aarjaneiro/planetmath huggingface dataset:
centered hexagonal number

centered hexagonal number

A

centered hexagonal number

, or

hex number

is a

figurate number

that represents a hexagon with a dot in the center and all other dots surrounding the center dot equidistantly. The centered hexagonal number for

n

is given by the formula

1

+

6

⁢

(

1

2

⁢

n

⁢

(

n

+

1

)

)

. In other words, the centered hexagonal number for

n

is the

triangular number

for

n

multiplied by 6, then add 1.

The first few centered hexagonal numbers are: 1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, 469, 547, 631, 721, 817, 919, … listed in A003215 of Sloane’s OEIS.

To find centered hexagonal numbers besides 1 that are also triangular numbers or squares, it is necessary to solve

Diophantine equations

. By solving the Diophantine equation

1

2

⁢

m

⁢

(

m

+

1

)

=

3

⁢

n

2

+

3

⁢

n

+

1

,

we learn that 91, 8911 and 873181 are numbers that are both centered hexagonal numbers and triangular numbers (they grow very large after that), while solving the Diophantine equation

m

2

=

3

⁢

n

2

+

3

⁢

n

+

1

,

we learn that 169 and 32761 are centered hexagonal numbers that are also squares.

The sum of the first

n

centered hexagonal numbers is

n

3

. The

difference

between

(

2

⁢

n

)

2

and the

n

th centered hexagonal number is a number of the form

n

2

+

3

⁢

n

-

1

, while the difference between

(

2

n

-

1

)

)

2

and the

n

th centered hexagonal number is an

oblong number

.

Title

centered hexagonal number

Canonical

name

CenteredHexagonalNumber

Date of creation

2013-03-22 16:34:35

Last modified on

2013-03-22 16:34:35

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

5

Author

PrimeFan (13766)

Entry type

Definition

Classification

msc 11D09

Synonym

hex number

Generated on Fri Feb 9 15:31:20 2018 by

LaTeXML
contraharmonic Diophantine equation

contraharmonic Diophantine equation

We call

contraharmonic Diophantine equation

the equation

u

2

+

v

2

=

(

u

+

v

)

⁢

c

(1)

of the three unknowns

u

,

v

,

c

required to get only

positive integer

values.  The equation expresses that

c

is the

contraharmonic
mean

of

u

and

v

.  As proved in the article
“

contraharmonic means and Pythagorean
hypotenuses

”, the supposition

u

≠

v

implies that the
number

c

must be the

hypotenuse

in a

Pythagorean triple

(

a

,

b

,

c

)

, and if particularly

u

<

v

, then

u

=

c

+

b

-

a

2

,

v

=

c

+

b

+

a

2

.

(2)

For getting the general solution of the quadratic

Diophantine
equation

(1), one can utilise the general

formulas

for
Pythagorean triples

a

=

l

⋅

(

m

2

-

n

2

)

,

b

=

l

⋅

2

⁢

m

⁢

n

,

c

=

l

⋅

(

m

2

+

n

2

)

(3)

where the parameters

l

,

m

,

n

are arbitrary positive
integers with

m

>

n

.  Using (3) in (2) one obtains the
result

{

u

1

=

l

⁢

(

m

2

-

m

⁢

n

)

,

u

2

=

l

⁢

(

n

2

+

m

⁢

n

)

,

v

=

l

⁢

(

m

2

+

m

⁢

n

)

,

c

=

l

⁢

(

m

2

+

n

2

)

,

(4)

in which

u

1

and

u

2

mean the alternative values for

u

gotten from (2) by swapping
the expressions of

a

and

b

in (3).

It’s clear that the contraharmonic Diophantine equation has an

infinite set

of solutions (4).  According to the

Proposition

6 of the article “

integer contraharmonic means

”, fixing e.g.
the variable

u

allows for the equation only a restricted
number of pertinent values

v

and

c

.  See also the
alternative expressions (1) and (2) in the article “

sums of
two squares

”.

Title

contraharmonic Diophantine equation

Canonical name

ContraharmonicDiophantineEquation

Date of creation

2013-11-19 21:49:13

Last modified on

2013-11-19 21:49:13

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

5

Author

pahio (2872)

Entry type

Derivation

Classification

msc 11D09

Classification

msc 11D45

Generated on Fri Feb 9 15:31:23 2018 by

LaTeXML
derivation of Pythagorean triples

derivation of Pythagorean triples

For finding all

positive

solutions of the

Diophantine equation

x

2

+

y

2

=

z

2

(1)

we first can determine such triples

x

,

y

,

z

which are

coprime

.  When these are then multiplied by all

positive integers

, one obtains all positive solutions.

