integer contraharmonic means

integer contraharmonic means

Let

u

and

v

be

positive integers

.  There exist nontrivial cases where their

contraharmonic mean

c

:=

u

2

+

v

2

u

+

v

(1)

is an

integer

, too.  For example, the values

u

=

3

,

v

=

15

have the contraharmonic mean

c

=

13

.  The only “

trivial

cases” are those with

u

=

v

,  when

c

=

u

=

v

.

u

2

3

3

4

4

5

5

6

6

6

6

7

7

8

8

8

9

9

…

v

6

6

15

12

28

20

45

12

18

30

66

42

91

24

56

120

18

45

…

c

5

5

13

10

25

17

41

10

15

26

61

37

85

20

50

113

15

39

…

The nontrivial

integer contraharmonic means

form Sloane’s

sequence

http://oeis.org/search?q=A146984&

language

=english&go=Search

A146984.

Proposition

1.

For any value of

u

>

2

, there are at least two

greater

values of

v

such that

c

∈

ℤ

.

Proof.

One has the

identities

u

2

+

(

(

u

-

1

)

⁢

u

)

2

u

+

(

u

-

1

)

⁢

u

=

u

2

-

2

⁢

u

+

2

,

(2)

u

2

+

(

(

2

⁢

u

-

1

)

⁢

u

)

2

u

+

(

2

⁢

u

-

1

)

⁢

u

=

2

⁢

u

2

-

2

⁢

u

+

1

,

(3)

the right hand sides of which are positive integers and different for

u

≠

1

.  The value

u

=

2

is an exception, since it has only

v

=

6

with which its contraharmonic mean is an integer.

In (2) and (3), the value of

v

is a

multiple

of

u

, but it needs not be always so in to

c

be an integer, e.g. we have

u

=

12

,

v

=

20

,

c

=

17

.

Proposition 2.

For all

u

>

1

, a necessary condition for

c

∈

ℤ

is that

gcd

⁡

(

u

,

v

)

>

1

.

Proof.

Suppose that we have positive integers

u

,

v

such that

gcd

⁡

(

u

,

v

)

=

1

.  Then as well,

gcd

⁡

(

u

+

v

,

u

⁢

v

)

=

1

,  since otherwise both

u

+

v

and

u

⁢

v

would be

divisible

by a prime

p

, and thus also one of the factors (

http://

planetmath

.org/

Product

)

u

and

v

of

u

⁢

v

would be divisible by

p

; then however

p

∣

u

+

v

would imply that

p

∣

u

and

p

∣

v

, whence we would have

gcd

⁡

(

u

,

v

)

≧

p

.  Consequently, we must have

gcd

⁡

(

u

+

v

,

u

⁢

v

)

=

1

.

We make the additional supposition that

u

2

+

v

2

u

+

v

is an integer, i.e. that

u

2

+

v

2

=

(

u

+

v

)

2

-

2

⁢

u

⁢

v

is divisible by

u

+

v

.  Therefore also

2

⁢

u

⁢

v

is divisible by this sum.  But because

gcd

⁡

(

u

+

v

,

u

⁢

v

)

=

1

, the factor 2 must be divisible by

u

+

v

, which is at least 2.  Thus

u

=

v

=

1

.

The

conclusion

is, that only the “most trivial case”

u

=

v

=

1

allows that

gcd

⁡

(

u

,

v

)

=

1

.  This settles the proof.

Proposition 3.

If

u

is an

odd prime

number, then (2) and (3) are the only possibilities enabling integer contraharmonic means.

Proof.

Let

u

be a

positive

odd prime.  The values

v

=

(

u

-

1

)

⁢

u

and

v

=

(

2

⁢

u

-

1

)

⁢

u

do always.  As for other possible values of

v

, according to the Proposition 2, they must be multiples of the

prime number

u

:

v

=

n

⁢

u

,

n

∈

ℤ

Now

ℤ

∋

u

2

+

v

2

u

+

v

=

(

n

2

+

1

)

⁢

u

n

+

1

,

and since

u

is prime, either

u

∣

n

+

1

or

n

+

1

∣

n

2

+

1

.

