aarjaneiro/planetmath huggingface dataset:
Euler product

Euler product

If

f

is a

multiplicative function

, then

∑

n

=

1

∞

f

⁢

(

n

)

=

∏

p

⁢

is prime

(

1

+

f

⁢

(

p

)

+

f

⁢

(

p

2

)

+

⋯

)

(1)

provided the sum on the left

converges absolutely

. The product
on the right is called the

Euler product

for the sum on the
left.

Proof of (

1

).

Expand partial products on right of (

1

) to
obtain by

fundamental theorem of arithmetic

∏

p

<

y

(

1

+

f

⁢

(

p

)

+

f

⁢

(

p

2

)

+

⋯

)

=

∑

k

1

f

⁢

(

p

1

k

1

)

⁢

∑

k

2

f

⁢

(

p

2

k

2

)

⁢

⋯

⁢

∑

k

t

f

⁢

(

p

t

k

t

)

=

∑

k

1

,

k

2

,

…

,

k

t

f

⁢

(

p

1

k

1

)

⁢

f

⁢

(

p

2

k

2

)

⁢

⋯

⁢

f

⁢

(

p

t

k

t

)

=

∑

k

1

,

k

2

,

…

,

k

t

f

⁢

(

p

1

k

1

⁢

p

2

k

2

⁢

⋯

⁢

p

t

k

t

)

=

∑

P

+

⁢

(

n

)

<

y

f

⁢

(

n

)

where

p

1

,

p

2

,

…

,

p

t

are all the primes between

1

and

y

, and

P

+

⁢

(

n

)

denotes the largest prime factor of

n

. Since
every natural number less than

y

has no factors exceeding

y

we have
that

|

∑

n

=

1

∞

f

⁢

(

n

)

-

∑

P

+

⁢

(

n

)

<

y

f

⁢

(

n

)

|

≤

∑

n

=

y

∞

|

f

⁢

(

n

)

|

which tends to zero as

y

→

∞

.
∎

Examples

•

If the

function

f

is defined on prime powers by

f

⁢

(

p

k

)

=

1

/

p

k

for all

p

<

x

and

f

⁢

(

p

k

)

=

0

for all

p

≥

x

,
then allows one to estimate

∏

p

<

x

(

1

+

1

/

(

p

-

1

)

)

∏

p

<

x

(

1

+

1

p

-

1

)

=

∏

p

<

x

(

1

+

1

p

+

1

p

2

+

⋯

)

=

∑

P

+

⁢

(

n

)

<

x

1

n

>

∑

n

<

x

1

n

>

ln

⁡

x

.

One of the consequences of this formula is that there are
infinitely many primes.

•

The

Riemann zeta function

is defined by the means of the
series

ζ

⁢

(

s

)

=

∑

n

=

1

∞

n

-

s

for

ℜ

⁡

s

>

1

.

Since the series converges absolutely, the Euler product for the

zeta function

is

ζ

⁢

(

s

)

=

∏

p

1

1

-

p

-

s

for

ℜ

⁡

s

>

1

.

If we set

s

=

2

, then on the one hand

ζ

⁢

(

s

)

=

∑

n

1

/

n

2

is

π

2

/

6

(proof is
here (

http://planetmath.org/ValueOfTheRiemannZetaFunctionAtS2

)), an

irrational number

, and on the other hand

ζ

⁢

(

2

)

is a product of

rational functions

of primes. This yields yet another

proof of infinitude of
primes

.

Title

Euler product

Canonical name

EulerProduct

Date of creation

2013-03-22 14:10:58

Last modified on

2013-03-22 14:10:58

Owner

bbukh (348)

Last modified by

bbukh (348)

Numerical id

8

Author

bbukh (348)

Entry type

Definition

Classification

msc 11A05

Classification

msc 11A51

Related topic

MultiplicativeFunction

Related topic

RiemannZetaFunction

Generated on Fri Feb 9 15:09:51 2018 by

LaTeXML
every even integer greater than 46 is the sum of two abundant numbers

every even integer greater than 46 is the sum of two abundant numbers

Theorem

.

