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Chapter 10: Congruences, Powers, and Euler’s Theorem
Practice HW p. 74 # 2, 3a, Additional Web Exercises
In this section, we state and prove Euler’s Theorem and look at its benefits.
Euler-Phi Function
Given an integer m, the Euler-Phi function, denoted by  (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}
For example,  (30)  8 since 1, 7, 11, 13, 17, 19, 23, and 29 are the eight integers
between 1 and 30 that are relatively prime to 30, that is, where
1  gcd(1, 30)  gcd( 7, 30)  gcd(11, 30)  gcd(13, 30)  gcd(17, 30)  gcd(19, 30)  gcd( 23, 30)  gcd( 29, 30)
Also,  (7)  6 since 1, 2, 3, 4, 5, 6, are the six integers between 1 and 7 that are
relatively prime to 7, that is, where
1  gcd(1, 7)  gcd( 2, 7)  gcd( 3, 7)  gcd( 4, 7)  gcd( 5, 7)  gcd( 6, 7)
Fact: For a prime p,  ( p )  p  1 .
The Euler Phi Function can be used to generalize Fermat’s Little Theorem as follows.
Theorem: Euler’s Formula. Let a be any integer. If gcd( a, m)  1 , then
a ( m)  1 (mod m)
Proof: Consider the integers b1 , b2 a, b3a,  , b ( m) a modulo m, where
b1  b2  b3    b ( m )  m are the list of integers between 1 and m that are relatively
prime to m, that is, gcd( bi , m)  1 . We first claim that for each integer of the  (m) distinct
integers in the list b1 , b2 a, b3a,  , b ( m) a , there is a distinct integer in the list of integers
b1 , b2 , b3 ,  , b ( m ) modulo m that is in its congruence class. Since gcd( a, m)  1 , m is
relatively prime to each number is the list b1 , b2 a, b3a,  , b ( m) a . Suppose that two of the
integers in the list b1 , b2 a, b3a,  , b ( m) a are in the same congruence class, that is, suppose
that
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b j a  bk a (mod m) for 1  k  j   ( m) .
Then by definition of congruence, m | (b j a  bk a ) or m | (b j b k )a . Since gcd( a, m )  1 by
assumption, then m | (b j  bk ) . But this is impossible since | b j  bk | m . Thus
b j a  bk a (mod m) only when b j  bk . Thus b1 , b2 a , b3a,  , b ( m) a are distinct modulo
m. But there are only  (m) distinct non-zero integers modulo m relatively prime to m,
namely b1 , b2 , b3 ,  , b ( m ) . Hence, b1 , b2 a, b3a,  , b ( m) a and b1 , b2 , b3 ,  , b ( m ) are the
same integers, possibly appearing in a different order. Hence, there products would give an
integer in the same congruence class, that is
b1  b2 a  b3a  b ( m ) a  (1  2  3 b ( m ) ) (mod m)
or
a ( m) B  B (mod m) ,
where B  b1  b2  b3  b ( m ) .Since gcd( B,  ( m ))  1 (if this were not true, then it can be
shown that gcd( bi ,  (m))  1 for some 1  i   ( m) ), then B has a multiplicative inverse.
Multiplying both sides by B 1 gives
a (m) BB 1  B B 1 (mod m) ,
which produces the result
a (m)  1(mod m)
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Note that since gcd( 5,7)  1 and  (7)  6 , then Euler’s Formula says
5 ( 7 )  1 (mod 7)
56  1 (mod 7)
Similarly, since gcd(11,30)  1 and  (30)  8 , then Euler’s Formula says
11 ( 30)  1 (mod 30)
118  1 (mod 30)
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Example 1: Use Euler’s Formula to find a where a  1142 (mod 30) .
Solution:
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