aarjaneiro/planetmath huggingface dataset:
proof of a corollary to Euler-Fermat theorem

proof of a corollary to Euler-Fermat theorem

This is an easy consequence of

Euler-Fermat theorem

:

Let

d

i

,

m

i

be defined as in the parent entry. Then

gcd

⁡

(

a

,

m

s

)

=

1

and Euler’s

theorem

implies:

a

ϕ

⁢

(

m

s

)

≡

1

mod

m

s

Note also that each of

d

s

-

1

,

…

,

d

0

divides

a

, so

∏

d

i

divides

a

s

, so

∏

d

i

divides

a

ϕ

⁢

(

m

s

)

+

s

-

a

s

. Also,

gcd

⁡

(

∏

d

i

,

m

s

)

=

1

and

m

s

⋅

∏

d

i

=

m

. Therefore:

a

ϕ

⁢

(

m

s

)

+

s

≡

a

s

mod

m

which is what the corollary claimed.

Title

proof of a corollary to Euler-Fermat theorem

Canonical

name

ProofOfACorollaryToEulerFermatTheorem

Date of creation

2013-03-22 14:23:20

Last modified on

2013-03-22 14:23:20

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

5

Author

alozano (2414)

Entry type

Proof

Classification

msc 11-00

Generated on Fri Feb 9 15:07:59 2018 by

LaTeXML
proof of Euler-Fermat theorem using Lagrange’s theorem

proof of Euler-Fermat theorem using Lagrange’s theorem

Theorem

.

Given

a

,

n

∈

Z

,

a

ϕ

⁢

(

n

)

≡

1

(

mod

n

)

when

𝑔𝑐𝑑

⁢

(

a

,

n

)

=

1

, where

ϕ

is the

Euler totient function

.

Proof.

We will make use of Lagrange’s

Theorem

: Let

G

be a

finite group

and let

H

be a

subgroup

of

G

. Then the order of

H

divides the order of

G

.

Let

G

=

(

ℤ

/

n

⁢

ℤ

)

×

and let

H

be the

multiplicative

subgroup of

G

generated by

a

(so

H

=

{

1

,

a

,

a

2

,

…

}

). The fact that

gcd

⁡

(

a

,

n

)

=

1

ensures that

a

∈

G

. Notice that the order of

H

,

h

=

|

H

|

is also the order of

a

, i.e. the smallest

natural number

n

>

1

such that

a

n

is the

identity

in

G

, i.e.

a

h

≡

1

mod

n

. Also, recall that the order of

G

=

(

ℤ

/

n

⁢

ℤ

)

×

is

ϕ

⁢

(

n

)

, where

ϕ

is the Euler

ϕ

function

.

By Lagrange’s theorem

h

∣

|

G

|

=

ϕ

(

n

)

, so

ϕ

⁢

(

n

)

=

h

⋅

m

for some

m

. Thus:

a

ϕ

⁢

(

n

)

=

(

a

h

)

m

≡

1

m

≡

1

mod

n

as claimed.
∎

Title

proof of Euler-Fermat theorem

using Lagrange’s theorem

Canonical name

ProofOfEulerFermatTheoremUsingLagrangesTheorem

Date of creation

2013-03-22 14:24:03

Last modified on

2013-03-22 14:24:03

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

4

Author

alozano (2414)

Entry type

Proof

Classification

msc 11-00

Related topic

LagrangesTheorem

Related topic

FermatsLittleTheorem

Related topic

FermatsTheoremProof

Generated on Fri Feb 9 15:08:01 2018 by

LaTeXML
proof of Fermat’s little theorem using Lagrange’s theorem

proof of Fermat’s little theorem using Lagrange’s theorem

Theorem

.

If

a

,

p

∈

Z

with

p

a prime and

p

∤

a

, then

a

p

-

1

≡

1

(

mod

p

)

.

Proof.

We will make use of Lagrange’s

Theorem

: Let

G

be a

finite group

and let

H

be a

subgroup

of

G

. Then the order of

H

divides the order of

G

.

Let

G

=

(

ℤ

/

p

⁢

ℤ

)

×

and let

H

be the multiplicative subgroup of

G

generated by

a

(so

H

=

{

1

,

a

,

a

2

,

…

}

). Notice that the order of

H

,

h

=

|

H

|

is also the order of

a

, i.e. the smallest

natural number

n

>

1

such that

a

n

is the

identity

in

G

, i.e.

a

h

≡

1

mod

p

.

