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 APM  Vol.10 No.10 , October 2020
Super Congruences Involving Alternating Harmonic Sums
Abstract: Let p be an odd prime, the harmonic congruence such as , and many different variations and generalizations have been studied intensively. In this note, we consider the congruences involving the combination of alternating harmonic sums,  where PP denotes the set of positive integers which are prime to p. And we establish the combinational congruences involving alternating harmonic sums for positive integer n=3,4,5.

Keywords:

1. Introduction

At the beginning of the 21th century, Zhao (Cf. [1] ) first announced the following curious congruence involving multiple harmonic sums for any odd prime p > 3 ,

i + j + k = p 1 i j k 2 B p 3 ( mod p ) , (1)

which holds when p = 3 evidently. Here, Bernoulli numbers B k are defined by the recursive relation:

i = 0 n ( n + 1 i ) B i = 0 , n 1.

A simple proof of (1) was presented in [2]. This congruence has been generalized along several directions. First, Zhou and Cai [3] established the following harmonic congruence for prime p > 3 and integer n p 2

l 1 + l 2 + + l n = p 1 l 1 l 2 l n ( ( n 1 ) ! B p n ( mod p ) , if 2 | n , n ( n ! ) 2 ( n + 1 ) p B p n 1 ( mod p 2 ) , if 2 | n . (2)

Later, Xia and Cai [4] generalized (1) to

i + j + k = p 1 i j k 12 B p 3 p 3 3 B 2 p 4 p 4 ( mod p 2 ) ,

where p > 5 is a prime.

Recently, Wang and Cai [5] proved for every prime p 3 and positive integer r,

i + j + k = p r i , j , k P p 1 i j k 2 p r 1 B p 3 ( mod p r ) , (3)

where P p denotes the set of positive integers which are prime to p.

Let n = 2 or 4, for every positive integer r n 2 and prime p > n , Zhao [6] extended (3) to

i 1 + i 2 + + i n = p r i 1 , i 2 , , i n P p 1 i 1 i 2 i n n ! n + 1 p r B p n 1 ( mod p r + 1 ) . (4)

For any prime p > 5 and integer r > 1 , Wang [7] proved that

i 1 + i 2 + + i 5 = p r i 1 , i 2 , , i 5 P p 1 i 1 i 2 i 5 5 ! 6 p r 1 B p 5 ( mod p r ) .

We consider the following alternating harmonic sums

i 1 + i 2 + + i n = p r i 1 , i 2 , , i n P p σ 1 i 1 σ 2 i 2 σ n i n i 1 i 2 i n ,

where σ i { 1 , 1 } , i = 1 , 2 , , n . Given n, we only need to consider the following alternating harmonic sums,

i 1 + i 2 + + i n = p r i 1 , i 2 , , i n P p ( 1 ) i 1 i 1 i 2 i n , i 1 + i 2 + + i n = p r i 1 , i 2 , , i n P p ( 1 ) i 1 + i 2 i 1 i 2 i n , , i 1 + i 2 + + i n = p r i 1 , i 2 , , i n P p ( 1 ) i 1 + i 2 + + i [ n 2 ] i 1 i 2 i n

where [ x ] denotes the largest integer less than or equal to x.

In this paper, we consider the congruences involving the combination of alternating harmonic sums,

i 1 + i 2 + + i n = p r i 1 , i 2 , , i n P p ( 1 ) i 1 i 1 i 2 i n , i 1 + i 2 + + i n = p r i 1 , i 2 , , i n P p ( 1 ) i 1 + i 2 i 1 i 2 i n , , i 1 + i 2 + + i n = p r i 1 , i 2 , , i n P p ( 1 ) i 1 + i 2 + + i [ n 2 ] i 1 i 2 i n .

We obtain the following theorems. Among them, Theorem 1 and Theorem 2 have been proved by Wang [8] using different method.

Theorem 1. Let p be an odd prime and r a positive integer, then

i + j + k = 2 p r i , j , k P p ( 1 ) i i j k p r 1 B p 3 ( mod p r ) .

Remark 1. There is no solution ( i , j , k ) for the equation i + j + k = 2 p r with i , j , k P 2 p .

Theorem 2. Let p be an odd prime and r a positive integer, then

i + j + k = p r i , j , k P p ( 1 ) i i j k 1 2 p r 1 B p 3 ( mod p r ) .

