Notes on the Chinese Remainder Theorem
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Chapter 4: Congruences Chinese Remainder Theorem
By the end of this section you will be able to Theorem prove the Chinese Remainder Theorem apply this theorem to solve simultaneous linear congruences
Up to now we have solved a single linear congruence such as ax b c mod n In this section we examine solving a set of simultaneous linear equations. How do we solve these? First we look at an example and then develop a general method. Example 1 Find the value of x x which satisfies both the following equations: x 1 mod 5 1 x
4 mod 7
2
Solution We need to find a value of x x such that equations (1) and (2) are true. Let us first use brute force to resolve these equations. Creating a table of values: x 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 mod 5 1 2 3 4 0 1 2 3 4 0 2 3 4 0 1 mod 7 1 2 3 4 5 6 0 1 2 3 5 6 0 1 4 By looking at the table we have x 11 will satisfy both the above equations (1) and (2) because 11 1 mod 5 , 11 4 mod 7 Of course we can apply brute force for simple integer values. However we need a systematic approach to solve these because x because x may be a large number and we just do not have the time to create a table table for large x large x.. Say Say if if x 701 then this would mean we would have to create a table with more than 701 columns. What does the first equation x 1 mod 5 in the above above example example mean? mean? Means that x 1 5k for some k . Re-arranging this we have x 1 5k Similarly the other equation x 4 mod 7 means x 4 7c for some c and x 4 7c . Equating these two equations, x 1 5k and x 4 7c , gives x 1 5k 4 7c k
3 7c
5 Since k i k is an int intege eger we need 3 7c to be a multiple of 5. If c 1 then 3 7c 3 7 10 10 and k 2. 5 Substituting these, c 1 into x 4 7c gives x 4 7 c 4 7 11 Putting k 2 into x 1 5k also yields x 11 which is our solution from the t he above example.
Notes on the Chinese Remainder Theorem
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Example 2 Solve the simultaneous equations: x 31 mod 49 x
6 mod 20
Solution From these equations we have x 31 49k
x 31 49 k
x 6 20c x 6 20c where k and c are integers. Equating these last two equations gives 25 49k 6 20c 31 49 k c 20 Since we want integer solutions so we try values of k such that the numerator 25 49k is a multiple of 20. Hence we trial multiples of 5 for k , that is k 5, 10, 15, . Note that k 5, 10 does not give a multiple of 20 but 15 does because 25 49 15 c 38 20 Substituting c 38 into x 6 20c gives x 6 20 38 766 . Check that this x 766 satisfies both the given equations: x 31 mod 49 x
6 mod 20
Example 3 Find an integer x such that when divided by 2, 3 and 5 the remainder is 1. Solution We can write these as congruences: x 1 mod 2 x x
mod 3 1 mod 5 1
What does this mean? Means that x 1 2k , x 1 3c and x 1 5m where k , c and m are integers This means that x 1 is a multiple of 2, 3 and 5. Which number is a multiple of 2, 3 and 5? Since gcd 2, 3, 5 1 so the smallest number which is a multiple of 2, 3 and 5 is 2 3 5 30 This means that x 1 is a multiple of 30 or x 1 30n x 30n 1 The general solution of the given three equations is 30n 1 : x 30 1 31, x 30 2 1 61, x 30 3 1 91, You may check that each of these solutions satisfies the given equations: x 1 mod 2 , x 1 mod 3 , x 1 mod 5 Well under what circumstances do we have a solution to simultaneous congruences?
