Ecuatia lui Bernoulli

8/13/2019 Ecuatia lui Bernoulli

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The paper will appear in Discrete Mathematics.CONGRUENCES INVOLVING BERNOULLI POLYNOMIALS

Zhi-Hong Sun

Department of Mathematics, Huaiyin Teachers College,Huaian, Jiangsu 223001, P.R. China

E-mail: [email protected]: http://www.hytc.edu.cn/xsjl/szh

Let {Bn(x)} be the Bernoulli polynomials. In the paper we establish some

congruences for Bj(x) (mod pn), where p is an odd prime and x is a rational p-integer.

Such congruences are concerned with the properties ofp-regular functions, the congru-ences for h(

sp) (mod p) (s = 3, 5, 8, 12) and the sum

kr (mod m)

p

k

, where h(d) is

the class number of the quadratic field (

d) of discriminant d and p-regular func-tions are those functions f such that f(k) (k = 0, 1, . . . ) are rational p-integers and n

k=0

n

k

(1)kf(k) 0 (mod pn) for n = 1, 2, 3, . . . We also establish many congru-ences for Euler numbers.

MSC: Primary 11B68, Secondary 11A07, 11R29.

Keywords: Congruence, Bernoulli polynomial, p-regular function, class number, Eulernumber

1. Introduction.

The Bernoulli numbers{Bn} and Bernoulli polynomials{Bn(x)} are defined by

B0= 1,n1k=0

n

k

Bk = 0 (n2) and Bn(x) =

nk=0

n

k

Bkx

nk (n0).

The Euler numbers{En} and Euler polynomials{En(x)} are defined by

2et

e2t + 1=

n=0

Entn

n! (|t|<

2) and

2ext

et + 1=

n=0

En(x)tn

n! (|t|< ),

which are equivalent to (see [MOS])

E0 = 1, E2n1= 0,n

r=0

2n

2r

E2r = 0 (n1)

and

(1.1) En(x) +n

r=0

n

r

Er(x) = 2x

n (n0).1


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It is well known that([MOS])

(1.2)

En(x) = 1

2n

nr=0

n

r

(2x 1)nrEr

= 2n+ 1

Bn+1(x) 2n+1Bn+1x2.LetZ and N be the set of integers and the set of positive integers respectively. Let

[x] be the integral part ofxand{x} be the fractional part ofx. Ifm, sN and pisan odd prime not dividing m, in Section 2 we show that

(1)s mp

p1k=1

ksp(mod m)

p

k

Bp1 (s1)pm Bp1 spm (modp) if 2|m,12

(1)[ (s1)pm ]Ep2

(s1)pm

(1)[ spm]Ep2 spm (mod p) if 2m.For a discriminant d let h(d) be the class number of the quadratic field Q(

d) (Q

is the set of rational numbers). Ifp >3 is a prime of the form 4m+3, it is well knownthat (cf. [IR])

(1.3) h(p) 2Bp+12

(mod p).

Ifp is a prime of the form 4m+ 1, according to [Er] we have

(1.4) 2h(4p)Ep12

(mod p).

Let ( an) be the Kronecker symbol. For odd primes p, in Section 3 we establish thefollowing congruences:

h(8p)Ep12

14

(mod p);

h(3p) 4

p

3Bp+1

2 1

3(mod p) forp1 (mod 4);

h(12p)8p3

Bp+12

112

(mod p) for p7, 11, 23 (mod 24);

h(5p) 8Bp+12

15

(mod p) forp11, 19 (mod 20).

For m N let Zm be the set of rational numbers whose denominator is coprimetom. For a prime p, in [S5] the author introduced the notion ofp-regular functions.If f(k) Zp for any nonnegative integers k and

nk=0

nk

(1)kf(k) 0 (mod pn)

2


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for all n N, then f is called a p-regular function. Iff is a p-regular function andk,m,n,tN, in Section 4 we show that

(1.5) f(ktpm1)n1

r=0

(1)n1rk 1 rn

1

r

k

rf(rtpm1) (mod pmn),

which was annouced by the author in [S5, (2.4)]. We also show that

(1.6) f(kpm1)(1 kpm1)f(0) +kpm1f(1) (mod pm+1) for p >2.

Let p be a prime, x Zp and let b be a nonnegative integer. Lettp be theleast nonnegative residue oft modulopand x = (x+ xp)/p. From [S4, Theorem3.1] we know that f(k) = p(pBk(p1)+b(x) pk(p1)+bBk(p1)+b(x)) is a pregularfunction. If p 1 b, in [S5] the author showed that f(k) = (Bk(p1)+b(x)

pk(p1)+b1

Bk(p1)+b(x

))/(k(p 1) + b) is also a pregular function. Using such re-sults in [S4, S5] and (1.5), in Section 5 we obtain general congruences for pBk(ps)+b(x),pBk(ps)+b, (mod p

sn), where k,n, s N, is Eulers totient function and is aDirichlet character modulo a positive integer. As a consequence of (1.6), if 2|b and

p 1b, we have

Bk(ps)+b

k(ps) +b (1 kps1)(1 pb1) Bb

b +kps1

Bp1+bp 1 +b (mod p

s+1).

In Section 6 we establish some congruences fornk=0

nk(1)

kpBk(p1)+b(x) mod-

ulopn+1

, where pis an odd prime,nN,xZp andb is a nonnegative integer.Let p be an odd prime and b {0, 2, 4, . . . }. In Section 7 we show that f(k) =(1(1) p12 pk(p1)+b)Ek(p1)+b is apregular function. Using this and (1.5) we givecongruences for Ek(pm)+b (mod p

mn), where k, mN. By (1.6) we have

Ek(pm)+b(1 kpm1)(1 (1)p12 pb)Eb+kp

m1Ep1+b (mod pm+1).

We also show that f(k) =E2k+b is a 2regular function and

E2mkt+bn1r=0

(1)n1rk 1 rn 1 rkrE2mrt+b (mod 2mn+n),

wherek,m,n,tN and N is given by 21 n


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Theorem 2.1. Letp, mN andk, rZ withk0. Then

p1x=0

xr(mod m)

xk = mk

k+ 1

Bk+1

pm

+r p

m

Bk+1

rm

and

p1x=0

xr(mod m)

(1)xrm xk =mk

2

(1)[ rpm ]Ek

pm

+r p

m

(1)[ rm ]Ek

rm

.

Proof. For any real number t and nonnegative integer n it is well known that (cf.[MOS])

(2.1) Bn(t+ 1) Bn(t) =ntn1 (n= 0) and En(t+ 1) +En(t) = 2tn.

Hence, for xZ we have

Bk+1

x+ 1m

+r x 1

m

Bk+1

xm

+r x

m

=

Bk+1

x+1m +

rxm

1m Bk+1 xm + rxm = 0 if m x r,Bk+1

x+1m +

m1m

Bk+1 xm= (k+ 1) xmk ifm|x r.Thus

Bk+1

pm

+

r pm

Bk+1

rm

=

p1x=0

Bk+1

x+ 1m

+r x 1

m

Bk+1

xm

+r x

m

=k+ 1

mk

p1x=0

xr(mod m)

xk.

Similarly, ifxZ, by (2.1) we have

(1)[ rx1m ]Ekx+ 1

m +

r x 1m

(1)[ rxm ]Ek

xm

+r x

m

=

(1)[ rxm ]Ekx+1m + rxm} 1m Ek xm + rxm = 0ifmx r,

(1) rxm 1Ekx+1m +

m1m

(1) rxm Ek xm=(1) rxm 2( xm)kifm|x r.

4


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Thus

(1)[ rpm ]Ekp

m+r p

m

(1)[ rm ]Ek

rm

=

p1

x=0(1)[ rx1m ]Ek

x+ 1

m +

r x 1

m (1)[ rxm ]Ek

xm

+r x

m

= 2mk

p1x=0

xr(mod m)

(1)xrm xk.

This completes the proof.

Corollary 2.1. Let p be an odd prime and k {0, 1, . . . , p2}. Let r Z andmN withpm. Then

p1x=0

xr(mod m)

xk mk

k+ 1

Bk+1

r pm

Bk+1

rm

(mod p)

and

p1x=0

xr(mod m)

(1)xrm xk

mk

2(1)

[ rpm

]Ekr pm

(1)[ rm

]Ekrm(mod p).

Proof. Ifx1, x2 Zp and x1 x2 (mod p), by [S5, Lemma 3.1] and [S3, Lemma3.3] we have

(2.2) Bk+1(x1) Bk+1(x2)

k+ 1 x1 x2

p pBk0 (mod p)

and

(2.3) Ek(x1)Ek(x2) (mod p).

Thus the result follows from Theorem 2.1.Remark 2.1 Putting k = p 2 in Corollary 2.1 and then applying Fermats littletheorem we see that ifpis an odd prime not dividing m, then

(2.4)

p1x=1

xr(mod m)

1

x 1

m

Bp1

r pm

Bp1

rm

(mod p)

5


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and

(2.5)

p1x=1

xr(mod m)

(1)xrm 1x

12m

(1)[ rpm ]Ep2r p

m

(1)[ rm ]Ep2rm

(mod p).

