homework solutions introductory modern algebra, spring...

Homework Solutions

Introductory Modern Algebra, Spring 2018

[Chapter 1, Problem 1] We verify the three properties:

(Reflexive) Since ab = ba, we have that (a, b) ∼ (b, a).

(Symmetric) If (a, b) ∼ (c, d), then ad = bc, so that cb = da and (c, d) ∼ (a, b).

(Transitive) If (a, b) ∼ (c, d) and (c, d) ∼ (e, f), then ad = bc and cf = de, so that multiplyingthe equations, we get that adcf = bcde, and since c 6= 0 and d 6= 0, we can divide by them, and getthat af = be, and (a, b) ∼ (e, f).

[Chapter 1, Problem 2] To check your work, we will list which properties fail:

(i) No, Transitive property fails.

(ii) Yes.

(ii) No, Reflexive property fails.

(iv) No, Reflexive and Transitive properties fail.

(v) No, Symmetric property fails.

[Chapter 1, Problem 3] Suppose that a, b ∈ S and x ∈ [a] and x ∈ [b]. Then x ∼ a and x ∼ b; bythe Symmetric property, a ∼ x and x ∼ b, so that by the Transitive property, a ∼ b.

Now, we will show that [a] ⊆ [b]: Take any y ∈ [a], so that y ∼ a. Then since a ∼ b, theTransitive property gives that y ∼ b, so that y ∈ [b].

Similarly, it can be shown that [b] ⊆ [a], so that [a] = [b].

[Chapter 2, Problem 2] We proceed by mathematical induction on integers n ≥ 1.

• Basis case: If n = 1, the left-hand side equals 12 = 1, and the right-hand side equals(1·22

)2=

12 = 1 as well.

• Inductive hypothesis: Assume that, for some integer n ≥ 1, that

13 + 23 + 33 + · · ·+ n3 =

(n(n+ 1)

2

)2

.

MATH 558, Spring 2018 Homework Solutions 2

Then

13 + 23 + 33 + · · ·+ (n+ 1)3 =

(n(n+ 1))

2

)2

+ (n+ 1)3 by the inductive hypothesis

=n2(n+ 1)2

4+

4(n+ 1)3

4

=n2(n+ 1)2 + (4n+ 4)(n+ 1)2

4

=(n2 + 4n+ 4)(n+ 1)2

4

=(n+ 2)2(n+ 1)2

4

=

((n+ 1)(n+ 2)

2

)2

.

Therefore, by the Principle of Mathematical Induction, the statement holds for all integers n ≥ 1.


• Basis case: If n = 2, the left-hand side equals 1+2 = 3, and the right-hand side equals 22−1 = 3as well.

• Inductive hypothesis: Assume that, for some integer n ≥ 2, that

1 + 2 + 22 + 23 + · · ·+ 2n−1 = 2n − 1.

Then

1 + 2 + 22 + 23 + · · ·+ 2n = (2n − 1) + 2n by the inductive hypothesis

= 2 · 2n − 1

= 2n+1 − 1.

Therefore, by the Principle of Mathematical Induction, the statement holds for all integers n > 1.


• Basis case: If n = 2, then for any real number x, we have that x2 − 1 = (x− 1)(x+ 1), showingthe desired property.

• Inductive hypothesis: Assume that for some fixed integer n ≥ 0, and any real number x, that

xn − 1 = (x− 1)(xn−1 + xn−2 + · · ·+ x+ 1).

Then

(x− 1)(xn + xn−1 + xn−2 · · ·+ x+ 1) = (x− 1)xn + (x− 1)(xn−1 + xn−2 · · ·+ x+ 1)

= (xn+1 − xn) + (xn − 1) by the inductive hypothesis

= xn+1 − 1,

and by the Principle of Mathematical Induction, the statement holds for all integers n ≥ 2 and allreal numbers x.


[Chapter 2, Problem 8] We prove the statement by induction on integers n ≥ 4.

• Basis case: If n = 4, we have that 4! = 24 and 24 = 16, so that 4! > 24.

• Inductive hypothesis: Assume that, for some n ≥ 4, that n! > 2n.