By Lagrange’s theorem

h

∣

|

G

|

=

p

-

1

, so

p

-

1

=

h

⋅

m

for some

m

. Thus:

a

p

-

1

=

(

a

h

)

m

≡

1

m

≡

1

mod

p

as claimed.
∎

Title

proof of Fermat’s little theorem using Lagrange’s theorem

Canonical name

ProofOfFermatsLittleTheoremUsingLagrangesTheorem

Date of creation

2013-03-22 14:23:53

Last modified on

2013-03-22 14:23:53

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

4

Author

alozano (2414)

Entry type

Proof

Classification

msc 11-00

Related topic

LagrangesTheorem

Generated on Fri Feb 9 15:08:03 2018 by

LaTeXML
proof of properties of primitive roots

proof of properties of primitive roots

The material in the main article is conveniently recast in terms of the groups

(

ℤ

/

m

⁢

ℤ

)

×

, the

multiplicative group

of units in

ℤ

/

m

⁢

ℤ

. Note that the order of this group is exactly

ϕ

⁢

(

m

)

where

ϕ

is the

Euler phi function

. Then saying that an

integer

g

is a

primitive root

of

m

is

equivalent

to saying that the

residue class

of

g

(

mod

m

)

generates

(

ℤ

/

m

⁢

ℤ

)

×

.

Proof.

(of

Theorem

):

The proof of the theorem is an immediate

consequence

of the

structure

of

(

ℤ

/

m

⁢

ℤ

)

×

as an

abelian group

;

(

ℤ

/

m

⁢

ℤ

)

×

is cyclic precisely for

m

=

2

,

4

,

p

k

, or

2

⁢

p

k

.
∎

Proof.

(of

Proposition

):

1.

Restated, this says that if the residue class of

g

(

mod

m

)

generates

(

ℤ

/

m

⁢

ℤ

)

×

, then the set

{

1

,

g

,

g

2

,

…

,

g

ϕ

⁢

(

m

)

}

is a

complete set

of representatives for

(

ℤ

/

m

⁢

ℤ

)

×

; this is obvious.

2.

Restated, this says that

g

(

mod

m

)

generates

(

ℤ

/

m

⁢

ℤ

)

×

if and only if

g

has exact order

m

, which is also obvious.

3.

If

g

(

mod

m

)

generates

(

ℤ

/

m

⁢

ℤ

)

×

, then

g

has exact order

ϕ

⁢

(

m

)

and thus

g

s

=

g

t

(

mod

m

)

if and only if

g

s

-

t

=

1

if and only if

ϕ

(

m

)

∣

s

-

t

.

4.

Suppose

g

generates

(

ℤ

/

m

⁢

ℤ

)

×

. Then

(

g

k

)

r

=

1

if and only if

g

k

⁢

r

=

1

if and only if

ϕ

(

m

)

∣

k

r

. Clearly we can choose

r

<

ϕ

⁢

(

m

)

if and only if

gcd

⁡

(

k

,

ϕ

⁢

(

m

)

)

>

1

.

5.

This follows immediately from (4).∎

Title

proof of properties of primitive roots

Canonical name

ProofOfPropertiesOfPrimitiveRoots

Date of creation

2013-03-22 18:43:48

Last modified on

2013-03-22 18:43:48

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

4

Author

rm50 (10146)

Entry type

Proof

Classification

msc 11-00

Generated on Fri Feb 9 15:08:05 2018 by

LaTeXML
proof of Pythagorean triples

proof of Pythagorean triples

Suppose that

a

2

+

b

2

=

1

where

a

,

b

∈

ℚ

.

a

2

+

b

2

=

N

ℚ

⁢

(

i

)

/

ℚ

⁢

(

a

+

b

⁢

i

)

(here

N

is the norm), so

a

2

+

b

2

=

1

if and only if

N

ℚ

⁢

(

i

)

/

ℚ

⁢

(

a

+

b

⁢

i

)

=

1

.