Here (2.4) and (2.5) are due to my brother Z.W. Sun. See [Su2, Theorem 2.1]. Inspiredby his work, the author established Theorem 2.1 and Corollary 2.1 in 1991.

Corollary 2.2. Letp be an odd prime. Letk {0, 1, . . . , p 2} andm, sN withpm. Then

(1)kk+ 1

Bk+1

(s 1)pm

Bk+1

spm

(s1)p

m


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Corollary 2.3. Letp be a prime.

(i) (Karpinski[K, UW]) Ifp3 (mod 8), then(p3)/4x=1

xp

= 0.

(ii) (Karpinski[K, UW]) Ifp5 (mod 8), then[p/6]

x=1 xp = 0.

(iii) (Berndt[B, UW]) Ifp5 (mod 24), then(p5)/12

x=1

xp

= 0.

Proof. By Corollary 2.2 and the known fact B2n+1 = 0, for mN with pm wehave

(2.8)

[p/m]x=1

xp

[p/m]x=1

xp12 (1)

p12

p+12

B p+1

2 B p+1

2

pm

2B p+1

2

pm

(mod p) ifp1 (mod 4),

2Bp+12 + 2B

p+12 p

m

(mod p) ifp3 (mod 4).It is well known that B2n(

34 ) = B2n(

14 ) = ( 122n1)B2n/24n1. Thus, if p

3 (mod 8), by (2.8) we see that

p34

x=1

xp

2Bp+1

2+ 2B p+1

2

34

=

1

2p1

1 2p12 Bp+12

2Bp+12

1 2

p

2

B p+1

2= 0 (mod p).

Asp34 p34x=1 xp p34 , we must have(p3)/4x=1 (xp ) = 0. This proves (i).Now we consider (ii). For n {0, 1, 2, . . . } and m N it is well known that (cf.

[IR], [MOS])

(2.9) Bn(1 x) = (1)nBn(x) andm1k=0

Bn

x+

k

m

= m1nBn(mx).

Thus

Bp+12

12n

+B p+1

2

12n

+1

2

= 2

p12 Bp+1

2

1n

and so

(2.10) B p+12

12n

2

p

Bp+1

2

1n

(1) p+12 Bp+1

2

n 12n

(mod p).

Since p5 (mod 8), taking n= 3 in (2.10) we find

(2.11) B p+12

16

Bp+1

2

13

+B p+1

2

13

= 0 (mod p).

7


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This together with (2.8) and (2.9) yields

[p/6]x=1

xp

2Bp+1

2

p6

=2

p3

Bp+1

2

16

0 (modp).

As|[p/6]x=1 xp| [p6 ] we have[p/6]x=1 xp= 0. This proves (ii).Finally we consider (iii). Assume p5 (mod 24). By (2.10) and (2.11) we have

Bp+12

112

2

p

Bp+1

2

16

+B p+1

2

512

B p+1

2

512

(mod p).

On the other hand, by (2.9) we have

B p+12

112

+B p+1

2

512

= 3

p12 Bp+1

2

14

Bp+1

2

912

3

p

B p+1

2

14

(1) p+12 B p+1

2

14

= 0 (mod p).Thus B p+1

2

112

B p+12

512

0 (mod p). Now applying (2.8) we see that[p/12]x=1

xp

2Bp+1

2

p12

=2Bp+1

2

512

0 (mod p).

This yields (iii) and so the corollary is proved.

Corollary 2.4. Supposep, q,mN, nZ, gcd(p, m) = 1 andqm. ForrZ letAr(m, p) be the least positive solution of the congruencepxr (mod m). Then

r: Ar

(m, p)

q, rZ,

n

r

p

1

n= pq+nm

n

m.

Proof. Using Theorem 2.1 we see thatr: Ar(m, p)q, rZ,nrp 1 n=

qx=1

p1nr=n

rpx (mod m)

1 =

qx=1

p1s=0

spx+n (mod m)

1

=

qx=1

B1 p

m+px+n p

m

B1

px+nm

=

qx=1

pm

+

p(x 1) +nm

px+nm

=pq

m+ n

m

pq+n

m

=

pq+n

m

pq+nm

n

m n

m

=pq+n

m

n

m

.

This proves the corollary.8


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Theorem 2.2. Letm, sN and letp be an odd prime not dividingm. Then

(1)s mp

p1k=1

ksp(mod m)

p

k

(s1)pm


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and

(1)[ (s1)pm ]Ep2 (s 1)p

m

(1)[ spm]Ep2

spm

2

(s1)pm


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Proof. From [UW, p. 58] we know that

(3.2) h(8p) = 2p1a=1

a1(mod 4)

8pa

.

Thus applying Corollary 2.1 in the case r= 1, m = 4 and k= p12 we see that

h(8p) = 2p1a=0

a1(mod 4)

2a

ap

2

p1a=0

a1(mod 4)

(1) a14 a p12

4p12

(1)[ 1p4 ]Ep12

1 p4

Ep1

2

14

(mod p).

Since E2n(0) = 22n+1 (B2n+1 22n+1B2n+1) = 0 by (1.2), we see that

Ep12

1 p

4

=

E2n(0) = 0 if p = 4n+ 1,E2n1(

12 ) = 2

12nE2n1= 0 if p = 4n 1.Thus

h(8p)4 p12 Ep12

14

Ep1

2

14

(mod p).

LetSn= 4nEn(

14 ). Now we show that Sn= S

n forn0. By (1.1) we have

4nSn+n

k=0

n

k

4kSk = 2 4n and so Sn+

nk=0

n

k

4nkSk = 2.

That is, Sn = 1n1

k=0

nk

22n2k1Sk. Since S

0 = S0 = 1 we see that S

n = Sn.

That is,

(3.3) Sn= 4nEn

14

.

HenceSp12

= 4p12 Ep1

2( 14 )h(8p) (mod p).This proves the theorem.

Corollary 3.1. Letp be an odd prime. ThenpSp12

.

Proof. From (3.2) we have 1 < h(8p)< p. Thus the result follows from Theorem

3.1.

Remark 3.1Since Sn = 4nEn(

14 ), by (1.2) and the binomial inversion formula we

have

(3.4) Sn=

nr=0

n

r

(1)nr2rEr and

nr=0

n

r

Sr = 2

nEn.

11


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Theorem 3.2. Letp be a prime greater than3.(i) Ifp1 (mod 4), then

h(3p)4Bp+1

2

13


4Bp+12 13 (mod p) ifp5 (mod 12).

(ii) Ifp3 (mod 4), then

h(12p)

8B p+12

112


8Bp+12

112


8B p+12

112

+ 8Bp+1

2(mod p) ifp19 (mod 24)

and

h(5p)8B p+1

2( 15 ) (mod p) ifp11, 19 (mod 20),

8Bp+12

( 15 ) + 4Bp+12(mod p) ifp

3, 7 (mod 20).

.

Proof. We first assume p1 (mod 4). From [UW, p. 40] or [B] we have

h(3p) = 2[p/3]x=1

px

.

Thus applying (2.8), (2.9) and the quadratic reciprocity law we see that

h(

3p) = 2

[p/3]

x=1

x

p

4Bp+12

p

3=

4p

3Bp+12

1

3 (mod p).

This proves (i).Now let us consider (ii). Assume p3 (mod 4). From [UW, pp. 3-5] we have

h(12p) =

4 p

12


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Thus

h(12p)

8B p+12

512

B p+12

16


8

B p+12

112

B p+1

2 16


8B p+12 13 B p+12 112 (mod p) ifp19 (mod 24).By (2.10) we have

Bp+12

112

2

p

Bp+1

2

16

B p+1

2

512

(mod p).

Thus, if p 7 (mod 24), then h(12p) 8(B p+12

16

B p+12

512

) 8B p+1

2

112

(mod p). It is well known that ([GS])

B2n1

3= 312n

1

2 B

2n and B

2n1

6= (212n

1)(312n

1)

2 B

2n.

Thus

B p+12

13

=

1

2

3

p12 1

Bp+1

2 1

2

3p

1

Bp+1

2(mod p)

and

Bp+12

16

=

(2p12 1)(3 p12 1)

2 Bp+1

2 1

2

2p

1

3p

1

Bp+1

2(modp).

Ifp

11 (mod 12), then 3p = 1 and so B p+12 16 0 (mod p). Hence h(12p)8Bp+12

112

(mod p). If p 19 (mod 24), then 3p =1 and so Bp+12 13

Bp+12

(mod p). Thus h(12p)8(B p+12

112

+B p+1

2) (mod p).

Finally we consider h(5p) (mod p). From [UW, p. 40] or [B] we have

h(5p) = 2

p

5


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From (2.9) we see that

Bp+12

+ 2Bp+12

15

+ 2B p+1

2

25

=

4k=0

B p+12

k5

= 5

p12 Bp+1

2

and so

B p+12

15

+B p+1

2

25

1

2

p5

1

B p+1

2(mod p).

Thus

h(5p) 4p

5

2Bp+1

2

15

+

1

2

1

p5

Bp+1

2

=

8Bp+12

15

(mod p) ifp11, 19 (mod 20),

8Bp+12

15

+ 4Bp+1

2(mod p) ifp3, 7 (mod 20).

The proof is now complete.When d is a negative discriminant, it is known that 1 h(d) < p. Thus, from

Theorem 3.2 we deduce the following result.