Then

(n+ 1)! = (n+ 1) · n!

> (n+ 1) · 2n by the inductive hypothesis

≥ 2 · 2n = 2n+1. since n ≥ 4, so certainly, n+ 1 ≥ 2

[Chapter 2, Problem 9] We proceed by mathematical induction on integers n > 4.

• Basis case (n = 5): 22·5 = 210 = 1024 > 256 = 44.

• Inductive hypothesis: Assume that, for some integer n ≥ 5, that 22n > n4.

Then

22(n+1) = 22n+2

= 22n · 22

> n4 · 4 by the inductive hypothesis

=≥ (n+ 1)4,

where the last equality holds since

4n4 ≥ (n+ 1)4 ⇐⇒ 2n2 ≥ (n+ 1)2 ⇐⇒√

2n ≥ n+ 1

⇐⇒ (√

2− 1)n ≥ 1 ⇐⇒ n ≥ 1/(√

2− 1) ≈ 2.41.

Since n ≥ 5, we certainly have that n ≥ 1/(√

2− 1) as well.

[Chapter 2, Problem 13] We proceed by induction on integers n ≥ 1.

• Basis case: We need to check that 24 divides 16− 16 = 0, which is true since 0 = 24 · 0.

• Inductive hypothesis: Suppose that for 24 divides 16n− 16, so 24k = 16n− 16 for some k ∈ Z.

We want to show that this implies that 24 also divides 16n+1 − 16. However, note that

16n+1 − 16 = 16(16n − 1)

= 16(16n − 1− 15 + 15)

= 16(16n − 16 + 15)

= 16(24k + 15) by our inductive hypothesis

= 16(24k) + 240

= 24(16k + 10).

Since we have written as 16n+1 − 16 as an integer multiple of 24, we are done.

[Chapter 2, Problem 14] We proceed by induction on n ≥ 1 to show that 5 | (8n − 3n).

• Basis case (n = 1): Since 81 − 31 = 5 = 5 · 1, we have that 5 | (81 − 31).


• Inductive hypothesis: Assume that for some integer n ≥ 1, 5 | (8n− 3n), so that 8n− 3n = 5kfor some integer k.

Then we know that 8n = 3n − 5k, so that

8n+1 − 3n+1 = 8 · 8n − 3n+1

= 8 · (3n − 5k)− 3n+1 by the inductive hypothesis

= 8 · 3n − 40k − 3 · 3n

= 5 · 3n − 40k

= 5(3n − 40k).

Therefore, since 3n−40k is an integer assuming that k is an integer, we know that 5 | (8n+1−3n+1).We conclude that 5 | (8n − 3n) for all integers n ≥ 1 by the Principle of Mathematical Induction.

[Chapter 2, Problem 15] We proceed by induction on n ≥ 1 to show that 5 | (34n − 1).

• Basis case (n = 1): we have that 5 | 80 = (34 − 1) since 80 = 5 · 16.

• Inductive hypothesis: Assume that for some integer n ≥ 1, 5 | (34n − 1), so that 34n − 1 = 5kfor some integer k.

Then 34n = 5k + 1, and

34(n+1) − 1 = 34n+4 − 1

= 34n · 34 − 1

= 81(5k + 1)− 1 by the inductive hypothesis

= 5(81k) + 81− 1

= 5(81k + 16).

Hence 5 | (34(n+1) − 1) as 81k + 16 is an integer (since k is one).We conclude that 5 | (34n−1) for all integers n ≥ 1 by the Principle of Mathematical Induction.

[Chapter 2, Problem 20] We claim that the minimum number of moves for n disks, n ≥ 1, is2n − 1. We proceed by induction on n ≥ 1.

• Basis case: If n = 1, then we need only move the disk from one pole to another, taking 21−1 = 1move. Certainly the disk cannot be moved in zero moves, so this number is minimal.

• Inductive hypothesis: Suppose that the minimum number of moves to stack n disks, n ≥ 1,from one pole to another is 2n−1.