Let

(

x

,

y

,

z

)

be a solution of the mentioned kind.  Then the numbers are

pairwise coprime

, since by (1), a common

divisor

of two of them is also a common divisor of the third.  Especially,

x

and

y

cannot both be even.  Neither can they both be odd, since because the square of any

odd number

is

≡

1

(

mod

4

)

, the equation (1) would imply an impossible

congruence

2

≡

z

2

(

mod

4

)

.  Accordingly, one of the numbers, e.g.

x

, is even and the other,

y

, odd.

Write (1) to the form

x

2

=

(

z

+

y

)

⁢

(

z

-

y

)

.

(2)

Now, both factors (

http://planetmath.org/

Product

) on the right hand side are even, whence one may denote

z

+

y

=

:

2

u

,

z

-

y

=

:

2

v

(3)

giving

z

=

u

+

v

,

y

=

u

-

v

,

(4)

and thus (2) reads

x

2

=

4

⁢

u

⁢

v

.

(5)

Because

z

and

y

are coprime and

z

>

y

>

0

,  one can infer from (4) and (3) that also

u

and

v

must be coprime and

u

>

v

>

0

.  Therefore, it follows from (5) that

u

=

m

2

,

v

=

n

2

where

m

and

n

are coprime and

m

>

n

>

0

.  Thus, (5) and (4) yield

x

=

2

⁢

m

⁢

n

,

y

=

m

2

-

n

2

,

z

=

m

2

+

n

2

.

(6)

Here, one of

m

and

n

is odd and the other even, since

y

is odd.

By substituting the

expressions

(6) to the equation (1), one sees that it is satisfied by arbitrary values of

m

and

n

.  If

m

and

n

have all the

properties

stated above, then

x

,

y

,

z

are positive integers and, as one may deduce from two first of the equations (6), the numbers

x

and

y

and thus all three numbers are coprime.

Thus one has proved the

Theorem

.

All coprime positive solutions

x

,

y

,

z

,  and only them, are gotten when one substitutes for

m

and

n

to the formulae (6) all possible coprime value pairs, from which always one is odd and the other even and

m

>

n

.

References

1

K. Väisälä

:

Lukuteorian ja korkeamman algebran alkeet

.  Tiedekirjasto No. 17. Kustannusosakeyhtiö Otava, Helsinki (1950).

Title

derivation of Pythagorean triples

Canonical

name

DerivationOfPythagoreanTriples

Date of creation

2013-03-22 18:34:40

Last modified on

2013-03-22 18:34:40

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Derivation

Classification

msc 11D09

Classification

msc 11A05

Related topic

LinearFormulasForPythagoreanTriples

Related topic

ContraharmonicMeansAndPythagoreanHypotenuses

Generated on Fri Feb 9 15:31:30 2018 by

LaTeXML
hexagonal number

hexagonal number

A

hexagonal number

is a

figurate number

that represents a hexagon. The hexagonal number for

n

is given by the formula

n

⁢

(

2

⁢

n

-

1

)

. The first few hexagonal numbers are
1, 6, 15, 28, 45, 66, 91, 120, 153, 190, 231, 276, 325, 378, 435, 496, 561, 630, 703, 780, 861, 946, etc., listed in A000384 of Sloane’s OEIS. Like a

triangular number

, the base 10

digital root

of a hexagonal number can only be 1, 3, 6 or 9. Every hexagonal number is a triangular number, but not every triangular number is a hexagonal number.

Any

integer

greater than 1791 can be expressed as a sum of at most four hexagonal numbers, a fact proven by

Adrien-Marie Legendre

in 1830.

Hexagonal numbers should not be confused with

centered hexagonal numbers

, which model the standard packaging of Vienna sausages.

Title

hexagonal number

Canonical name

HexagonalNumber

Date of creation

2013-03-22 17:50:51

Last modified on

2013-03-22 17:50:51

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

4

Author

PrimeFan (13766)

Entry type

Definition

Classification

msc 11D09

Generated on Fri Feb 9 15:31:32 2018 by

LaTeXML
Pell’s equation and simple continued fractions

Pell’s equation and simple continued fractions

Theorem 1

.