Theorem 3. Let p 5 be a prime and r a positive integer, then

4 i 1 + i 2 + i 3 + i 4 = 2 p r i 1 , i 2 , i 3 , i 4 P p ( 1 ) i 1 i 1 i 2 i 3 i 4 + 3 i 1 + i 2 + i 3 + i 4 = 2 p r i 1 , i 2 , i 3 , i 4 P p ( 1 ) i 1 + i 2 i 1 i 2 i 3 i 4 ( 216 5 p B p 5 ( m o d p 2 ) , if r = 1, 36 5 p r B p 5 ( m o d p r + 1 ) , if r > 1.

Theorem 4. Let p > 5 be a prime and r a positive integer, then

i 1 + i 2 + i 3 + i 4 + i 5 = 2 p r i 1 , i 2 , i 3 , i 4 , i 5 P p ( 1 ) i 1 i 1 i 2 i 3 i 4 i 5 + 2 i 1 + i 2 + i 3 + i 4 + i 5 = 2 p r i 1 , i 2 , i 3 , i 4 , i 5 P p ( 1 ) i 1 + i 2 i 1 i 2 i 3 i 4 i 5 ( 12 B p 5 ( m o d p ) , if r = 1, 6 p r 1 B p 5 ( m o d p r ) , if r > 1.

2. Preliminaries

In order to prove the theorems, we need the following lemmas.

Lemma 1 ( [5] ) Let p be an odd prime and r, m positive integers, then

i + j + k = m p r i , j , k P p 1 i j k 2 m p r 1 B p 3 ( mod p r ) .

Lemma 2. Let p be an odd prime and r, m positive integers, then

i + j + k = m p r i , j , k P p 1 i j k = 6 m p r 1 j < l m p r j , l , l j P p 1 j l .

Proof. It is easy to see that

i + j + k = m p r i , j , k P p 1 i j k = 1 m p r i + j + k = m p r i , j , k P p i + j + k i j k = 3 m p r i + j + k = m p r i , j , k P p 1 i j .

Let l = j + k , then 1 j < l m p r and j , l , l j P p . By symmetry, we have

3 m p r i + j + k = m p r i , j , k P p 1 i j = 3 m p r i + j < m p r i , j , l P p 1 l i + j i j = 6 m p r 1 j < l m p r j , l , l j P p 1 j l .

This completes the proof of Lemma 2. ¨

Lemma 3. Let p > 4 be a prime and r, m positive integers, then

i 1 + i 2 + i 3 + i 4 = m p r i 1 , i 2 , i 3 , i 4 P p 1 i 1 i 2 i 3 i 4 = 24 m p r 1 u 1 < u 2 < u 3 m p r u 1 , u 3 , u 2 u 1 , u 3 u 2 P p 1 u 1 u 2 u 3 .

Proof. The proof of Lemma 3 is similar to the proof of Lemma 2. ¨

Lemma 4 ( [3] ) Let r , α 1 , , α n be positive integers, r = α 1 + + α n p 3 , then

1 l 1 , , l n p 1 l i l j , i j 1 l 1 α 1 l 2 α 2 l n α n ( ( 1 ) n ( n 1 ) ! r ( r + 1 ) 2 ( r + 2 ) B p r 2 p 2 ( m o d p 3 ) , if 2 | r , ( 1 ) n 1 ( n 1 ) ! r r + 1 B p r 1 p ( m o d p 2 ) , if 2 | r .

Lemma 5 ( [7] ). Let p be an odd prime, and α 1 , , α n positive integers, where r = α 1 + + α n p 3 , then

1 l 1 , , l n 2 p l i l j , l i P p 1 l 1 α 1 l 2 α 2 l n α n ( ( 1 ) n ( n 1 ) ! 2 r ( r + 1 ) r + 2 B p r 2 p 2 ( m o d p 3 ) , if 2 | r , ( 1 ) n 1 ( n 1 ) ! 2 r r + 1 B p r 1 p ( m o d p 2 ) , if 2 | r .

Lemma 6. Let p > 4 be a prime, then

i 1 + i 2 + i 3 + i 4 = 2 p i 1 , i 2 , i 3 , i 4 P p 1 i 1 i 2 i 3 i 4 240 5 p B p 5 ( m o d p 2 ) .