Corollary 3.2. Letp be a prime.(i) Ifp1 (mod 4), thenBp+1

2( 13 )0 (mod p).

(ii) Ifp7, 11, 23 (mod 24), thenBp+12

( 112 )0 (mod p).(iii) Ifp11, 19 (mod 20), thenBp+1

2( 15 )0 (mod p).

Remark 3.2Forn= 0, 1, . . . it is well known that nk=0 nk 1nk+1 Bk(x) = xn. From

this we deduce that ifmN and an= mnBn( 1m), thennk=0 n+1k mnkak =n + 1.4. p-regular functions.

For a prime p, in [S5] the author introduced the notion ofp-regular functions. Iff(k) is a complex number congruent to an algebraic integer modulo p for any givennonnegative integer k and

nk=0

nk

(1)kf(k) 0 (mod pn) for all n N, then f

is called a p-regular function. Iff and g are p-regular functions, in [S5] the authorshowed that f g is also a p-regular function. Thus we see thatp-regular functionsform a ring. In the section we discuss further properties ofp-regular functions.

Suppose nN andk {0, 1, . . . , n}. Let s(n, k) be the unsigned Stirling numberof the first kind and S(n, k) be the Stirling number of the second kind defined by

x(x 1) (x n+ 1) =n

k=0

(1)nks(n, k)xk

and

xn =

nk=0

S(n, k)x(x 1) (x k+ 1).14


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For our convenience we also define s(n, k) = S(n, k) = 0 for k > n. FormN it iswell known that

(4.1)n

r=0

n

r

(1)nrrm =n!S(m, n)

In particular, taking m = n we have the following Eulers identity

(4.2)n

r=0

n

r

(1)nrrn =n!.

Lemma 4.1. Letx, d be variables, m, nN and iZ withi0. Thenn

r=0

n

r

(1)nr

rx+d

m

ri

=

n!

m!

mj=ni

mk=j

kj

(1)mk

s(m, k)dkj

S(i+j, n)xj

.

In particular we have

nr=0

n

r

(1)nr

rx

m

ri =

n!

m!

mj=ni

(1)mjs(m, j)S(i+j, n)xj .

Proof. Since

m!rx+d

m = (rx+d)(rx+d 1) (rx+d m+ 1)=

mk=0

(1)mks(m, k)(rx+d)k

=

mk=0

(1)mks(m, k)k

j=0

k

j

(rx)jdkj

=mj=0

mk=j

k

j

(1)mks(m, k)dkj

rjxj ,

we have

nr=0

n

r

(1)nr

rx+d

m

ri

= 1

m!

mj=0

mk=j

k

j

(1)mks(m, k)dkj

xj

nr=0

n

r

(1)nrri+j .

Now applying (4.1) we obtain the result.15


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Lemma 4.2. Letp be a prime andm, nN. Thenm!s(n, m)

n! pnm Zp and m!S(n, m)

n! pnm Zp.

Moreover, ifm < n, we have

m!s(n, m)

n! pnm m!S(n, m)

n! pnm 0 (mod p) forp > 2

andm!s(n, m)

n! 2nm

m

n m

(mod 2).

Proof. It is well known that

(ex 1)mm!

=

n=m

S(n, m)xn

n!.

Thus, applying the multinomial theorem we see that

(ex 1)m = k=1

xk

k!

m=

n=m

k1+k2++kn=mk1+2k2++nkn=n

m!

k1!k2! kn!n

r=1

1

r!kr

xn

and so

(4.3) S(n, m) = k1+k2++kn=mk1+2k2++nkn=n

n!

1!k1k1!2!k2k2!

n!knkn!.

Hence

m!S(n, m)

n! pnm =


(k1+k2+ +kn)!k1!k2! kn!

nr=1

pr1r!

kr.

From [S5, pp. 196-197] we also have

(4.4) s(n, m) = k1+k2++kn=mk1+2k2++nkn=n

n!

1k1k1!2k2k2! nknkn!

and

m!s(n, m)

n! pnm =


(k1+k2+ +kn)!k1!k2! kn!

nr=1

pr1r

kr.

16


17/47

It is known that (k1+ +kn)!/(k1! kn!)Z. ForrN we know that ifp r!(thatis p | r! but p+1 r!), then =i=1 rpi rp. Thus pr1/r, pr1/r!Zp. For

p >2 we see that pr1/rpr1/r!0 (mod p) for r >1. Hence the result followsfrom the above. For p = 2 we see that 2r1/r0 (mod 2) forr >2. Thus

m!s(n, m)

n! 2nm

k1+k2=mk1+2k2=n

(k1+k2)!

k1!k2! = m

n m

(mod 2).

Summarizing the above we prove the lemma.From Lemma 4.1 we have the following identities, which are generalizations of

Eulers identity.

Theorem 4.1. Letx, d be variables andm, nN.(i) Ifmn, then

n

r=0

n

r

(1)nr

rx+d

m

rnm =

n!

m!xm.

In particular, whenm= n we haven

r=0

n

r

(1)nr

rx+d

n

= xn.

(ii) Ifmn+ 1, thenn

r=0

n

r

(1)nr

rx+d

m

rn+1m =

n!

m!

n(n+ 1)2

xm m(m 1 2d)2

xm1

.

In particular, whenm= n+ 1 we haven

r=0

nr

(1)nrrx+d

n+ 1

=

d+

n(x

1)

2

xn

.

Proof. Observe thats(m, m) = 1 and S(n, n) = 1. Putting i = n m in Lemma4.1 we obtain (i). By (4.3) and (4.4) we have

s(n, n 1) =S(n, n 1) =n(n 1)/2 for n= 2, 3, 4, . . .Thus applying Lemma 4.1 we see that ifmn+ 1, then

nr=0

n

r

(1)nr

rx+d

m

rn+1m

= n!

m!

m

j=m1

m

k=j

kj(1)

mks(m, k)dkjS(n+ 1 m+j, n)xj

= n!

m!

S(n+ 1, n)xm +

mk=m1

k

m 1

(1)mks(m, k)dk(m1)xm1

= n!

m!

n(n+ 1)2

xm +

dm m(m 1)2

xm1

.

This yields (ii) and so the theorem is proved.17


18/47

Corollary 4.1. Let p be an odd prime, m Z and d {0, 1, . . . , p1}. Thenmp m (mod p) and

mp mp

p1k=1

1

k

km+dp

+m

dk=1

1

k (mod p).

Proof. From Theorem 4.1(i) we have

mp =

pk=0

p

k

(1)pk

km+d

p

=

mp+d

p

+

p1k=1

p

k

(1)pk

km+d

p

.

Asp1

k=11k 0 (modp), we see that

mp+d

p = (mp+d)(mp+d 1) (mp+d p+ 1)

p!

= mp

p (mp+ 1) (mp+d)((m 1)p+d+ 1) ((m 1)p+p 1)

(p 1)!

m

1 +mpd

k=1

1

k+ (m 1)p

p1k=d+1

1

k

m

1 +mpd

k=1

1

k (m 1)p

dk=1

1

k

=m1 +pd

k=1

1

k (mod p2).

Letrkbe the least nonnegative residue ofkm+dmodulop. Fork {1, 2, . . . , p1}we see that

p

k

=

p(p 1) (p k+ 1)k!

(1)k1

k p(mod p2).

Thus,p1k=1

p

k

(1)pk

km+d

p

p1

k=1

p

k(km+d)(km+d 1) (km+d p+ 1)

p!

=p

p1k=1

1

k km+d rk

p 1

(p 1)!p1i=0i=rk

(km+d i)

pp1k=1

1

k km+d rk

p =p

p1k=1

1

k

km+dp

(modp2).

18


19/47

Now putting all the above together we obtain the result.Remark 4.1 In the case d = 0, Corollary 4.1 was first found by Lerch [Ler]. For adifferent proof of Lerchs result, see [S5].

Theorem 4.2. Letp be a prime. Let f be ap-regular function. Supposem, n Nandd, tZ withd, t0. Then

nr=0

n

r

(1)rf(pm1rt+d)0 (mod pmn).

Moreover, ifAk =pkk

r=0

kr

(1)rf(r), then

n

r=0

n

r

(1)rf(pm1rt+d)

pmntnAn (mod pmn+1) ifp >2 orm= 1,

2mntnn

r=0

nr

Ar+n (mod 2

mn+1) ifp= 2 andm2.

Proof. Since f is a p-regular function, we have AkZp for k0. Set

a0=A0 and ai = (1)inr=i

s(r, i)pr

r!Ar for i= 1, 2, . . . , n.

Aspr/r!

Zp and Ar

Zp we have a0, . . . , an

Zp. From [S5, p. 197] we have

f(k)ni=0

aiki (mod pn+1) for k= 0, 1, 2, . . . .

Thus applying (4.1) and (4.2) we see that

nr=0

n

r

(1)rf(rt+d)

nr=0

n

r

(1)r

ni=0

ai(rt+d)i

=

nr=0

nr

(1)r(antnrn +bn1rn1 + +b1r+b0)

=an(t)nn! = (1)ns(n, n)pn

n!An (t)nn!