In the former case

n

+

1

=

k

⁢

u

,  one obtains

c

=

(

n

2

+

1

)

⁢

u

n

+

1

=

(

k

2

⁢

u

2

-

2

⁢

k

⁢

u

+

2

)

⁢

u

k

⁢

u

=

k

⁢

u

2

-

2

+

2

k

,

which is an integer only for

k

=

1

and

k

=

2

, corresponding (2) and (3).

In the latter case, there must be a prime number

p

dividing both

n

+

1

and

n

2

+

1

, whence

p

∤

n

.  The equation

n

2

+

1

=

(

n

+

1

)

2

-

2

⁢

n

then implies that

p

∣

2

n

.  So we must have

p

∣

2

,  i.e. necessarily

p

=

2

.  Moreover, if we had

4

∣

n

+

1

and

4

∣

n

2

+

1

,  then we could write

n

+

1

=

4

⁢

m

,  and thus

n

2

+

1

=

(

4

⁢

m

-

1

)

2

+

1

=

16

⁢

m

2

-

8

⁢

m

+

2

≢

0

(

mod

4

)

,

which is impossible.  We infer, that now

gcd

⁡

(

n

+

1

,

n

2

+

1

)

=

2

,  and in any case

gcd

⁡

(

n

+

1

,

n

2

+

1

)

≦

2

.

Nevertheless, since

n

+

1

≧

3

and

n

+

1

∣

n

2

+

1

,  we should have

gcd

⁡

(

n

+

1

,

n

2

+

1

)

≧

3

.  The

contradiction

means that the latter case is not possible, and the Proposition 3 has been proved.

Proposition 4.

If

(

u

1

,

v

,

c

)

is a nontrivial solution of (1) with

u

1

<

c

<

v

,  then there is always another nontrivial solution

(

u

2

,

v

,

c

)

with

u

2

<

v

.  On the contrary, if

(

u

,

v

1

,

c

)

is a nontrivial solution of (1) with

u

<

c

<

v

1

,  there exists no different solution

(

u

,

v

2

,

c

)

.

For example, there are the solutions

(

2

,

6

,

5

)

and

(

3

,

6

,

5

)

;

(

5

,

20

,

17

)

and

(

12

,

20

,

17

)

.

Proof.

The

Diophantine equation

(1) may be written

u

2

-

c

⁢

u

+

(

v

2

-

c

⁢

v

)

=

0

,

(4)

whence

u

=

c

±

c

2

+

4

⁢

c

⁢

v

-

4

⁢

v

2

2

,

(5)

and the

discriminant

of (4) must be nonnogative because of the

existence

of the real root (

http://planetmath.org/Equation

)

u

1

.  But if it were zero, i.e. if the equation

c

2

+

4

⁢

c

⁢

v

-

4

⁢

v

2

=

0

were true, this would imply for

v

the

irrational

value

1

2

⁢

(

1

+

2

)

⁢

c

.  Thus the discriminant must be positive, and then also the smaller root

u

of (4) gotten with “

-

” in front of the

square root

is positive, since we can rewrite it

c

-

c

2

+

4

⁢

c

⁢

v

-

4

⁢

v

2

2

=

c

2

-

(

c

2

+

4

⁢

c

⁢

v

-

4

⁢

v

2

)

2

⁢

(

c

+

c

2

+

4

⁢

c

⁢

v

-

4

⁢

v

2

)

=

2

⁢

(

v

-

c

)

⁢

v

c

+

c

2

+

4

⁢

c

⁢

v

-

4

⁢

v

2

and the

numerator

is positive because

v

>

c

.  Thus, when the discriminant of the equation (4) is positive, the equation has always two distinct positive roots

u

.  When one of the roots (

u

1

) is an integer, the other is an integer, too, because in the numerator of (5) the sum and the

difference

of two integers are simultaneously even.  It follows the existence of

u

2

, distinct from

u

1

.

