aarjaneiro/planetmath huggingface dataset:
theorem on multiples of abundant numbers

theorem on multiples of abundant numbers

Theorem. The product

n

⁢

m

of an

abundant number

n

and any

integer

m

>

0

is also an abundant number, regardless of the

abundance

or

deficiency

of

m

.

Proof. Choose an abundant number

n

with

k

divisors

d

1

,

…

,

d

k

(where the divisors are sorted in

ascending order

and

d

1

=

1

,

d

k

=

n

) that add up to

2

⁢

n

+

a

, where

a

>

0

is the abundance of

n

. For maximum flair, set

a

=

1

, the bare minimum for abundance (that is, a

quasiperfect number

). Next, for

m

choose a spectacularly

deficient number

such that

gcd

⁡

(

m

,

n

)

=

1

, preferably some large prime number. If we choose a

prime number

, its divisors will only add up to

m

+

1

. However, the divisors of

n

⁢

m

will include each

d

i

⁢

m

, where

d

i

is a divisor of

n

and

0

<

i

≤

k

. Therefore, the divisors of

n

⁢

m

will add up to

∑

i

=

1

k

d

i

+

∑

i

=

1

k

d

i

⁢

m

=

2

⁢

n

⁢

m

+

a

⁢

(

m

+

1

)

+

2

⁢

n

.

It now becomes obvious that by insisting that

m

and

n

be

coprime

we are guaranteeing that if

m

is itself prime, it will bring at least

k

new divisors to the table. But what if

gcd

⁡

(

m

,

n

)

>

1

, or in the most extreme case,

m

=

n

? In such a case, we just can’t use the same formula for the sum of divisors of

n

⁢

m

that we used when

m

and

n

were coprime, as that would count some divisors twice. However,

m

=

n

still brings new divisors to the table, and those new divisors add up to

∑

i

=

2

k

d

i

⁢

d

k

=

2

⁢

n

2

+

2

⁢

a

2

+

a

.

Having proven these extreme cases, it is obvious that

n

⁢

m

will be abundant in other cases, such as

m

being a

composite

deficient number, a

perfect number

, an abundant number sharing some but not all

prime factors

with

n

, etc.

Title

theorem on multiples of abundant numbers

Canonical name

TheoremOnMultiplesOfAbundantNumbers

Date of creation

2013-03-22 16:05:46

Last modified on

2013-03-22 16:05:46

Owner

CompositeFan (12809)

Last modified by

CompositeFan (12809)

Numerical id

15

Author

CompositeFan (12809)

Entry type

Theorem

Classification

msc 11A05

Related topic

APositiveMultipleOfAnAbundantNumberIsAbundant

Generated on Fri Feb 9 15:10:52 2018 by

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theorem on sums of two squares by Fermat

theorem on sums of two squares by Fermat

Suppose that an

odd prime

number

p

can be written as the sum

p

=

a

2

+

b

2

where

a

and

b

are

integers

.  Then they have to be

coprime

.
We will show that

p

is of the form

4

⁢

n

+

1

.

Since

p

∤

b

,  the

congruence

b

⁢

b

1

≡

1

(

mod

p

)

has a solution

b

1

, whence

0

≡

p

⁢

b

1

2

=

(

a

⁢

b

1

)

2

+

(

b

⁢

b

1

)

2

≡

(

a

⁢

b

1

)

2

+

1

(

mod

p

)

,

and thus

(

a

⁢

b

1

)

2

≡

-

1

(

mod

p

)

.

Consequently, the

Legendre symbol

(

-

1

p

)

is

+

1

, i.e.

(

-

1

)

p

-

1

2

=

1

.

Therefore, we must have

p

=

4

⁢

n

+

1

(1)

where

n

is a

positive integer

.

Euler has first proved the following

theorem

presented by
Fermat and containing also the

converse

of the above claim.

Theorem
(Thue’s lemma (

http://planetmath.org/ThuesLemma

)).