=pntnAn (mod pn+1),

whereb0, b1, . . . , bn1Zp. Thus the result is true form= 1.19


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Now assume m2. By the binomial inversion formula we have f(k) =ks=0 ks(p)sAs.Thus

n

r=0

n

r

(1)rf(pm1rt)

=

nr=0

n

r

(1)r

pm1rtk=0

pm1rt

k

(p)kAk

=

pm1ntk=0

(p)kAkn

r=0

n

r

(1)r

pm1rt

k

=

pm1ntk=n

(p)kAk (1)n n!k!

kj=n

(1)kjs(k, j)S(j, n)pm1tj (by Lemma 4.1)

=

pm1ntk=n

(p)n(1)kAkk

j=n(1)kj

s(k, j)j!

k! pkj

S(j, n)n!

j! pjn pm1tj

=Antnpmn +

pm1ntk=n+1

(p)n(1)kAk (1)kns(k, n)n!

k! pkn p(m1)ntn

+k

j=n+1

(1)kjs(k, j)j!k!

pkj S(j, n)n!j!

pjn (pm1t)j

.

By Lemma 4.2, for j, k, nN we haves(k, j)j!

k! pkj Zp and S(j, n)n!

j! pjn Zp.Hence, by the above, Lemma 4.2 and the fact (m 1)(n + 1 ) + nmn + 1 we obtain

nr=0

n

r

(1)rf(pm1rt)

pmntn

An+

pm1ntk=n+1

s(k, n)n!

k! pknAk

pmntnAn (mod p

mn+1) ifp >2,

2mntn2m

1ntk=n

nkn

Ak = 2

mntnn

r=0

nr

Ar+n (mod 2

mn+1) ifp = 2.

Thus the result holds for d= 0.Now suppose g(r) =f(r+d). By the previous argument,

nr=0

n

r

(1)rg(r)pnAn (mod pn+1).

20


21/47

Thus g is also a p-regular function. Note thatn

r=0

n

r

(1)rf(pm1rt+d) =

nr=0

n

r

(1)rg(pm1rt).

By the above we see that the result is also true for d >0. The proof is now complete.

Theorem 4.3. Let p be a prime, k,m,n,t N and d {0, 1, 2, . . . }. Let f be ap-regular function. Then

f(ktpm1 +d)n1r=0

(1)n1r

k 1 rn 1 r

k

r

f(rtpm1 +d) (mod pmn).

Moreover, settingAs = pss

r=0

sr

(1)rf(r) we then have

f(ktpm1 +d) n1r=0

(1)n1r

k 1 rn 1 r

k

r

f(rtpm1 +d)

pmnkn(t)

nAn (mod pmn+1) ifp >2 orm= 1,

2mnkn

(t)nnr=0 nrAr+n (mod 2mn+1) ifp= 2 andm2.

Proof. From [S4, Lemma 2.1] we know that for any function F,

(4.5)

F(k) =

n1r=0

(1)n1r

k 1 rn 1 r

k

r

F(r)

+k

r=n

k

r

(1)r

rs=0

r

s

(1)sF(s),

where the second sum vanishes when k < n.Now taking F(k) =f(ktpm1 +d) we obtain

f(ktpm1 +d) =n1r=0

(1)n1r

k 1 rn 1 r

k

r

f(rtpm1 +d)

+k

r=n

k

r

(1)r

rs=0

r

s

(1)sf(stpm1 +d).

By Theorem 4.2 we havek

r=n

k

r

(1)r

rs=0

r

s

(1)sf(stpm1 +d)

(1)n

kn

ns=0

ns

(1)sf(stpm1 +d)

kn

pmn(t)nAn (modpmn+1) ifp >2 or m = 1,

kn

2mn(t)nnr=0 nrAr+n (mod 2mn+1) ifp = 2 and m2.

Now combining the above we prove the theorem.Puttingn = 1, 2, 3 and d= 0 in Theorem 4.3 we deduce the following result.

21


22/47

Corollary 4.2. Letp be a prime, k,m, tN. Letf be ap-regular function. Then(i) ([S5, Corollary 2.1]) f(kpm1)f(0) (mod pm).(ii) f(ktpm1)kf(tpm1) (k 1)f(0) (mod p2m).(iii) We have

f(ktpm1)k(k 1)2

f(2tpm1) k(k 2)f(tpm1)

+(k 1)(k 2)

2 f(0) (mod p3m).

(iv) We have

f(kpm1)

f(0) k(f(0) f(1))pm1 (mod pm+1) ifp > 2 orm= 1,f(0) 2m2k(f(2) 4f(1) + 3f(0)) (mod 2m+1) ifp= 2 andm2.

Theorem 4.4. Letp be a prime and letf be ap-regular function. LetnN.(i) Ford, xZp andm {0, 1, . . . , n 1} we have

nk=0

n

k

(1)k

kx+d

m

f(k)0 (mod pnm).

(ii) We have

n

k=1

n

k(1)kf(k 1) f(pn1 1) (mod pn).

Proof. From [S5, Theorem 2.1] we know that there are a0, a1, . . . , anm1 Zsuch that

f(k)anm1knm1 + +a1k+a0 (mod pnm) for k= 0, 1, 2, . . .

Thus applying Lemma 4.1 and (4.1) we have

n

k=0

n

k

(1)k

kx+d

m

f(k)

n

k=0

n

k

(1)k

kx+d

m

nm1i=0

aiki

=nm1i=0

ai

nk=0

n

k

(1)k

kx+d

m

ki = 0 (mod pnm).

This proves (i).22


23/47

Now we consider (ii). By [S5, Theorem 2.1] there are a0, a1, . . . , an1 Zp suchthats!as/p

s Zp (s= 0, 1, . . . , n 1) and

f(k)an1kn1 + +a1k+a0 (mod pn) for k= 0, 1, 2, . . .

Note that ps1

/s!Zp for sN. We then have a1 an10 (modp). Letan1(k 1)n1 + +a1(k 1) +a0 = bn1kn1 + +b1k+b0.

Then clearlyb1 bn10 (mod p) and

f(k 1)bn1kn1 + +b1k+b0 (mod pn) for k= 1, 2, 3, . . .

Thusf(pn1 1)bn1(pn1)n1 + +b1pn1 +b0b0 (mod pn).

Hence, applying (4.1) we haven

k=1

n

k

(1)kf(k 1)

nk=1

n

k

(1)k(bn1kn1 + +b1k+b0)

=

n1i=1

bi

nk=0

n

k

(1)kki +b0

nk=1

n

k

(1)k

=b0 f(pn1 1) (mod pn).

So the theorem is proved.

5. Congruences for pBk(pm)+b(x) and pBk(pm)+b, (mod pmn).For given prime p and t Zp we recall thattp denotes the least nonnegative

residue oftmodulo p.

Theorem 5.1. Letp be a prime, and k,m,n,t,bZ withm, n1 andk,b, t0.LetxZp andx = (x+ xp)/p. Then

pBkt(pm)+b(x) pkt(pm)+bBkt(pm)+b(x

)

n1

r=0(1)n1r

k 1 rn 1 r

k

r

pBrt(pm)+b(x) prt(p

m)+bBrt(pm)+b(x)

(b,n,p)kn

(t)npmn1 (mod pmn) ifp >2 orm= 1,

0 (mod 2mn) ifp= 2 andm2,

where

(b,n,p) =

1 ifp= 2 andn {1, 2, 4, 6, . . . }or ifp >2, p 1|b andp 1|n,

0 otherwise.23


24/47

Proof. From [S4, Theorem 3.1] we know that

nk=0

n

k

(1)k

pBk(p1)+b(x) pk(p1)+bBk(p1)+b(x)

pn1(b,n,p) (mod pn).

Set f(k) =ppBk(p1)+b(x) pk(p1)+bBk(p1)+b(x)

. Then

nk=0

nk

(1)kf(k)

(b,n,p)pn (mod pn+1).Thusf is ap-regular function. Hence appealing to Theorem4.3 we have

f(ktpm1) n1r=0

(1)n1r

k 1 rn 1 r

k

r

f(rtpm1)

pmnkn

(t)n(b,n,p) (mod pmn+1) ifp >2 or m = 1,

2mn

kn

(t)n

nr=0

nr

(b, n+r, 2) (mod 2mn+1) ifp = 2 and m2.

Note that(b, n+r, 2) =

1 ifn+r {1, 2, 4, 6, . . . },0 ifn+r {3, 5, 7, . . . }.

We then have

nr=0

n

r

(b, n+r, 2)

=

(b, 1, 2) +(b, 2, 2) = 1 + 10 (mod 2) ifn = 1,n

r=02|n+r

nr= 2

n1 0 (mod 2) if n >1.

Thusf(ktpm1)

p

n1r=0

(1)n1r

k 1 rn 1 r

k

r

f(rtpm1)

p

pmn1kn

(t)n(b,n,p) (modpmn) ifp >2 orm= 1,

0 (mod 2mn) ifp= 2 and m2.This is the result.

