Congruences - Prof Still

Congruences

The theory of congruences was developed at the beginning of thenineteenth century by Carl Friedrich Gauss. A congruence is nothing morethan a statement about divisibility.

Definition 1. Let m 6= 0 where m is an integer. If m|(a− b) we say that ais congruent to b modulo m, and we write a ≡ b(mod m). If m 6 |(a − b),we write a 6≡ b(mod m).

Example 1.

3 ≡ −6(mod 9) 12 6≡ 0(mod 7)15 ≡ 0(mod 5) 63 6≡ 24(mod 2)1 ≡ 17(mod 4) 4! 6≡ −1(mod 4)

Chapter 2: Congruences 1

Theorem 1. Let a, b, c, d ∈ Z.

1. The following statements are equivalent:

(a) a ≡ b(mod m)(b) b ≡ a(mod m)(c) a− b ≡ 0(mod m)

2. If a ≡ b(mod m) and b ≡ c(mod m) then a ≡ c(mod m).

3. If a ≡ b(mod m) and c ≡ d(mod m) then a+ c ≡ b+ d(mod m).

4. If a ≡ b(mod m) and c ≡ d(mod m) then ac ≡ bd(mod m).

5. If a ≡ b(mod m) and d|m, d > 0 then a ≡ b(mod d).

6. If a ≡ b(mod m) then ac ≡ bc(mod mc) for c > 0.


Illustration 1.

1. 7 ≡ 32(mod 5)⇔ 32 ≡ 7(mod 5)⇔ 32− 7 ≡ 0(mod 5).

2. 7 ≡ 32(mod 5) and 32 ≡ 2(mod 5)⇒ 7 ≡ 2(mod 5).

3. 2 ≡ 17(mod 3) and 5 ≡ −4(mod 3)⇒ 2 + 5 ≡ 17 + (−4)(mod 3).

4. 21 ≡ 1(mod 4) and 3 ≡ −9(mod 4)⇒ 63 ≡ −9(mod 4)

5. 6 ≡ 12(mod 6) and 2|6⇒ 6 ≡ 12(mod 2)

6. 4 ≡ 10(mod 6)⇒ 12 ≡ 30(mod 18)


Theorem 2. Let f be a polynomial with integral coefficients. If a ≡b(mod m) then f(a) ≡ f(b)(mod m).

Example 2. Let f(x) = x3−2x2 +5. Since 3 ≡ −1(mod 4) we must havef(3) ≡ f(−1)(mod 4).

f(3) = 33 − 2(32) + 5 = 27− 18 + 5 = 14

f(−1) = (−1)3 − 2(−1)2 + 5 = −1− 2 + 5 = 2

=⇒ 14 ≡ 2(mod 4)


Theorem 3.

1. ax ≡ ay(mod m) if and only if x ≡ y(

mod m(a,m)

).

2. If ax ≡ ay(mod m) and (a,m) = 1 then x ≡ y(mod m).

3. x ≡ y(mod mi) if and only if x ≡ y(mod [m1,m2, . . . ,mk]).

Illustration 2.

1. 32 ≡ 8(mod 8)⇔ 8 ≡ 2(mod 2) where a = 4 and m = 8.

2. 10 ≡ 30(mod 2)⇔ 2 ≡ 6(mod 2) where a = 5 and m = 2.

3. 62 ≡ 72(mod 10)⇔ 62 ≡ 72(mod 5) and 62 ≡ 72(mod 2).


Definition 2. If x ≡ r(mod m) then r is called a residue of x modulo m.A set {x1, x2, . . . xm} is called a complete residue system modulo m iffor all integer y there exists exactly one xi such that y ≡ xi(mod m).

Illustration 3.

Let y ∈ Z =⇒ y = mq + r, 0 ≤ r < m

=⇒ y − r = mq

=⇒ y ≡ r(mod m)

Hence, for all integer y there exists exactly one xi such that y ≡xi(mod m) and {0, 1, 2, . . . ,m− 1} constitute a complete residue systemmodulo m.


Let m = 10 then S = {0, 1, 2, . . . , 9} is a complete residue systemmodulo 10. If y = 299 then 299 ≡ 9(mod 10). However, there areinfinitely many complete residue systems modulo m. For this particularexample, S′ = {1, 2, . . . , 10} is another complete residue system modulo10.