An odd prime

p

is
uniquely expressible as

sum of two squares

of integers if and
only if it

satisfies

(1) with an integer value of

n

.

The theorem implies easily the

Corollary.

If all odd prime factors of a positive
integer are

congruent

to 1 modulo 4 then the integer is a sum
of two squares. (Cf. the proof of the parent article and the article
“

prime factors

of Pythagorean hypotenuses (

http://planetmath.org/primefactorsofpythagoreanhypotenuses

)”.)

Title

theorem on sums of two squares by Fermat

Canonical name

TheoremOnSumsOfTwoSquaresByFermat

Date of creation

2014-10-25 17:44:02

Last modified on

2014-10-25 17:44:02

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

13

Author

pahio (2872)

Entry type

Theorem

Classification

msc 11A05

Classification

msc 11A41

Classification

msc 11A67

Classification

msc 11E25

Generated on Fri Feb 9 15:10:54 2018 by

LaTeXML
unitary perfect number

unitary perfect number

A

unitary perfect number

is an

integer

which is the sum of its

positive

proper

unitary divisors

, not including the number itself. (A

divisor

d

of a number

n

is a unitary divisor if

d

and

n

d

share no common factors.) Some

perfect numbers

are not unitary perfect numbers, and some unitary perfect numbers are not

regular

perfect numbers.

The first few unitary perfect numbers are 6, 60, 90, 87360, 146361946186458562560000 (

sequence

A002827 in the OEIS).

There are no odd unitary perfect numbers. This follows since one has

2

⁢

d

*

(

n

)

dividing the sum of the unitary divisors of an

odd number

(where

d

*

(

n

)

is the number of distinct

prime divisors

of

n

). One gets this because the sum of all the unitary divisors is a

multiplicative function

and one has the sum of the unitary divisors of a power of a prime

p

a

is

p

a

+

1

which is even for all

odd primes

p

. Therefore, an odd unitary perfect number must have only one

distinct prime factor

, and it is not hard to show that a power of prime cannot be a unitary perfect number, since there are not enough divisors. It’s not known whether or not there are infinitely many unitary perfect numbers.

Title

unitary perfect number

Canonical name

UnitaryPerfectNumber

Date of creation

2013-03-22 16:19:29

Last modified on

2013-03-22 16:19:29

Owner

CompositeFan (12809)

Last modified by

CompositeFan (12809)

Numerical id

4

Author

CompositeFan (12809)

Entry type

Definition

Classification

msc 11A05

Generated on Fri Feb 9 15:10:56 2018 by

LaTeXML
untouchable number

untouchable number

An

untouchable number

is an

integer

n

for which there exists no integer

m

such that

(

∑

d

|

m

d

)

-

m

=

n

,

, thus

n

can’t be ”touched” by the sum of

proper divisors

of any other integer. Paul Erdős proved that there are infinitely many untouchable numbers.

Obviously no

perfect number

can be an untouchable number. Neither can any integer of the form

p

+

1

, where

p

is a

prime number

. What is not so obvious is whether 5 is the only odd untouchable number, and the related question of whether 2 and 5 are the only prime untouchable numbers.

References

1

P. Erdős, Über die Zahlen der Form

σ

⁢

(

n

)

-

n

und

n

-

ϕ

⁢

(

n

)

.

Elem. Math.

28

(1973), 83–86.

Title

untouchable number

Canonical name

UntouchableNumber

Date of creation

2013-03-22 15:51:52

Last modified on

2013-03-22 15:51:52

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

8

Author

PrimeFan (13766)

Entry type

Definition

Classification

msc 11A05

Generated on Fri Feb 9 15:10:57 2018 by

LaTeXML
Anton’s congruence

Anton’s congruence

For every

n

∈

ℕ

(

n

⁢

!

¯

)

p

stands for the

product

of numbers
between

1

and

n

which are not

divisible

by a given prime

p

. And we set

(

0

⁢

!

¯

)

p

=

1

.