ℚ

⁢

(

i

)

is cyclic over

ℚ

with

Galois group

isomorphic

to

ℤ

/

2

⁢

ℤ

, so by Hilbert’s

Theorem

90, there is some element

s

+

t

⁢

i

∈

ℚ

⁢

(

i

)

such that

a

+

b

⁢

i

=

s

+

t

⁢

i

σ

⁢

(

s

+

t

⁢

i

)

=

s

+

t

⁢

i

s

-

t

⁢

i

=

s

2

-

t

2

+

2

⁢

s

⁢

t

⁢

i

s

2

+

t

2

so that

a

=

s

2

-

t

2

s

2

+

t

2

,

b

=

2

⁢

s

⁢

t

s

2

+

t

2

Now, given any

integer

right triangle

p

,

q

,

r

with

p

2

+

q

2

=

r

2

, we have

(

p

r

)

2

+

(

q

r

)

2

=

1

where

p

/

r

,

q

/

r

∈

ℚ

, so for some

s

,

t

∈

ℚ

,

p

r

=

s

2

-

t

2

s

2

+

t

2

,

q

r

=

2

⁢

s

⁢

t

s

2

+

t

2

Clearing

fractions

on the

right hand side

of these equations by multiplying numerator and denominator by the square of the least common

multiple

of the denominators of

s

,

t

, we get

p

r

=

m

2

-

n

2

m

2

+

n

2

,

q

r

=

2

⁢

m

⁢

n

m

2

+

n

2

for

m

,

n

∈

ℤ

. Thus for some

d

∈

ℤ

,

p

=

d

⁢

(

m

2

-

n

2

)

,

q

=

2

⁢

m

⁢

n

⁢

d

,

r

=

d

⁢

(

m

2

+

n

2

)

Title

proof of Pythagorean triples

Canonical

name

ProofOfPythagoreanTriples1

Date of creation

2013-03-22 17:44:34

Last modified on

2013-03-22 17:44:34

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

5

Author

rm50 (10146)

Entry type

Proof

Classification

msc 11-00

Generated on Fri Feb 9 15:08:09 2018 by

LaTeXML
proof of Wilson’s theorem

proof of Wilson’s theorem

We first show that, if

p

is a prime, then

(

p

-

1

)

!

≡

-

1

(

mod

p

)

.

Since

p

is prime,

ℤ

p

is a field and thus,

pairing

off each

element

with its

inverse

in the

product

(

p

-

1

)

!

=

∏

x

=

1

p

-

1

x

,

we are left with the elements which are their own inverses (i.e. which

satisfy

the equation

x

2

≡

1

(

mod

p

)

),

1

and

-

1

, only. Consequently,

(

p

-

1

)

!

≡

-

1

(

mod

p

)

.

To prove that the condition is necessary, suppose that

(

p

-

1

)

!

≡

-

1

(

mod

p

)

and that

p

is not a prime. The case

p

=

1

is

trivial

. Since

p

is composite, it has a divisor

k

such that

1

<

k

<

p

, and we have

(

p

-

1

)

!

≡

-

1

(

mod

k

)

. However, since

k

⩽

p

-

1

, it divides

(

p

-

1

)

!

and thus

(

p

-

1

)

!

≡

0

(

mod

k

)

,

a

contradiction

.

Title

proof of Wilson’s

theorem

Canonical name

ProofOfWilsonsTheorem

Date of creation

2013-03-22 12:09:09

Last modified on

2013-03-22 12:09:09

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

9

Author

CWoo (3771)

Entry type

Proof

Classification

msc 11-00

Generated on Fri Feb 9 15:08:11 2018 by

LaTeXML
proof of Wilson’s theorem result

proof of Wilson’s theorem result

\PMlinkescapephrase

set of primes

We denote by

ℙ

the set of primes and by

x

¯

the

multiplicative inverse

of

x

in

ℤ

p

.

Theorem

(Generalisation of Wilson’s Theorem).

For all integers

1

≤

k

≤

p

-

1

,

p

∈

P

⇔

(

p

-

k

)

!

(

k

-

1

)

!

≡

(

-

1

)

k

(

mod

p

)

Proof.

If

p

is a prime, then:

(

p

-

k

)

!

≡

(

p

-

1

)

!

⁢

(

p

-

1

)

¯

⁢

⋯

⁢

(

p

-

k

+

1

)

¯

≡

(

p

-

1

)

!

⁢

(

-

1

)

¯

⁢

⋯

⁢

(

1

-

k

)

¯

=

=

(

p

-

1

)

!

⁢

(

-

1

)

k

-

1

⁢

(

k

-

1

)

!

¯

(

mod

p

)

,

and since

(

p

-

1

)

!