Every sufficiently large

even integer

n

>

46

can be expressed as the sum of

abundant numbers

a

and

b

thus:

a

+

b

=

n

.

Proof.

First we rewrite

n

=

2

⁢

x

(where

x

is some

positive integer

) as

n

=

20

⁢

m

+

r

, where

r

satisfies

n

≡

r

mod

20

and

m

=

n

-

r

20

. If

r

=

0

, then we’re done, we can simply set

a

=

20

and

b

=

20

⁢

(

m

-

1

)

(or vice versa if preferred), thanks to the

theorem on multiples of abundant numbers

.

The other nine possible values of

r

are almost as easy to dispose of:

If

r

=

2

, then assign

a

=

20

⁢

(

m

-

2

)

and

b

=

42

. This works for

m

>

2

.

If

r

=

4

, then set

a

=

20

⁢

(

m

-

1

)

and

b

=

24

. This works for

m

>

1

.

If

r

=

6

, then

a

=

20

⁢

(

m

-

3

)

and

b

=

66

. This works for

m

>

3

.

If

r

=

8

, then

a

=

20

⁢

(

m

-

2

)

and

b

=

48

.

If

r

=

10

, then

a

=

20

⁢

(

m

-

1

)

and

b

=

30

.

If

r

=

12

, then

a

=

20

⁢

m

and

b

=

r

.

If

r

=

14

, then

a

=

20

⁢

(

m

-

5

)

and

b

=

114

. This works for

m

>

4

.

If

r

=

16

, then

a

=

20

⁢

(

m

-

1

)

and

b

=

36

.

If

r

=

18

, then

a

=

20

⁢

m

and

b

=

r

.

This leaves us 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 26, 28, 34, 46, 54, 74 and 94 to concern ourselves with. We can get rid of 54, 74 and 94 by simply subtracting 24 from each of them. 46 is the largest even value we can’t remove from this list. Thus it’s proven that all even

n

>

46

can be expressed as the sum of a pair of abundant numbers.
∎

We wish to generalize this to odd integers. Given the fact that the smallest odd abundant number is 945, by ”sufficiently large” we will have to mean an

integer

greater than that, and an empirical proof such as the one given above for even integers might have to use the second odd abundant number, 1575. This would give us almost eight hundred possible values of

r

to worry about, making a more abstract method appear more attractive.

Title

every even integer greater than 46 is the sum of two abundant numbers

Canonical name

EveryEvenIntegerGreaterThan46IsTheSumOfTwoAbundantNumbers

Date of creation

2013-03-22 16:06:17

Last modified on

2013-03-22 16:06:17

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

10

Author

PrimeFan (13766)

Entry type

Theorem

Classification

msc 11A05

Generated on Fri Feb 9 15:09:53 2018 by

LaTeXML
every even integer greater than 70 is the sum of two abundant numbers in more than one way

every even integer greater than 70 is the sum of two abundant numbers in more than one way

Theorem

Every even

n

>

70

can be expressed as

n

=

a

+

b

, with both

a

and

b

abundant numbers

, in more than one way. Due to the commutative property of

addition

, swaps of

a

and

b

are not counted as separate ways.

Proof.

To prove this it is enough to find just two ways for each even

n

>

70

, though of course there are plenty more ways as the numbers get larger, purely for our convenience we’ll seek to choose the smallest values for

b

possible. Since every

multiple

of a

perfect number

is an abundant number, and 6 is a perfect number, it follows that every multiple of 6 is abundant, and it is small enough a

modulus

that reviewing all possible cases should not prove tiresome.