Corollary 5.1. Let p be a prime, and k,m, b Z with k, m 1 and b 0. LetxZp andx

= (x+ xp)/p. Supposep >2 orm >1. Then

pBk(pm)+b(x)

3 (mod 4) ifp= m = 2, k= 1 andb= 0,

pBb(x) pbBb(x) (mod pm) otherwise.Proof. Putting n= t = 1 in Theorem 5.1 we see that

pBk(pm)+b(x) pk(pm)+bBk(pm)+b(x

)pBb(x) pbBb(x) (mod pm).24


25/47

Ifp = m = 2, k = 1 and b = 0, then pBk(pm)+b(x) = 2B2(x) = 2(x2 x+ 16 )

3 (mod 4). Otherwise, we havek(pm) + bm + 1 and sopk(pm)+bBk(pm)+b(x)0 (mod pm). Thus the result follows from the above.

In the case p >2, Corollary 5.1 has been proved by the author in [S4].Let be a primitive Dirichlet character of conductor m. The generalized Bernoulli

number Bn, is defined bymr=1

(r)tert

emt 1 =n=0

Bn,tn

n!.

Let0 be the trivial character. It is well known that (see [W])

B1,0 = 1

2, Bn,0 =Bn(n= 1) and Bn,= mn1

mr=1

(r)Bn r

m

.

If is nontrivial and nN, then clearlymr=1(r) = 0 and so

Bn,n

=mn1mr=1

(r)Bn( rm ) Bn

n +

Bnn

= mn1

mr=1

(r)Bn( rm) Bn

n .

When p is a prime with pm, by [S4, Lemma 2.3] we have (Bn(rm ) Bn)/n Zp.

Thus Bn,/nis congruent to an algebraic integer modulo p.

Lemma 5.1. Letp be a prime and letb be a nonnegative integer.(i) ([S5, Theorem 3.2], [Y2]) Ifp1b,xZp andx = (x+xp)/p, thenf(k) =

(Bk(p1)+b(x) pk(p1)+b1Bk(p1)+b(x))/(k(p 1) +b) is apregular function.(ii) ([S5, (3.1), Theorem 3.1 and Remark 3.1]) If a, b N andp a, then f(k) =

(1

pk(p1)+b1)(ak(p1)+b

1)Bk(p1)+b/(k(p

1) +b) is ap-regular function.

(iii) ([Y3, Theorem 4.2], [Y1, p. 216], [F], [S5, Lemma 8.1(a)]) If b, m N, p mand is a nontrivial primitive Dirichlet character of conductor m, then f(k) =(1 (p) pk(p1)+b1)Bk(p1)+b,/(k(p 1) +b) is apregular function.

(iv) ([S5, Lemma 8.1(b)]) If m N, p m and is a nontrivial Dirichlet char-acter of conductorm, thenf(k) = (1 (p)pk(p1)+b1)pBk(p1)+b, is apregularfunction.

From Lemma 5.1 and Theorem 4.3 we deduce the following theorem.

Theorem 5.2. Letp be a prime, k,n,s, tN andb {0, 1, 2, . . . }.(i) Ifp 1b, xZp andx = (x+ xp)/p, then

Bktps1(p1)+b(x) pktps1

(p1)+b1Bktps1(p1)+b(x)ktps1(p 1) +b

n1r=0

(1)n1r

k 1 rn 1 r

k

r

Brtps1(p1)+b(x) prtps1(p1)+b1Brtps1(p1)+b(x

)

rtps1(p 1) +b (mod psn).

25


26/47

(ii) Ifa, bN andpa, then

1 pktps1(p1)+b1aktps1(p1)+b 1 Bktps1(p1)+b

ktps1(p 1) +b

n1r=0

(1)n1rk 1 rn 1 rkr1 prtps1(p1)+b1 artps1(p1)+b 1 Brtps1(p1)+b


(iii) If b, m N, p m and is a nontrivial primitive Dirichlet character ofconductorm, then

(1 (p)pktps1(p1)+b1)Bktps1(p1)+b,ktps1(p

1) +b

n1r=0

(1)n1r

k 1 rn 1 r

k

r

(1 (p)prtps1(p1)+b1)Brtps1(p1)+b,


(iv) IfmN, p m and is a nontrivial Dirichlet character of conductorm, then

1 (p)pktps1(p1)+b1

pBktps1(p1)+b,

n1r=0

(1)n1rk 1 rn 1 rkr 1 (p)prtps1(p1)+b1pBrtps1(p1)+b, (mod psn).

Remark 5.1 Theorem 5.2 can be viewed as generalizations of some congruencesin [S5]. In the case n = 1, Theorem 5.2(i) was given by Eie and Ong [EO], andindependently by the author in [S5, p. 204]. In the case s = t = 1, Theorem 5.2(i)was announced by the author in [S4] and proved in [S5], and Theorem 5.2(iii) (in thecase p 1 b) and Theorem 5.2(iv) were also given in [S5]. When n = 1, Theorem5.2(iii) was given in [W, p. 141].

Combining Lemma 5.1 and Corollary 4.2(iv) we obtain the following result.

Theorem 5.3. Letp be an odd prime, k, sN andb {0, 1, 2, . . . }.(i) Ifp 1b, xZp andx = (x+ xp)/p, then

Bk(ps)+b(x)

k(ps) +b (1 kps1) Bb(x) p

b1Bb(x)

b +kps1

Bp1+b(x)

p 1 +b (mod ps+1).

26


27/47

(ii) If b, m N, p m and is a nontrivial primitive Dirichlet character ofconductorm, then

Bk(ps)+b,

k(ps) +b(1 kps1)

1 (p)pb1

Bb,

b +kps1

Bp1+b,

p 1 +b (mod ps+1).

(iii) IfmN, p m and is a nontrivial Dirichlet character of conductorm, then(1 (p)pk(ps)+b1)pBk(ps)+b,(1 kps1)1 (p)pb1pBb,

+kps1

1 (p)pp2+bpBp1+b, (mod ps+1).Corollary 5.2. Letp be an odd prime andk,s, bN with2|b andp 1b. Then

Bk(ps)+bk(ps) +b

(1 kps1)(1 pb1) Bbb

+kps1 Bp1+bp

1 +b

(mod ps+1).

Theorem 5.4. Letp be a prime, a, nN andpa.(i) There are integersb0, b1, , bn1 such that

1 pk(p1)1ak(p1) 1 Bk(p1)

k(p 1)bn1kn1 + +b1k+b0 (mod pn) for k= 1, 2, 3, . . .

(ii) Ifp >2 orn >2, then

nk=1

nk

(1)k

(1 pk(p1)1

)(ak(p1)

1)Bk(p1)

k(p 1) 1

a(p

n)

pn (mod pn

).

Proof. Suppose bN. From Lemma 5.1(ii) we know that

f(k) =

1 pk(p1)+b1

ak(p1)+b 1 Bk(p1)+b

k(p 1) +bis a p-regular function. Hence taking b = p 1 and applying [S5, Theorem 2.1] weknow that there exist integers a0, a1, . . . , an1 such that

1 p(k+1)(p1)1a(k+1)(p1)

1 B(k+1)(p1)

(k+ 1)(p 1)an1kn1 + +a1k+a0 (modpn) for k= 0, 1, 2, . . .That is,

1 pk(p1)1ak(p1) 1 Bk(p1)

k(p 1)an1(k 1)n1 + +a1(k 1) +a0 (mod pn) for k= 1, 2, 3, . . .

27


28/47

On setting

an1(k 1)n1 + +a1(k 1) +a0=bn1kn1 + +b1k+b0we obtain (i).

Now we consider (ii). Supposep >2 or n >2. Since f(k) is a p-regular function,

by Theorem 4.4(ii) we have

nk=1

n

k

(1)k(1 p(k1)(p1)+b1)(a(k1)(p1)+b 1) B(k1)(p1)+b

(k 1)(p 1) +b

(1 p(pn11)(p1)+b1)(a(pn11)(p1)+b 1) B(pn11)(p1)+b(pn1 1)(p 1) +b (mod p

n).

Substitutingbbyp 1 +b we see that for b0,

(5.1)

n

k=1

n

k(1)k(1 pk(p1)+b1)(ak(p1)+b 1) Bk(p1)+b

k(p

1) +b

(1 p(pn)+b1)(a(pn)+b 1)B(pn)+b(pn) +b

(mod pn).

By Corollary 5.1 we have pB(pn)p 1 (mod pn). Thus taking b= 0 in (5.1) andnoting that (pn)n+ 1 we obtain

nk=1

n

k

(1)k(1 pk(p1)1)(ak(p1) 1)Bk(p1)

k(p 1)

(1 p(pn)1)(a(pn) 1) B(pn)(pn)

=(1 p(pn)1) a(pn) 1

pn pB(pn)

p 1 a(p

n) 1pn

(mod pn).

This completes the proof of the theorem.

6. Congruences forn

k=0

nk

(1)kpBk(p1)+b(x) (mod pn+1).

ForaN and bZ we define (a|b) = 1 or 0 according as a|b orab.Lemma 6.1. Letp be an odd prime andnN. Then

n

s=1sn+1 (mod p1)

n

s (p 1|n) (mod p).Proof. Let n0 {1, 2, . . . , p 1} be such that nn0 (modp 1). Since Glaisher

(see [D]) it is well known that

ns=0

sr (mod p1)

n

s

n0s=0

sr (mod p1)

n0s

(mod p) for rZ.