≡

-

1

(

mod

p

)

(Wilson’s

Theorem

, simply pair up each number — except

p

-

1

and

1

, the only numbers in

ℤ

p

which are their own

inverses

— with its inverse), the first

implication

follows.

Now, if

p

∣

(

p

-

1

)

!

(

k

-

1

)

!

-

(

-

1

)

k

, then

p

∈

ℙ

as the opposite would mean that

p

=

a

⁢

b

, for some integers

1

<

a

,

b

<

p

, and so

p

would not be relatively prime to

(

p

-

1

)

!

⁢

(

k

-

1

)

!

as the initial

hypothesis

implies.
∎

Title

proof of Wilson’s theorem result

Canonical

name

ProofOfWilsonsTheoremResult

Date of creation

2013-03-22 15:07:08

Last modified on

2013-03-22 15:07:08

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

15

Author

CWoo (3771)

Entry type

Proof

Classification

msc 11-00

Generated on Fri Feb 9 15:08:13 2018 by

LaTeXML
properties of primitive roots

properties of primitive roots

Definition

.

Let

m

>

1

be an

integer

. An integer

g

is said to be a

primitive root

of

m

if

gcd

⁡

(

g

,

m

)

=

1

and the

multiplicative order

of

g

is exactly

ϕ

⁢

(

m

)

, where

ϕ

is the

Euler phi function

. In other words,

g

ϕ

⁢

(

m

)

≡

1

mod

m

and

g

n

≠

1

mod

m

for any

n

<

ϕ

⁢

(

m

)

.

Theorem

.

An integer

m

>

1

has a primitive root if and only if

m

is

2

,

4

,

p

k

or

2

⁢

p

k

for some

k

≥

1

. In particular, every prime has a primitive root.

Proposition

.

Let

m

>

1

be an integer.

1.

If

g

is a primitive root of

m

then the set

{

1

,

g

,

g

2

,

…

,

g

ϕ

⁢

(

m

)

}

is a

complete set

of representatives for

(

ℤ

/

m

⁢

ℤ

)

×

.

2.

If

gcd

⁡

(

g

,

m

)

=

1

then

g

is a primitive root of

m

if and only if

g

ϕ

⁢

(

m

)

/

q

≠

1

mod

m

for every

prime divisor

q

of

ϕ

⁢

(

m

)

.

3.

If

g

is a primitive root of

m

, then

g

s

≡

g

t

mod

m

if and only if

s

≡

t

mod

ϕ

⁢

(

m

)

. Thus

g

s

≡

1

mod

m

if and only if

ϕ

⁢

(

m

)

divides

s

.

4.

If

g

is a primitive root of

m

, then

g

k

is a primitive root of

m

if and only if

gcd

⁡

(

k

,

ϕ

⁢

(

m

)

)

=

1

.

5.

If

m

has a primitive root then

m

has exactly

ϕ

⁢

(

ϕ

⁢

(

m

)

)

incongruent primitive roots.

Title

properties of primitive roots

Canonical name

PropertiesOfPrimitiveRoots

Date of creation

2013-03-22 16:20:47

Last modified on

2013-03-22 16:20:47

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

4

Author

alozano (2414)

Entry type

Theorem

Classification

msc 11-00

Generated on Fri Feb 9 15:08:15 2018 by

LaTeXML
properties of the index of an integer with respect to a primitive root

properties of the index of an integer with respect to a primitive root

Definition

.

Let

m

>

1

be an

integer

such that the integer

g

is a

primitive root

for

m

. Suppose

a

is another integer relatively prime to

g

. The index of

a

(to base

g

) is the smallest

positive integer

n

such that

g

n

≡

a

mod

m

, and it is denoted by

ind

⁡

a

or

ind

g

⁡

a

.

Proposition

.

Suppose

g

is a primitive root of

m

.

1.

ind

⁡

1

≡

0

mod

ϕ

⁢

(

m

)

;

ind

⁡

g

≡

1

mod

ϕ

⁢

(

m

)

, where

ϕ

is the

Euler phi function

.

2.

a

≡

b

mod

m

if and only if

ind

⁡

a

≡

ind

⁡

b

mod

ϕ

⁢

(

m

)

.

3.

ind

⁡

(

a

⁢

b

)

≡

ind

⁡

a

+

ind

⁡

b

mod

ϕ

⁢

(

m

)

.