Consider the process of moving a stack of n + 1 disks from one pole to another. Because thebiggest disk must move from its place, there must be a point in the process when all the n smallerdisks must be on a separate pole (since one pole must be empty for the biggest disk to move to).This takes 2n−1 moves (minimally) by the inductive hypothesis. Then, we must move the big diskover to the empty pole, taking one move. Finally, we must move the n smaller disks to the polethat now has the largest disk, which takes 2n − 1 moves (minimally) by the inductive hypothesis,(since, ignoring the big disk, we are reduced to the same problem with n disks) .

Therefore, the minimal number of moves is (2n − 1) + 1 + (2n − 1) = 2 · 2n − 1 = 2n+1 − 1.

[Chapter 2, Problem 22] We proceed by induction on n ≥ 1.


• Basis case: If n = 1, then n = 12 is a square of 1.

• Inductive hypothesis: Suppose that for some fixed integer n ≥ 1, that one of n, n + 1, n +2, . . . , 2n is a perfect square.

We aim to show that one of

n+ 1, n+ 2, . . . , 2(n+ 1) = 2n+ 2

is a perfect square.Now, if n is not a square, we know by the inductive hypothesis that one of n+ 1, n+ 2, . . . , 2n

is a square of an integer, and we’re done. Otherwise, n = k2 for some integer k, and so the perfectsquare (k + 1)2 is at most 2n+ 2 = 2k2 + 2 if and only if

k2 + 2k + 1 ≤ 2k2 + 2 ⇐⇒ k2 − 2k + 1 = (k − 1)2 ≥ 0,

which is always true. Therefore, n = k2 < (k + 1)2 ≤ 2n + 2, and (k + 1)2 is a perfect squarebetween n+ 1 and 2n+ 2.

Therefore, by the Principle of Mathematical Induction, the statement holds for all integersn ≥ 1.

[Chapter 2, Problem 23] It is true that “We may assume by induction that in the set L of then Model T’s to the left of all had the same exterior color, and similarly that in the set R of the nModel T’s to the right all had the same exterior color.” However, if n = 1 so that n+ 1 = 2, thenthere is no overlap between the two sets, L and R: they each only have one element! Therefore,the subsequent sentence is not true: we cannot conclude that all n+ 1 Model T’s have the samecolor.

[Chapter 2, Problem 26] We proceed by induction on n ≥ 2.

• Basis case: If n = 2, then n is prime.

• Inductive hypothesis: Suppose that for some fixed integer n ≥ 2, then either n is prime orfactors into a product of primes.

If n + 1 is prime, we’re done. Otherwise, n = ab, where 2 ≤ a, b ≤ n, so that by the inductivehypothesis, a and b factor into products of primes p1p2 · · · pr and q1q2 · · · qs, respectively, Then

n+ 1 = p1p2 · · · prq1q2 · · · qs

also factors into a product of primes.

[Chapter 2, Problem 27] We proceed by induction on n ≥ 3, since a triangle is the polygon withthe fewest sides.

• Basis case: The smallest possible number of sides of a convex polygon is n = 3, a triangle; weknow that the sum of its interior angles is 180 = 180 · (3− 2) degrees.

• Inductive hypothesis: Fix a natural number n ≥ 3 and assume that for any convex k-gon,3 ≤ k ≤ n, the sum of its interior angles is 180 · (k − 2).

Consider a convex (n+1)-gon. If we draw a chord between two vertices of the polygon, skippingone in-between, our polygon is the union of a triangle and an n-gon. Thus, its interior angles sumto

180 + 180 · (n− 2) = 180(n− 1) = 180 ((n+ 1)− 2) .


[Chapter 2, Problem 30] We proceed by induction on n ≥ 1 tp show that an <(53

)n.

• Basis case: a1 = 1 < 53 and a2 = 1 < 5

3 (both are necessary!).

• Inductive hypothesis: Fix a natural number n ≥ 2, and assume that for all k satisfying

1 ≤ k ≤ n, ak <(53

)k.

Then an+1 = an + an−1

<

(5

3

)n+

(5

3

)n−1by our inductive hypothesis

=

(5

3

)n−1·(

5

3+ 1

)=

(5

3

)n−1· 8

3

<

(5

3

)n−1·(

5

3

)2

as

(5

3

)2

=25

9>

8

3

=

(5

3

)n+1

.