If one solves (1) for

v

, the smaller root

c

-

c

2

+

4

⁢

c

⁢

u

-

4

⁢

u

2

2

=

2

⁢

(

u

-

c

)

⁢

u

c

+

c

2

+

4

⁢

c

⁢

u

-

4

⁢

u

2

is negative.  Thus there cannot be any

(

u

,

v

2

,

c

)

.

Proposition 5.

When the contraharmonic mean of two different positive integers

u

and

v

is an integer, their sum is never

squarefree

.

Proof.

By Proposition 2 we have

gcd

(

u

,

v

)

=

:

d

>

1

.

Denote

u

=

u

′

⁢

d

,

v

=

v

′

⁢

d

,

when

gcd

⁡

(

u

′

,

v

′

)

=

1

.  Then

c

=

(

u

′

⁣

2

+

v

′

⁣

2

)

⁢

d

u

′

+

v

′

,

whence

(

u

′

+

v

′

)

⁢

c

=

(

u

′

⁣

2

+

v

′

⁣

2

)

⁢

d

≡

[

(

u

′

+

v

′

)

2

-

2

⁢

u

′

⁢

v

′

]

⁢

d

.

(6)

If

p

is any odd prime factor of

u

′

+

v

′

, the last equation implies that

p

∤

u

′

,

p

∤

v

′

,

p

∤

[

]

,

and consequently

p

∣

d

.  Thus we see that

p

2

∣

(

u

′

+

v

′

)

d

=

u

+

v

.

This means that the sum

u

+

v

is not squarefree.  The same result is easily got also in the case that

u

and

v

both are even.

Note 1.

Cf.

u

+

v

=

c

+

b

in

2

∘

of
the proof of
this

theorem

(

http://planetmath.org/ContraharmonicMeansAndPythagoreanHypotenuses

)
and the Note 4 of

http://planetmath.org/node/138

this entry.

Proposition 6.

For each integer

u

>

0

there are only a finite number of solutions

(

u

,

v

,

c

)

of the Diophantine equation (1).  The number does not exceed

u

-

1

.

Proof.

The expression of the contraharmonic mean in (1) may be edited as follows:

c

=

(

u

+

v

)

2

-

2

⁢

u

⁢

v

u

+

v

=

u

+

v

-

2

⁢

u

⁢

(

u

+

v

-

u

)

u

+

v

=

v

-

u

+

2

⁢

u

2

u

+

v

In to

c

be an integer, the

quotient

w

:=

2

⁢

u

2

u

+

v

must be integer; rewriting this last equation as

v

=

2

⁢

u

2

w

-

u

(7)

we infer that

w

has to be a

http://planetmath.org/node/923

divisor

of

2

⁢

u

2

(apparently

1

≦

w

<

u

for getting values of

v

greater than

u

).  The amount of such divisors is quite restricted, not more than

u

-

1

, and consequently there is only a finite number of suitable values of

v

.

Note 2.

The equation (7) explains the result of Proposition 1 (

w

=

1

,

w

=

2

).  As well, if

u

is an odd prime number, then the only factors of

2

⁢

u

2

less than

u

are 1 and 2, and for these the equation (7) gives the values

v

:=

(

2

⁢

u

-

1

)

⁢

u

and

v

:=

(

u

-

1

)

⁢

u

which explains Proposition 3.

References

1

J. Pahikkala

: “On contraharmonic mean and

Pythagorean triples

”.  –

Elemente der Mathematik

65

:2 (2010).

Title

integer contraharmonic means

Canonical name

IntegerContraharmonicMeans

Date of creation

2013-12-04 10:25:44

Last modified on

2013-12-04 10:25:44

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

45

Author

pahio (2872)

Entry type

Topic

Classification

msc 11Z05

Classification

msc 11D45

Classification

msc 11D09

Classification

msc 11A05

Synonym

integer contraharmonic means of integers

Related topic

ComparisonOfPythagoreanMeans

Related topic

DivisibilityInRings

Related topic

Gcd

Defines

contraharmonic integer

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