Proof. By Lemma 3, we have

i 1 + i 2 + i 3 + i 4 = 2 p i 1 , i 2 , i 3 , i 4 P p 1 i 1 i 2 i 3 i 4 = 24 2 p 1 u 1 < u 2 < u 3 2 p u 1 , u 3 , u 2 u 1 , u 3 u 2 P p 1 u 1 u 2 u 3 . (5)

It is easy to see that

1 u 1 < u 2 < u 3 2 p u 1 , u 3 , u 2 u 1 , u 3 u 2 P p 1 u 1 u 2 u 3 = 1 u 1 < u 2 < u 3 2 p u 1 , u 2 , u 3 , u 2 u 1 , u 3 u 2 P p 1 u 1 u 2 u 3 + 1 u 1 < p < u 3 2 p u 1 , u 3 P p 1 u 1 p u 3 .

By Lemma 4, we have

1 u 1 < p < u 3 2 p u 1 , u 3 P p 1 u 1 p u 3 = 1 p 1 u 1 < p 1 u 1 p < u 3 < 2 p 1 u 3 0 ( mod p 3 ) . (6)

Hence

1 u 1 < u 2 < u 3 2 p u 1 , u 3 , u 2 u 1 , u 3 u 2 P p 1 u 1 u 2 u 3 1 u 1 < u 2 < u 3 2 p u 1 , u 2 , u 3 , u 2 u 1 , u 3 u 2 P p 1 u 1 u 2 u 3 1 u 1 < u 2 < u 3 2 p u 1 , u 2 , u 3 P p 1 u 1 u 2 u 3 1 u 1 < u 1 + p < u 3 2 p u 1 , u 3 P p 1 u 1 ( u 1 + p ) u 3 1 u 1 < u 2 < u 2 + p 2 p u 1 , u 2 P p 1 u 1 u 2 ( u 2 + p ) ( mod p 3 ) .

Replace u 3 = u 2 + p , then

1 u 1 < u 1 + p < u 3 2 p u 1 , u 3 P p 1 u 1 ( u 1 + p ) u 3 = 1 u 1 < u 1 + p < u 2 + p 2 p u 1 , u 2 P p 1 u 1 ( u 1 + p ) ( u 2 + p ) 1 u 1 < u 2 < p 1 u 1 2 u 2 ( 1 p u 1 + p 2 u 1 2 ) ( 1 p u 2 + p 2 u 2 2 ) ( mod p 3 ) .

and

1 u 1 < u 2 < u 2 + p 2 p u 1 , u 2 P p 1 u 1 u 2 ( u 2 + p ) 1 u 1 < u 2 < p 1 u 1 u 2 2 ( 1 p u 2 + p 2 u 2 2 ) ( mod p 3 ) .

Thus

1 u 1 < u 2 < u 3 2 p u 1 , u 3 , u 2 u 1 , u 3 u 2 P p 1 u 1 u 2 u 3 1 u 1 < u 2 < u 3 2 p u 1 , u 2 , u 3 P p 1 u 1 u 2 u 3 1 u 1 < u 2 < p ( 1 u 1 2 u 2 + 1 u 1 u 2 2 p ( 1 u 1 3 u 2 + 1 u 1 2 u 2 2 + 1 u 1 u 2 3 ) + p 2 ( 1 u 1 4 u 2 + 1 u 1 3 u 2 + 1 u 1 2 u 2 3 + 1 u 1 1 u 2 4 ) 1 3 ! 1 u 1 , u 2 , u 3 2 p u i u j , u i P p 1 u 1 u 2 u 3 1 u 1 , u 2 < p ( 1 u 1 2 u 2 p ( 1 u 1 3 u 2 + 1 2 1 u 1 2 u 2 2 ) + p 2 ( 1 u 1 4 u 2 + 1 u 1 3 u 2 ) ) ( mod p 3 ) . (7)

Using Lemma 5 in the first sum of the right hand in (7) and using Lemma 4 in the second sum, we have

1 u 1 < u 2 < u 3 2 p u 1 , u 3 , u 2 u 1 , u 3 u 2 P p 1 u 1 u 2 u 3 1 3 ! ( 1 ) 3 ( 3 1 ) ! 24 5 B p 5 p 2 ( 1 ) 2 12 10 B p 5 p 2 + 3 p 2 ( 4 5 B p 5 p ) p 2 30 14 B p 7 p 2 20 5 p 2 B p 5 ( m o d p 3 ) . (8)

Combining (5) with (8), we complete the proof of Lemma 6. ¨

Lemma 7. Let p > 4 be a prime and r > 1 a positive integer, then

i 1 + i 2 + i 3 + i 4 = 2 p r i 1 , i 2 , i 3 , i 4 P p 1 i 1 i 2 i 3 i 4 48 5 p r B p 5 ( mod p r + 1 ) .