4.

ind

⁡

a

k

≡

k

⁢

ind

⁡

a

mod

ϕ

⁢

(

m

)

for any

k

≥

0

.

Title

properties of the index of an integer with respect to a primitive root

Canonical name

PropertiesOfTheIndexOfAnIntegerWithRespectToAPrimitiveRoot

Date of creation

2013-03-22 16:20:52

Last modified on

2013-03-22 16:20:52

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

4

Author

alozano (2414)

Entry type

Theorem

Classification

msc 11-00

Generated on Fri Feb 9 15:08:17 2018 by

LaTeXML
properties of the Legendre symbol

properties of the Legendre symbol

Let

p

be an

odd prime

and let

a

be an arbitrary

integer

. Let

(

a

p

)

be the

Legendre symbol

of

a

modulo

p

. Then:

Proposition

.

The following properties are satisfied:

1.

If

a

≡

b

mod

p

then

(

a

p

)

=

(

b

p

)

.

2.

If

a

≠

0

mod

p

then

(

a

2

p

)

=

1

.

3.

If

a

≠

0

mod

p

and

b

∈

ℤ

then

(

a

2

⁢

b

p

)

=

(

b

p

)

.

4.

(

a

p

)

⁢

(

b

p

)

=

(

a

⁢

b

p

)

.

Proof.

The first three properties are immediate from the definition of the Legendre symbol. Remember that

(

a

/

p

)

is

1

if

x

2

≡

a

mod

p

has solutions, the value is

-

1

if there are no solutions, and equals

0

if

a

≡

0

mod

p

.

The fourth property is a

consequence

of Euler’s criterion. Indeed,

(

a

p

)

≡

a

(

p

-

1

)

/

2

,

(

b

p

)

≡

b

(

p

-

1

)

/

2

,

and

⁢

(

a

⁢

b

p

)

≡

(

a

⁢

b

)

(

p

-

1

)

/

2

mod

p

.

It is clear then that

(

a

/

p

)

⁢

(

b

/

p

)

≡

(

a

⁢

b

/

p

)

mod

p

. Since
the numbers involved are all

±

1

or

0

, the

congruence

also
holds with equality in

ℤ

.
∎

Remark

.

Property (4) is somewhat surprising because, in particular, it says that the

product

of two

quadratic non-residues

modulo

p

is a

quadratic residue

modulo

p

, which is not at all obvious.

Title

properties of the Legendre symbol

Canonical name

PropertiesOfTheLegendreSymbol

Date of creation

2013-03-22 16:17:52

Last modified on

2013-03-22 16:17:52

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

4

Author

alozano (2414)

Entry type

Theorem

Classification

msc 11-00

Related topic

EulersCriterion

Generated on Fri Feb 9 15:08:19 2018 by

LaTeXML
properties of the multiplicative order of an integer

properties of the multiplicative order of an integer

Definition

.

Let

m

>

1

be an

integer

and let

a

be another integer relatively prime to

m

. The order of

a

modulo

m

(or the

multiplicative order

of

a

mod

m

) is the smallest

positive integer

n

such that

a

n

≡

1

mod

m

. The order is sometimes denoted by

ord

⁡

a

or

ord

m

⁡

a

.

Proposition

.

Let

m

be a positive integer and suppose that

(

a

,

m

)

=

1

.

1.

a

s

≡

1

mod

m

if and only if

ord

⁡

a

divides

s

. In particular,

ord

⁡

a

divides

ϕ

⁢

(

m

)

, where

ϕ

is the

Euler phi function

.

2.

a

s

≡

a

t

mod

m

if and only if

s

≡

t

mod

ord

⁡

a

.

3.

If

ord

⁡

a

=

d

then

ord

⁡

a

k

=

d

gcd

⁡

(

k

,

d

)

for any

k

≥

1

.

4.

If

ord

⁡

a

=

d

and

e

is a

positive

divisor

of

d

then

a

d

/

e

has exact order

e

.

5.

Suppose

ord

⁡

a

=

h

and

ord

⁡

b

=

k

with

gcd

⁡

(

h

,

k

)

=

1

. Then

ord

⁡

(

a

⁢

b

)

=

h

⁢

k

.

