values of n for which φ⁢(n)=τ⁢(n)

values of

n

for which

φ

⁢

(

n

)

=

τ

⁢

(

n

)

Within this entry, we use the following notation:

•

ℕ

denotes the

natural numbers

(

positive integers

)

•

n

∈

ℕ

•

p

denotes a prime

•

k

∈

ℕ

∪

{

0

}

•

φ

denotes the

Euler phi function

•

τ

denotes the

divisor function

•

∣

denotes divides

•

∥

denotes exactly divides

Within this entry, we will determine all values of

n

for which

φ

⁢

(

n

)

=

τ

⁢

(

n

)

.

Define

γ

:

ℕ

→

ℚ

by

γ

⁢

(

n

)

=

τ

⁢

(

n

)

φ

⁢

(

n

)

.

Note that

γ

is a

multiplicative function

since both

φ

and

τ

are. Thus, we will initially focus on the values of

γ

at prime powers. We will need specific values of

γ

⁢

(

p

k

)

. These are calculated below.

γ

⁢

(

1

)

=

1

1

=

1

γ

⁢

(

2

)

=

2

1

=

2

γ

⁢

(

4

)

=

3

2

γ

⁢

(

8

)

=

4

4

=

1

γ

⁢

(

16

)

=

5

8

γ

⁢

(

32

)

=

6

16

=

3

8

γ

⁢

(

3

)

=

2

2

=

1

γ

⁢

(

9

)

=

3

6

=

1

2

γ

⁢

(

5

)

=

2

4

=

1

2

Note that

γ

⁢

(

p

k

)

=

τ

⁢

(

p

k

)

φ

⁢

(

p

k

)

=

k

+

1

p

k

-

1

⁢

(

p

-

1

)

.

If

p

is fixed, we can extend this to a

continuous function

Γ

p

:

ℝ

→

ℝ

defined by

Γ

p

⁢

(

x

)

=

x

+

1

p

x

-

1

⁢

(

p

-

1

)

.

We investigate the

derivative

(

http://planetmath.org/Derivative

) of

Γ

p

for

x

≥

1

:

Γ

p

′

⁢

(

x

)

=

1

p

-

1

⋅

p

x

-

1

-

(

x

+

1

)

⁢

p

x

-

1

⁢

ln

⁡

p

(

p

x

-

1

)

2

=

1

-

(

x

+

1

)

⁢

ln

⁡

p

p

x

-

1

⁢

(

p

-

1

)

<

1

-

2

⁢

ln

⁡

2

p

x

-

1

⁢

(

p

-

1

)

<

0

.

Thus, for

p

fixed and

k

≥

1

,

γ

⁢

(

p

k

)

is a

strictly decreasing function

of

k

.

On the other hand, from the equation

γ

⁢

(

p

k

)

=

k

+

1

p

k

-

1

⁢

(

p

-

1

)

,

it is clear that, if

k

is fixed,

γ

⁢

(

p

k

)

is a strictly decreasing function of

p

.

Thus, we have proven the following:

Lemma 1

.

Let

p

be a prime and

k

be a

nonnegative integer

with

p

k

∉

{

1

,

2

,

3

,

4

,

8

,

16

}

. Then

γ

⁢

(

p

k

)

≤

1

2

with equality holding if and only if

p

k

∈

{

5

,

9

}

.

This lemma has an immediate

consequence

:

Lemma 2

.

Let

m

be an odd natural number. Then

γ

⁢

(

m

)

=

1

⁢

or

⁢

γ

⁢

(

m

)

≤

1

2

.

Moreover,

γ

⁢

(

m

)

=

1

if and only if

m

=

1

or

m

=

3

.

Now we will examine the general case. Let

φ

⁢

(

n

)

=

τ

⁢

(

n

)

. Then

γ

⁢

(

n

)

=

1

.

Suppose that

4

∥

n

. Let

m

be an odd natural number with

n

=

4

⁢

m

. Thus,

1

=

γ

⁢

(

n

)

=

γ

⁢

(

4

⁢

m

)

=

γ

⁢

(

4

)

⁢

γ

⁢

(

m

)

=

3

2

⁢

γ

⁢

(

m

)

.

Therefore,

γ

⁢

(

m

)

=

2

3

,

which contradicts the second lemma. Hence,

4

∦

n

.