[Chapter 2, Problem 31] We claim that the number of compositions of a natural number n ≥ 1is c(n) = 2n−1.

• Basis case: There is only one composition of n = 1 “1” itself.

• Inductive hypothesis: Assume that for some n ≥ 1 that for every k satisfying 1 ≤ k ≤ n,

c(k) = 2k−1.

Now consider the compositions of n+ 1. We will first show that

c(n+ 1) = c(n) + c(n− 1) + . . .+ c(2) + c(1) + 1 :

Take any composition of n+ 1. Then either:

(1) The number of summands in the composition is one, in which case the composition must besimply “(n+ 1),” or

(2) The number of summands is at least two.

In the second case, if we “tear off” the first summand of the composition, we get a compositionof some k, 1 ≤ k ≤ n (since the summand is at least one). In fact, we obtain every composition ofevery k between 1 and n using this method, since “(n+ 1− k)+(the composition of k)” is, in fact,a composition of n+ 1. Moreover, we don’t get any repeats in this process, since if two “torn off”compositions are the same, they must be compositions of the same number k, 1 ≤ k ≤ n, and thenthe number we tore off must have been n + 1− k (so, in fact, the entire initial compositions werethe same).

Thus, case (2) gives us c(1) + c(2) + . . .+ c(n) compositions of n+ 1, and we must add the oneextra from case (1) to get the full list.


Thus, we have that c(n + 1) = c(n) + c(n− 1) + . . . + c(2) + c(1) + 1, which, by the inductivehypothesis equals

2n−1 + 2n−2 + . . .+ 21 + 20 + 1,

which equals 2n since it is the sum of a geometric series.

[Chapter 2, Problem 32] We will show that there is no rational number ba whose square is 2:

By way of contradiction, suppose that a < b are natural numbers and 2 =(ba

)2= b2

a2. Then

b2 = 2a2, and b2 is even, which means that b is even. So b = 2c for some natural number c. Inparticular, we note that c < b. Substituting, we get that b2 = (2c)2 = 4c2, so that 2a2 = 4c2, anda2 = 2c2. This means, in particular, that c < a.

Now, this means that a2 is even, so that a is even, and a = 2d for some natural number d < a.Then we have that 2c2 = a2 = 4d2, so that c2 = 2d2 and d < c. Then c2 is even, and so c is alsoeven. Thus, c = 2e for some natural number e < c.

For clarity, let’s rename a1 = b, a2 = a, a3 = c, a4 = d, a5 = e. We can continue the processdescribed above, obtaining an infinite chain of natural numbers a1 < a2 < a3 < · · · , which cannotexist by the Principle of infinite descent. Thus, we could not have that b2 = 2a2 to begin with, sothere is no such rational number whose square is 2.

Alternatively, collect all of the an into a set S. By the Well-ordering principle, the set S has aleast element, ak. However, we showed above that ak+1 < ak, a contradiction!

[Chapter 2, Problem 33] Suppose that P (n) is a statement that makes sense for all n ≥ n0.Suppose that

(a) P (n0) is true, and

(b) for any n ≥ n0, if P (n) is true, then P (n+ 1) is true.

Let S be the set of n ≥ n0 for which P (n) is false. Suppose that S is not empty. Then by theWell-ordering principle, the set S contains a least element; call it m. We note that m ≥ n0 andP (m) is false, moreover, since P (n0) is true, m > n0 (that is, m − 1 ≥ n0). Since m is the leastnatural number n ≥ n0 for which P (n) is false, P (m− 1) must be true. But m− 1 ≥ n0 and using(b), since P (m− 1) is true, this means that P (m) is true, a contradiction. Thus, S must be empty,and P (n) must be true for every n ≥ n0.

[Chapter 2, Problem 34] Suppose that the Well-ordering Principle holds. By way of contradiction,suppose that there exists an infinite descending chain of natural numbers

a1 > a2 > a3 > · · · .