Let

d

be a

positive integer

which is not a

perfect square

, and let

(

x

,

y

)

be
a solution of

x

2

-

d

⁢

y

2

=

1

. Then

x

y

is a

convergent

in the

simple
continued fraction

expansion of

d

.

Proof.

Suppose we have a non-trivial solution

x

,

y

of Pell’s equation, i.e.

y

≠

0

. Let

x

,

y

both be positive integers. From

(

x

y

)

2

=

d

+

1

y

2

we see that

(

x

y

)

2

>

d

, hence

x

y

>

d

. So we get

|

x

y

-

d

|

=

1

y

2

⁢

(

x

y

+

d

)

<

1

y

2

⁢

(

2

⁢

d

)

<

1

2

⁢

y

2

.

This implies that

x

y

is a convergent of the

continued fraction

of

d

.
∎

Title

Pell’s equation and simple continued fractions

Canonical name

PellsEquationAndSimpleContinuedFractions

Date of creation

2013-03-22 13:21:04

Last modified on

2013-03-22 13:21:04

Owner

Thomas Heye (1234)

Last modified by

Thomas Heye (1234)

Numerical id

9

Author

Thomas Heye (1234)

Entry type

Theorem

Classification

msc 11D09

Generated on Fri Feb 9 15:31:37 2018 by

LaTeXML
Pythagorean triangle

Pythagorean triangle

The side lengths of any

right triangle

satisfy the equation of the

Pythagorean theorem

,
but if they are

integers

then the triangle is a

Pythagorean triangle

.

The side lengths are said to form a

Pythagorean triple

.  They are always different
integers, the smallest of them being at least 3.

Any Pythagorean triangle has the

property

that the

hypotenuse

is
the

contraharmonic mean

c

=

u

2

+

v

2

u

+

v

(1)

and one

cathetus

is the

harmonic mean

h

=

2

⁢

u

⁢

v

u

+

v

(2)

of a certain pair of distinct

positive integers

u

,

v

; the
other cathetus is simply

|

u

-

v

|

.

If there is given the value of

c

as the length of the
hypotenuse and a compatible value

h

as the length of one
cathetus, the pair of equations (1) and (2) does not determine
the numbers

u

and

v

uniquely (cf. the

Proposition

4 in the
entry

integer contraharmonic means

).  For example, if

c

=

61

and

h

=

11

, then the equations give for

(

u

,

v

)

either

(

6

,

66

)

or

(

55

,

66

)

.

As for the hypotenuse and (1), the proof is found in [1] and also
in the

PlanetMath

article

contraharmonic means and Pythagorean
hypotenuses

.  The contraharmonic and the harmonic mean of two
integers are simultaneously integers (see
this article (

http://planetmath.org/IntegerHarmonicMeans

)).  The above
claim concerning the catheti of the Pythagorean triangle is
evident from the identity

(

2

⁢

u

⁢

v

u

+

v

)

2

+

|

u

-

v

|

2

=

(

u

2

+

v

2

u

+

v

)

2

.

If the catheti of a Pythagorean triangle are

a

and

b

,
then the values of the

parameters

u

and

v

determined by
the equations (1) and (2) are

c

+

b

±

a

2

(3)

as one instantly sees by substituting them into the equations.

References

1

J. Pahikkala

: “On contraharmonic mean and Pythagorean triples”.  –

Elemente der Mathematik

65

:2 (2010).

Title

Pythagorean triangle

Canonical

name

PythagoreanTriangle

Date of creation

2013-11-23 11:53:13

Last modified on

2013-11-23 11:53:13

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

15

Author

pahio (2872)

Entry type

Result

Classification

msc 11D09

Classification

msc 51M05

Generated on Fri Feb 9 15:31:47 2018 by

LaTeXML
test for hexagonal numbers

test for hexagonal numbers

Given an arbitrary

positive integer

n

, one can determine whether or not it is a

hexagonal number

by calculating

x

=

1

+

1

+

8

⁢

n

4

.

If

x

is an

integer

, (that is,

x

∈

ℤ

), then

n

is a hexagonal number. If

n

is a

triangular number

but not a hexagonal number, then

x

=

y

2

, with

y

being an

odd integer

. In all other cases,

x

will be an

irrational number

.

It will suffice to work out an example of each. We choose 1729, 277770576188160 and 1900221452387519291741168640. With the exception of 1729, these numbers have been deliberately chosen so that they would be so big that in Sloane’s OEIS they would be crowded out by much smaller numbers.

