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Chapter 8: Congruences
Practice HW p. 62 # 1, 2, 3, 5, 6, Additional Web Exercises
In this section, we look at the fundamental concept of modular arithmetic, which is used
in a variety of applications in number theory.
Modular Arithmetic
Definition: Given two integers a, b and a positive integer m , we say that a is congruent
to b modulo m, written
a  b (mod m)
if m | (a  b) . The number m is called the modulus of the congruence.
Example 1: Explain why 23  3 (mod 5), 48  12 (mod 6) but 20  3 (mod 4) .
Solution: 23  3 (mod 5) and 48  12 (mod 6) since 5 | (23  3) or 5 | 20 and 6 | (48  12)
or 6 | 36 . However, 20  3 (mod 4) since 4 | (20  3) or 4 | 17 .
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Theorem: a  b (mod m) if and only if a  b  km for some integer k
Proof:
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Fact: Computationally, b (mod m) gives the integer remainder of b  m . We say that
a  b (mod m) if a and b produce the same integer remainder upon division by m.
For example, 23  8 (mod 5) since both 23 and 8 produce are remainder of 3 when
divided by 5, that is 23 (mod 5)  3 and 8 (mod 5)  3 . We can write 23  8  3 (mod 5) .
Note: When performing modular arithmetic computationally, the remainder r should
never be negative (this fact comes from the division algorithm that when computing
b  qm  r for b  m that 0  r  m ). Hence, when finding the remainder for b (mod m) ,
look for the nearest integer that m divides that is less than b.
Example 2: Compare computing 23 (mod 9) with  23 (mod 9) .
Solution:
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Doing Modular Arithmetic For Larger Numbers With A Calculator
To do modular arithmetic with a calculator, we use the fact from the division algorithm
that
b  qm  r ,
and solve for the remainder to obtain
r  b  qm .
We put this result in division tableau format as follows:
m
q
b
 qm
b  qm  r
Take Floor of Quotient (largest integer
less than calculator value of b  m ).
(1)
Remainder
Example 3: Compute 1024 (mod 37)
Solution:
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Example 4: Compute 500234 (mod 10301)
500234
 48.6 . The largest integer less than 48.6
10301
is 48. Hence, we assign q = floor(48.6) = 48. If we let b= 500234 and m = 10301 in (2),
then
48
10301 500234 .
Solution: Using a calculator, we obtain
 494448
5786
The remainder of the division is r = 5786. Hence, 500234 (mod 10301)  5786 .
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Example 5: Compute  3071 (mod 107)
 3071
 28.7 . The largest integer less than
107
 28.7 is  29 . Hence, we assign q = floor(-28.7) = -29. If we let b= -3071 and m = 107
in (2), then
Solution: Using a calculator, we obtain
 29
 3071
107

  3103
32
Thus,  3071 (mod 107)  32
Maple Commands for Doing Modular Arithmetic
Compute 1024 (mod 37)
> 1024 mod 37;
Compute 500234 (mod 10301)
> 500234 mod 10301;
Compute  3071 (mod 107)
> -3071 mod 107;
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Generalization of Modular Arithmetic
Fact: The common remainder of two numbers have when they are divided can be used to
define a congruence class. The remainder r will be the smallest positive integer in the
congruence class. Suppose r is the remainder of x divided by m, that is
x  r (mod m).
Theorem 1 says that then
x  r  km , where k is an integer.
Example 6: Find all elements of the congruence class x  4 (mod 7) .
Solution:
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Example 7 Find congruence class 2 modulo 7.
Solution:
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Note: For x  b (mod 7) , the set of distinct congruence classes are 0 , 1 , 2 , 3 , 4 , 5 ,
and 6 . This partitions the integers Z into disjoint subsets.