28


29/47

From [S1] we know that

ns=0

sr (mod p1)

n

s

=

ns=0

snr (mod p1)

n

s

.

Thusn

s=0sn+1 (mod p1)

n

s

=

ns=0

s1 (mod p1)

n

s

n0s=0

sp2 (mod p1)

n0s

=

p 1 1 (mod p) ifn0 = p 1,1 (mod p) ifn0 = p 2,0 (mod p) ifn0 < p 2.

Hence

ns=1

sn+1 (mod p1)

ns

=

ns=0

sn+1 (mod p1)

ns

(p1|n+1) (p1|n) (modp).

This proves the lemma.

Proposition 6.1. Letp be an odd prime, nN andxZp. Letb be a nonnegativeinteger. Then

nk=0

n

k

(1)k

pBk(p1)+b(x) pk(p1)+bBk(p1)+b

x+ xpp

p1j=0

j=xp

(x+j)bnpnBn

(x+j)p (x+j)p(p 1)

+pn(b,n,p) (mod pn+1),

where

(b,n,p) =

(n b)T n ifp 1|b andp 1|n,(n b)T ifp 1b andp 1|n,b n ifp 1|b andp 1|n+ 1,0 otherwise

and

T =p1j=0

j=xp

(x+j)p1+b (x+j)bp

.

Proof. Let

Sn=

nk=0

n

k

(1)k


x+ xpp

.

29


30/47

From [S4, p.157] we know that

Sn=

n(p1)+br=0

prBr

p1j=0

j=xp

nk=0

n

k

(1)k

k(p 1) +b

r

(x+j)k(p1)+br.

By [S5, p.199] we know that for any functionsf and g we have

(6.1)

nk=0

n

k

(1)kf(k)g(k)

=n

s=0

n

s

nsi=0

n s

i

(1)if(i+s)

sj=0

s

j

(1)jg(j).

Now taking f(k) =k(p1)+b

r

and g(k) = ak(p1)+br (a= 0) in (6.1) we obtainn

k=0

n

k

(1)k

k(p 1) +b

r

ak(p1)+br

=n

s=0

n

s

nsi=0

n s

i

(1)i

(i+s)(p 1) +b

r

sj=0

s

j

(1)jaj(p1)+br

=n

s=0

n

s

abr(1 ap1)s

nsi=0

n s

i

(1)i

i(p 1) +s(p 1) +b

r

.

Thus applying the above and Lemma 4.1 we have

Sn=

n(p1)+br=0

prBr

p1j=0

j=xp

ns=0

n

s

(x+j)br

1 (x+j)p1s

nsi=0

n s

i

(1)i

i(p 1) +s(p 1) +b

r

=

p1j=0

j=xp

ns=0

ns1 (x+j)

p1

psn(p1)+b

r=nspr+sBr (x+j)br

nsi=0

n s

i

(1)i

i(p 1) +s(p 1) +b

r

.

SincepBrZp and sopr+sBr0 (mod pn+1) forrn s + 2, by Theorem 4.1 we30


31/47

have

n(p1)+br=ns

(x+j)brpr+sBr

nsi=0

n s

i

(1)i

i(p 1) +s(p 1) +b

r

(x+j)b(ns)pnBnsnsi=0

n s

i

(1)ii(p 1) +s(p 1) +b

n s

+ (x+j)b(ns+1)pn+1Bns+1

nsi=0

n s

i

(1)i

i(p 1) +s(p 1) +b

n s+ 1

= (x+j)b(ns)pnBns (1 p)ns + (x+j)b(ns+1)pn+1Bns+1 (s(p 1) +b+ (n s)(p 2)/2)(1 p)ns

(x+j)b(ns)(1 p)nspnBns+ (x+j)b(ns+1)(b

n)pn+1Bns+1 (mod p

n+1).

Thus,

Snp1j=0

j=xp

ns=0

n

s

1 (x+j)p1p

s(x+j)bn+s(1 p)nspnBns

+ (x+j)bn+s1(b n)pn+1Bns+1

=

p1j=0

j=xp

(x+j)bn(1 p)npnn

s=0

n

s

1 (x+j)p1p

x+j1 p

sBns

+

p1j=0

j=xp

ns=0

n

s

1 (x+j)p1p

s(x+j)bn+s1(b n)pn+1Bns+1

p1j=0

j=xp

(x+j)bn(1 p)npnBn(xj) +p1j=0

j=xp

ns=0

p1|ns+1

n

s

1 (x+j)p1p

s

(x+j)bn+s1(n b)pn (modpn+1),

where

xj = (x+j)p (x+j)

p(p 1) .

In the last step we use the facts

Bn(t) =

ns=0

n

s

tsBns and pBk (p 1|k) (mod p) (k1).

31


32/47

ForaZ, using Lemma 6.1 and Fermats little theorem we see thatn

s=0sn+1 (mod p1)

n

s

as =

ns=1

sn+1 (mod p1)

n

s

as +(p 1|n+ 1)

an+1n

s=1sn+1 (mod p1)

n

s

+(p 1|n+ 1)

(p 1|n)an+1 +(p 1|n+ 1)

=

an+1 a (modp) ifp 1|n,1 (mod p) ifp 1|n+ 1,0 (mod p) ifp 1n and p 1n+ 1.

We also note that (see [S5, (5.1)])

(6.2)

p1j=0

j=xp

(x+j)b p1r=1

rb (p 1|b) (mod p).

Thus

p1j=0

j=xp

ns=0

p1|ns+1

n

s

1 (x+j)p1p

s(x+j)bn+s1(n b)pn

pn(n b) p1j=0

j=xp

(x+j)bn

s=0sn+1 (mod p1)

ns

1 (x+j)p1

p

s

pn(n b)p1j=0

j=xp

(x+j)b((x+j)p1 1)/p(mod pn+1)

ifp 1|n,

pn(n b)p1

j=0j=xp(x+j)b (p 1|b)(n b)pn (mod pn+1)

ifp 1|n+ 1,0 (mod pn+1) ifp 1n and p 1n+ 1.

On the other hand, fortZp we have Bn(t) BnZp (cf. [S4, Lemma 2.3]) and so

(np)pnBn(xj) npn+1Bn

npn (mod pn+1) ifp 1|n,0 (mod pn+1) ifp 1n.

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33/47

Thus applying (6.2) we get

p1j=0

j=xp

(x+j)bn (np)pnBn(xj)

p1j=0

j=xp

(x+j)b npn npn(p 1|b) (modpn+1) ifp 1|n,

0 (mod pn+1) ifp 1n.

Hence, by the above and the fact (1 p)n 1 np(mod p2) we obtainp1

j=0j=xp

(x+j)bn(1 p)npnBn(xj) p1

j=0j=xp

(x+j)bnpnBn(xj)

p1j=0

j=xp

(x+j)bn (np)pnBn(xj)

npn (mod pn+1) ifp 1|b and p 1|n,

0 (mod pn+1) ifp 1b or p 1n.

Now combining the above we see that

Snp1j=0

j=xp

(x+j)bnpnBn(xj)

npn + (n b)pnT (mod pn+1) ifp 1|b andp 1|n,pn(n b)T (modpn+1) ifp 1b andp 1|n,pn(b n) (mod pn+1) ifp 1|b andp 1|n+ 1,0 (mod pn+1) otherwise.

This is the result.Remark 6.1 When p= 2, b1 and n2, setting (b,n,p) =b n we can showthat the result of Proposition 6.1 is also true.Theorem 6.1. Letp be a prime greater than3, xZp, nN, n0, 1 (mod p 1)andb {0, 1, 2, . . . }. Letn0 be given bynn0 (modp1)andn0 {2, 3, . . . , p2}.Set

Sn=

nk=0

n

k

(1)k


x+ xpp

.

33


34/47

Then

Sn

nn0

Sn0pn0 + (n+2)b2pn (mod pn+1) ifp 1|b andp 1|n+ 1,

nn0

Sn0pn0pn (mod pn+1) ifp 1b orp 1n+ 1.

Proof. Sincep 1 n we know that Bn/nZp. FortZp, by [S4, Lemma 2.3]we have (Bn(t) Bn)/nZp. Thus

Bn(t)

n =

Bn(t) Bnn

+Bn

n Zp.

Asn0, 1 (mod p 1), by [S5, Corollary 3.1] we have

Bn(t)

n Bn0(t) p

n01Bn0

(t+ tp)/p

n0 Bn0(t)

n0(mod p).

Set xj = ((x+j)p (x+j))/(p(p 1)). Then xj Zp. ThusBn(xj)/n Zp andBn(xj)/nBn0(xj)/n0 (mod p). From Proposition 6.1 and the above we see that

Snpn

p1j=0

j=xp

(x+j)bnBn(xj) + (b n)(p 1|b)(p 1|n+ 1)

np1j=0

j=xp

(x+j)bn0Bn0(xj)

n0+ (b n)(p 1|b)(p 1|n+ 1) (mod p)

and so

Sn0pn0

n0p1j=0

j=xp

(x+j)bn0Bn0(xj)

n0+ (b n0)(p 1|b)(p 1|n+ 1) (mod p).