The

corollary

below generalizes a result first found by Anton, Stickelberger,
and Hensel:

Let

N

0

be the least non-negative

residue

of

n

(

mod

p

s

)

where

p

is a

prime number

and

n

∈

ℕ

. Then

(

n

⁢

!

¯

)

p

≡

(

±

1

)

⌊

n

/

p

s

⌋

⋅

(

N

0

⁢

!

¯

)

p

(

mod

p

s

)

.

Proof.

We write each

r

in the product below as

i

⁢

p

s

+

j

to get

(

n

⁢

!

¯

)

p

=

∏

1

≤

r

≤

n

p

s

÷̸

r

r

=

(

∏

0

≤

i

≤

⌊

n

/

p

s

⌋

-

1

1

≤

j

<

p

s

p

s

÷̸

j

i

⁢

p

s

+

j

)

⁢

(

∏

i

=

⌊

n

/

p

s

⌋

1

≤

j

≤

N

0

p

s

÷̸

j

i

⁢

p

s

+

j

)

≡

∏

i

=

0

⌊

n

/

p

s

⌋

-

1

∏

1

≤

j

<

p

s

p

s

÷̸

j

j

⋅

∏

j

=

1

p

s

÷̸

j

N

0

j

)

≡

(

p

s

⁢

!

¯

)

p

⌊

n

/

p

s

⌋

⋅

(

N

0

⁢

!

¯

)

p

(

mod

p

s

)

.

From Wilson’s

theorem

for prime powers it follows that

(

n

⁢

!

¯

)

p

≡

{

(

N

0

⁢

!

¯

)

p

⁢

if

p

=

2

,

s

≥

3

(

-

1

)

⌊

n

/

p

s

⌋

⋅

(

N

0

⁢

!

¯

)

p

otherwise.

(

mod

p

s

)

.

∎

Title

Anton’s

congruence

Canonical

name

AntonsCongruence

Date of creation

2013-03-22 13:22:49

Last modified on

2013-03-22 13:22:49

Owner

Thomas Heye (1234)

Last modified by

Thomas Heye (1234)

Numerical id

10

Author

Thomas Heye (1234)

Entry type

Theorem

Classification

msc 11A07

Related topic

Factorial

Generated on Fri Feb 9 15:11:00 2018 by

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calculating the Jacobi symbol

calculating the Jacobi symbol

To

calculate

the

Jacobi symbol

(

a

m

)

for

positive integers

a

,

m

,

m

odd, we apply the quadratic reciprocity law and the fact that

(

a

m

)

=

(

b

m

)

if

a

≡

b

mod

m

. So if

gcd

⁡

(

a

,

m

)

=

1

(otherwise the Jacobi symbol is

0

) and

a

>

m

choose

b

≡

a

mod

m

with

b

<

m

. Then

(

a

m

)

=

(

b

m

)

. To apply the quadratic reciprocity law, which basically allows us to exchange

b

and

m

, we need to make

b

odd if it is even. Writing

b

=

2

s

⁢

c

with odd

c

does the trick since

(

b

m

)

=

(

2

m

)

s

⁢

(

c

m

)

.

Using the fact that

(

2

m

)

=

(

-

1

)

m

2

-

1

8

(1)

we compute

(

2

m

)

to be

1

if

m

≡

±

1

mod

8

and

-

1

if

m

≡

±

3

mod

8

.
Now we apply the quadratic reciprocity law:

(

c

m

)

=

(

-

1

)

(

c

-

1

)

⁢

(

m

-

1

)

4

⁢

(

m

c

)

.

The factor

(

-

1

)

(

c

-

1

)

⁢

(

m

-

1

)

4

is equal to

-

1

if

c

,

m

≡

3

mod

4

and

1

if at least one of the numbers

c

,

m

is

congruent

to

1

modulo

4

. At this point we can start thw whole procedure over again for

(

m

c

)

.

Eventually

we can apply the equation

(

-

1

m

)

=

(

-

1

)

m

-

1

2

, which is equal to

1

if

m

≡

1

mod

4

and

-

1

otherwise.

