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Chapter 11: Euler’s Phi Function
Practice HW p. 80 # 1, 2a, 3, 5 (Additional Web Exercises)
In this chapter, we want to look at how to compute efficiently the number  (m) , which
 (m) , is the number of integers between 1 and m that are relatively prime to m. That is,
 (m) # {a : 1  a  m  1 and gcd( a, m)  1}
We would like to have a method of computing  (m) when m is larger. The next theorems
describe some efficient ways of doing this.
Theorem 1: If p is a prime number, then  ( p )  p  1 .
Proof:
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Example 1: Compute  (11) and  (97) .
Solution:
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Theorem 2: If p is a prime number, then  ( p k )  p k  p k 1 .
Proof: The only divisors of p k are 1 and the powers of p less than k, that is p i | p k for
1  i  k . Let a be an integer where 1  a  p k . If gcd( a, p k )  1 , then a must have a
factor of p , that is p | a . Thus, a number a is not relatively prime ( gcd( a, p k )  1) when
p | a . Thus
 ( p k )  p k  #{a : 1  a  p k and p | a}
If p | a , then a  kp for some integer k. The multiples of p between 1 and p k are
(1) p, 2 p, 3 p, , ( p k 1  1) p, p k 1  p  p k ,
which is p k 1 multiples of p. Hence, #{a : 1  a  p k and p | a}  p k 1 and thus
 ( p k )  p k  p k 1 .
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Example 2: Compute  (64) .
Solution:
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We next state and prove a lemma that will be useful in a later result.
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Lemma: For integers a, m, and n, gcd( a, mn)  1 if and only if gcd( a, m )  1 and
gcd( a, n )  1 .
Proof:  Assume gcd( a, mn)  1 and suppose gcd( a, m )  d  1 . Then d | a and d | m .
Hence d | mn . This contradicts the fact that gcd( a, mn)  1 . A similar argument can be
made if we assume gcd( a, n )  d  1 . Thus gcd( a, m )  1 and gcd( a, n )  1
 Now assume gcd( a, m )  1 and gcd( a, n )  1 and suppose gcd( a, mn)  d  1 . Then
d | a and d | mn . By the Fundamental Theorem of Arithmetic, d must have a prime divisor
p where p | d . Thus, p | a and p | mn . By the prime divisibility property, that says that
p | m or p | n . If p | m , since p | a this contradicts the fact that gcd( a, m )  1 . Similarly, if
p | n , since p | a this contradicts the fact that gcd( a, n )  1 . In either case, we have a
contradiction and hence gcd( a, mn)  1 .
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For example, if a = 5, m = 6, and c = 7, the lemma implies
gcd( 5,6)  1, gcd( 5,7)  1  gcd( 5,6  7)  gcd( 5,42)  1
We next use the previous lemma to prove a fundamental result.
Theorem 3: For two positive integers m and n, if the gcd( m, n)  1 , then  (mn)   (m)   (n) .
Proof: We rearrange the integers between 1 and mn into an n  n array with n rows and m
columns.
1
2

r

m 1
m2

mr
 2m
2m  1
2m  2

2m  r
 3m





m

( n  1)m  1 ( n  1)m  2  ( n  1)m  r  nm
By Lemma 5,
 (mn)  # {a : gcd( a, mn)  1}  # {a : gcd( a, m)  1 and gcd( a, n)  1}
Consider the r th column of the array. Each entry in the r th column when paired with m,
has the same greatest common divisor.
Continued on Next Page
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That is, if 0  i, j  n  1 , gcd( im  r, m)  gcd( jm  r, m)  1 (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,  (m) of the columns will have the
elements in the array that are relatively prime to m.
Now, consider the  (m) columns. We must now show that each of these columns has
 (n ) elements relatively prime to n, thus giving a total of  (m) (n ) total elements
relatively prime to both m and n and hence m n by the previous lemma. Consider the
r th column