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Fact: Given x  b (mod m) , Z can be partitioned into distinct congruence classes of the
form
_____ 

0, 1, 2, 3,  m  1


Facts about Congruences
1. If a  b (mod m) , then a  t  b  t (mod m ) and at  bt (mod m) for any integer t.
Proof:
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2. If a1  b1 (mod m) and a2  b2 (mod m) , then a1  a2  b1  b2 (mod m) ,
a1  a2  b1  b2 (mod m) and a1a2  b1b2 (mod m) .
Proof:
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Note: Computationally, in mod m arithmetic, we have
[a1  a2    ak ](mod m)  [a1 (mod m)  a2 (mod m)   a k (mod m)] (mod m)
[a1  a2 ak ](mod m)  [a1 (mod m)  a2 (mod m)ak (mod m)] (mod m)
For example, we can by simplify 50 mod 3 as
Now, consider the problem of solving ax  c, a  0
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Solving Equations Involving Congruences
We want to consider the problem of solving the linear congruence equation
ax  c (mod m)
for x.
For example, consider the problem of solving the linear congruence
4 x  2(mod 10)
Note that we can see that x  3 is a solution since
x  3  4(3)  12 mod 10  2
Also, it can be seen that x  13 is a solution since
x  13  4(13)  52 mod 10  2
However, x  3 and x  13 are in the same congruence class modulo 10, since both
3 mod 10  3 and 13 mod 10  3 .
We want solution representations that are not in the same congruence class. These
solutions are called incongruent solutions.
Fact: One way to find all of the incongruent solutions when solving a congruence for x in
mod m arithmetic, set x  0, 1, 2, 3, , m  1 and find the values of x that satisfies the
congruence. This is known as the method of brute force for finding the incongruent solutions.
Example 8: Find the incongruent solutions to 4 x  2(mod 10) .
Solution:
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However, the brute force method becomes less practical for finding incongruent solutions
to congruences with larger moduli, like for example, solving the congruence
1000 x  2000 (mod 1076)
The next theorem deals with this issue.
Theorem: Linear Congruence Theorem. Let a, c, and m be integers where m  1 and let
g  gcd( a, m). For
ax  c (mod m)
*
i.)
ii.)
If g | c , then * has no solutions.
If g | c , then * has exactly g incongruent solutions. To find these g incongruent
solutions , we first using the Euclidean Algorithm remainder solution process to
find a solution ( x , y ) to the equation
ax  my  g  gcd( a, m)
Then a complete set of incongruent solutions is given by
c x m 
x  
 k (mod m),
g 
 g
k  0, 1, 2, , g  1
Proof: To prove part i, suppose there is a solution to ax  c (mod m) when g | c . Since
ax  c (mod m) , then by the definition of congruence m | ( ax  c ) , or ax  c  tm or
ax  tm  c
Since g  gcd( a, m) , g | a and g | m . Hence g | ( ax  tm) , which implies g | c . This is a
contradiction. Thus, when, g | c , * has no solutions.
To prove part ii, assume g | c . Since g  gcd( a, m) , there exist integers x and y to the
equation
ax  my  g
c
We multiply both sides of this equation by the integer
(recall g | c ) to get
g
 cx 
 cy 
a    m   c
 g
g
Continued on Next Page
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Hence,
 cx 
 cy 
a    c  m  
 g
 g
  cx  
 cx 
which says m | a   c or by definition of congruence a    c (mod m) . Hence
 g
  g 
cx
is a solution to *. To find the other g  1 incongruent solutions, suppose x is
x0 
g
another solution to *. Then
ax  ax0  c (mod m)
or
ax  ax0 (mod m) .
Then
m | (ax  ax0 )
or
ax  ax0  tm
a ( x  x0 )  tm
t is an integer
Since g  gcd( a, m) , g | a and g | m and hence are integers. We divide both sides of the
previous relationship by g.
a ( x  x0 )
m
t .
g
g
m a ( x  x0 )
m a
m a
or
|
| ( x  x0 ) . We claim gcd( , )  1 , for if not, there exists
g
g
g g
g g
a
m
m
a
a common divisor d  1 where d |
and d | . Thus
 ud and  vd for integers
g
g
g
g
u and v. Solving for m and a, we get m  u (dg ) and a  v (dg ) . Hence, dg | m and
dg | a , which contradicts that gcd( a, m)  g ( dg  g ).