Example:

We try to calculate

(

107

23

)

. Since

gcd

⁡

(

107

,

23

)

=

1

and

107

≡

15

mod

23

we find

(

107

23

)

=

(

15

23

)

. Since both

15

and

23

are congruent

3

modulo

4

we have

(

15

23

)

=

-

(

23

15

)

=

-

(

8

15

)

=

-

(

2

15

)

.

This can be evaluated using equation (

1

) and we find

(

107

23

)

=

-

1

.

Title

calculating the Jacobi symbol

Canonical name

CalculatingTheJacobiSymbol

Date of creation

2013-03-22 14:55:05

Last modified on

2013-03-22 14:55:05

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

4

Author

mathwizard (128)

Entry type

Algorithm

Classification

msc 11A07

Classification

msc 11A15

Classification

msc 11Y99

Generated on Fri Feb 9 15:11:02 2018 by

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conditional congruences

conditional congruences

Consider

congruences

(

http://planetmath.org/Congruences

) of the form

f

⁢

(

x

)

:=

a

n

⁢

x

n

+

a

n

-

1

⁢

x

n

-

1

+

…

+

a

0

≡

0

(

mod

m

)

(1)

where the coefficients

a

i

and

m

are

rational integers

.

Solving

the congruence means finding all the integer values of

x

which satisfy (1).

•

If

a

i

≡

0

(

mod

m

)

for all

i

’s, the congruence is satisfied by each

integer

, in which case the congruence is identical (cf. the

formal congruence

).  Therefore one can assume that at least

a

n

≢

0

(

mod

m

)

,

since one would otherwise have

a

n

⁢

x

n

≡

0

(

mod

m

)

and the first term could be left out of (1).  Now, we say that the

degree of the congruence

(1) is

n

.

•

If

x

=

x

0

is a solution of (1) and

x

1

≡

x

0

(

mod

m

)

,  then by the properties of congruences (

http://planetmath.org/Congruences

),

f

⁢

(

x

1

)

≡

f

⁢

(

x

0

)

≡

0

(

mod

m

)

,

and thus also

x

=

x

1

is a solution.  Therefore, one regards as different

roots

of a congruence modulo

m

only such values of

x

which are incongruent modulo

m

.

•

One can think that the congruence (1) has as many roots as is found in a

complete residue system

modulo

m

.

Title

conditional congruences

Canonical name

ConditionalCongruences

Date of creation

2013-03-22 18:52:23

Last modified on

2013-03-22 18:52:23

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Topic

Classification

msc 11A07

Classification

msc 11A05

Related topic

LinearCongruence

Related topic

QuadraticCongruence

Defines

degree of congruence

Defines

root of congruence

Defines

root

Generated on Fri Feb 9 15:11:05 2018 by

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factorial modulo prime powers

factorial modulo prime powers

For

n

∈

ℕ

and any

prime number

p

,

(

n

!

)

p

is the

product

of numbers

1

≤

m

≤

n

, where

p

|̸

m

.

For

natural numbers

n

,

s

and a given prime number

p

, we have the

congruence

n

!

p

∑

i

=

0

d

⌊

n

p

i

⌋

≡

∏

i

=

0

d

(

(

±

1

)

⌊

n

p

s

+

i

⌋

⁢

(

N

i

!

)

p

)

(

mod

p

s

)

,

where

N

i

is the least non-negative

residue

of

⌊

n

p

i

⌋

(

mod

p

s

)

. Here

d

+

1

denotes the number of digits in the

representation

of

n

in base

p

. More precisely,

±

1

is

-

1

unless

p

=

2

,

s

≥

3

.

Proof.

Let

i

≥

0

. Then the set of numbers between 1 and

⌊

n

p

i

⌋

is

{

k

⁢

p

,

k

≥

1

,

k

≤

⌊

n

p

i

+

1

⌋

}

.

This is true for every

integer

i

with

p

i

+

1

≤

n

. So we have

⌊

n

p

i

⌋

!