Notes on the Chinese Remainder Theorem
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We need to prove the result for general congruences. The proof below is l ifted from Burton’s Number Theory book page 79. Chinese Remainder Theorem. Let n1 , n2 , n3 , , nr be positive integers such that gcd ni ,
nj
1
for i j
Then the simultaneous linear congruences x a1 mod n1 x
a2 mod n2
x
ar mod nr
has a solution satisfying all these equations. Moreover the solution is unique modulo n1n2 n3 nr . How do we prove this result? We need to prove two things – (1) existence of solution and (2) uniqueness of solution Proof . (1) Existence Let n n1n2 n3 nr . For each integer k 1, 2, 3, , r let N k
n
n1n2 n3 nk 1nk nk 1 nr
nk
nk
Cancelling out nk
n 1n2n3 nk 1nk 1 nr
This means that N k is the product of all the moduli ni with the modulus nk missing. We are given that gcd ni ,
nj
1
for i j which implies that gcd N k ,
nk
1
Why? Because by problem 20(a) on page 25 we have: If gcd a, b
gcd
a,
c
1
then gcd a,
bc
1
Consider the linear congruence N k x
1
mod
nk
mod
nk
Does this have any solutions? Since gcd N k , nk 1 the linear congruence N k x
1
has a unique solution. Why? Because by Theorem (4.7) on page 76 we have: The linear congruence ax
b
mod n
has a solution
g
gcd a,
n divides b. If
g b then the linear congruence has exactly g solutions. Let xk be the unique solution of N k x N k xk
1
1
mod
mod
nk which means that
nk
(†)
We construct a solution which satisfies all the given simultaneous linear congruences. The solution we consider is x ' a1 N1 x1 a2 N 2 x 2 a 3N 3x 3 a r N r x r Let us see if this solution x ' satisfies the first given linear congruence:
Notes on the Chinese Remainder Theorem
x
4 a1 mod n1
Taking the solution to modulus n1 gives x ' a1 N1 x1 a2 N 2 x 2 a 3N 3x 3 a r N r x r mod n1 By the above definition of N 2 , N 3 , N 4 ,
, N r
(*)
these numbers are multiplies of n1 .
Therefore a2 N 2 x2
0 mod n1 , a3 N 3x 3
0
mod
Substituting this into (*) we have x ' a1 N1 x1 0 0 0 By the above (†) we have N1 x1 x ' a1 N1 x1
mod
1
mod
n1 , , a r N r x r 0 mod n1
mod
a1N 1x1
n1
n1 . Substituting this into the above
n1 gives x ' a1 N1 x1
a1
1 a1 mod
Hence x ' satisfies the first simultaneous congruence x
a1
n1
mod
n1 . Similarly we can
show that the solution constructed satisfies the remaining congruences. (2) Uniqueness Suppose there is another solution y ' which satisfies the given linear congruences. This means we have x ' ak y ' mod nk where k 1, 2, 3, , r From this congruence x ' y ' mod nk we have n1
x ' y ' ,
n2
x ' y ' ,
n3
x ' y ' ,
, nr
Remember we are given that the n’s are pairwise prime - gcd ni ,
x ' y ' nj
1
for i j .
Applying Corollary 2 on page 23: If a
c and b c with gcd a, b
1
then ab c .
to the above list gives
n1n2n3 nr x ' y ' This means that x '
y ' mod n1n2 n3 nr
This completes our proof. ■ The proof gives us a systematic approach on how to construct the solution of any given linear simultaneous congruences. In the proof the solution was x ' a1 N1 x1 a2 N 2 x 2 a 3N 3x 3 a r N r x r We use this to solve the remaining examples. Example 4 Solve the following simultaneous linear congruences: x 1 mod 3 , x 2 mod 5 , x
3
Solution Using the constructed solution in the proof we have x ' a1 N1 x1 a2 N 2 x 2 a 3N 3x 3 Each of N ’s is given by
n nk
where n 3 5 7 105 . Hence
mod 7 (*)
Notes on the Chinese Remainder Theorem
N1
3 5 7 3
35,
N2
5
3 5 7
5
21 and N 3
3 5 7 7
15
The a’s are given to us: a1
1,
a2
2 and a3
3
We need to find the xk ’s which are given by N k xk 35 x1
1
mod 3 ,
By trial and error we have x1
1
21 x2
2 , x2
1
1
mod 5
and x3
1.