Proof. The proof of Lemma 7 is similar to the proof method of (4) in [6]. ¨

Lemma 8 ( [7] ). Let p > 5 be a prime and r , m positive integers, ( m , p ) = 1 , then

i 1 + i 2 + + i 5 = m p i 1 , i 2 , , i 5 P p 1 i 1 i 2 i 3 i 4 i 5 ( 4 ( 5 m + m 3 ) B p 5 ( m o d p ) , if r = 1, 20 m p r 1 B p 5 ( m o d p r ) , if r > 1.

Lemma 9. Let p > 5 be a prime and r , m positive integers, then

i 1 + i 2 + + i 5 = m p r i 1 , i 2 , , i 5 P p 1 i 1 i 2 i 3 i 4 i 5 = 120 m p r 1 u 1 < u 2 < u 3 < u 4 m p r u 1 , u 4 , u 2 u 1 , u 3 u 2 , u 4 u 3 P p 1 u 1 u 2 u 3 u 4 .

Proof. The proof of Lemma 9 is similar to the proof of Lemma 2. ¨

3. Proofs of the Theorems

Proof of Theorem 1. It is easy to see that

i + j + k = 2 p r i , j , k P p ( 1 ) i i j k = 1 2 p r i + j + k = 2 p r i , j , k P p ( 1 ) i ( i + j + k ) i j k = 1 2 p r i + j + k = 2 p r i , j , k P p ( ( 1 ) i j k + 2 ( 1 ) i i j ) . (9)

Let l = j + k ,then 1 j < l 2 p r and j , l , l j P p , hence

i + j + k = 2 p r i , j , k P p ( 1 ) i j k = 1 j < l 2 p r j , l , l j P p 1 l ( 1 ) l ( j + k ) j k = 1 j < l 2 p r j , l , l j P p 2 ( 1 ) l j l . (10)

Let l = i + j , then 1 i < l 2 p r and i , l , l i P p , hence

i + j + k = 2 p r i , j , k P p 2 ( 1 ) i i j = 1 i < l 2 p r i , l , l i P p 1 l 2 ( 1 ) i ( i + j ) i j = 1 i < l 2 p r i , l , l i P p ( 2 ( 1 ) i j l + 2 ( 1 ) i i l ) . (11)

Noting that i = l j , ( 1 ) l j = ( 1 ) l + j and we rename l to l , then

1 i < l 2 p r i , l , l i P p 2 ( 1 ) i j l = 1 j < l 2 p r j , l , l j P p 2 ( 1 ) j + l j l . (12)

Rename i to j and l to l , then

1 i < l 2 p r i , l , l i P p 2 ( 1 ) i i l = 1 j < l 2 p r j , l , l j P p 2 ( 1 ) j j l . (13)

Combining (9)-(13), we have

i + j + k = 2 p r i , j , k P p ( 1 ) i i j k = 1 p r 1 j < l 2 p r j , l , l j P p ( ( 1 ) l j l + ( 1 ) j + l j l + ( 1 ) j j l ) = 1 p r 1 j < l 2 p r j , l , l j P p ( 1 + ( 1 ) l ) ( 1 + ( 1 ) j ) j l 1 p r 1 j < l 2 p r j , l , l j P p 1 j l = 1 p r 1 j < l 2 p r j , l , l j P p , j even , l even 4 j l 1 p r 1 j < l 2 p r j , l , l j P p 1 j l . (14)

Let j = 2 j , l = 2 l in the first sum of (14) and noting that

1 j < l p r j , l , l j P p 1 j l = 1 j < l p r j , l , l j P p 1 j l ,

(14) is equal to

i + j + k = 2 p r i , j , k P p ( 1 ) i i j k = 1 p r 1 j < l p r j , l , l j P p 1 j l 1 p r 1 j < l 2 p r j , l , l j P p 1 j l . (15)

By Lemma 1, Lemma 2 and (15), we obtain

i + j + k = 2 p r i , j , k P p ( 1 ) i i j k = 1 p r 1 j < l p r j , l , l j P p 1 j l 1 p r 1 j < l 2 p r j , l , l j P p 1 j l p r 1 B p 3 ( mod p r ) .