Title

properties of the multiplicative order of an integer

Canonical name

PropertiesOfTheMultiplicativeOrderOfAnInteger

Date of creation

2013-03-22 16:20:44

Last modified on

2013-03-22 16:20:44

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

4

Author

alozano (2414)

Entry type

Theorem

Classification

msc 11-00

Classification

msc 13M05

Classification

msc 13-00

Generated on Fri Feb 9 15:08:21 2018 by

LaTeXML
Pythagorean n-tuple

Pythagorean n-tuple

Let

n

≥

3

be an

integer

. A

Pythagorean n-tuple

is an n-tuple of integers

(

x

1

,

…

,

x

n

)

for which

∑

i

=

1

n

-

1

x

i

2

=

x

n

2

.

It is said to be if

gcd

⁢

(

x

1

,

…

,

x

n

)

=

1

.

If

n

=

3

the 3-tuple is called a

Pythagorean triplet

or

Pythagorean triple

.

Title

Pythagorean n-tuple

Canonical name

PythagoreanNtuple

Date of creation

2013-03-22 16:55:58

Last modified on

2013-03-22 16:55:58

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

7

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 11-00

Defines

primitive Pythagorean n-tuple

Generated on Fri Feb 9 15:08:22 2018 by

LaTeXML
Ramanujan’s formula for pi

Ramanujan’s formula for pi

Around

1910

, Ramanujan proved the following formula:

Theorem

.

The following series

converges

and the sum equals

1

π

:

1

π

=

2

⁢

2

9801

⁢

∑

n

=

0

∞

(

4

⁢

n

)

!

⁢

(

1103

+

26390

⁢

n

)

(

n

!

)

4

⁢

396

4

⁢

n

.

Needless to say, the

convergence

is extremely fast. For example, if we only use the term

n

=

0

we obtain the following approximation:

π

≈

9801

2

⋅

1103

⋅

2

=

3.14159273001

⁢

…

and the error is (in

absolute value

) equal to

0.0000000764235

⁢

…

In

1985

, William Gosper used this formula to calculate the first 17 million digits of

π

.

Another similar formula can be easily obtained from the

power series

of

arctan

⁡

x

. Although the convergence is good, it is not as impressive as in Ramanujan’s formula:

π

=

2

⁢

3

⁢

∑

n

=

0

∞

(

-

1

)

n

(

2

⁢

n

+

1

)

⁢

3

n

.

Title

Ramanujan’s formula for pi

Canonical name

RamanujansFormulaForPi

Date of creation

2013-03-22 15:53:41

Last modified on

2013-03-22 15:53:41

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

7

Author

alozano (2414)

Entry type

Theorem

Classification

msc 11-00

Classification

msc 51-00

Related topic

CyclometricFunctions

Generated on Fri Feb 9 15:08:26 2018 by

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rational points on one dimensional sphere

rational points on one dimensional sphere

Let

𝕊

1

=

{

(

x

,

y

)

∈

ℝ

2

|

x

2

+

y

2

=

1

}

be a one dimensional sphere.

We will denote by

𝕊

ℚ

1

=

{

(

x

,

y

)

∈

ℚ

2

|

(

x

,

y

)

∈

𝕊

1

}

the

rational

sphere. We shall try to describe

𝕊

ℚ

1

in terms of

Pythagorean triplets

.

Theorem

.

Let

(

x

,

y

)

∈

ℝ

2

. Then

(

x

,

y

)

∈

𝕊

ℚ

1

if and only if there exists a Pythagorean triplet

a

,

b

,

c

∈

ℤ

(i.e.

|

a

|

,

|

b

|

,

|

c

|

∈

ℕ

is a Pythagorean triplet) such that

x

,

y

are of the form

a

c

and

b

c

.

Proof.

,,

⇐

” If (for example)

x

=

a

c

and

y

=

b

c

for a Pythagorean triplet

a

,

b

,

c

∈

ℤ

, then we have

x

2

+

y

2

=

a

2

c

2

+

b

2

c

2

=

a

2

+

b

2

c

2

=

c

2

c

2

=

1

and thus

(

x

,

y

)

∈

𝕊

ℚ

1

.

,,

⇒

” Assume that

(

x

,

y

)

∈

𝕊

ℚ

1

. Then

x

=

p

q

for some

p

,

q

∈

ℤ

. It follows, that

1

=

x

2

+

y

2

=

p

2

q

2

+

y

2

and this is if and only if

y

=

q

2

-

p

2

q

(up to a sign of course). Therefore

y

∈

ℚ

if and only if

q

2

-

p

2

=

n

is an

integer

. In this case we have

x

=

p

q

,

y

=

n

q

.

