Math 373/578, Spring 2013 due: Thursday, February 7, 2013

Homework 2 Solutions (fixed)

Section 1.4

Written problems

[1] (5 pts) p53, Problem 1.27

Without using the fact that every integer has a unique factorization into primes, prove that if gcd(a,b) = 1 and if

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a|bc , then

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a|c .

proof: Let

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a,b,c ∈ Z . Suppose

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gcd(a,b) =1 and

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a|bc .

Since

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gcd(a,b) =1, then

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∃u,v ∈ Z such that

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au + bv =1. (1)

Since

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a|bc , then

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∃d ∈ Z such that

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ad = bc . (2)

Multiplying both sides of (2) by v:

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adv = avc . (3)

Rearranging (1):

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bv =1−au and substituting into (3):

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adv = (1−au)c ⇒ adv = c − cau ⇒ adv + cau = c ⇒ a(dv + cu) = c

Since

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d,c ,u,v ∈ Z , then

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dv + cu ∈ Z , so

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a|c .

Section 1.5

Computational problems

[2] (5 pts) p53, Problem 1.30

For each of the following primes p and numbers a, compute

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a−1(mod p) in two different ways.

(a)

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p = 47, a =11

(1) >> [d,u,v]=gcd(11,47)

d = so,

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11−1 ≡ −17≡ 30(mod 47)

1

u =

-17

v =

4

(2) >> powermod(11,45,47)

ans =

30

Math 373/578, Spring 2013 due: Thursday, February 7, 2013

(b)

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p = 587, a = 345

(1) >> [d,u,v]=gcd(345,587)

d = so,

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345−1 ≡114(mod 587)

1

u =

114

v =

-67

(2) >> powermod(345,585,587)

ans =

114

(c)

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p =104801, a = 78467

(1) >> [d,u,v]=gcd(78467,104801)

d = so,

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78467−1 ≡1763(mod104801)

1

u =

1763

v =

-1320

(2) >> powermod(78467,104799,104801)

ans =

1763

[3] (5 pts) p53, Problem 1.32, parts (a), (b), & (c)

Recall that g is a primitive root modulo p if the powers of g give all the nonzero elements of

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Fp .

(a) For which of the following primes is 2 a primitive root modulo p?

Math 373/578, Spring 2013 due: Thursday, February 7, 2013

(i)

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p = 7

>> X=[]; for x=1:6; y=powermod(2,x,7); X=[X,y];end

>> X

X =

2 4 1 2 4 1 so 2 is not a primitive root mod7

(ii)

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p =13

>> X=[]; for x=1:12; y=powermod(2,x,13); X=[X,y];end

>> X

X =

Columns 1 through 9

2 4 8 3 6 12 11 9 5

Columns 10 through 12

10 7 1 so 2 is a primitive root mod13

(iii)

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p =19

>> X=[]; for x=1:18; y=powermod(2,x,19); X=[X,y];end

>> X

X = so 2 is a primitive root mod19

Columns 1 through 9

2 4 8 16 13 7 14 9 18

Columns 10 through 18

17 15 11 3 6 12 5 10 1

(iv)

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p = 23

>> X=[]; for x=1:22; y=powermod(2,x,23); X=[X,y];end

>> X

Math 373/578, Spring 2013 due: Thursday, February 7, 2013

X =

Columns 1 through 9

2 4 8 16 9 18 13 3 6

Columns 10 through 18

12 1 2 4 8 16 9 18 13

Columns 19 through 22

3 6 12 1 so 2 is not a primitive root mod23

(b) For which of the following primes is 3 a primitive root modulo p?

(i)

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p = 5

>> X=[]; for x=1:4; y=powermod(3,x,5); X=[X,y];end

>> X

X =

3 4 2 1 so 3 is a primitive root mod5

(ii)

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p = 7

>> X=[]; for x=1:6; y=powermod(3,x,7); X=[X,y];end

>> X

X =

3 2 6 4 5 1 so 3 is a primitive root mod7

(iii)

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p =11

>> X=[]; for x=1:10; y=powermod(3,x,11); X=[X,y];end

>> X so 3 is not a primitive root mod11

X =

3 9 5 4 1 3 9 5 4 1 1

Math 373/578, Spring 2013 due: Thursday, February 7, 2013

(iv)

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p =17

>> X=[]; for x=1:16; y=powermod(3,x,17); X=[X,y];end

>> X

X =

Columns 1 through 12

3 9 10 13 5 15 11 16 14 8 7 4

Columns 13 through 16

12 2 6 1 so 3 is a primitive root mod17

(c) Find a primitive root for each of the following primes.