This completes the proof of Theorem 1. ¨

Proof of Theorem 2. For every triple ( i , j , k ) of positive integers which satisfies i + j + k = 2 p r , i , j , k P p , we take it to 3 cases.

Cases 1. Let A ( p r ) = { ( i , j , k ) | 1 i , j , k p r 1 and i , j , k P p } . ( i , j , k ) ( p r i , p r j , p r k ) is a bijection between the solutions of i + j + k = 2 p r , ( i , j , k ) A ( p r ) and i + j + k = p r , i , j , k P p , we have

i + j + k = 2 p r ( i , j , k ) A ( p r ) ( 1 ) i i j k i + j + k = p r i , j , k P p ( 1 ) p r i ( p r i ) ( p r j ) ( p r k ) i + j + k = p r i , j , k P p ( 1 ) i i j k ( mod p r ) . (16)

Cases 2. Let B ( p r ) = { ( i , j , k ) | p r + 1 i 2 p r 1,1 j , k p r 1 and i , j , k P p } . ( i , j , k ) ( p r + i , j , k ) is a bijection between the solutions of i + j + k = 2 p r , ( i , j , k ) B ( p r ) and i + j + k = p r , i , j , k P p , we have

i + j + k = 2 p r ( i , j , k ) B ( p r ) ( 1 ) i i j k i + j + k = p r i , j , k P p ( 1 ) p r + i ( p r + i ) j k i + j + k = p r i , j , k P p ( 1 ) i i j k ( mod p r ) . (17)

Cases 3. Let

C ( p r ) = { ( i , j , k ) | p r + 1 j 2 p r 1,1 i , k p r 1 and i , j , k P p }

and

D ( p r ) = { ( i , j , k ) | p r + 1 k 2 p r 1,1 i , j p r 1 and i , j , k P p } .

( i , j , k ) ( i , p r + j , k ) in the former and ( i , j , k ) ( i , j , p r + k ) in the later are the bijections between the solutions of i + j + k = 2 p r , ( i , j , k ) C ( p r ) or ( i , j , k ) D ( p r ) and i + j + k = p r , i , j , k P p , we have

i + j + k = 2 p r ( i , j , k ) C ( p r ) ( 1 ) i i j k + i + j + k = 2 p r ( i , j , k ) D ( p r ) ( 1 ) i i j k i + j + k = p r i , j , k P p ( 1 ) i i ( p r + j ) k + i + j + k = p r i , j , k P p ( 1 ) i i j ( p r + k ) 2 i + j + k = p r i , j , k P p ( 1 ) i i j k ( mod p r ) . (18)

Combining (16)-(18), we have

i + j + k = 2 p r i , j , k P p ( 1 ) i i j k = i + j + k = 2 p r ( i , j , k ) A ( p r ) ( 1 ) i i j k + i + j + k = 2 p r ( i , j , k ) B ( p r ) ( 1 ) i i j k + i + j + k = 2 p r ( i , j , k ) C ( p r ) ( 1 ) i i j k + i + j + k = 2 p r ( i , j , k ) D ( p r ) ( 1 ) i i j k 2 i + j + k = p r i , j , k P p ( 1 ) i i j k ( mod p r ) .

By Theorem 1, we complete the proof of Theorem 2. ¨

Proof of Theorem 3. By symmetry, it is easy to see that

i 1 + i 2 + i 3 + i 4 = 2 p r i 1 , i 2 , i 3 , i 4 P p ( 1 ) i 1 i 1 i 2 i 3 i 4 = 1 2 p r i 1 + i 2 + i 3 + i 4 = 2 p r i 1 , i 2 , i 3 , i 4 P p ( 1 ) i 1 ( i 1 + i 2 + i 3 + i 4 ) i 1 i 2 i 3 i 4 = 1 2 p r i 1 + i 2 + i 3 + i 4 = 2 p r i 1 , i 2 , i 3 , i 4 P p ( ( 1 ) i 1 i 2 i 3 i 4 + 3 ( 1 ) i 1 i 1 i 3 i 4 ) . (19)