⌊

n

p

i

+

1

⌋

⁢

p

⌊

n

p

i

+

1

⌋

=

(

⌊

n

p

i

⌋

!

)

p

.

(1)

Multiplying all terms with

0

≤

i

≤

d

, where

d

is the largest power of

p

not greater than

n

, the statement follows from the

generalization

of
Anton’s congruence.
∎

Title

factorial modulo prime powers

Canonical name

FactorialModuloPrimePowers

Date of creation

2013-03-22 13:22:52

Last modified on

2013-03-22 13:22:52

Owner

Thomas Heye (1234)

Last modified by

Thomas Heye (1234)

Numerical id

7

Author

Thomas Heye (1234)

Entry type

Result

Classification

msc 11A07

Generated on Fri Feb 9 15:11:07 2018 by

LaTeXML
Hensel’s lemma for integers

Hensel’s lemma for integers

Let

f

⁢

(

x

)

be a

polynomial

with

integer

coefficients

,

p

a

prime number

, and

n

a

positive integer

.  Assume that an integer

a

(and naturally its whole

residue class

modulo

p

n

)

satisfies

the

congruence

f

⁢

(

x

)

≡

0

(

mod

p

n

)

.

(1)

The solution

x

=

a

of (1) may be refined in its residue class modulo

p

n

to a solution

x

=

a

+

r

⁢

p

n

of the congruence

f

⁢

(

x

)

≡

0

(

mod

p

n

+

1

)

.

(2)

This refinement is unique modulo

p

n

+

1

iff

f

′

⁢

(

a

)

≢

0

(

mod

p

)

.

Proof.

Now we have

f

⁢

(

a

)

=

s

⁢

p

n

.  We have to find an

r

such that

f

⁢

(

a

+

r

⁢

p

n

)

≡

0

(

mod

p

n

+

1

)

.

The

short Taylor theorem

requires that

f

⁢

(

a

+

r

⁢

p

n

)

≡

f

⁢

(

a

)

+

r

⁢

f

′

⁢

(

a

)

⁢

p

n

(

mod

r

2

⁢

p

2

⁢

n

)

where

2

⁢

n

≥

n

+

1

, whence this congruence can be simplified to

s

⁢

p

n

+

r

⁢

f

′

⁢

(

a

)

⁢

p

n

≡

0

(

mod

p

n

+

1

)

.

Thus the integer

r

must satisfy the

linear congruence

s

+

r

⁢

f

′

⁢

(

a

)

≡

0

(

mod

p

)

.

When

f

′

⁢

(

a

)

≢

0

,  this congruence has a unique solution modulo

p

(see linear congruence); thus we have the refinement

a

′

=

a

+

r

⁢

p

n

which is unique modulo

p

n

+

1

.

When

f

′

⁢

(

a

)

≡

0

and

s

≢

0

(

mod

p

)

,  the congruence evidently is impossible.

In the case

f

′

⁢

(

a

)

≡

s

≡

0

(

mod

p

)

the congruence (2) is identically true in the residue class of

a

modulo

p

n

.   □

References

1

Peter Hackman

:

Elementary Number Theory

. HHH

Productions

, Linköping (2009).

Title

Hensel’s lemma for integers

Canonical name

HenselsLemmaForIntegers

Date of creation

2013-04-08 19:35:26

Last modified on

2013-04-08 19:35:26

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Definition

Classification

msc 11A07

Generated on Fri Feb 9 15:11:12 2018 by

LaTeXML
inductive proof of Fermat’s little theorem proof

inductive proof of Fermat’s little theorem proof

We will show

a

p

≡

a

(

mod

p

)

with

p

prime. The

equivalent

statement

a

p

-

1

≡

1

(

mod

p

)

when

p

does not divide

a

follows by cancelling

a

both sides (which can be done since then

a

,

p

are

coprime

).