Let {x1, x2, . . . , xm} be a complete residue system modulo m.

If x′1 ≡ x1(mod m)

x′2 ≡ x2(mod m)

...

x′m ≡ xm(mod m)

Then {x′1, x′2, . . . , x′m} is another complete residue system modulo m.


Remark: If C is any complete residue system modulo m then

1. C has exactly m elements

2. The elements of C are pairwise incongruent modulo m.

Theorem 4. If x ≡ y(mod m) then (x,m) = (y,m).

Illustration 4. 12 ≡ 32(mod 10) then (12, 10) = (32, 10) = 2.


Definition 3. A reduced residue system modulo m is a set S of integersri such that

1. (ri,m) = 1 for all ri ∈ S

2. ri 6≡ rj(mod m) if ri 6= rj and

3. for each integer y, if (y,m) = 1 then y ≡ ri(mod m) for some ri ∈ S.


Example 3. We know that C = {1, 2, 3, 4, 5, 6, 7, 8, 9} is a completeresidue system modulo 9. If we delete from this set every integer xwith (x, 9) > 1, we obtain the set S = {1, 2, 4, 5, 7, 8}. From definition3,

1. (ri, 9) = 1 for all ri ∈ S,

2. since S ⊆ C and C is a complete residue system modulo m, then allits elements are pairwise incongruent modulo 9,

3. let y ∈ Z such that (y, 9) = 1. By the definition of a complete residuesystem, there exists xi ∈ C such that y ≡ xi(mod 9). By theorem 4,(xi,m) = (y,m) = 1

Hence, S is a reduced residue system modulo 9.


Theorem 5. The number φ(m) is equal to the number of positive integersnot exceeding m and relatively prime to m.

Theorem 6. Let (a,m) = 1. Let A = {x1, x2, . . . , xk} and aA ={ax1, ax2, . . . , axk}.

1. If A is complete residue system modulo m then aA is complete residuesystem modulo m.

2. If A is reduced residue system modulo m then aA is reduced residuesystem modulo m.


Illustration 5.

1. Let A = {1, 2, 3, 4, 5} and let a = 2. Then aA = {2, 4, 6, 8, 10}.Observe that

1 ≡ 6(mod 5)

2 ≡ 2(mod 5)

3 ≡ 8(mod 5)

4 ≡ 4(mod 5)

5 ≡ 10(mod 5)

Since A is a complete residue system modulo 5 then aA is also acomplete residue system modulo 5.

2. Let S = {1, 2, 4, 5, 7, 8} and let a = 4. Then aS = {4, 8, 16, 20, 28, 32}.


Observe that

1 ≡ 28(mod 9)

2 ≡ 20(mod 9)

4 ≡ 4(mod 9)

5 ≡ 32(mod 9)

7 ≡ 16(mod 9)

8 ≡ 8(mod 9)

Since S is a reduced residue system modulo 5 then aS is also a reducedresidue system modulo 9. Also, φ(9) = 6.


Theorem 7. (Euler) If (a,m) = 1 then aφ(m) ≡ 1(mod m).

Illustration 6. Let a = 2 and m = 9. From illustration 5 part (2),φ(9) = 6. Hence, 1 ≡ 26(mod 9).

Theorem 8. (Fermat) Let p denote a prime. If p 6 |a then ap−1 ≡1(mod p).

Illustration 7. Let p = 29, since 29 6 |6 then 628 ≡ 1(mod 29).

Theorem 9. If (a,m) = 1 then ax ≡ b(mod m) has a solution x = x0. Ifx = x1 is any other solution, then x1 ≡ x0(mod m).

Example 4. Given 3x ≡ 5(mod 11), x0 = 3φ(11)−1 ·5 is a solution. Hence,39 · 5 ≡ 5(mod 11) since φ(11) = 10.

Theorem 10. (Wilson) If p is a prime then (p− 1)! ≡ −1(mod p).


Illustration 8. Let p = 5 then 4! ≡ −1(mod 5) and if p = 101 then100! ≡ −1(mod 101)

Theorem 11. Let p denote a prime. Then x2 ≡ −1(mod p) has solutionsif and only if p = 2 or p ≡ 1(mod 4).

Illustration 9. Let p = 17 then x =

8∏j=1

j = 40320 is a solution to x2 ≡

−1(mod 17).


