Elementary Number Theory, 7th
Edition by Kenneth H. Rosen
Complete Chapter Solutions Manual
are included (Ch 1 to 15)
** Immediate Download
** Swift Response
** All Chapters included
,Table of Contents are given below
1.The Integers
2.Integer Representations and Operations
3.Greatest Common Divisors
4.Prime Numbers
5.Congruences
6.Applications of Congruences
7.Some Special Congruences
8.Arithmetic Functions
9.Cryptography
10.Primitive Roots
11.Applications of Primitive Roots and the Order of an Integer
12.Quadratic Residues
13.Decimal Fractions and Continued Fractions
14.Nonlinear Diophantine Equations and Elliptic Curves
15.The Gaussian Integers
,Solutions Manual organized in reverse order, with the last chapter displayed first, to ensure that all
chapters are included in this document. (Complete Chapters included Ch15-1)
CHAPTER 15
The Gaussian Integers
15.1. Gaussian Integers and Gaussian Primes
15.1.1. a. First, (2 + i)(2 + i) = 4 + 2i + 2i + i2 = 4 + 4i − 1 = 3 + 4i. Then we have (2 + i)2 (3 + i) =
(3 + 4i)(3 + i) = 9 + 12i + 3i + 4i2 = 9 + 15i − 4 = 5 + 15i.
b. First, (2 − 3i)(2 − 3i) = 4 − 6i − 6i + 9i2 = 4 − 12i − 9 = −5 − 12i. Then we have (2 − 3i)3 =
(−5 − 12i)(2 − 3i) = −10 − 24i + 15i + 36i2 = −10 − 9i − 36 = −46 − 9i.
c. First, (−i + 3)(−i + 3) = i2 − 3i − 3i + 9 = −1 − 6i + 9 = 8 − 6i. Next, −i(−i + 3) = i2 − 3i = −1 − 3i.
Finally, we have −i(−i + 3)3 = (8 − 6i)(−1 − 3i) = −8 + 6i − 24i + 18i2 = −8 − 18i − 18 = −26 − 18i.
15.1.2. a. First we compute (−1+i)(1+i) = −1+i−i+i2 = −2. Then we have (−1+i)3 (1+i)3 = (−2)3 = −8.
b. First, (3 − i)(3 − i) = 9 − 6i + i2 = 8 − 6i, so that (3 + 2i)(3 − i)2 = (3 + 2i)(8 − 6i) = 24 + 16i −
18i − 12i2 = 36 − 2i.
c. By the binomial theorem, we have (5 − i)3 = 53 − 3 · 52 i + 3 · 5i2 − i3 = 125 − 75i − 15 + i =
110 − 74i. Also (2 + i)(2 + i) = 4 + 4i + i2 = 3 + 4i. Therefore, (2 + i)2 (5 − i)3 = (3 + 4i)(110 − 74i) =
330 + 440i − 222i − 296i2 = 626 + 218i.
β 5 + 5i (5 + 5i)(2 + i) 5 + 15i
15.1.3. a. We evaluate the fraction = = = = 1 + 3i, which is a Gaussian in-
α 2−i (2 − i)(2 + i) 5
teger. Therefore, α divides β, since α(1 + 3i) = β.
8 8(1 + i) 8+i
b. We evaluate the fraction = = = 4 + 4i, which is a Gaussian integer. There-
1−i (1 − i)(1 + i) 2
fore 8 = (1 − i)(4 + 4i) and so α divides β.
c. Since N (α) = N (5) = 25 and N (β) = N (2 + 3i) = 4 + 9 = 13, we observe that 25 ∤ 13. Therefore, α
can not divide β.
26 26(3 − 2i) 78 − 52i
d. We evaluate the fraction = = = 6 − 4i, which is a Gaussian inte-
3 + 2i (3 + 2i)(3 − 2i) 13
ger. Therefore, α divides β.
15.1.4. a. Since N (α) = N (3) = 9 and N (β) = N (4 + 7i) = 16 + 49 = 65, we observe that 9 ∤ 65, and so α can
not divide β.
15 15(2 − i) 30 − 15i
b. We evaluate the fraction = = = 6 − 3i, which is a Gaussian integer.
2+i (2 + i)(2 − i) 5
Therefore, α divides β.
30 + 6i (5 + i)(5 − 3i) 28 − 10i
c. We evaluate the fraction =6 =6 , which is not a Gaussian integer.
5 + 3i (5 + 3i)(5 − 3i) 34
Therefore, α does not divide β.
269
, 270 Section 15.1
274 274(11 − 4i) 274(11 − 4i)
d. We evaluate the fraction = = = 2(11 − 4i) = 22 − 8i, which
11 + 4i (11 + 4i)(11 − 4i) 137
is a Gaussian integer. Therefore, α divides β.