Thus

Snpn

nn0

Sn0pn0

(b n0)(p 1|b)(p 1|n+ 1)+ (b n)(p 1|b)(p 1|n+ 1)

= n

n0 Sn0

pn0+b

1 n

n0

(p 1|b)(p 1|n+ 1)

nn0

Sn0pn0

+b

1 +n

2

(p 1|b)(p 1|n+ 1) (mod p).

This proves the theorem.34


35/47

Theorem 6.2. Let p be an odd prime, x Zp, b, n Z with n 1 and b 0. Ifp|n andp 1n, then

n

k=0

n

k

(1)k


x+ xp

p

bpn (mod pn+1) ifp 1|b andp 1|n+ 1,0 (mod pn+1) ifp 1b orp 1n+ 1.

Proof. As p1 n and p| n, for t Zp we see that Bn(t)/n Zp and soBn(t) =nBn(t)/n0 (modp). Thus the result follows from Proposition 6.1.Theorem 6.3. Letp be an odd prime, nN andb {0, 2, 4, . . . }. Ifp(p 1)| n,then

nk=0

n

k

(1)k(1 pk(p1)+b1)pBk(p1)+b

pn1

2pn (mod pn+1) ifp

1|

b,

0 (mod pn+1) ifp 1b.Proof. From Proposition 6.1 we see that

nk=0

n

k

(1)k(1 pk(p1)+b1)pBk(p1)+b

p1j=1

jbnpnBn

jp jp(p 1)

bT pn (modpn+1),

where

T =

p1j=1

jp1+b jbp

.

For p >3 and mN, from [S5, (5.1)] we havep1j=1

jm pBm+ p2

2mBm1+

p3

6m(m 1)Bm2 (mod p3).

Ifm4 is even, then Bm1 = 0 andpBm2Zp. Thus

(6.3)

p1j=1j

m

pBm (mod p2

) for m= 2, 4, 6, . . .

Hence

T

pBp1+bpBbp (mod p) ifp > 3 and b >0,

pBp1(p1)p (mod p) ifp > 3 and b = 0,

22+b2b

3 = 2b (1)b = 1 3B223 (mod 3) ifp= 3.

35


36/47

Ifp >3 and b = k(p 1) for some kN, by [S4, Corollary 4.2] we have

(6.4) pBb = pBk(p1)kpBp1 (k 1)(p 1) (mod p2)

and

pBp1+b=pB(k+1)(p1)(k+ 1)pBp1 k(p 1) (mod p2).Thus

T pBp1+bpBbp

pBp1 (p 1)p

(mod p).

Ifp >3 and p 1b, by Kummers congruences we have

Bp1+bp 1 +b

Bbb

(mod p) and so Bp1+b(b 1) Bbb

(mod p).

Thus

T pBp1+bpBbp

b 1b

Bb Bb=Bbb

(mod p).

Summarizing the above we have

(6.5) T

pBp1(p1)p (modp) ifp 1|b,

Bbb (mod p) ifp 1b.

Asp(p 1)|n, from Corollary 5.1 we have pBn(x)p 1 (mod p2) for xZp.Note that jn 1 (mod p2) forj = 1, 2, . . . , p 1. Combining the above we obtain

nk=0

n

k

(1)k(1 pk(p1)+b1)pBk(p1)+b

p1j=1

jbnpn1 pBn jp j

p(p 1)

bT pn

p1j=1

jbpn1(p 1) bT pn (mod pn+1).

From (6.3) and (6.4) we see that

p1j=1

jb

pBb bp1pBp1 ( bp1 1)(p 1) (mod p2)ifp >3, b >0 and p 1|b,

pBb (mod p2) ifp >3 and p 1b,

p 1 (mod p2) ifp >3 and b = 0,1 + (1 + 3)

b2 2 + 3b2 2 + 6b (mod 9) ifp= 3.

36


37/47

That is,

p1

j=1jb

bp1 (pBp1 (p 1)) +p 1 (mod p2) ifp 1|b,pBb (mod p2) ifp

1b.

Hence

nk=0

n

k

(1)k(1 pk(p1)+b1)pBk(p1)+b

pn1(p 1)p1j=1

jb bT pn

pn1

(b(pBp1 (p 1)) + (p 1)2

) pn1

b(pBp1 (p 1))=pn1(p 1)2 pn1 2pn (mod pn+1) ifp 1|b,

pn1(p 1) pBb bpn (Bbb ) =pn+1Bb0 (mod pn+1) ifp 1b.

This completes the proof.

Theorem 6.4. Letp be a prime greater than3, xZp, nN, n0, 1 (mod p 1)andb {0, 1, 2, . . . }. Letn0 be given bynn0 (modp1)andn0 {2, 3, . . . , p2}.Let

f(k) =pBk(p1)+b(x) pk(p1)+bBk(p1)+b

x+ xpp

.

Then fork= 0, 1, 2, . . . we have

f(k)n1r=0

(1)n1r

k 1 rn 1 r

k

r

f(r) +

n

n0n0

s=0

n0s

(1)sf(s)

pn0

k

n

(p)n

+(p 1|n+ 1)(p 1|b)(n+ 2)b

2 k

n k

n+ 1 (p)n (modpn+1).

Proof. From [S4, Theorem 3.1] we have

mk=0

m

k

(1)kf(k)pm1(p 1|m)(p 1|b) (mod pm) for mN.

37


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Thus applying [S4, Lemma 2.1], Theorem 6.1 and the above we see that

f(k) n1r=0

(1)n1r

k 1 rn 1 r

k

r

f(r)

=

kr=n

kr

(1)rr

s=0

rs

(1)sf(s)

k

n

(1)n

ns=0

n

s

(1)sf(s) +

k

n+ 1

(1)n+1

n+1s=0

n+ 1

s

(1)sf(s)

k

n

(1)npn

nn0

n0

s=0

n0s

(1)sf(s)

pn0+

(n+ 2)b

2 (p 1|n+ 1)(p 1|b)

+

k

n+ 1

(1)n+1pn(p 1|n+ 1)(p 1|b) (mod pn+1).

This yields the result.

Corollary 6.1. Letk, nN.(i) Ifn2 (mod 4), then

(5 54k)B4kn1r=0

(1)n1r

k 1 rn 1 r

k

r

(5 54r)B4r

+ 3n

k

n

5n (mod 5n+1)

and

(5 54k+2)B4k+2n1r=0

(1)n1rk 1 rn 1 rkr(5 54r+2)B4r+2 n

k

n

5n (mod 5n+1).

(ii) Ifn3 (mod 4), then

(5 54k)B4kn1r=0

(1)n1r

k 1 rn 1 r

k

r

(5 54r)B4r

+ k

n+ 15n (mod 5n+1)

and

(5 54k+2)B4k+2n1r=0

(1)n1r

k 1 rn 1 r

k

r

(5 54r+2)B4r+2

+n

k

n

5n (mod 5n+1).

38


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40/47

Theorem 7.1. Letp be an odd prime, k,m,n,tN andb {0, 2, 4, . . . }. Then

1 (1) p12 pktpm1(p1)+bEktpm1(p1)+b

n1

r=0

(

1)n1r

k 1 r

n 1 rk

r1

(

1)

p12 prtp

m1(p1)+b Ertpm1(p1)+b (mod pmn).

Puttingn = 1, 2, 3 and t= 1 in Theorem 7.1 we obtain the following result.

Corollary 7.1. Letp be an odd prime, k, mN andb {0, 2, 4, . . . }. Then(i) ([C, p. 131]) Ek(pm)+b

1 (1) p12 pbEb (mod pm).

(ii) Ek(pm)+bkE(pm)+b (k 1)

1 (1) p12 pbEb (mod p2m).(iii) We have

Ek(pm)+b

k(k 1)2

E2(pm)+b

k(k

2)1 (1)

p12 p(p

m)+bE(pm)+b+

(k 1)(k 2)2

1 (1)p12 pbEb (mod p3m).

From Lemma 7.1 and Corollary 4.2(iv) we have:

Theorem 7.2. Letp be an odd prime, k, mN andb {0, 2, 4, . . . }. Then

Ek(pm)+b(1 kpm1)(1 (1)p12 pb)Eb+kp

m1Ep1+b (mod pm+1).

Corollary 7.2. Letp be an odd prime andk, mN. Then

Ek(pm)

kpm1Ep1 (mod pm+1) ifp1 (mod 4),2 +kpm1(Ep1 2) (mod pm+1) ifp3 (mod 4).

From [S5, Theorem 2.1] and Lemma 7.1 we have:

Theorem 7.3. Letp be an odd prime, nN andb {0, 2, 4, . . . }. Then there areintegersa0, a1, . . . , an1 such that

(1 (1) p12 pk(p1)+b)Ek(p1)+ban1kn1 + +a1k+a0 (mod pn)for every k = 0, 1, 2, . . . Moreover, if p n, then a0, a1, . . . , an1 (mod pn) areuniquely determined.