Hence,
Now, since
m a
m a
m
| ( x  x0 ) and gcd( , )  1 , this implies that
| ( x  x0 ) .
g g
g g
g
Hence, we have
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x  x0  k
m
g
or
m
k )(mod m)
g
cx m
x  (  k )(mod m)
g g
x  ( x0 
Now, that taking k  0, 1, 2,  , g  1 produces g incongruent solutions. Note, if k  g , a
previous value would be repeated, since if k  g  i for any 0  i  g  1 would give
x(
cx m
cx
m
cx
m
cx m
 ( g  i ))(mod m)  (  m  i )(mod m)  (  0  i ) mod m  (  i )(mod m)
g g
g
g
g
g
g g
Since 0  i  g  1 , this value will have been repeated. This completes the proof.
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Example 9: Use the Linear Congruence Theorem to find all of the incongruent solutions
to 12 x  6 (mod 18)
Solution:
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Example 10: Use the Linear Congruence Theorem to find all of the incongruent
solutions to 76 x  50 (mod 176)
Solution:
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Example 11: Use the Linear Congruence Theorem to find all of the incongruent
solutions to 1000 x  2000 (mod 1076)
Solution:
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Corollary to the Linear Congruence Theorem: The congruence ax  1 (mod m) has a
solution for x (note that x  a 1 is the multiplicative inverse of a modulo m if
gcd( a, m )  1 ). The solution is unique for 1  x  m .
Proof:
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Solving Congruences of Higher Degree
Solving congruences of higher degrees are generally more difficult than linear
congruences. If the modulus m is small enough, the brute force method will suffice.
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Example 12: Find the incongruent solutions for x 2  x  0 (mod 6)
Solution:
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Example 13: Find the incongruent solutions for x 2  3 (mod 8)
Solution:
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However, the brute force method becomes more inefficient the larger the modulus. In our
examples, we saw that x 2  x  0 (mod 6) has 4 solutions and that x 2  3 (mod 8) has
none. Is there any way we can predict the number of incongruent solutions beforehand?
The next theorem partially answers this question.
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Theorem: Polynomial Roots Mod p Theorem. Let p be a prime and let
f ( x )  a0 x d  a1 x d 1    a0
be a polynomial of degree d  1 with integer coefficients and with p | a0 . Then the
congruence
f ( x )  0 (mod p )
has at most d incongruent solutions.
Proof: Suppose there at least one polynomial where p does not divide the leading
coefficient that has more distinct incongruent solutions then its degree. Let d be a
polynomial of least degree for these polynomials given by
f ( x )  a0 x d  a1 x d 1    a0
Suppose r1 , r2 , rd , rd 1 be these incongruent solutions. For each incongruent solution
rk of r1 , r2 , rd , rd 1 , we know that
f (rk )  0 (mod p)
Since r1 is a zero, by the factor theorem for polynomials,
f ( x)  ( x  r1 ) g( x)
where g ( x ) is of degree d  1 , that is, g ( x ) has the form
g ( x )  a0 x d 1  b1 x d  2    bd 1
For the other incongruent solutions rk  r1 of f ( x ) , we have
f (rk )  (rk  r1 ) g (rk )  0 (mod p)
Since rk  r1 (mod p ) , it follows from the prime divisibility property that
g ( rk )  0 (mod p) . Hence, r2 , r3 , rd , rd 1 are incongruent solutions to g ( x ) , which
says that g ( x ) has d incongruent solutions, which is a contradiction that f ( x ) is the
polynomial of least degree that has more incongruent solutions than its degree. Thus, the
result holds.
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Example 14: Find the incongruent solutions for x 2  9 (mod 61)
Solution:
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