Solution of Congruences

In this section, we shall let f(x) denote a polynomial in x with integralcoefficients. If f(u) ≡ 0(mod m), we shall say that u is a solution ofthe congruence f(x) ≡ 0(mod m). We know that if v ≡ u(mod m) thenf(v) ≡ 0(mod m) by theorem 2. Hence there are infinitely many solutions.However, we shall count solutions in a different way. If v ≡ u(mod m), weshall count v as a solution distinct from u. We shall define the number ofsolutions of the congruence f(x) ≡ 0(mod m) to be the number of pairwiseincongruent solutions modulo m.


Definition 4. Let x1, x2, . . . , xm form a complete residue system modulom. The number of solutions of the congruence f(x) ≡ 0(mod m) is thenumber of xi such that f(xi) ≡ 0(mod m)

Example 5. Given 3x ≡ 0(mod 6) and C = {1, 2, 3, 4, 5, 6} a completeresidue system modulo 6. 2, 4 and 6 are solutions which implies thenumber of solutions is 3.

Definition 5. Let f(x) = a0 + a1x+ a2x2 + . . .+ anx

n. If an 6≡ 0(mod m)then f(x) ≡ 0(mod m) is of degree n.

Example 6.

1. f(x) = 3x− 8 ≡ 0(mod 6) is of degree 1 since 3 6≡ 0(mod 6).

2. 2 − 5x + 6x2 − 12x3 ≡ 0(mod 4) is of degree 2 since 12 ≡ 0(mod 4)and 6 6≡ 0(mod 4).


Theorem 12. If d|m, d > 0, and if u is a solution of f(x) ≡ 0(mod m),then u is a solution of f(x) ≡ 0(mod d).

Any congruence of degree 1 can be written in the form ax ≡ b(mod m),a 6≡ 0(mod m). From theorem 9, there is exactly one solution if (a,m) = 1.The next theorem treats the general case of a congruence of degree 1.

Theorem 13. Let (a,m) = g. Then ax ≡ b(mod m) has no solution ifg 6 |b. If g|b it has g solutions given by

x ≡ b

gx0 + t

m

g(mod m) t = 0, 1, 2, . . . , g − 1

where x0 is any solution ofa

gx ≡ 1

(mod

m

g

).


Illustration 10.

1. 4x ≡ 7(mod 6) has no solution since (4, 6) = 2 and 2 6 |7.

2. 4x ≡ 8(mod 6) has 2 solutions since (4, 6) = 2 and 2|8. The 2solutions are x1 = 4x0 and x2 = 4x0 + 3 where x0 is a solution to4

2x ≡ 1

(mod

6

2

). Hence, x0 = 2φ(3)−1 = 2. Thus, x1 = 8 and

x2 = 11 where 4(8) ≡ 8(mod 6) and 4(11) ≡ 8(mod 6).


The Chinese Remainder Theorem

A very rich man owns a great number of cars. If he will divide them asevenly as possible among his 12 friends, there will be remainder of 4, if hewill divide them among his 7 children, there will be remainder of 2. Whatis the minimum number of cars that the rich man could possibly have?

Problems similar to this can be solved by the following theorem:

Theorem 14. The Chinese Remainder Theorem. Let m1,m2, . . . ,mk

denote k positive integers that are relatively prime in pairs andlet a1, a2, . . . , ak denote any k integers. The congruences x ≡ai(mod mi), i = 1, 2, . . . , k, have common solutions. Any two solutionsare congruent modulo m1,m2, . . . ,mk.


Illustration 11. For the problem above, x ≡ 4(mod 12) and x ≡ 2(mod 7).

Let m = 84. To find b1, we solve84

12b1 ≡ 1(mod 12) and to find b2, we

solve84

7b2 ≡ 1(mod 7).

7b1 ≡ 1(mod 12) =⇒ b1 = 7φ(12)−1 = 73 ≡ 7(mod 12)

=⇒ b1 = 7

12b2 ≡ 1(mod 7) =⇒ b2 = 12φ(7)−1 = 125 ≡ 55(mod 7) ≡ 3(mod 7)

=⇒ b2 = 3


Hence,

x ≡ 7(7)a1 + 12a2(mod 84)

≡ 49(4) + 36(2)(mod 84)

≡ 268(mod 84)

≡ 16(mod 84)

Therefore, the minimum number of cars that the rich man couldpossible have is 16.