As examples, we have

(1 + 32k)E2k 12k+ 2 (mod 33),(7.2)(1 54k)E4k 750k3 + 1375k2 620k (mod 55),(7.3)(1 54k+2)E4k+21000k3 + 1500k2 + 540k+ 24 (mod 55).(7.4)

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Theorem 7.4. LetnN andb {0, 2, 4, . . . }. Supposen N and2n1 n 1

and thusf(k) =E2k+b is a2regular function.Proof. Suppose nN and 2n1 n n, thus n < nand hence 2n n n+ 1. Therefore, forn 3 we have

nk=0

nk

(1)kE2k+b

0 (mod 2n+1). AsE0 E2 = 1 (1) = 2 and E0 2E2+E4 = 1 2(1) + 5 = 8,applying (7.6) and the above we see that E

bE

b+2 E

0E

2 = 2 (mod 8) and

Eb 2Eb+2+Eb+40 (mod 8).So the result follows.Theorem 7.5. Suppose k,m,n,tN andb {0, 2, 4, . . . }. ForsN letsN begiven by2s1 s


44/47

From Corollary 7.3 and the proof of Theorem 4.2 we know that

nr=0

n

r

(1)rE22m1rt+b

=Antn 2mn +2m1ntr=n+1

(2)n(1)rAr (1)rns(r, n)n!r!

2rn 2(m1)ntn

+r

j=n+1

(1)rjs(r, j)j!r!

2rj S(j, n)n!j!

2jn (2m1t)j

.

By Lemma 4.2, for n + 1jr we haves(r, j)j!

r! 2rj ,

S(j, n)n!

j! 2jn Z2 and s(r, n)n!

r! 2rn

n

r n

(mod 2).

As 2n+1n+1

|Ar forr

n+ 1, by the above we obtain

(7.7)

nr=0

n

r

(1)rE2mrt+b2mnAntn 2mntnen (mod 2mn+n+1n+1)

and so

(7.8)n

r=0

n

r

(1)rE2mrt+b0 (mod 2mn+nn).

Forrn + 1 we havemr+ r rm(n + 1 ) + n + 1 n+1mn + n + 2 n+1.Thus, ifrn+ 1, by (7.8) we have

(7.9)r

s=0

r

s

(1)sE2mst+b0 (mod 2mn+n+2n+1).

By (4.5) we have

E2mkt+b =

n1r=0

(1)n1r

k 1 rn 1 r

k

r

E2mrt+b

+k

r=n

k

r

(1)r

r

s=0

r

s

(1)sE2mst+b.

Hence, applying (7.9) we obtain

(7.10)

E2mkt+bn1r=0

(1)n1r

k 1 rn 1 r

k

r

E2mrt+b

k

n

(1)n

ns=0

n

s

(1)sE2mst+b (mod 2mn+n+2n+1).

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45/47

In view of (7.7), we get

E2mkt+bn1r=0

(1)n1r

k 1 rn 1 r

k

r

E2mrt+b

+k

n

(1)n

2mn

tn

en (mod 2mn+n+1n+1

).

Now assume m2. Then (m 1)(n+ 1) +nmn+ 1. From the above we seethat

nr=0

n

r

(1)rE2mrt+b

2mnAntn +2m1ntr=n+1

(2)n(1)rAr(1)rns(r, n)n!

r! 2rn 2(m1)ntn

2mntnAn+ 2m1ntr=n+1

nr n

Ar2mntn n+2

r=n

nr n

Ar

2mntn

en+nen+1+

n

2

en+2

(mod 2mn+n+2n+1).

This together with (7.10) yields the remaining result. Hence the proof is complete.As 2nn |en and n + 1 n+1n n, by Theorem 7.5 we have:

Corollary 7.4. Let k,m,n,t N and b {0, 2, 4, . . . }. Let N be given by21 n


46/47

Corollary 7.6. Letk, mN, m2 andb {0, 2, 4, . . . }. Then

E2mk+bkE2m+b (k 1)Eb+ 22mk(k 1) (mod 22m+2).

Taking m = 2 and b = 0, 2 in Corollary 7.6 we get:

Corollary 7.7. ForkN we have

E4k

4k+ 1 (mod 64) ifk0, 1 (mod 4),4k+ 33 (mod 64) ifk2, 3 (mod 4)

and

E4k+2

4k 1 (mod 64) ifk0, 1 (mod 4),4k 33 (mod 64) ifk2, 3 (mod 4).

Corollary 7.8. Letk, mN, m2 andb {0, 2, 4, . . . }. Letk = 0 or1 accordingas4k 3 or4|k 3. Then

E2mk+b

k

2

E2m+1+b k(k 2)E2m+b+

k 1

2

Eb+ 2

3m+1k (mod 23m+2).

Proof. Observe thate3 = 10, e4 = 104, e5 = 1816 andk3

k (mod 2). Takingn= 3 and t = 1 in Theorem 7.5 we obtain the result.

Taking m = 2, b = 0, 2 in Corollary 7.8 and noting that E8 105 (mod 256),E10 89 (mod 256) we deduce:Corollary 7.9. LetkN andk = 0 or1 according as4k 3 or4|k 3. Then

E4k48k244k+1+128k (mod 256)andE4k+216k276k1+128k (mod 256).

Remark 7.2 Let{Sn} be given by (3.1). From Remark 3.1 we know that (1)kSkis a 2-regular function and hence f(k) = (1)k+bSk+b is also a 2-regular function,where b {0, 1, 2, . . . }. Thus, by Corollary 4.2, for m 2, k 1 and b 0 we haveS2m1k+bSb(mod 2m) andS2m1k+bSb2m2k(Sb+2+4Sb+1+3Sb) (mod 2m+1).

References

[B] B.C. Berndt, Classical theorems on quadratic residues, Enseign. Math. 22 (1976), 261-304.

[C] K.W. Chen, Congruences for Euler numbers, Fibonacci Quart. 42 (2004), 128-140.[D] L.E. Dickson, History of the Theory of Numbers, Vol.I, Chelsea Publ., 1999, p. 271.[EO] M. Eie and Y.L. Ong, A generalization of Kummers congruences, Abh. Math. Sem. Univ.

Hamburg 67 (1997), 149-157.[Er] R. Ernvall, A congruence on Euler numbers, Amer. Math. Monthly 89 (1982), 431.[F] G.J. Fox, Kummer congruences for expressions involving generalized Bernoulli polynomials,

J. Theor. Nombres Bordeaux 14 (2002), 187-204.[GS] A. Granville and Z.W. Sun, Values of Bernoulli polynomials, Pacific J. Math. 172 (1996),

117-137.

46


47/47

[IR] K. Ireland and M. Rosen,A Classical Introduction to Modern Number Theory(2nd edition),Springer, New York, 1990, pp. 238,248.

[K] L.C. Karpinski, Uber die Verteilung der quadratischen Reste, J. Reine Angew. Math. 127(1904), 1-19.

[L] E. Lehmer, On congruences involving Bernoulli numbers and the quotients of Fermat andWilson, Ann. Math. 39 (1938), 350-360.

[Ler] M. Lerch, Zur Theorie des Fermatschen Quotienten ap1

1p =q(a), Math. Ann. 60 (1905),

471-490.[MOS] W. Magnus, F. Oberhettinger and R.P. Soni, Formulas and Theorems for the Special Func-

tions of Mathematical Physics (3rd edition), Springer, New York, 1966, pp. 25-32.[St] M.A. Stern, Zur Theorie der Eulerschen Zahlen, J. Reine Angew. Math. 79 (1875), 67-98.

[S1] Z.H. Sun, Combinatorial sumn

k=0kr(mod m)

n

k

and its applications in number theory I, J.

Nanjing Univ. Math. Biquarterly 9 (1992), 227-240.

[S2] Z.H. Sun, Combinatorial sumn

k=0kr(mod m)

n

k

and its applications in number theory II, J.

Nanjing Univ. Math. Biquarterly 10 (1993), 105-118.

[S3] Z.H. Sun, Combinatorial sum

kr(mod m)

n

k

and its applications in number theory III, J.

Nanjing Univ. Math. Biquarterly 12 (1995), 90-102.[S4] Z.H. Sun, Congruences for Bernoulli numbers and Bernoulli polynomials, Discrete Math.

163 (1997), 153-163.[S5] Z.H. Sun, Congruences concerning Bernoulli numbers and Bernoulli polynomials, Discrete

Appl. Math. 105 (2000), 193-223.[SS] Z.H. Sun and Z.W. Sun, Fibonacci numbers and Fermats last theorem, Acta Arith. 60

(1992), 371-388.[Su1] Z.W. Sun, On the sum

kr(mod m)

n

k

and related congruences, Israel J. Math. 128 (2002),

135-156.

[Su2] Z.W. Sun, Binomial coefficients and quadratic fields, Proc. Amer. Math. Soc. 134 (2006),2213-2222.

[UW] J. Urbanowicz and K.S. Williams,Congruences for L-Functions, Kluwer Academic Publish-ers, Dordrecht, Boston, London, 2000, pp. 3-8, 28, 40, 55.

[W] L.C. Washington, Introduction to Cyclotomic Fields, Springer, New York, 1982, pp. 30-31,141.

[Y1] P.T. Young, Congruences for Bernoulli, Euler, and Stirling numbers, J. Number Theory 78(1999), 204-227.

[Y2] P.T. Young,Kummer congruences for values of Bernoulli and Euler polynomials, Acta Arith.

99 (2001), 277-288.

[Y3] P.T. Young,Degenerate andn-adic versions of Kummers congruences for values of Bernoullipolynomials, Discrete Math. 285 (2004), 289-296.

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