15.1.5. Since a Gaussian integer must be of the form
-20
a + bi with a and b rational integers, then for a
Gaussian integer α to be divisible by 4 + 3i, we
must have α = (4 + 3i)(a + bi) = (4a − 3b) +
-10
(4b + 3a)i and this gives us a formula for all
Gaussian integers divisible by 4 + 3i in terms
of rational integers a and b. To the right is a
-20
-10
10
20
0
0
display of the pattern of this set in the plane.
10
20
15.1.6. Since a Gaussian integer must be of the form 10
a + bi with a and b rational integers, then for a
Gaussian integer α to be divisible by 4 − i, we
must have α = (4 − i)(a + bi) = (4a + b) + (4b −
5
a)i and this gives us a formula for all Gaussian
integers divisible by 4 − i in terms of rational
integers a and b. To the right is a display of the
pattern of this set in the plane. –10 –5 5 10
–5
–10
15.1.7. Since α|β and β|γ, there exist Gaussian integers µ and ν such that µα = β and νβ = γ. Since the prod-
uct of Gaussian integers is again a Gaussian integer, we have that νµ is also a Gaussian integer. Then
γ = νβ = νµα and so α|γ.
15.1.8. Since γ | α and γ | β there exist Gaussian integers ρ and σ such that α = ργ and β = σγ. Then we
have µα + νβ = µργ + νσγ = (µρ + νσ)γ. Since (µρ + νσ) is a Gaussian integer, we have γ | (µα + νβ).
15.1.9. Consider the equation x5 = x or x5 − x = 0. The left side factors over the Gaussian integers as x(x −
1)(x + 1)(x − i)(x + i) = 0, so the solutions of the equation are 0, 1, −1, i, and −i. Since this includes all
of the units for the Gaussian integers, this proves the result.
15.1.10. If α is an associate of α = a + bi then we must have α = ϵα where ϵ is a unit, so there are 4 cases to
consider. If ϵ = 1, we have a − bi = a + bi and so b = 0 and α = a is a rational integer. If ϵ = −1, we
have a − bi = −a − bi and so a = 0 and α = bi is a pure imaginary number. If ϵ = i, we have a − bi =
i(a + bi) = −b + ai from which we deduce a = −b, so α is of the form a − ai = a(1 − i). If ϵ = −i, we have
a − bi = −i(a + bi) = b − ai from which we deduce a = b, so α is of the form a + ai = a(1 + i). Therefore
if α is an associate of its conjugate it must be of the one of the forms a, ai, a(1 − i), a(1 + i), where a is a
rational integer.
Copyright © 2023 Pearson Education, Inc.
Edition by Kenneth H. Rosen
Complete Chapter Solutions Manual
are included (Ch 1 to 15)
** Immediate Download
** Swift Response
** All Chapters included
,Table of Contents are given below
1.The Integers
2.Integer Representations and Operations
3.Greatest Common Divisors
4.Prime Numbers
5.Congruences
6.Applications of Congruences
7.Some Special Congruences
8.Arithmetic Functions
9.Cryptography
10.Primitive Roots
11.Applications of Primitive Roots and the Order of an Integer
12.Quadratic Residues
13.Decimal Fractions and Continued Fractions
14.Nonlinear Diophantine Equations and Elliptic Curves
15.The Gaussian Integers
,Solutions Manual organized in reverse order, with the last chapter displayed first, to ensure that all
chapters are included in this document. (Complete Chapters included Ch15-1)
CHAPTER 15
The Gaussian Integers
15.1. Gaussian Integers and Gaussian Primes
15.1.1. a. First, (2 + i)(2 + i) = 4 + 2i + 2i + i2 = 4 + 4i − 1 = 3 + 4i. Then we have (2 + i)2 (3 + i) =
(3 + 4i)(3 + i) = 9 + 12i + 3i + 4i2 = 9 + 15i − 4 = 5 + 15i.
b. First, (2 − 3i)(2 − 3i) = 4 − 6i − 6i + 9i2 = 4 − 12i − 9 = −5 − 12i. Then we have (2 − 3i)3 =
(−5 − 12i)(2 − 3i) = −10 − 24i + 15i + 36i2 = −10 − 9i − 36 = −46 − 9i.
c. First, (−i + 3)(−i + 3) = i2 − 3i − 3i + 9 = −1 − 6i + 9 = 8 − 6i. Next, −i(−i + 3) = i2 − 3i = −1 − 3i.
Finally, we have −i(−i + 3)3 = (8 − 6i)(−1 − 3i) = −8 + 6i − 24i + 18i2 = −8 − 18i − 18 = −26 − 18i.
15.1.2. a. First we compute (−1+i)(1+i) = −1+i−i+i2 = −2. Then we have (−1+i)3 (1+i)3 = (−2)3 = −8.
b. First, (3 − i)(3 − i) = 9 − 6i + i2 = 8 − 6i, so that (3 + 2i)(3 − i)2 = (3 + 2i)(8 − 6i) = 24 + 16i −
18i − 12i2 = 36 − 2i.
c. By the binomial theorem, we have (5 − i)3 = 53 − 3 · 52 i + 3 · 5i2 − i3 = 125 − 75i − 15 + i =
110 − 74i. Also (2 + i)(2 + i) = 4 + 4i + i2 = 3 + 4i. Therefore, (2 + i)2 (5 − i)3 = (3 + 4i)(110 − 74i) =
330 + 440i − 222i − 296i2 = 626 + 218i.