1729 has many properties, and it is relevant here to note that is 12-gonal and 24-gonal. But is it hexagonal?

x

=

1

+

1

+

8

×

1729

4

=

1

+

13833

4

=

1

+

3

⁢

1537

4

≈

29.653443675868988254

Though 1729 is a

figurate number

in a few different ways, hexagonal is not among them.

Given

n

set to 277770576188160,

x

=

23569921

2

. Clearly this is a

rational number

, but not an integer. This tells us that 277770576188160 is a triangular number but not a hexagonal number.

Given 1900221452387519291741168640, our

x

is 30823866178560, an integer. If we plug in this integer into the

formula

for hexagonal numbers, we should get 1900221452387519291741168640 back, confirming that this number is in fact the 30823866178560th hexagonal number.

Title

test for hexagonal numbers

Canonical name

TestForHexagonalNumbers

Date of creation

2013-03-22 17:50:54

Last modified on

2013-03-22 17:50:54

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

6

Author

PrimeFan (13766)

Entry type

Result

Classification

msc 11D09

Generated on Fri Feb 9 15:31:48 2018 by

LaTeXML
Beal conjecture

Beal conjecture

The

Beal conjecture

:

Let

A

,

B

,

C

,

x

,

y

,

z

be nonzero

integers

such that

x

,

y

, and

z

are all

≥

3

, and

A

x

+

B

y

=

C

z

(1)

Then

A

,

B

, and

C

(or any two of them) are not relatively prime.

It is clear that the famous statement known as Fermat’s Last

Theorem

would follow from this stronger claim.

Solutions of equation (1) are not very scarce. One parametric solution is

[

a

⁢

(

a

m

+

b

m

)

]

m

+

[

b

⁢

(

a

m

+

b

m

)

]

m

=

(

a

m

+

b

m

)

m

+

1

for

m

≥

3

, and

a

,

b

such that the are nonzero.
But computerized searching brings
forth quite a few additional solutions, such as:

3

3

+

6

3

=

3

5

3

9

+

54

3

=

3

11

3

6

+

18

3

=

3

8

7

6

+

7

7

=

98

3

27

4

+

162

3

=

9

7

211

3

+

3165

3

=

422

4

386

3

+

4825

3

=

579

4

307

3

+

614

4

=

5219

3

5400

3

+

90

4

=

630

4

217

3

+

5642

3

=

651

4

271

3

+

813

4

=

7588

3

602

3

+

903

4

=

8729

3

624

3

+

14352

3

=

312

5

1862

3

+

57722

3

=

3724

4

2246

3

+

4492

4

=

74118

3

1838

3

+

97414

3

=

5514

4

Mysteriously, the summands have a common factor

>

1

in each instance.

Dan Vanderkam has verified the Beal conjecture for all values of all six

variables

up to 1000, and he provides source code for anyone who wants to repeat the verification for himself. A 64-bit machine is required. See http://www.owlnet.rice.edu/ danvk/beal.html

This

conjecture

is “wanted in Texas, dead or alive”. For the details,
plus some additional , see

http://www.math.unt.edu/ mauldin/beal.html

Mauldin.

Title

Beal conjecture

Canonical

name

BealConjecture

Date of creation

2013-03-22 13:16:53

Last modified on

2013-03-22 13:16:53

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

16

Author

mathcam (2727)

Entry type

Conjecture

Classification

msc 11D41

Synonym

Beal’s conjecture

Generated on Fri Feb 9 15:31:54 2018 by

LaTeXML
Fermat’s last theorem (analytic form of)

Fermat’s last theorem (analytic form of)

Let

x

,

y

,

z

be positive real numbers.

For each

positive integer

r

, let

a

r

=

(

x

r

+

y

r

)

/

r

!

and

b

r

=

z

r

/

r

!

.

For

s

divisible

by 4, let

A

s

=

a

2

-

a

4

+

a

6

-

⋯

+

a

s

-

2

-

a

s

,

B

s

=

b

2

-

b

4

+

b

6

-

⋯

+

b

s

-

2

-

b

s

.

Then Fermat’s last

theorem

is

equivalent

(by

elementary

means) to:

Theorem

If

a

n

=

b

n

for some

odd integer

n

>

2

, then either

(i)

A

N

>

0

for some

N

>

x

,

y

,

or

(ii)

B

M

>

0

for some

M

>

z

.