Let u 3 = i 2 + i 3 + i 4 in the first sum of the last equation in (19), then i 1 = 2 p r u 3 , (19) equals to

= 1 2 p r [ u 3 = i 2 + i 3 + i 4 < 2 p r u 3 , i 2 , i 3 , i 4 P p ( 1 ) 2 p r u 3 ( i 2 + i 3 + i 4 ) i 2 i 3 i 4 u 3 + 3 u 3 = i 1 + i 3 + i 4 < 2 p r u 3 , i 1 , i 3 , i 4 P p ( 1 ) i 1 ( i 1 + i 3 + i 4 ) i 1 i 3 i 4 u 3 ] = 1 2 p r [ 3 u 3 = i 2 + i 3 + i 4 < 2 p r u 3 , i 2 , i 3 , i 4 P p ( 1 ) u 3 i 3 i 4 u 3 + 3 u 3 = i 1 + i 3 + i 4 < 2 p r u 3 , i 1 , i 3 , i 4 P p ( 1 ) i 1 i 3 i 4 u 3 + 6 u 3 = i 1 + i 3 + i 4 < 2 p r u 3 , i 1 , i 3 , i 4 P p ( 1 ) i 1 i 1 i 3 u 3 ] . (20)

Let u 2 = i 3 + i 4 in the second sum of the last equation in (20), since u 3 = i 1 + i 3 + i 4 , then i 1 = u 3 u 2 , (20) equals to

Similarly, we have

i 1 + i 2 + i 3 + i 4 = 2 p r i 1 , i 2 , i 3 , i 4 P p ( 1 ) i 1 + i 2 i 1 i 2 i 3 i 4 = 4 p r 0 < u 1 < u 2 < u 3 < 2 p r u 1 , u 3 , u 3 u 2 , u 2 u 1 P p ( 1 ) u 2 + ( 1 ) u 1 + u 2 + u 3 + ( 1 ) u 1 + u 3 u 1 u 2 u 3 .

Hence

4 i 1 + i 2 + i 3 + i 4 = 2 p r i 1 , i 2 , i 3 , i 4 P p ( 1 ) i 1 i 1 i 2 i 3 i 4 + 3 i 1 + i 2 + i 3 + i 4 = 2 p r i 1 , i 2 , i 3 , i 4 P p ( 1 ) i 1 + i 2 i 1 i 2 i 3 i 4 = 12 p r 0 < u 1 < u 2 < u 3 < 2 p r u 1 , u 3 , u 3 u 2 , u 2 u 1 P p [ 1 + ( 1 ) u 1 ] [ 1 + ( 1 ) u 2 ] [ 1 + ( 1 ) u 3 ] 1 u 1 u 2 u 3 = 12 p r 0 < u 1 < u 2 < u 3 < p r u 1 , u 3 , u 3 u 2 , u 2 u 1 P p 1 u 1 u 2 u 3 12 p r 0 < u 1 < u 2 < u 3 < 2 p r u 1 , u 3 , u 3 u 2 , u 2 u 1 P p 1 u 1 u 2 u 3 .

By Lemma 3, we have

4 i 1 + i 2 + i 3 + i 4 = 2 p r i 1 , i 2 , i 3 , i 4 P p ( 1 ) i 1 i 1 i 2 i 3 i 4 + 3 i 1 + i 2 + i 3 + i 4 = 2 p r i 1 , i 2 , i 3 , i 4 P p ( 1 ) i 1 + i 2 i 1 i 2 i 3 i 4 = 1 2 i 1 + i 2 + i 3 + i 4 = 2 p r i 1 , i 2 , i 3 , i 4 P p 1 i 1 i 2 i 3 i 4 i 1 + i 2 + i 3 + i 4 = 2 p r i 1 , i 2 , i 3 , i 4 P p 1 i 1 i 2 i 3 i 4 .

By (2) and Lemma 6, we have

4 i 1 + i 2 + i 3 + i 4 = 2 p i 1 , i 2 , i 3 , i 4 P p ( 1 ) i 1 i 1 i 2 i 3 i 4 + 3 i 1 + i 2 + i 3 + i 4 = 2 p i 1 , i 2 , i 3 , i 4 P p ( 1 ) i 1 + i 2 i 1 i 2 i 3 i 4 24 5 p B p 5 + 240 5 p B p 5 216 5 p B p 5 ( mod p 2 ) .