Note, that

q

2

-

p

2

=

n

2

and thus

q

2

=

n

2

+

p

2

,

so

n

,

p

,

q

∈

ℤ

is a Pythagorean triplet, which

completes

the proof.

□

Corollary.

The rational sphere

𝕊

ℚ

1

is an

infinite set

.

Proof.

Let

m

,

n

∈

ℕ

be

natural numbers

such that

n

is fixed and even. Let

m

run through primes. Then (due to the theorem in parent entry)

2

⁢

m

⁢

n

,

m

2

-

n

2

,

m

2

+

n

2

∈

ℕ

is a Pythagorean triplet.
Let

x

m

=

2

⁢

m

⁢

n

m

2

+

n

2

=

2

⁢

n

m

+

n

2

m

.

Ir follows from the theorem, that there exists

y

m

∈

ℚ

such that

(

x

m

,

y

m

)

∈

𝕊

ℚ

1

. It is easy to see, that

x

m

=

x

m

′

if and only if

m

=

m

′

and thus we generated infinitely many

rational points

on sphere. This completes the proof.

□

Title

rational points on one dimensional sphere

Canonical name

RationalPointsOnOneDimensionalSphere

Date of creation

2013-03-22 19:07:49

Last modified on

2013-03-22 19:07:49

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

7

Author

joking (16130)

Entry type

Definition

Classification

msc 11-00

Related topic

RationalSineAndCosine

Generated on Fri Feb 9 15:08:28 2018 by

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residue systems

residue systems

Definition.

Let

m

be a

positive integer

.  A

complete residue system

modulo

m

is a set

{

a

0

,

a

1

,

…

,

a

m

-

1

}

of

integers

containing one and only one representant from every

residue class

modulo

m

.  Thus the numbers

a

ν

may be ordered such that for all

ν

=

0

,

1

,

…

,

m

-

1

,  they satisfy

a

ν

≡

ν

(

mod

m

)

.

The

least nonnegative

remainders

modulo

m

, i.e. the numbers

0

,

1

,

…

,

m

-

1

form a complete residue system modulo

m

(see long division).  If

m

is odd, then also the

absolutely least remainders

-

m

-

1

2

,

-

m

-

3

2

,

…

,

-

2

,

-

1

,

0

,

1

,

2

,

…

,

m

-

3

2

,

m

-

1

2

offer a complete residue system modulo

m

.

Theorem 1.

If

gcd

⁡

(

a

,

m

)

=

1

and

b

is an arbitrary integer, then the numbers

0

⁢

a

+

b

,

1

⁢

a

+

b

,

2

⁢

a

+

b

,

…

,

(

m

-

1

)

⁢

a

+

b

form a complete residue system modulo

m

.

One may speak also of a

reduced residue system

modulo

m

, which contains only one representant from each

prime residue class

modulo

m

.  Such a system consists of

φ

⁢

(

m

)

integers, where

φ

means the Euler’s

totient function

.

Remark.

A set of integers forms a reduced residue system modulo

m

if and only if

1

∘

it contains

φ

⁢

(

m

)

numbers,

2

∘

its numbers are

coprime

with

m

,

3

∘

its numbers are incongruent modulo

m

.

Theorem 2.

If

gcd

⁡

(

a

,

m

)

=

1

and

{

k

1

,

k

2

,

…

,

k

φ

⁢

(

m

)

}

is a reduced residue system modulo

m

, then also the numbers

k

1

⁢

a

,

k

2

⁢

a

,

…

,

k

φ

⁢

(

a

)

⁢

a

form a reduced residue system modulo

m

.

References

1

K. Väisälä:

Lukuteorian ja korkeamman algebran alkeet

.  Otava, Helsinki (1950).

Title

residue systems

Canonical name

ResidueSystems

Date of creation

2013-03-22 16:57:08

Last modified on

2013-03-22 16:57:08

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

12

Author

pahio (2872)

Entry type

Definition

Classification

msc 11-00

Related topic

PrimeResidueClass

Related topic

Totative

Related topic

MathbbZ_n

Defines

complete residue system

Defines

reduced residue system

Defines

least nonnegative remainder

Defines

absolutely least remainder

Generated on Fri Feb 9 15:08:30 2018 by

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