The the set of all elements of the sequence, S = {ak | k ≥ 1} consists of natural numbers, isnonempty (since, for example, a1 ∈ S), so must have a least element an for some n ≥ 1. Butan+1 < an and an+1 ∈ S, a contradiction. We conclude that there is no infinite descending chainof natural numbers.

Now suppose that there is no infinite descending chain of natural numbers. By way of contra-diction, suppose that the Well-ordering Principle fails, meaning that there is some nonempty setT consisting of natural numbers that has no least element. Since T is nonempty, there is someb1 ∈ T , and since b1 cannot be the least element in T , there is some b2 ∈ T less than b1. Likewise,


b2 cannot be the least element of T , so there is b3 ∈ T less than b2. Following this process, weobtain a chain of natural numbers

b1 > b2 > b3 > · · · ,contradicting the fact that no such sequence exists. We conclude that Well-ordering is valid.

[Chapter 2, Problem 35] Suppose that S is a nonempty set of integers, each strictly less than N .Suppose, by way of contradiction, that S does not have a maximal element, so that for any

fixed x ∈ S, there is another element y ∈ S such that y > x.If N ≤ 1, then every element of S is negative. If we consider a new set P = {−a | a ∈ S}, then

P consists of natural numbers. This set is nonempty since S is not empty. Thus, P has a leastelement, b. It is easy to see that −b is the maximal element of S.

On the other hand, consider the case that N > 1. Using the hint, consider the new set

T = {n ∈ N | n ≥ a for all a ∈ S}.

In this case, N − 1 ∈ T since N − 1 > 0, and by the definitions of S and T . (Alternatively, N ∈ Tas well.) In particular, T is not empty. Therefore, by the Well-ordering principle, T has a leastelement; call it M .

If M = 0, then all elements of S are negative, we can use the first argument.Otherwise, since M is the least element, M − 1 is a natural number that is not in T , meaning

that there must be an element x ∈ S for which x > M − 1. But by our assumption that S has nomaximal element, there is another element y > x (so y ≥ x + 1) in S; in particular, y > M . Thiscontradicts the fact that M ∈ T , and we conclude that S must have a maximal element.

[Chapter 3, Problem 3] Since d | a and d | b, a = dk and b = dj for some integers k, j. Then

r = b− aq = dj − (dk)q = d(j − kq),

and since j − kq is an integer, this means that d | r.

[Chapter 3, Problem 4] Since m is the least common multiple of a and b, and c is a commonmultiple, we have that m ≤ c. Using the Division Algorithm, we write

c = m · q + r,

where 0 ≤ r < m.Now, a | m and a | c, so by Problem #3, a | r. By the same reasoning, we have that r is also a

multiple of b. Thus, r is a common multiple of a and b. However, r < c, and c is the least commonmultiple of a and b. Thus, r must be zero.

Thus, c = m · q, and m | c.

[Chapter 3, Problem 5] Let d be the least element of J , and write d = ax+ by for some integersx and y. Since a = a · 1 + b · 0 and b = a · 0 + b · 1, both a and b are in J , so that d ≤ a and d ≤ b.

Using the Division Algorithm, write a = dq + r where 0 ≤ r < d (these bounds on r will beimportant soon). Solving for r in this expression shows that

r = a− dq= a− (ax+ by)q (subbing in expression for d)

= a(1− xq) + b(−yq)= ax′ + by′ (just renaming coefficients of a and b).


This shows that if r 6= 0, then r ∈ J . However, this is impossible as r < d and d is the minimalelement of J . We conclude that r = 0, so that a = dq.

[Chapter 3, Problem 19] Final answers: (i) 5 (ii) 3 (iii) 1

[Chapter 3, Problem 23] First say that n > 0. Then using the Euclidean Algorithm, we obtainthe sequence of equalities

n+ 1 = n · 1 + 1

n = 1 · n+ 0.