β 5 + 5i (5 + 5i)(2 + i) 5 + 15i
15.1.3. a. We evaluate the fraction = = = = 1 + 3i, which is a Gaussian in-
α 2−i (2 − i)(2 + i) 5
teger. Therefore, α divides β, since α(1 + 3i) = β.
8 8(1 + i) 8+i
b. We evaluate the fraction = = = 4 + 4i, which is a Gaussian integer. There-
1−i (1 − i)(1 + i) 2
fore 8 = (1 − i)(4 + 4i) and so α divides β.
c. Since N (α) = N (5) = 25 and N (β) = N (2 + 3i) = 4 + 9 = 13, we observe that 25 ∤ 13. Therefore, α
can not divide β.
26 26(3 − 2i) 78 − 52i
d. We evaluate the fraction = = = 6 − 4i, which is a Gaussian inte-
3 + 2i (3 + 2i)(3 − 2i) 13
ger. Therefore, α divides β.
15.1.4. a. Since N (α) = N (3) = 9 and N (β) = N (4 + 7i) = 16 + 49 = 65, we observe that 9 ∤ 65, and so α can
not divide β.
15 15(2 − i) 30 − 15i
b. We evaluate the fraction = = = 6 − 3i, which is a Gaussian integer.
2+i (2 + i)(2 − i) 5
Therefore, α divides β.
30 + 6i (5 + i)(5 − 3i) 28 − 10i
c. We evaluate the fraction =6 =6 , which is not a Gaussian integer.
5 + 3i (5 + 3i)(5 − 3i) 34
Therefore, α does not divide β.
269
, 270 Section 15.1
274 274(11 − 4i) 274(11 − 4i)
d. We evaluate the fraction = = = 2(11 − 4i) = 22 − 8i, which
11 + 4i (11 + 4i)(11 − 4i) 137
is a Gaussian integer. Therefore, α divides β.
15.1.5. Since a Gaussian integer must be of the form
-20
a + bi with a and b rational integers, then for a
Gaussian integer α to be divisible by 4 + 3i, we
must have α = (4 + 3i)(a + bi) = (4a − 3b) +
-10
(4b + 3a)i and this gives us a formula for all
Gaussian integers divisible by 4 + 3i in terms
of rational integers a and b. To the right is a
-20
-10
10
20
0
0
display of the pattern of this set in the plane.
10
20
15.1.6. Since a Gaussian integer must be of the form 10
a + bi with a and b rational integers, then for a
Gaussian integer α to be divisible by 4 − i, we
must have α = (4 − i)(a + bi) = (4a + b) + (4b −
5
a)i and this gives us a formula for all Gaussian
integers divisible by 4 − i in terms of rational
integers a and b. To the right is a display of the
pattern of this set in the plane. –10 –5 5 10
–5
–10
15.1.7. Since α|β and β|γ, there exist Gaussian integers µ and ν such that µα = β and νβ = γ. Since the prod-
uct of Gaussian integers is again a Gaussian integer, we have that νµ is also a Gaussian integer. Then
γ = νβ = νµα and so α|γ.
15.1.8. Since γ | α and γ | β there exist Gaussian integers ρ and σ such that α = ργ and β = σγ. Then we
have µα + νβ = µργ + νσγ = (µρ + νσ)γ. Since (µρ + νσ) is a Gaussian integer, we have γ | (µα + νβ).
15.1.9. Consider the equation x5 = x or x5 − x = 0. The left side factors over the Gaussian integers as x(x −
1)(x + 1)(x − i)(x + i) = 0, so the solutions of the equation are 0, 1, −1, i, and −i. Since this includes all
of the units for the Gaussian integers, this proves the result.
15.1.10. If α is an associate of α = a + bi then we must have α = ϵα where ϵ is a unit, so there are 4 cases to
consider. If ϵ = 1, we have a − bi = a + bi and so b = 0 and α = a is a rational integer. If ϵ = −1, we
have a − bi = −a − bi and so a = 0 and α = bi is a pure imaginary number. If ϵ = i, we have a − bi =
i(a + bi) = −b + ai from which we deduce a = −b, so α is of the form a − ai = a(1 − i). If ϵ = −i, we have
a − bi = −i(a + bi) = b − ai from which we deduce a = b, so α is of the form a + ai = a(1 + i). Therefore
if α is an associate of its conjugate it must be of the one of the forms a, ai, a(1 − i), a(1 + i), where a is a
rational integer.
Copyright © 2023 Pearson Education, Inc.