2m  r  n total entries



( n  1)m  r 
r
mr
Claim: In modulo n arithmetic, all entries in this column are just a rearrangement of
1, 2, 3,, n  1, that is, each entry in the column is in a the same distinct congruence class
as one of the integers 1, 2, 3,, n  1 . For if not, there would exist integers 0  i, j  n  1
such that
jm  r  im  r (mod n)
This implies that
n | ( j  i )m
Since by the Theorem assumption, gcd( m, n)  1 , this implies that n | ( j  i ) which is
impossible since | j  i | n . Hence, since the r th column elements
r, m  r, 2m  r,, (n  1)m  r and 1, 2, 3,, n  1 are the same elements with respect to
congruence in modulo n arithmetic and there are  (n ) elements relatively prime to n in list
1, 2, 3,, n  1, there are  (n ) elements in the r th column r, m  r, 2m  r,, (n  1)m  r
relatively prime to n. Thus there are  (m) columns in the array containing the elements
relatively prime to n, and in each of these  (m) columns  (n ) entries relatively prime to n.
This gives a total of  ( m)   ( n ) total elements that are relatively prime to both m and n.
Since, by the previous lemma, these are the same elements relatively prime to m  n , we have
the result  ( mn)   ( m)   ( n )
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Example 3: Compute  (24) .
Solution:
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Theorem 4: If m has the prime factorization m  p1k1  p 2k2  prkr , then
 (m)   ( p1k1  p2k2  prkr )   ( p1k1 )   ( p2k2 ) ( prkr )
 ( p k1  p1k1 1 )( p k2  p2k2 1 ) ( p kr  prkr 1 )
.
Proof: We can prove this result using mathematical induction. For the trivial case, that is,
if r  1 , then using Theorem 13.6 we have
 (m)   ( p1k1 )  ( p1k1  p1k11 ) .
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 m  p1k1  p2k2  prkr  prkr 11 . Noting that
gcd( p1k1  p2k2  prkr , prkr 11 )  1, we have
 ( m)   (( p1k1  p2k 2  prk r )  prkr 11 )
  ( p1k1  p2k 2  prk r )   ( prkr 11 )
(Use previous Theorem 3)
 ( p k1  p1k1 1 )( p k 2  p2k 2 1 )  ( p k r  prk r 1 )   ( prkr 11 )
(By the induction hypothesis )
 ( p k1  p1k1 1 )( p k 2  p2k 2 1 )  ( p k r  prk r 1 )( prkr 11  prkr 11 1 )
(Use previous Theorem 2)
Hence, by the principle of mathematical induction, the result holds.
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Corollary 1: If p and q are primes where p  q , then  ( pq)   ( p) (q)  ( p  1)( q  1) .
Proof:
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Example 4: Compute  (35) .
Solution:
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Example 5: Compute  (360) .
Solution:
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Example 6: Compute  (1575) .
Solution:
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Euler Phi Function with 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  (35) .
> phi(35);
Compute  (360) .
> phi(360);
Compute  (1575) .
> phi(1575);
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Chinese Remainder Theorem
Theorem 5: Chinese Remainder Theorem. The system of linear congruences
x  a1 (mod m1 )
x  a2 (mod m2 )

*
x  ak (mod mk )
where gcd( mi , m j )  1 (Moduli are pairwise relatively prime) can be solved for an integer
x modulus M  m1  m2  mk . Moreover, if y is another solution to these congruences, then
y  x (mod M ) .
Proof: Let M i 
M
for 1, 2, 3,, k .
mi
Then gcd( M i , mi )  1 (Proof Exercise). By the Linear Congruence Theorem, there exists
an integer y i where
M i yi  1 (mod mi )
(Note y i is the multiplicative inverse of M i mod mi
Also, M i  0 (mod m j ) whenever i  j . (Proof Exercise)
k
Let x   ai M i yi (mod M )  (a1M 1 y1  a2 M 2 y2    ak M k yk ) (mod M )
i 1
Claim that x satisfies every linear congruence. To show, that the jth arbitrary congruence
with modulus m j . Then
x  ( a1M 1 y1  a2 M 2 y 2   a j M j y j    ak M k y k (mod m j )
 a j M j y j (mod m j )
 aj
Continued on Next Page
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To show there are no other incongruent solutions, suppose y is another solution to *. Then
for all 1  j  k , y  a j (mod m j ) . Since x  a j (mod m j ) , it follows that
x  y (mod m j )
Thus, m j | ( x  y ) . Since gcd( mi , m j )  1 when i  j , then it follows that
M  m1  m2  mk | ( x  y )
Hence, M | ( x  y ) and x  y (mod M ) . This completes the proof.
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Chinese Remainder Theorem Summary
To solve the system of linear congruences
x  a1 (mod m1 )
x  a2 (mod m2 )

gcd( mi , m j )  1
x  ak (mod mk )
We compute
k
x   ai M i yi (mod M )  (a1M 1 y1  a2 M 2 y2    ak M k yk ) (mod M )
i 1
where




each ai comes from the right had side of the given congruences
M  m1  m2  mk ,
M
Mi 
for 1, 2, 3,, k ,
mi
yi is the multiplicative inverse of M i mod mi , that is, yi solves the congruence
M i yi  1 (mod mi )
for 1, 2, 3,, k .
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Example 7: Use the Chinese Remainder Theorem to solve the system of congruences
x  2(mod 3) , x  4(mod 5) , x  6(mod 19)
Solution:
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Using the Chinese Remainder Theorem in Maple
To solve x  2(mod 3) , x  4(mod 5) , x  6(mod 19)
> chrem( [2, 4, 6], [3, 5, 19] );