For a proof that these theorems are equivalent see:

proof of equivalence of Fermat’s Last Theorem to its

analytic

form

Title

Fermat’s last theorem (analytic form of)

Canonical name

FermatsLastTheoremanalyticFormOf

Date of creation

2013-03-22 16:17:34

Last modified on

2013-03-22 16:17:34

Owner

whm22 (2009)

Last modified by

whm22 (2009)

Numerical id

8

Author

whm22 (2009)

Entry type

Theorem

Classification

msc 11D41

Generated on Fri Feb 9 15:31:56 2018 by

LaTeXML
proof of equivalence of Fermat’s Last Theorem to its analytic form

proof of equivalence of Fermat’s Last Theorem to its analytic form

Consider the

Taylor expansion

of the cosine function. We have

lim

s

→

∞

⁢

(

A

s

)

=

2

-

cos

⁢

x

-

cos

⁢

y

and

lim

s

→

∞

⁢

(

B

s

)

=

1

-

cos

⁢

z

.

For

r

>

x

,

y

the

sequence

a

r

is decreasing as the

denominator

grows faster than the numerator.
Hence for

s

>

x

,

y

the sequence

A

s

is increasing as

A

s

+

4

=

A

s

+

a

s

+

2

-

a

s

+

4

and

a

s

+

2

>

a

s

+

4

. So if

A

N

>

0

for some

N

>

x

,

y

, we have

2

-

cos

⁢

x

-

cos

⁢

y

>

0

.

Conversely

if no such

N

exists then

A

s

≤

0

for

s

>

x

,

y

, so its limit,

2

-

cos

⁢

x

-

cos

⁢

y

, is also less than or equal to

0

. However as this
expression cannot be

negative

we would have

2

-

cos

⁢

x

-

cos

⁢

y

=

0

.

Similarly for

r

>

z

the sequence

b

r

is decreasing, and for

s

>

z

the sequence

B

s

is
increasing. So if

B

M

>

0

for some

M

>

z

we have

1

-

cos

⁢

z

>

0

.
Conversely if no such

M

exists then

1

-

cos

⁢

z

≤

0

. However as this
expression cannot be negative we would have

1

-

cos

⁢

z

=

0

.

Note that

2

-

cos

⁢

x

-

cos

⁢

y

=

0

precisely when

x

,

y

∈

2

⁢

π

⁢

ℤ

. Also

1

-

cos

⁢

z

=

0

precisely when

z

∈

2

⁢

π

⁢

ℤ

.

So the form of the

theorem

may be read:

If for positive reals

x

,

y

,

z

we have

x

n

+

y

n

=

z

n

for some

odd integer

n

>

2

, then either

x

or

y

not in

2

⁢

π

⁢

ℤ

or

z

not in

2

⁢

π

⁢

ℤ

.

Clearly this only fails if for

positive integers

a

,

b

,

c

and some odd

n

>

2

, we have

(

2

⁢

π

⁢

a

)

n

+

(

2

⁢

π

⁢

b

)

n

=

(

2

⁢

π

⁢

c

)

n

.

Dividing through by

(

2

⁢

π

)

n

we see that

a

n

+

b

n

=

c

n

.

Conversely suppose we have non-zero

integers

satisfying

a

n

+

b

n

=

c

n

for some

n

>

2

. If

n

=

4

⁢

k

we have

(

a

k

)

4

+

(

b

k

)

4

=

(

c

k

)

4

, contradicting
example of Fermat’s last theorem. Hence if

n

is even we may replace

a

,

b

,

c

with

a

2

,

b

2

,

c

2

and

n

with

n

/

2

, which will be odd and greater than 1 (and hence greater than 2 as it is odd). So without loss of generality we may
assume

n

odd.

Finally replace

a

,

b

,

c

with their

absolute values

and if
reorder to obtain a positive
integer solution. This would be a counterexample to the
form of the theorem as stated above.

Title

proof of equivalence of Fermat’s Last Theorem to its

analytic

form

Canonical name

ProofOfEquivalenceOfFermatsLastTheoremToItsAnalyticForm

Date of creation

2013-03-22 16:19:04

Last modified on

2013-03-22 16:19:04

Owner

whm22 (2009)

Last modified by

whm22 (2009)

Numerical id

7

Author

whm22 (2009)

Entry type

Proof

Classification

msc 11D41

Generated on Fri Feb 9 15:31:58 2018 by

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Catalan’s conjecture

Catalan’s conjecture

The successive

positive integers

8 and 9 are

integer

powers of positive integers (

2

3

and

3

2

), with exponents greater than 1.