mod
nk :
and 15 x3
1
mod 7
Substituting these numbers into (*)
gives x ' a1 N1x1 a2 N 2 x 2 a 3N 3x 3 Hence 157
52
mod
1 35 2 2 211 3 15 1 157
105 . The least positive integer which satisfies the given
congruences is x 52 . Check that this is correct by substituting into the given linear congruences. Example 5 Solve the following simultaneous linear congruences: 2 x 1 mod 5 , 3 x 9 mod 6 , 4 x Solution This time we do not have x
1
mod 7
mod m but ax ? mod m . How do we solve these? We convert them into x ? mod m by multiplying by an appropriate factor. Multiply the first linear congruence 2 x 1 mod 5 by 3 gives 6 x x 3 mod 5 We can simplify the second 3 x 9 mod 6 by dividing by 3 because gcd 3, 6 3 : x 3 mod 2 1 mod 2 We multiply the third 4 x 1 mod 7 by 2: 8 x x 2 mod 7 ?
We solve the equivalent system: x 3 mod 5 , x
1
mod 2 , x
2 mod 7
The solution is given by x a1 N1 x1 a2 N 2 x 2 a 3N 3x 3
(*)
Now using the method described in the proof we have n 3 5 7 105 Hence 3 5 7 N 1 21 5 3 5 7 N 2 35 3 3 5 7 N 3 15 7 The a’s are a1 3 , a2 1 and a3 2 . We need to find the xk ’s which are given by N k xk
1
mod
nk :
Notes on the Chinese Remainder Theorem
21 x1
1
mod 5 ,
35 x2
The solutions by trial and error are x1
1
6
mod 2
1 , x2 1
and 15 x3
and x3
1
mod 7
1.
Putting all these numbers into (*) gives: x a1 N1 x1 a2 N 2 x 2 a 3N 3x 3
21 1 1 35 1 2 15 1 128 105 . The least positive integer is x 23 . Check this solution 3
We have x 128 23 mod
by making sure each of the given congruences is satisfied. Example 6 Solve the following simultaneous linear congruences: 2 x 1 mod 5 , 3x 9 mod 6 , 4 x 1 mod 7 , 5 x Solution This time we do not have x
9
mod
11
mod m but ax ? mod m . How do we solve these? We convert them into x ? mod m by multiplying by an appropriate factor. Multiply the first linear congruence 2 x 1 mod 5 by 3 gives 6 x x 3 mod 5 We can simplify the second 3 x 9 mod 6 to x 3 mod 2 1 mod 2 We multiply the third 4 x 1 mod 7 by 2: 8 x x 2 mod 7 Lastly we multiply the last given congruence 5 x 9 mod 11 by 9: 45 x x 81 4 mod 11 We can rewrite this as x 4 mod 11 . ?
We solve the equivalent system: x 3 mod 5 , x 1 mod 2 , x
2 mod 7 and x
The solution is given by x a1 N1 x1 a 2 N 2x 2 a 3N 3x 3 a 4N 4x 4
1
mod
4 mod 11 (*)
Now using the method described in the proof we have n 5 2 7 11 770 Hence 2 5 7 11 N 1 154 5 2 5 7 11 N 2 385 2 2 5 7 11 N 3 110 7 2 5 7 11 N 4 70 11 The a’s are a1 3 , a2 1 , a3 2 and a4 4 . We need to find the xk ’s which are given by N k xk
nk :
Notes on the Chinese Remainder Theorem
154 x1
1
mod 5 ,
385 x2
1
7
mod 2 , 110 x3
The solutions by trial and error are x1
4 , x2
1
mod 7
1 , x3 3
and x4
and 70 x4
1
mod
3.
Putting all these numbers into (*) gives: x a1 N1x1 a2 N 2x 2 a 3N 3x 3 a 4N 4x 4
154 4 1 385 1 2 110 3 4 70 3 3733 We have x 3733 653 mod 770 . The least positive integer is x 653 . 3
Check that this solution is correct. Summary For simultaneous linear congruences we can apply the Chinese Remainder Theorem to resolve for the unknown.
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