Now we recall the function φ(n) which is defined for all positive integersn. In general, any function whose domain is the set Z+ is called a numericalfunction. A numerical function f : Z+ −→ Z is said to be multiplicative iff(mn) = f(m)f(n) for all pairs of relatively prime integers m and n. Thenumerical function f is called totally multiplicative if f(mn) = f(m)f(n)for all pairs of integers m and n.


Theorem 15. The function φ(n) is multiplicative.

Illustration 12.

1. Since (6, 7) = 1, φ(6) = 2 and φ(7) = 6, then φ(42) = φ(6)φ(7) = 12.

2. Since (2, 4) = 2, φ(2) = 1 and φ(4) = 2, then φ(8) 6= φ(2)φ(4) sinceφ(8) = 4 6= 2.

Let A be any finite set and let xi be a real number for each i ∈ A. Weshall denote by ∏

i∈A

xi

the product of all the numbers xi where i ranges over all elements of thefinite set A. In particular, if A = {1, 2, . . . , n} then the same product will


be denoted byn∏i=1

xi.

In case A = ∅, then the product will be defined to be equal to 1. If A isthe set of all prime divisors of n ∈ Z, then we shall write∏

p|n

xp for∏p∈A

xp.

Example 7. If A = {1, 2, 3, 4} then∏i∈A

xi = x1 · x2 · x3 · x4. If n = 10

then A = {2, 5} and∏p|n

xp = x2 · x5.


Primitive Roots

Definition 6. Let m denote a positive integer and a any integer suchthat (a,m) = 1. Let h be the smallest positive integer such that ah ≡1(mod m). We say that the order of a modulo m is h, or that abelongs to the exponent h modulo m.

Lemma 16. If a has order h(mod m), then the positive integers k suchthat ak ≡ 1(mod m) are precisely those for which h|k.

Corollary 16.1. If (a,m) = 1, then the order of a modulo m divides φ(m)

Lemma 17. If a has order h modulo m, then ak has orderh

(h, k)modulo

m.

Lemma 18. If a has order h(mod m), b has order k(mod m) , and if(h, k) = 1, then ab has order hk(mod m).


Definition 7. If g belongs to the exponent φ(m) modulo m, then g is calleda primitive root modulo m

In algebraic language, this definition can be stated: If the order ofg modulo m is φ(m), then the multiplicative group of reduced residuesmodulo mis a cyclic group generated by the element g.

In view of Lemma 16, the number a is a solution of the congruencexk ≡ 1(mod m) if and only if the order of a(mod m) divides k. In onespecial case, where xd ≡ 1(mod p) with d|(p− 1), the number of solutionsis d.

If p is prime and k|(p − 1), then there are precisely k residue classesa(mod p) such that the order of a modulo p is a divisor of k. If k happensto be a prime power, we can then determine the exact number of residuesa(mod p) of order k.


Lemma 19. Let p and q be primes, and suppose that qα|(p − 1), whereα ≥ 1. Then there are precisely qα − qα−1 residue classes a(mod p) oforder qα.

Theorem 20. If p is a prime, then there exist φ(p − 1) primitive rootsmodulo p.

Definition 8. If (a, p) = 1 and xn ≡ a(mod p) has a solution, then a iscalled an nth power residue modulo p.

Theorem 21. Let a, b and m > 0 be given integers, and let g = (a,m).The congruence ax ≡ b(mod m) has a solution if and only if g|b. If thiscondition is met then the solution form an arithmetic progression withcommon difference m/g giving g solutions (mod m).


Theorem 22. If p is a prime and (a, p) = 1, then the congruence xn ≡a(mod p) has (n,p-1) solutions or no solution according as

a(p−1)/(n,p−1) ≡ 1(mod p)

or not.

Lemma 23. Let p be a prime number. Then x2 ≡ 1(mod p) if and onlyif x ≡ ±1(mod p)

Corollary 23.1. (Euler’s Criterion) If p is an odd prime and (a, p) =1, then x2 ≡ a(mod p) has two solutions or no solution according asa(p−1)/2 ≡ 1 or ≡ −1(mod p).


Congruences - Prof Still

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