If

n

=

6

⁢

m

and

m

>

5

, the two desired pairs are

a

=

6

⁢

(

m

-

2

)

,

b

=

12

, and

a

=

6

⁢

(

m

-

3

)

,

b

=

18

. This leaves us the cases

n

=

6

⁢

m

+

2

and

n

=

6

⁢

m

+

4

to concern ourselves with.

If

n

≡

2

mod

6

and

m

>

10

then the pairs are are

a

=

6

⁢

(

m

-

3

)

,

b

=

20

, and

a

=

6

⁢

(

m

-

9

)

,

b

=

56

.

If

n

≡

4

mod

6

and

m

>

12

then the pairs are

a

=

6

⁢

(

m

-

6

)

,

b

=

40

, and

a

=

6

⁢

(

m

-

11

)

,

b

=

70

.

The

lower bounds

of

m

have been chosen to ensure the

formulas

give distinct pairs of abundant numbers and never the perfect number 6 itself, but its multiples. These values of

m

correspond to the values of

n

36, 68, 82. To

complete

the proof we are left with the special case of

n

=

76

to examine on its own. Ignoring the bounds for

m

, the formulas above give us 76 = 40 + 36, a valid pair, and 76 = 70 + 6, which is not a pair of abundant numbers. But there is one other pair, 56 + 20, of which neither

a

nor

b

is a multiple of 6.
∎

The special case of 76 shows that there are solutions that don’t use multiples of 6. These become more readily available as the numbers get larger.

Title

every

even integer

greater than 70 is the sum of two abundant numbers in more than one way

Canonical name

EveryEvenIntegerGreaterThan70IsTheSumOfTwoAbundantNumbersInMoreThanOneWay

Date of creation

2013-03-22 17:44:14

Last modified on

2013-03-22 17:44:14

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

5

Author

PrimeFan (13766)

Entry type

Theorem

Classification

msc 11A05

Generated on Fri Feb 9 15:09:55 2018 by

LaTeXML
every sufficiently large even integer can be expressed as the sum of a pair of abundant numbers

every sufficiently large even integer can be expressed as the sum of a pair of abundant numbers

Theorem 1

.

If

n

>

1540539

, then

n

=

a

+

b

, where

a

and

b

are

abundant numbers

.

Proof.

Note that both

20

and

81081

are abundant numbers.
Furthermore, we have

81081

=

4054

⋅

20

+

1

.
If

n

is a

multiple

of

20

, then

n

-

20

is also
a multiple of

20

hence, as a multiple of an abundant
number, is also abundant, so we may choose

a

=

20

and

b

=

n

-

20

. Otherwise, write

n

=

20

⁢

m

+

k

where

m

and

k

are

positive

and

k

<

20

. Note that,
since

n

>

1540539

and

k

<

20

, it follows that

m

>

77026

>

4054

⁢

k

, hence we have

n

=

20

⁢

(

m

-

4054

⁢

k

)

+

81081

⁢

k

.

Since positive multiples of abundant numbers are
abundant, we may set

a

=

20

⁢

(

m

-

4054

⁢

k

)

and

b

=

81081

⁢

k

.
∎

Title

every sufficiently large

even integer

can be expressed as the sum of a pair of abundant numbers

Canonical name

EverySufficientlyLargeEvenIntegerCanBeExpressedAsTheSumOfAPairOfAbundantNumbers

Date of creation

2013-03-22 16:46:58

Last modified on

2013-03-22 16:46:58

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

9

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 11A05

Generated on Fri Feb 9 15:09:57 2018 by

LaTeXML
examples of smooth numbers

examples of smooth numbers

If a number is

k

-smooth, it is also a

smooth number

for smaller

k

. So, to give just a handful of

examples of smooth numbers

for a few different values of

k

, the following table only goes up to 100, and for

k

>

2

, only those numbers not already listed for smaller

k

are given. Also, the

k

have been limited to

prime numbers

. Obviously the 2-smooth numbers are the

powers of two

.

