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**Congruence of Integers**

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**Definition: Let n (>1) be a positive integer.**

For x, yZ, we say that x is congruent to y modulo n if and only if n | x y, denoted by x y (mod n).

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**Note: x y (mod n) n |x y x y = nq for some qZ x = y + nq**

x and y yield the same remainder when each divided by n.

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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.

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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).

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**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,……}

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**Theorem 2. If a b (mod n) and xZ,**

then a + x b + x (mod n) and ax bx (mod n). Pf:

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Theorem 3. Suppose a b (mod n), c d (mod n). Then a+c b+d (mod n) and acbd (mod n). Pf:

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

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**Use the Euclidean Algorithm to find a solution of ax b (mod n)**

when (a, n) = 1.

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**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=203+3, 20=36+2, 3=21+1. Thus 1 =

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Ex2. Solve 20x 14 (mod 63)

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

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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.

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

Pf:

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**Thm5. Moreover, any two solutions of ax b (mod n) in Z are congruent modulo n.**

Pf:

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**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).

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