By (4) and Lemma 7, if r 2 , then

4 i 1 + i 2 + i 3 + i 4 = 2 p r i 1 , i 2 , i 3 , i 4 P p ( 1 ) i 1 i 1 i 2 i 3 i 4 + 3 i 1 + i 2 + i 3 + i 4 = 2 p r i 1 , i 2 , i 3 , i 4 P p ( 1 ) i 1 + i 2 i 1 i 2 i 3 i 4 12 5 p r B p 5 + 48 5 p r B p 5 36 5 p r B p 5 ( mod p r + 1 ) .

This completes the proof of Theorem 3. ¨

Proof of Theorem 4. Similar to the proofs of Theorem 1 and Theorem 3, we have

i 1 + i 2 + i 3 + i 4 + i 5 = 2 p r i 1 , i 2 , i 3 , i 4 , i 5 P p ( 1 ) i 1 i 1 i 2 i 3 i 4 i 5 = 12 p r 0 < u 1 < u 2 < u 3 < u 4 < 2 p r u 1 , u 4 , u 2 u 1 , u 3 u 2 , u 4 u 3 P p 1 u 1 u 2 u 3 u 4 [ ( 1 ) u 1 + ( 1 ) u 4 + ( 1 ) u 1 + u 2 + ( 1 ) ) u 2 + u 3 + ( 1 ) u 3 + u 4 ]

and

i 1 + i 2 + i 3 + i 4 + i 5 = 2 p r i 1 , i 2 , i 3 , i 4 , i 5 P p ( 1 ) i 1 + i 2 i 1 i 2 i 3 i 4 i 5 = 6 p r 0 < u 1 < u 2 < u 3 < u 4 < 2 p r u 1 , u 4 , u 2 u 1 , u 3 u 2 , u 4 u 3 P p 1 u 1 u 2 u 3 u 4 [ ( 1 ) u 2 + ( 1 ) u 3 + ( 1 ) u 1 + u 3 + ( 1 ) u 1 + u 4 + ( 1 ) u 2 + u 4 + ( 1 ) u 1 + u 2 + u 3 + ( 1 ) u 1 + u 2 + u 4 + ( 1 ) u 1 + u 3 + u 4 + ( 1 ) u 2 + u 3 + u 4 + ( 1 ) u 1 + u 2 + u 3 + u 4 ]

Hence

i 1 + i 2 + i 3 + i 4 + i 5 = 2 p r i 1 , i 2 , i 3 , i 4 , i 5 P p ( 1 ) i 1 i 1 i 2 i 3 i 4 i 5 + 2 i 1 + i 2 + i 3 + i 4 + i 5 = 2 p r i 1 , i 2 , i 3 , i 4 , i 5 P p ( 1 ) i 1 + i 2 i 1 i 2 i 3 i 4 i 5 = 12 p r 0 < u 1 < u 2 < u 3 < u 4 < 2 p r u 1 , u 4 , u 2 u 1 , u 3 u 2 , u 4 u 3 P p [ 1 + ( 1 ) u 1 ] [ 1 + ( 1 ) u 2 ] [ 1 + ( 1 ) u 3 ] [ 1 + ( 1 ) u 4 ] 1 u 1 u 2 u 3 u 4 = 12 p r 0 < u 1 < u 2 < u 3 < u 4 < p r u 1 , u 4 , u 2 u 1 , u 3 u 2 , u 4 u 3 P p 1 u 1 u 2 u 3 u 4 12 p r 0 < u 1 < u 2 < u 3 < u 4 < 2 p r u 1 , u 4 , u 2 u 1 , u 3 u 2 , u 4 u 3 P p 1 u 1 u 2 u 3 u 4 .

By Lemma 9, we have

i 1 + i 2 + i 3 + i 4 + i 5 = 2 p r i 1 , i 2 , i 3 , i 4 , i 5 P p ( 1 ) i 1 i 1 i 2 i 3 i 4 i 5 + 2 i 1 + i 2 + i 3 + i 4 + i 5 = 2 p r i 1 , i 2 , i 3 , i 4 , i 5 P p ( 1 ) i 1 + i 2 i 1 i 2 i 3 i 4 i 5 = 1 10 i 1 + i 2 + i 3 + i 4 + i 5 = p r i 1 , i 2 , i 3 , i 4 , i 5 P p 1 i 1 i 2 i 3 i 4 i 5 2 10 i 1 + i 2 + i 3 + i 4 + i 5 = 2 p r i 1 , i 2 , i 3 , i 4 , i 5 P p 1 i 1 i 2 i 3 i 4 i 5 .