2

1, 2, 4, 8, 16, 32, 64

3

3, 6, 9, 12, 18, 24, 27, 36, 48, 54, 64, 72, 81, 96

5

5, 10, 15, 20, 25, 30, 40, 45, 50, 60, 75, 80, 90, 100

7

7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 84, 98

11

11, 22, 33, 44, 55, 66, 77, 88, 99

13

13, 26, 39, 52, 65, 78, 91

17

17, 34, 51, 68, 85

19

19, 38, 57, 57, 76, 95

Title

examples of smooth numbers

Canonical name

ExamplesOfSmoothNumbers

Date of creation

2013-03-22 18:10:25

Last modified on

2013-03-22 18:10:25

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

4

Author

PrimeFan (13766)

Entry type

Example

Classification

msc 11A05

Generated on Fri Feb 9 15:10:00 2018 by

LaTeXML
fast Euclidean algorithm

fast Euclidean algorithm

Given two

polynomials

of degree

n

with coefficients from a field

K

, the straightforward Eucliean

Algorithm

uses

O

⁢

(

n

2

)

field

operations

to compute their

greatest common divisor

. The

Fast Euclidean Algorithm

computes the same GCD in

O

⁢

(

𝖬

⁢

(

n

)

⁢

log

⁡

(

n

)

)

field operations, where

𝖬

⁢

(

n

)

is the time to multiply two

n

-degree polynomials; with FFT

multiplication

the GCD can thus be computed in time

O

⁢

(

n

⁢

log

2

⁡

(

n

)

⁢

log

⁡

(

log

⁡

(

n

)

)

)

. The algorithm can also be used to compute any particular pair of coefficients from the Extended

Euclidean Algorithm

, although computing every pair of coefficients would involve

O

⁢

(

n

2

)

outputs and so the

efficiency

is not as helpful when all are needed.

The algorithm can be made to work over

ℤ

but it is very tricky. A newer version that is easier to understand was published by Damien Stehlé and Paul Zimmerman, “A Binary

Recursive

Gcd Algorithm.”

Here we sketch the algorithm over

K

⁢

[

x

]

. The basic idea is that the

quotients

q

i

computed by the Euclidean Algorithm can usually be computed by looking at only the first few coefficients of the polynomial; for example, if

A

⁢

(

x

)

=

a

n

⁢

x

n

+

a

n

-

1

⁢

x

n

-

1

+

…

+

a

0

,

B

⁢

(

x

)

=

b

n

-

1

⁢

x

n

-

1

+

…

+

b

0

then

q

⁢

u

⁢

o

⁢

(

A

⁢

(

x

)

,

B

⁢

(

x

)

)

=

a

n

b

n

-

1

⁢

x

+

b

n

-

1

⁢

a

n

-

1

-

a

n

⁢

b

n

-

2

b

n

-

1

2

With more detailed

analysis

, we can show that in fact a divide-and-conquer approach can be used to

calculate

the GCD. First, we remove the terms whose degree is

n

/

2

or less from both polynomials

A

and

B

. Then, we recursively compute their GCD and

Euclidean

coefficients. We then apply the Euclidean coefficients to

A

and

B

, and recursively

complete

the Euclidean Algorithm.

The full algorithm, and a comprehensive runtime analysis is given in “Modern

Computer Algebra

” by von zur Gathen and Gerhard.

Title

fast Euclidean algorithm

Canonical name

FastEuclideanAlgorithm

Date of creation

2013-03-22 14:28:52

Last modified on

2013-03-22 14:28:52

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

7

Author

mathcam (2727)

Entry type

Algorithm

Classification

msc 11A05

Synonym

Half-GCD Algorithm

Synonym

Lehmer’s Algorithm

Related topic

EuclidsAlgorithm

Generated on Fri Feb 9 15:10:02 2018 by

LaTeXML
formula for sum of divisors

formula for sum of divisors

If one knows the factorization of a number,
one can compute the sum of the

positive

divisors

of
that number without having to write down
all the divisors of that number. To do
this, one can use a formula which is obtained
by

summing

a

geometric series

.

