1

Chapter

11

:

Euler’s

Phi Function

Practice HW

p.

80

#

1

,

2

a

, 3, 5 (Additional Web Exercises)

In

this chapter, we want to look at how to compute efficiently the number

, which

, is the number of integers betwe

en 1 and

m

that are relatively prime to

m

. That is,

We would like to have a method of computing

when

m

is larger. The next theorems

describe some efficient ways of doing this.

Theorem

1

:

If

p

is a prime n

umber, then

.

Proof:

█

Example 1

:

Compute

and

.

Solution:

█

2

Theorem 2

:

If

p

is a prime number, then

.

Proof:

The only divisors of

are 1 and the powers of

p

less than

k

, that is

for

. Let

a

be an integer where

. If

, then

must have a

factor of

, that is

. Thus, a number

a

is not relatively prime (

when

. Th

us

If

, then

for some integer

k

. The multiples of

p

between 1 and

are

,

which is

multiples of

p

. Hence,

and thus

.

█

Example 2

:

Compute

.

Solution:

█

We next state

and prove a lemma that will be useful in a later result.

3

Lemma:

For integers

a, m,

and

n,

if and only if

and

.

Proof:

Assume

and suppose

. Then

and

.

Hence

. This contradicts the fa

ct that

. A similar argument can be

made if we assume

.

Thus

and

Now assume

and

and suppo

se

. Then

and

. By the Fundamental Theorem of Arithmetic,

d

must have a prime divisor

p

where

. Thus,

and

. By the p

rime divisibility property, that says that

or

. If

, since

this contradicts the fact that

. Similarly, if

, since

this contradicts the fact that

.

In either case, we have a

contradiction and hence

.

█

For example, if

a

= 5,

m =

6, and

c

= 7, the lemma implies

W

e next use the previous lemma to prove a fundamental result.

Theorem 3

:

For two positive integers

m

and

n

, if the

, then

.

Proof:

We rearrange the integers between 1 and

mn

into an

array with

n

rows and

m

columns.

By Lemma 5,

Consider the

column of the array. Each entry in the

column when paired with

m

,

has the same greatest common

divisor.

Continued on Next Page

4

That is, if

,

(This can be seen by

calculating both using the first step of the Euclidean Algorithm). Since all elements in

each column have the same greatest common

divisor,

of the columns will have the

elements in the array that are relatively prime to

m

.

Now, consider the

columns. We must now show that each of these columns has

elements relat

ively prime to

n

, thus giving a total of

total elements

relatively prime to both

m

and

n

and hence

by the previous lemma. Consider the

column

Claim: In modu

lo

n

arithmetic, all entries in this column are just a rearrangement of

, that is, each entry in the column is in a the same distinct congruence class

as one of the integers

. For if not, there would exist inte

gers

such that

This implies that

Since by the Theorem assumption,

, this implies that

which is

impossible since

. Hence, since the

column elements

and

are the same elements with respect to

congruence in modulo

n

arithmetic and there are

elements relatively prime to

n

i

n list

, there are

elements in the

column

relatively prime to

n

. Thus there are

columns in the array containing the elements

relatively p

rime to

n

, and in each of these

columns

entries relatively prime to

n

.

This gives a total of

total elements that are relatively prime to both

m

and

n.

Since, by the previous lemma, the

se are the same elements relatively prime to

, we have

the result

█

5

Example 3

:

Compute

.

Solution:

█

Theorem 4

:

If

m

has the

prime factorization

, then

.

Proof:

We can prove this result using mathematical induction. For the trivial case, that is,

if

, then using Theorem 13.6 we have

.

Now, assume the result is true if

m

is a product of

r

primes. We want to show the result is

true if

m

is a product of

r

+ 1 primes. Suppose

. Noting that

we have

Hence, by the princi

ple of mathematical induction, the result holds.

█

Corollary 1

:

If

p

and

q

are primes where

, then

.

Proof:

█

6

Example 4

:

Compute

.

Solution:

█

Example

5

:

Compute

.

Solution:

█

Example 6

:

Compute

.

Solution:

█

7

Euler Phi Function w

ith Maple

Note that the

numtheory

package must be loaded to the home directory using the

with

statement before the

phi

command can be used.

> with(numtheory):

Compute

.

> phi(35);

Compute

.

> phi(360);

Compute

.

> phi(1575);

8

Chinese Remainder Theorem

Theorem 5:

Chinese Remainder Theorem.

The system of linear congruences

*

where

(Moduli are pairwise relatively prime

) can be solved for an integer

x

modulus

. Moreover, if

y

is another solution to these congruences, then

.

Proof:

Let

for

.

Then

(Proof Exercise).

By the Linear Congruence Theorem, there exists

an integer

where

(Note

is the multiplicative inverse of

mod

Al

so,

whenever

. (Proof Exercise)

Let

Claim that

x

satisfies every linear congruence. To show, that the

j

th

arbitrary congruence

with modulus

. Then

Continued on Next Page

9

To show there are no other incongruent solutions, suppose

y

is another solution to *. Then

for all

,

. Since

, it follows that

Thus,

. Since

when

, then it follows that

Hence,

and

. This complete

s the proof.

█

Chinese Remainder Theorem Summary

To solve t

he system of linear congruences

We compute

where

each

comes from the right had side of the given c

ongruences

,

for

,

is the multiplicative inverse of

mod

, that is,

solves the congruence

for

.

10

Example 7:

Use the Chinese Remainder Theorem to solve

the system of congruences

,

,

Solution:

11

█

12

13

Using the Chinese Remainder Theorem in Maple

To solve

,

,

> chrem( [2, 4, 6], [3, 5, 19] );

## Comments 0

Log in to post a comment