(i)

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p = 23

>> X=[]; for x=1:22; y=powermod(5,x,23); X=[X,y];end

>> X

X = so 5 is a primitive root mod23

Columns 1 through 12

5 2 10 4 20 8 17 16 11 9 22 18

Columns 13 through 22

21 13 19 3 15 6 7 12 14 1

(ii)

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p = 29

>> X=[]; for x=1:28; y=powermod(2,x,29); X=[X,y];end

>> X

X = so 2 is a primitive root mod29

Columns 1 through 12

2 4 8 16 3 6 12 24 19 9 18 7

Columns 13 through 24

14 28 27 25 21 13 26 23 17 5 10 20

Columns 25 through 28

11 22 15 1

Math 373/578, Spring 2013 due: Thursday, February 7, 2013

(iii)

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p = 41

>> X=[]; for x=1:40; y=powermod(7,x,41); X=[X,y];end

>> X

X =

Columns 1 through 12

7 8 15 23 38 20 17 37 13 9 22 31

Columns 13 through 24

12 2 14 16 30 5 35 40 34 33 26 18

Columns 25 through 36

3 21 24 4 28 32 19 10 29 39 27 25

Columns 37 through 40

11 36 6 1 so 7 is a primitive root mod41

(iv)

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p = 43

>> X=[]; for x=1:42; y=powermod(3,x,43); X=[X,y];end

>> X

X =

Columns 1 through 12

3 9 27 38 28 41 37 25 32 10 30 4

Columns 13 through 24

12 36 22 23 26 35 19 14 42 40 34 16

Columns 25 through 36

5 15 2 6 18 11 33 13 39 31 7 21

Columns 37 through 42

20 17 8 24 29 1 so 3 is a primitive root mod43

[4] (5 pts) p54, Problem 1.34, part (b) [omit]

Math 373/578, Spring 2013 due: Thursday, February 7, 2013

For Math 578 students only:

[5] (10 pts) p54, Problem 1.34, part (a)

Let p be an odd prime number and let b be an integer with

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p / | b . Prove that either b has two square roots modulo p or else b has no square roots modulo p.

In other words, prove that the congruence

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X 2 ≡ b(mod p) has either two solutions or no solutions in

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Z p .

(What happens for p = 2? What happens if

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p|b ?)

proof: It suffices to show that if b has at least one square root modulo p, then b must have exactly two square roots modulo p.

Suppose that a is a square root of b modp. Since

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p / | b , then

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a / ≡ 0(mod p) . Therefore

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p / | a and thus

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gcd(a, p) =1.

Since

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a2 ≡ b(mod p), it follows that

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(−a)2 ≡ a2 ≡ b(mod p) and so –a must also be a square root of b (modp). We need to explain that

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a / ≡ −a(mod p) . If we had

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a ≡ −a(mod p) , then

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2a ≡ 0(mod p), so

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p|(2a). Since

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gcd(a, p) =1 and

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p|(2a), then

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p|2, which is contrary to p being prime. Hence,

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a / ≡ −a(mod p) , and so b has at least two square roots.

We need to show that b cannot have more than 2 square roots. Suppose that for some

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c ∈ Z p ,

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c / ≡ a(mod p) and

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c / ≡ −a(mod p) but

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c 2 ≡ b(mod p). Then we have

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a2 ≡ b ≡ c 2(mod p), or

equivalently,

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(a − c )(a + c ) ≡ a2 − c 2 ≡ b −b ≡ (mod p) . This means

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p|(a + c )(a − c ). If

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p|(a − c ), then

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c ≡ a(mod p) ; if

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p|(a + c ), then

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c ≡ −a(mod p) . Either way will lead to a contradiction to the assumptions that

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c / ≡ a(mod p) and

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c / ≡ −a(mod p) . Thus, b must have exactly two square roots, if it has at least one.