 (mod pr ). Proof. For any integer 0 ≤ k ≤ pr − 1, write k as k = a + pr−1 b, 0 ≤ a ≤ pr−1 − 1, 0 ≤ b ≤ p − 1. Then S(x, pr+1 ) − pS(x, pr ) =
r pX −1
=
r−1 p−1 X p X−1
k=0
(3)
b=0
r−1
pX −1 1 1 −p x + kp x + kp k=0
a=0
p−1 Xp
p−1 X p X−1 1 1 − x + (a + pr−1 b)p x + ap a=0 r−1
b=0
r−1
= −pr
b=1
X−1 a=0
b . (x + ap){x + (a + pr−1 b)p}
Because p−1 X b=0
b=
p(p − 1) ≡ 0 (mod p), 2
we have S(x, pr+1 ) − pS(x, pr ) ≡ 0 (mod pr+1 ). Note that S(x, p) = x−1 , by induction on r it’s easy to see that S(x, pr ) ≡ r−1 −1 p x (mod pr ).
A CURIOUS CONGRUENCE INVOLVING ALTERNATING HARMONIC SUMS
3
3. Proofs of the Theorems Proof of Theorem 1. It’s easy to see that pr −j−k
i
X
i+j+k=pr i,j,k∈Pp
(−1) = ijk
X
i+j+k=pr i,j,k∈Pp
X
≡
(4)
(−1) (pr − j − k)jk
j+k
