proof of congruence of Clausen and von Staudt

proof of congruence of Clausen and von Staudt

Theorem 0.1

For

m

≥

1

,

(

m

+

1

)

⁢

S

m

⁢

(

n

)

=

∑

k

=

0

m

(

m

+

1

k

)

⁢

B

k

⁢

n

m

+

1

-

k

,

where

S

m

⁢

(

n

)

=

1

m

+

2

m

+

…

+

(

n

-

1

)

m

.

Proof 1

In the equation

e

k

⁢

t

=

∑

m

=

0

∞

k

m

⁢

(

t

m

m

!

)

,

substitute

k

=

0

,

1

,

2

,

…

,

n

-

1

and add, obtaining,

∑

m

=

0

∞

S

m

⁢

(

n

)

⁢

t

m

m

!

=

e

n

⁢

t

-

1

e

t

-

1

=

e

n

⁢

t

-

1

t

⋅

t

e

t

-

1

=

∑

k

=

1

∞

n

k

⁢

t

k

-

1

k

!

⁢

∑

j

=

0

∞

B

j

⁢

t

j

j

!

,

since

t

e

t

-

1

=

∑

j

=

0

∞

B

j

⁢

t

j

j

!

.

Now by comparing coefficients of

t

m

and then multiplying by

(

m

+

1

)

!

, we obtain the result.

We will write this

identity

in the following form:

S

m

⁢

(

n

)

=

∑

k

=

0

m

(

m

k

)

⁢

B

m

-

k

⁢

n

k

+

1

k

+

1

(1)

=

B

m

⁢

n

+

(

m

1

)

⁢

B

m

-

1

⁢

n

2

2

+

…

+

n

m

+

1

m

+

1

(2)

This follows by replacing

(

m

+

1

k

)

by

m

+

1

m

-

k

+

1

⁢

(

m

k

)

and then switching

m

-

k

and

k

in the

theorem

.

Proposition 0.2

Let

p

be prime and

m

≥

1

. Then

p

⁢

B

m

is

p

-integral, and if

m

≥

2

is even, then

p

⁢

B

m

≡

S

m

⁢

(

p

)

(

mod

p

)

.

Proof 2

The first statement is

equivalent

to showing if

p

∣

b

then

p

2

∤

b

, where

b

is the

denominator

of

p

⁢

B

m

. This is clear for

p

⁢

B

1

=

-

p

/

2

. We proceed by

induction

. Suppose

m

>

1

and let

n

=

p

in (

2

). Since

S

m

⁢

(

p

)

∈

Z

, it suffices to prove that

(

m

k

)

⁢

(

p

⁢

B

m

-

k

)

⁢

p

k

k

+

1

is

p

-integral for

k

=

1

,

2

,

…

,

m

. By induction

p

⁢

B

m

-

k

is

p

-integral for

k

≥

1

, and

p

k

k

+

1

is

p

-integral since

k

+

1

≤

p

k

for all primes

p

. It follows that

p

⁢

B

m

is

p

-integral.
To establish the

congruence

, we need to show that if

k

≥

1

, then

(

m

k

)

⁢

(

p

⁢

B

m

-

k

)

⁢

p

k

k

+

1

≡

0

(

mod

p

)

.

For

k

≥

2

,

p

k

k

+

1

≡

0

(

mod

p

)

, since

k

+

1

<

p

k

.
For

k

=

1

, we have

m

2

⁢

(

p

⁢

B

m

-

1

)

⁢

p

≡

0

(

mod

p

)

,

since

m

is even. In fact, since

B

m

-

1

=

0

for

m

≥

4

even, it suffices to check it for

m

=

2

, which is obvious.

Lemma 1

Let

p

be prime. Then

S

m

⁢

(

p

)

≡

{

0

(

mod

p

)

,

if

p

-

1

∤

m

-

1

(

mod

p

)

,

if

p

-

1

∣

m

Proof 3

Let

g

be a

primitive root

modulo

p

. Then

S

m

⁢

(

p

)

=

1

m

+

2

m

+

…

+

(

p

-

1

)

m

≡

1

m

+

g

m

+

g

2

⁢

m

+

…

+

g

(

p

-

2

)

⁢

m

(

mod

p

)

Hence,

(

g

m

-

1

)

⁢

S

m

⁢

(

p

)

≡

g

m

⁢

(

p

-

1

)

-

1

≡

0

(

mod

p

)

.

If

p

-

1

∤

m

, then

g

m

≢

1

(

mod

p

)

, and

S

m

⁢

(

p

)

≡

0

(

mod

p

)

.
If

p

-

1

∣

m

, then

S

m

⁢

(

p

)

≡

1

+

1

+

…

+

1

≡

p

-

1

≡

-

1

(

mod

p

)

.

We are now ready to prove the congruence.

Proof 4

(Proof of von Staudt-Claussen congruence)

Assume

m

is even. Then by the

proposition

,

p

⁢

B

m

is

p

-integral and

p

⁢

B

m

≡

S

m

⁢

(

p

)

(

mod

p

)

. Therefore, by the lemma, if

p

-

1

∤

m

, then

B

m

is a

p

-

integer

and if

p

-

1

∣

m

, then

p

⁢

B

m

≡

-

1

(

mod

p

)

. Hence,

A

m

=

B

m

+

∑

p

-

1

∣

m

1

p

is a

p

-integer for all primes

p

. For if

q

is a prime, and

q

-

1

∤

m

, then

B

m

is

q

-integral and hence

A

m

is as well, since the sum contributes no

negative

power of

q

. Otherwise,

q

-

1

∣

m

and

A

m

=

B

m

+

1

q

+

∑

p

-

1

∣

m

p

≠

q

1

p

=

q

⁢

B

m

+

1

q

+

∑

p

-

1

∣

m

p

≠

q

1

p

≡

∑

p

-

1

∣

m

p

≠

q

1

p

(

mod

ℤ

)

,

which is clearly

q

-integral. Since

A

m

is

p

-integral for all primes

p

, it must be the case that

A

m

∈

Z

. That is,

B

m

≡

-

∑

p

⁢

p

⁢

r

⁢

i

⁢

m

⁢

e

p

-

1

∣

m

1

p

(

mod

ℤ

)

Title

proof of congruence of Clausen and von Staudt

Canonical name

ProofOfCongruenceOfClausenAndVonStaudt

Date of creation

2013-03-22 15:33:58

Last modified on

2013-03-22 15:33:58

Owner

slachter (11430)

Last modified by

slachter (11430)

Numerical id

5

Author

slachter (11430)

Entry type

Proof

Classification

msc 11B68

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