Theorem 1

.

Suppose that

n

is a

positive integer

whose factorization
into

prime factors

is

∏

i

=

1

k

p

i

m

i

,
where the

p

i

’s are distinct primes and the

multiplicities

m

i

are all at least

1

. Then
the sum of the divisors of

n

equals

∏

i

=

1

k

p

i

m

i

+

1

-

1

p

i

-

1

and the sum of the

proper divisors

of

n

equals

∏

i

=

1

k

p

i

m

i

+

1

-

1

p

i

-

1

-

∏

i

=

1

k

p

i

m

i

.

Proof.

A number will divide

n

if and only if prime
factors are also prime factors of

n

and
their multiplicity is less than to or equal
to their multiplicities in

n

. In other
words, a divisors

n

can be expressed
as

∏

i

=

1

k

p

i

μ

i

where

0

≤

μ

i

≤

m

i

. Then the sum over all divisors
becomes the sum over all possible choices
for the

μ

i

’s:

∑

d

∣

n

d

=

∑

0

≤

μ

i

≤

m

i

∏

i

=

1

k

p

i

μ

i

This sum may be expressed as a

multiple

sum like so:

∑

μ

1

=

0

m

1

∑

μ

2

=

0

m

2

⋯

⁢

∑

μ

k

=

0

m

k

∏

i

=

1

k

p

i

μ

i

This sum of products may be factored into
a product of sums:

∏

i

=

1

k

(

∑

μ

i

=

0

m

i

p

i

μ

i

)

Each of these sums is a geometric series;
hence we may use the formula for sum of a
geometric series to conclude

∑

d

∣

n

d

=

∏

i

=

1

k

p

i

m

i

+

1

-

1

p

i

-

1

.

If we want only proper divisors, we should
not include

n

in the sum, so we obtain
the formula for proper divisors by subtracting

n

from our formula.

∎

As an illustration, let us compute the sum
of the divisors of

1800

. The factorization
of our number is

2

3

⋅

3

2

⋅

5

2

.
Therefore, the sum of its divisors equals

(

2

4

-

1

2

-

1

)

⁢

(

3

3

-

1

3

-

1

)

⁢

(

5

3

-

1

5

-

1

)

=

15

⋅

26

⋅

124

2

⋅

4

=

6045

.

The sum of the proper divisors equals

6045

-

1800

=

4245

,  so we see that

1800

is an

abundant number

.

Title

formula for sum of divisors

Canonical name

FormulaForSumOfDivisors

Date of creation

2013-03-22 16:47:35

Last modified on

2013-03-22 16:47:35

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

14

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 11A05

Generated on Fri Feb 9 15:10:04 2018 by

LaTeXML
fundamental theorem of arithmetic

fundamental theorem of arithmetic

Each

positive integer

n

has a unique as a

product

n

=

∏

i

=

0

l

p

i

a

i

of

positive

powers of its distinct positive

prime divisors

p

i

. The

prime divisor

of

n

means a (

rational

)

prime number

dividing (

http://

planetmath

.org/

Divisibility

)

n

. A synonymous name is

prime factor

.

The of the prime divisors and for

n

=

1

is an empty product.

For some results it is useful to assume that

p

i

<

p

j

whenever

i

<

j

.

The FTA was the last of the fundamental theorems proven by C.F. Gauss. Gauss wrote his proof in “Discussions on

Arithmetic

” (

Disquisitiones Arithmeticae

) in 1801 formalizing

congruences

. Euclid and Greeks used prime properties of the FTA without rigorously proving its

existence

. It appears that the fundamentals of the FTA were used centuries before, and after, the Greeks within

Egyptian fraction

arithmetic.

Fibonacci

, for example, wrote in Egyptian fraction arithmetic, used three notations to detail

Euclidean

and medieval

factoring

methods.