By (2) and Lemma 8 (1), we have

i 1 + i 2 + i 3 + i 4 + i 5 = 2 p i 1 , i 2 , i 3 , i 4 , i 5 P p ( 1 ) i 1 i 1 i 2 i 3 i 4 i 5 + 2 i 1 + i 2 + i 3 + i 4 + i 5 = 2 p i 1 , i 2 , i 3 , i 4 , i 5 P p ( 1 ) i 1 + i 2 i 1 i 2 i 3 i 4 i 5 24 10 B p 5 + 144 10 B p 5 12 B p 5 ( mod p ) .

By Lemma 8 (2), if r 2 , then

i 1 + i 2 + i 3 + i 4 + i 5 = 2 p r i 1 , i 2 , i 3 , i 4 , i 5 P p ( 1 ) i 1 i 1 i 2 i 3 i 4 i 5 + 2 i 1 + i 2 + i 3 + i 4 + i 5 = 2 p r i 1 , i 2 , i 3 , i 4 , i 5 P p ( 1 ) i 1 + i 2 i 1 i 2 i 3 i 4 i 5 2 p r 1 B p 5 + 8 p r 1 B p 5 6 p r 1 B p 5 ( mod p r ) .

This completes the proof of Theorem 4. ¨

4. Conclusions

Let p be an odd prime and r , m positive integers, ( m , p ) = 1 , using Lemma 1 and Lemma 2, similar to the proof of Theorem 1, we can prove that

i + j + k = 2 m p r i , j , k P p ( 1 ) i i j k m p r 1 B p 3 ( mod p r ) .

In particular, if m = 1 , it becomes Theorem 1.

Let p be odd prime and r , m positive integers, ( m , p ) = 1 , similar to the proof of Theorem 2, we can prove that

i + j + k = m p r i , j , k P p ( 1 ) i i j k 1 2 i + j + k = 2 m p r i , j , k P p ( 1 ) i i j k m 2 p r 1 B p 3 ( mod p r ) .

In particular, if m = 1 , it becomes Theorem 2.

Let p > 4 be a prime and r , m positive integers, ( m , p ) = 1 , we can deduce the congruence ( m o d p r + 1 ) for

4 i 1 + i 2 + i 3 + i 4 = 2 m p r i 1 , i 2 , i 3 , i 4 P p ( 1 ) i 1 i 1 i 2 i 3 i 4 + 3 i 1 + i 2 + i 3 + i 4 = 2 m p r i 1 , i 2 , i 3 , i 4 P p ( 1 ) i 1 + i 2 i 1 i 2 i 3 i 4 .

Let p > 5 be a prime and r , m positive integers, ( m , p ) = 1 , we can deduce the congruence ( m o d p r ) for

i 1 + i 2 + i 3 + i 4 + i 5 = 2 m p r i 1 , i 2 , i 3 , i 4 , i 5 P p ( 1 ) i 1 i 1 i 2 i 3 i 4 i 5 + 2 i 1 + i 2 + i 3 + i 4 + i 5 = 2 m p r i 1 , i 2 , i 3 , i 4 , i 5 P p ( 1 ) i 1 + i 2 i 1 i 2 i 3 i 4 i 5 .

Similarly, we can consider the congruence ( m o d p r + 1 ) for

i 1 + i 2 + + i n = p r i 1 , i 2 , , i n P p σ 1 i 1 σ 2 i 2 σ n i n i 1 i 2 i n ,

where σ i { 1 , 1 } , i = 1 , 2 , , n , but it seems much more complicated.

Founding

This work is supported by the Natural Science Foundation of Zhejiang Province, Project (No. LY18A010016) and the National Natural Science Foundation of China, Project (No. 12071421).

Cite this paper: Shen, Z. and Cai, T. (2020) Super Congruences Involving Alternating Harmonic Sums. Advances in Pure Mathematics, 10, 611-622. doi: 10.4236/apm.2020.1010037.
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