When

a

=

1

, we have

1

p

≡

1

(

mod

p

)

Now assume the

theorem

holds for some

positive

a

and we want to prove the statement for

a

+

1

. We will have as a direct

consequence

that

a

p

≡

a

(

mod

p

)

Let’s examine

a

+

1

. By the

binomial theorem

, we have

(

a

+

1

)

p

≡

(

p

0

)

⁢

a

p

+

(

p

1

)

⁢

a

p

-

1

+

⋯

+

(

p

p

-

1

)

⁢

a

+

1

≡

a

+

p

⁢

a

p

-

1

+

p

⁢

(

p

-

1

)

2

⁢

a

p

-

2

+

⋯

+

p

⁢

a

+

1

≡

(

a

+

1

)

+

[

p

⁢

a

p

-

1

+

p

⁢

(

p

-

1

)

2

⁢

a

p

-

2

+

⋯

+

p

⁢

a

]

However, note that the entire bracketed term is

divisible

by

p

, since each element of it is divsible by

p

. Hence

(

a

+

1

)

p

≡

(

a

+

1

)

(

mod

p

)

Therefore by

induction

it follows that

a

p

≡

a

(

mod

p

)

for all

positive integers

a

.

It is easy to show that it also holds for

-

a

whenever it holds for

a

, so the statement works for all

integers

a

.

Title

inductive proof of Fermat’s little theorem proof

Canonical name

InductiveProofOfFermatsLittleTheoremProof

Date of creation

2013-03-22 11:47:46

Last modified on

2013-03-22 11:47:46

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

17

Author

mathcam (2727)

Entry type

Proof

Classification

msc 11A07

Related topic

FermatsTheoremProof

Generated on Fri Feb 9 15:11:15 2018 by

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Jacobi symbol

Jacobi symbol

The

Jacobi symbol

is a

generalization

of the

Legendre symbol

to all odd

positive integers

.

Let

n

be an odd positive integer, with

prime factorization

p

1

e

1

⁢

⋯

⁢

p

k

e

k

. Let

a

≥

0

be an

integer

. The

Jacobi symbol

(

a

n

)

is defined to be

(

a

n

)

=

∏

i

=

1

k

(

a

p

i

)

e

i

where

(

a

p

i

)

is the Legendre symbol of

a

and

p

i

.

A further generalization of the Legendre symbol, due to Kronecker, is the

Kronecker symbol

.

Title

Jacobi symbol

Canonical name

JacobiSymbol

Date of creation

2013-03-22 12:36:26

Last modified on

2013-03-22 12:36:26

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

9

Author

mathwizard (128)

Entry type

Definition

Classification

msc 11A07

Classification

msc 11A15

Related topic

LegendreSymbol

Related topic

KroneckerSymbol

Generated on Fri Feb 9 15:11:17 2018 by

LaTeXML
Kronecker symbol

Kronecker symbol

The

Kronecker symbol

is a

generalization

of the

Jacobi symbol

to all

integers

.

Let

n

be an integer, with

prime factorization

u

⋅

p

1

e

1

⁢

⋯

⁢

p

k

e

k

, where

u

is a unit and the

p

i

are primes. Let

a

≥

0

be an integer. The Kronecker symbol

(

a

n

)

is defined to be

(

a

n

)

=

(

a

u

)

⁢

∏

i

=

1

k

(

a

p

i

)

e

i

For odd

p

i

, the number

(

a

p

i

)

is simply the usual

Legendre symbol

. This leaves the case when

p

i

=

2

. We define

(

a

2

)

by

(

a

2

)

=

{

0

if

a

is even

1

if

a

is odd and

n

≡

1

or

n

≡

7

(

mod

8

)

-

1

if

a

is odd and

n

≡

3

or

n

≡

5

(

mod

8

)

Since it extends the Jacobi symbol, the quantity

(

a

u

)

is simply 1 when

u

=

1

. When

u

=

-

1

, we define it by

(

a

-

1

)

=

{

-

1

if

a

<

0

1

if

a

>

0

These

extensions

suffice to define the Kronecker symbol for all integer values

n

.