Title

fundamental theorem of arithmetic

Canonical name

FundamentalTheoremOfArithmetic

Date of creation

2013-03-22 11:46:03

Last modified on

2013-03-22 11:46:03

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

21

Author

CWoo (3771)

Entry type

Theorem

Classification

msc 11A05

Classification

msc 17B66

Classification

msc 17B45

Related topic

Divisibility

Related topic

UFD

Related topic

AnyNonzeroIntegerIsQuadraticResidue

Related topic

NumberTheory

Defines

prime divisor

Defines

prime factor

Generated on Fri Feb 9 15:10:06 2018 by

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greatest common divisor of several integers

greatest common divisor of several integers

Theorem

.

If the

greatest common divisor

of the nonzero

integers

a

1

,

a

2

,

…

,

a

n

is

d

, then there exist the integers

x

1

,

x

2

,

…

,

x

n

such that

d

=

x

1

⁢

a

1

+

x

2

⁢

a

2

+

…

+

x

n

⁢

a

n

.

(1)

Proof.

In the case

n

=

2

the

Euclidean algorithm

for two nonzero integers

a

,

b

always guarantees the integers

x

,

y

such that

gcd

⁡

(

a

,

b

)

=

x

⁢

a

+

y

⁢

b

.

(2)

Make the

induction hypothesis

that the theorem is true whenever

n

<

k

.

Since obviously

d

=

gcd

⁡

(

a

1

,

a

2

,

…

,

a

k

)

=

gcd

⁡

(

gcd

⁡

(

a

1

,

a

2

,

…

,

a

k

-

1

)

,

a

k

)

,

we may write by (2) that

d

=

x

⁢

gcd

⁡

(

a

1

,

a

2

,

…

,

a

k

-

1

)

+

y

⁢

a

k

and thus by the induction hypothesis that

d

=

x

⁢

(

x

1

⁢

a

1

+

x

2

⁢

a

2

+

…

+

x

k

-

1

⁢

a

k

-

1

)

+

y

⁢

a

k

.

Consequently, we have gotten an equation of the form (1), Q.E.D.

Note.

An analogous theorem concerns elements of any Bézout domain (

http://planetmath.org/BezoutDomain

).

Title

greatest common divisor of several integers

Canonical name

GreatestCommonDivisorOfSeveralIntegers

Date of creation

2013-03-22 19:20:14

Last modified on

2013-03-22 19:20:14

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

5

Author

pahio (2872)

Entry type

Theorem

Classification

msc 11A05

Synonym

GCD of several integers

Related topic

BezoutsLemma

Related topic

IdealDecompositionInDedekindDomain

Generated on Fri Feb 9 15:10:08 2018 by

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highly abundant number

highly abundant number

An

integer

n

is a

highly abundant number

if

σ

⁢

(

n

)

>

σ

⁢

(

m

)

for all

m

<

n

(with

σ

being the

sum of divisors function

). The highly abundant numbers less than 100 are 1, 2, 3, 4, 6, 8, 10, 12, 16, 18, 20, 24, 30, 36, 42, 48, 60, 72, 84, 90, 96 (see A002093 in Sloane’s OEIS). Highly abundant numbers are like

highly composite numbers

except the definition for the latter uses the

divisor function

τ

instead of

σ

. The highly abundant numbers grow much more slowly than the highly composite numbers.

Though the first eight

factorials

are highly abundant, not all factorials are highly abundant. Two examples: 360360 is more abudant than 362880; and 3492720, 3538080, 3598560, 3603600 are all more abundant than 3628800.