Catalan’s conjecture

(1844) said that there are no other such successive positive integers, i.e. that the only integer solution of the

Diophantine equation

x

m

-

y

n

=

1

with

x

>

1

,

y

>

1

,

m

>

1

,

n

>

1

is

x

=

n

=

3

,

y

=

m

=

2

.

It took more than 150 years before the conjecture was proven. Mihailescu gave in 2002 a proof in which he used the theory of

cyclotomic fields

and Galois modules.

For details, see e.g.

http://www.ams.org/journals/bull/2004-41-01/S0273-0979-03-00993-5/S0273-0979-03-00993-5.pdf

this article.

See also a related problem concerning the equation

x

y

=

y

x

(

http://planetmath.org/solutionsofxyyx

).

Title

Catalan’s conjecture

Canonical name

CatalansConjecture

Date of creation

2014-12-16 16:16:07

Last modified on

2014-12-16 16:16:07

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Conjecture

Classification

msc 11D45

Classification

msc 11D61

Synonym

Mihailescu’s theorem

Related topic

FermatsLastTheorem

Related topic

SolutionsOfXyYx

Generated on Fri Feb 9 15:32:14 2018 by

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perfect power

perfect power

The power

m

n

is called a

perfect power

if

m

and

n

are

integers

both

greater

than 1. The perfect powers form the

ascending order

sequence

(cf. Sloane’s

http://oeis.org/classic/A001597

A001597)

4

,

8

,

9

,

16

,

25

,

27

,

32

,

36

,

49

,

64

,

81

,

100

,

121

,

125

,

…

,

(1)

i.e.

2

2

,

2

3

,

3

2

,

2

4

=

4

2

,

5

2

,

3

3

,

2

5

,

6

2

,

7

2

,

2

6

=

4

3

=

8

2

,

3

4

=

9

2

,

10

2

,

11

2

,

5

3

,

…

S. S. Pillai has conjectured in 1945, that if the

i

th

member of the sequence (1) is denoted by

a

i

, then

lim inf

i

→

∞

⁡

(

a

i

+

1

-

a

i

)

=

∞

.

(2)

This does not necessarily that one had

lim

i

→

∞

⁡

(

a

i

+

1

-

a

i

)

=

∞

,  since there may always exist little

differences

a

i

+

1

-

a

i

arbitrarily far from the begin of the sequence (1).

The equation (2) is

equivalent

(

http://planetmath.org/Equivalent3

) to the

Pillai’s

conjecture

.

For any

positive integer

k

, the

Diophantine equation

x

m

-

y

n

=

k

has only a finite number of solutions

(

x

,

y

,

m

,

n

)

where the integers

x

,

y

,

m

,

n

all are greater than 1.

Pillai’s conjecture generalises the Catalan’s conjecture (

k

=

1

) in which the number of solutions is 1.

The series formed by the

inverse numbers

of the perfect powers

converges absolutely

, and its sum may be calculated easily:

∑

m

,

n

=

2

∞

1

m

n

=

∑

m

=

2

∞

∑

n

=

2

∞

1

m

n

=

∑

m

=

2

∞

∑

n

=

2

∞

1

m

2

⋅

1

m

n

-

2

=

∑

m

=

2

∞

1

m

2

⁢

∑

n

=

2

∞

1

m

n

-

2

=

∑

m

=

2

∞

1

m

2

⁢

∑

n

=

0

∞

(

1

m

)

n

=

∑

m

=

2

∞

1

m

2

⋅

1

1

-

1

m

=

∑

m

=

2

∞

1

m

⁢

(

m

-

1

)

=

∑

m

=

2

∞

(

1

m

-

1

-

1

m

)

The sum of this telescoping series (

http://planetmath.org/TelescopingSum

) is equal to 1.

References

1

S. S. Pillai

:  On

a

x

-

b

y

=

c

.  –

J. Indian math. Soc.

2

(1936).

Title

perfect power

Canonical

name

PerfectPower

Date of creation

2013-03-22 19:15:51

Last modified on

2013-03-22 19:15:51

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

11

Author

pahio (2872)

Entry type

Definition

Classification

msc 11D61

Classification

msc 11D45

Classification

msc 11B83

Synonym

Pillai’s conjecture

Related topic

PerfectSquare

Related topic

LimitInferior

Related topic

DoubleSeries

Related topic

SolutionsOfXyYx

Related topic

MarshallHallsConjecture

Generated on Fri Feb 9 15:32:18 2018 by

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