Title

Kronecker symbol

Canonical name

KroneckerSymbol

Date of creation

2013-03-22 14:33:21

Last modified on

2013-03-22 14:33:21

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

6

Author

mathwizard (128)

Entry type

Definition

Classification

msc 11A07

Classification

msc 11A15

Synonym

Kronecker-Jacobi symbol

Related topic

JacobiSymbol

Related topic

LegendreSymbol

Generated on Fri Feb 9 15:11:19 2018 by

LaTeXML
orderly number

orderly number

An

orderly number

is an

integer

n

such that there exists at least one other integer

k

such that each

divisor

of

n

from 1 to

d

τ

⁢

(

n

)

(with

τ

⁢

(

n

)

being the number of divisors

function

)

satisfies

the

congruences

d

i

≡

1

mod

k

,

d

i

≡

2

mod

k

through

d

i

≡

τ

⁢

(

n

)

mod

k

. For example, 20 is an orderly number, with

k

=

7

, since it has six divisors, 1, 2, 4, 5, 10, 20, and we can verify that

1

≡

1

mod

7

2

≡

2

mod

7

10

≡

3

mod

7

4

≡

4

mod

7

5

≡

5

mod

7

20

≡

6

mod

7

The orderly numbers less than 100 are 1, 2, 5, 7, 8, 9, 11, 12, 13, 17, 19, 20, 23, 27, 29, 31, 37, 38, 41, 43, 47, 52, 53, 57, 58, 59, 61, 67, 68, 71, 72, 73, 76, 79, 83, 87, 89, 97, listed in A167408 of

Neil Sloane

’s OEIS. With the exception of 3, all

prime numbers

are orderly.

Title

orderly number

Canonical name

OrderlyNumber

Date of creation

2013-03-22 19:05:28

Last modified on

2013-03-22 19:05:28

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

4

Author

PrimeFan (13766)

Entry type

Definition

Classification

msc 11A07

Generated on Fri Feb 9 15:11:21 2018 by

LaTeXML
proof of Euler-Fermat theorem

proof of Euler-Fermat theorem

Let

a

1

,

a

2

,

…

,

a

ϕ

⁢

(

n

)

be all

positive integers

less than

n

which are

coprime

to

n

. Since

gcd

⁢

(

a

,

n

)

=

1

, then the set

a

⁢

a

1

,

a

⁢

a

2

,

…

,

a

⁢

a

ϕ

⁢

(

n

)

are each

congruent

to one of the

integers

a

1

,

a

2

,

…

,

a

ϕ

⁢

(

n

)

in some order. Taking the

product

of these

congruences

, we get

(

a

⁢

a

1

)

⁢

(

a

⁢

a

2

)

⁢

⋯

⁢

(

a

⁢

a

ϕ

⁢

(

n

)

)

≡

a

1

⁢

a

2

⁢

⋯

⁢

a

ϕ

⁢

(

n

)

(

mod

n

)

hence

a

ϕ

⁢

(

n

)

⁢

(

a

1

⁢

a

2

⁢

⋯

⁢

a

ϕ

⁢

(

n

)

)

≡

a

1

⁢

a

2

⁢

⋯

⁢

a

ϕ

⁢

(

n

)

(

mod

n

)

.

Since

gcd

⁢

(

a

1

⁢

a

2

⁢

⋯

⁢

a

ϕ

⁢

(

n

)

,

n

)

=

1

, we can divide both sides by

a

1

⁢

a

2

⁢

⋯

⁢

a

ϕ

⁢

(

n

)

, and the desired result follows.

Title

proof of Euler-Fermat theorem

Canonical name

ProofOfEulerFermatTheorem

Date of creation

2013-03-22 11:47:57

Last modified on

2013-03-22 11:47:57

Owner

KimJ (5)

Last modified by

KimJ (5)

Numerical id

10

Author

KimJ (5)

Entry type

Proof

Classification

msc 11A07

Classification

msc 11A25

Generated on Fri Feb 9 15:11:25 2018 by

LaTeXML