Title

highly abundant number

Canonical name

HighlyAbundantNumber

Date of creation

2013-03-22 18:20:55

Last modified on

2013-03-22 18:20:55

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 15:10:10 2018 by

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hyperperfect number

hyperperfect number

A

hyperperfect number

n

for a given

k

is an

integer

such that

k

⁢

σ

⁢

(

n

)

=

-

1

+

k

+

(

k

+

1

)

⁢

n

, where

σ

⁢

(

x

)

is the

sum of divisors function

.

n

is then called

k

-hyperperfect. For example, 325 is 3-hyperperfect since its

divisors

(1, 5, 13, 25, 65, 325) add up to 434, and

3

×

434

=

-

1

+

3

+

(

3

+

1

)

⁢

325

=

1302

. Numbers that are 1-hyperperfect are by default called

perfect numbers

, since

1

⁢

σ

⁢

(

n

)

=

-

1

+

1

+

(

1

+

1

)

⁢

n

=

2

⁢

n

.

The 2-hyperperfect numbers are listed in A007593 of Sloane’s OEIS. As of 2007, the only known 3-hyperperfect number is 325. The two known 4-hyperperfect numbers are 1950625 and 1220640625, a

sequence

too short to list in the OEIS, and no 5-hyperperfect numbers are known to exist. The 6-hyperperfect numbers are listed in A028499.

References

1

Judson S. McCrainie, “A Study of Hyperperfect Numbers”

Journal of

Integer Sequences

3

(2000): 00.1.3

Title

hyperperfect number

Canonical name

HyperperfectNumber

Date of creation

2013-03-22 17:49:38

Last modified on

2013-03-22 17:49:38

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 15:10:11 2018 by

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incircle radius determined by Pythagorean triple

incircle radius determined by Pythagorean triple

If the sides of a

right triangle

are

integers

, then so is the radius of the

incircle

of this triangle.

For example, the incircle radius of the Egyptian triangle is 1.

Proof.

The sides of such a right triangle may be expressed by the integer

parametres

m

,

n

with

m

>

n

>

0

as

a

=

2

⁢

m

⁢

n

,

b

=

m

2

-

n

2

,

c

=

m

2

+

n

2

;

(1)

the radius of the incircle (

http://planetmath.org/Incircle

) is

r

=

2

⁢

A

a

+

b

+

c

,

(2)

where

A

is the area of the triangle.  Using (1) and (2) we obtain

r

=

2

⋅

2

⁢

m

⁢

n

⋅

(

m

2

-

n

2

)

/

2

2

⁢

m

⁢

n

+

(

m

2

-

n

2

)

+

(

m

2

+

n

2

)

=

2

⁢

m

⁢

n

⁢

(

m

+

n

)

⁢

(

m

-

n

)

2

⁢

m

⁢

(

m

+

n

)

=

(

m

-

n

)

⁢

n

,

which is a

positive integer

.

Remark.

The corresponding radius of the

circumcircle

need not to be integer, since by Thales’

theorem

, the radius is always half of the

hypotenuse

which may be odd (e.g. 5).

Title

incircle radius determined by Pythagorean triple

Canonical

name

IncircleRadiusDeterminedByPythagoreanTriple

Date of creation

2013-03-22 17:45:46

Last modified on

2013-03-22 17:45:46

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

14

Author

pahio (2872)

Entry type

Feature

Classification

msc 11A05

Synonym

incircle radius of right triangle

Related topic

Triangle

Related topic

PythagoreanTriple

Related topic

DifferenceOfSquares

Related topic

FirstPrimitivePythagoreanTriplets

Related topic

X4Y4z2HasNoSolutionsInPositiveIntegers

Generated on Fri Feb 9 15:10:13 2018 by

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Metrically Convex Space

Metrically Convex Space

Metrically Convex Space

A

metric space

(

M

,

d

)

is said to be metrically convex if for any
two points

x

,

y

∈

M

with

x

≠

y

there exists

z

∈

M

:

x

≠

z

≠

y

, such that

d

⁢

(

x

,

z

)

+

d

⁢

(

z

,

y

)

=

d

⁢

(

x

,

y

)

.

Generated on Fri Feb 9 15:10:14 2018 by

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