1. Congruence of Integers

2. Definition: Let n (>1) be a positive integer.
For x, yZ, we say that x is congruent to y modulo n
if and only if n | x y,
denoted by x  y (mod n).

3. Note: x  y (mod n)  n |x y  x  y = nq for some qZ  x = y + nq
 x and y yield the same remainder
when each divided by n.

4. Theorem1.
The relation of congruence modulo n is an equivalence relation on the set of integers.
Pf: Need to check
(i) reflexive,
(ii) symmetric and
(iii) transitive.

5. Note:
The equivalence classes for congruence modulo n form a partition of Z, i.e. they separate Z into mutually disjoint subsets.
These subsets are called congruence classes (or residue classes).

6. Ex1. When n = 4, we have 4 congruence classes as follow:
[0]={ … , 8, 4, 0, 4, 8, …..}
[1]={ …., 7, 3, 1, 5, 9,……}
[2]={ …., 6, 2, 2, 6, 10,……}
[3]={ …., 5, 1, 3, 7, 11,……}

7. Theorem 2. If a  b (mod n) and xZ,
then a + x  b + x (mod n) and ax  bx (mod n).
Pf:

8. Theorem 3.
Suppose a  b (mod n), c  d (mod n). Then a+c  b+d (mod n) and acbd (mod n).
Pf:

9. Theorem 4.
If ax  ay(mod n) and (a, n)=1, then x  y (mod n).
Pf:

10. Use the Euclidean Algorithm to find a solution of ax  b (mod n)
when (a, n) = 1.

11. Ex2. Consider the congruence equation 20x  14 (mod 63).
Since (20, 63) = 1,we can find two integers m, n such that 1 = 20m+63n.
Appling the Euclidean Algorithm, we have 63=203+3, 20=36+2, 3=21+1. Thus
1 =

12. Ex2. Solve 20x  14 (mod 63)

13. Ex2’. Use another way to find the solution of 20x  14 (mod 63).

14. Theorem 5.
If a and n are relatively prime, then the congruence equation ax  b (mod n) has an integer solution x.
Moreover, any two solutions in Z are congruent modulo n.

15. Thm5. If (a, n)=1, then the congruence ax  b (mod n) has an integer solution x.
Pf:

16. Thm5. Moreover, any two solutions of ax  b (mod n) in Z are congruent modulo n.
Pf:

17. Ex3. Consider the equation 24x  45 (mod 12).
Suppose that there is a solution, x0, of this congruence equation.
Then 24x0–45 is a multiple of 12.
It is impossible, because 24x0–45 is always an odd integer which cannot be a multiple of 12. Thus the equation 24x  45 (mod 12) has no solution.
This Example tells us that the condition in Theorem5 “(a, n) = 1” is a very important condition for getting a solution of
ax  b (mod n).