Suppose that

16

∣

n

. Let

m

be an odd natural number with

n

=

2

k

⁢

m

. Then

k

≥

4

. Thus,

1

=

γ

⁢

(

n

)

=

γ

⁢

(

2

k

⁢

m

)

=

γ

⁢

(

2

k

)

⁢

γ

⁢

(

m

)

≤

γ

⁢

(

16

)

⁢

γ

⁢

(

m

)

=

5

8

⁢

γ

⁢

(

m

)

.

Therefore,

γ

⁢

(

m

)

≥

8

5

,

which contradicts the second lemma. Hence,

16

∤

n

.

Now we deal with the cases that can actually occur.

•

Case I:

n

is odd

The second lemma immediately applies, yielding

n

=

1

or

n

=

3

.

•

Case II:

2

∥

n

and

3

∤

n

Let

m

be an odd natural number with

n

=

2

⁢

m

. Then

3

∤

m

and

1

=

γ

⁢

(

n

)

=

γ

⁢

(

2

⁢

m

)

=

γ

⁢

(

2

)

⁢

γ

⁢

(

m

)

=

2

⁢

γ

⁢

(

m

)

.

Thus,

γ

⁢

(

m

)

=

1

2

.

By the first lemma, for all

p

k

∥

m

with

k

>

0

,

γ

⁢

(

p

k

)

≤

1

2

with equality holding if and only if

p

k

=

5

. Therefore,

m

=

5

. Hence

n

=

10

.

•

Case III:

2

∥

n

and

3

∥

n

Let

m

be an odd natural number with

n

=

6

⁢

m

. Then

3

∤

m

and

1

=

γ

⁢

(

n

)

=

γ

⁢

(

6

⁢

m

)

=

γ

⁢

(

2

)

⁢

γ

⁢

(

3

)

⁢

γ

⁢

(

m

)

=

2

⁢

γ

⁢

(

m

)

.

Thus,

γ

⁢

(

m

)

=

1

2

.

By the first lemma, for all

p

k

∥

m

with

k

>

0

,

γ

⁢

(

p

k

)

≤

1

2

with equality holding if and only if

p

k

=

5

. Therefore,

m

=

5

. Hence

n

=

30

.

•

Case IV:

2

∥

n

and

9

∣

n

Let

m

be an odd natural number with

3

∤

m

such that

n

=

2

⋅

3

k

⁢

m

. Then

k

≥

2

and

1

=

γ

⁢

(

n

)

=

γ

⁢

(

2

⋅

3

k

⁢

m

)

=

γ

⁢

(

2

)

⁢

γ

⁢

(

3

k

)

⁢

γ

⁢

(

m

)

=

2

⁢

γ

⁢

(

3

k

)

⁢

γ

⁢

(

m

)

≤

2

⁢

γ

⁢

(

9

)

⁢

γ

⁢

(

m

)

=

γ

⁢

(

m

)

.

Since

3

∤

m

, the second lemma yields that

m

=

1

. Thus,

1

=

γ

⁢

(

n

)

=

γ

⁢

(

2

⋅

3

k

)

=

γ

⁢

(

2

)

⁢

γ

⁢

(

3

k

)

=

2

⁢

γ

⁢

(

3

k

)

.

Therefore,

γ

⁢

(

3

k

)

=

1

2

.

By the first lemma,

k

=

2

. Hence,

n

=

18

.

•

Case V:

8

∣

n

Recall that

16

∤

n

. Thus, there exists an odd natural number

m

with

n

=

8

⁢

m

. Then

1

=

γ

⁢

(

n

)

=

γ

⁢

(

8

⁢

m

)

=

γ

⁢

(

8

)

⁢

γ

⁢

(

m

)

=

γ

⁢

(

m

)

.

The second lemma yields that

m

=

1

or

m

=

3

. Hence,

n

=

8

or

n

=

24

.

It follows that

{

n

∈

ℕ

:

φ

⁢

(

n

)

=

τ

⁢

(

n

)

}

=

{

1

,

3

,

8

,

10

,

18

,

24

,

30

}

.

This list of numbers appears in the OEIS as

sequence

http://www.research.att.com/ njas/sequences/A020488

A020488.

Title

values of

n

for which

φ

⁢

(

n

)

=

τ

⁢

(

n

)

Canonical name

ValuesOfNForWhichvarphintaun

Date of creation

2013-03-22 18:03:48

Last modified on

2013-03-22 18:03:48

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

12

Author

Wkbj79 (1863)

Entry type

Feature

Classification

msc 11A25

Related topic

EulerPhifunction

Related topic

TauFunction

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