Thus, the last nonzero remainder is 1, which equals (n, n+ 1).The remaining cases are when n ≤ 0. If n = 0, then n + 1 = 1, and (0, 1) = 1. If n < 0,

then n + 1 is also negative. If we take the absolute value of both (i.e., −n and −(n + 1)) and dothe Euclidean Algorithm, we obtain the exact same calculation as the one above. Since greatestcommon divisors do not depend on signs (positive or negative), (n, n+ 1) = (−n,−(n+ 1)) = 1.

In all cases, (n, n+ 1) = 1, and the two are relatively prime.Alternatively, suppose, by way of contradiction, that n and n + 1 are not coprime. Then they

have a common factor d, and n = kd, n+ 1 = jd for some integers k and j. Then 1 = (n+ 1)−n =jd− kd = (j − k)d.

[Chapter 3, Problem 24] Assuming that a, b > 0, if a | b, then since a | a (a = a · 1), we knowthat a is a common divisor of a and b. Moreover, (a, b) ≤ a, so (a, b) must equal a.

[Chapter 3, Problem 25] Suppose that a and b have a common factor d ≥ 1. Then a = kd andb = jd for some integers k and j. Then

1 = ar + bs = (kd)r + (jd)s = d(kr + ds),

so that d | 1, which means that d = 1, and their only common factor is one.

[Chapter 3, Problem 28] Let d = (c, b). Then d | c and d | b, and c = dk for some integer k.Moreover, since c | a, a = c` = (dk)` = d(k`) for some integer `, so d | a. But then d is a commonmultiple of a and b, so 1 = (a, b) ≥ d, which forces d = 1.

[Chapter 3, Problem 29] Consider three consecutive integers, and call them n, n+ 1, and n+ 2.First say that they are non-negative, meaning that n ≥ 0. We notice that a number is divisible by3 if and only if the remainder used in the Division Algorithm using the number and 3 is 0. First,do the Division Algorithm with n and 3:

n = 3q + r, with q ≥ 0 and 0 ≤ r < 3

There are three cases: either r = 0, 1, or 2.First, say that r = 0, so that n is divisible by 3. Then

n+ 1 = 3q + 1, and

n+ 2 = 3q + 2,


where both 1 and 2 satisfy the constraint of the remainder in the Division Algorithm, meaning thatthey are non-negative, and less than 3. Since these remainders are nonzero, n + 1 and n + 2 arenot divisible by 3.

The second case is when r = 1, so that n = 3q + 1 and n is not divisible by 3. Here we obtainthat

n+ 1 = 3q + 2, and

n+ 2 = 3q + 3 = 3(q + 1) + 0,

noticing that we needed to change the quotient in the last implementation of the Division Algorithmto q + 1 in order to obtain a remainder less than 3. Thus, we see that n + 1 is not divisible by 3,but n+ 2 is.

Finally, suppose that r = 2. Then

n+ 1 = 3q + 3 = 3(q + 1) + 0, and

n+ 2 = 3q + 4 = 3(q + 1) + 1,

and n+ 1 is divisible by 3, but n+ 2 is not.Now, if all of n, n+1, and n+2 are non-positive, then we can negate each, and the same process

works (since a number is a divisor of an integer m if and only if it is a divisor of −m). The onlyremaining cases are when n = −1 and −2, which we can do by hand.

[Chapter 3, Problem 33] To check your work, the greatest common divisors are as follows: (i) 3(ii) 17 (iii) 257.

[Chapter 3, Problem 34] (i) 1 (ii) 1 (iii) 5 (iv) 77.

[Chapter 3, Problem 36] . Since e | a and e | b, then a = ek and b = ej for some integers j, k.Then ar + bs = (ek)r + (ej)s = e(kr + bs), and kr + bs is an integer, so e | (ar + bs) regardless ofthe choice of integers r, s.

[Chapter 3, Problem 37] To prove this, it is enough to show that every common divisor of b anda is a common divisor of a and r, and vice versa. This implies that the set of common divisors ofb and a equals that of a and r, so their greatest common divisor must also coincide.

First, take a common divisor b and a, call it c. Then b = ck and a = cj for some k, j ∈ Z. Ifa | r, then c is a common divisor of a and r. Indeed, r = b− aq = ck − (cj)q = c(k − jq), and c | rsince k − jq ∈ Z.

On the other hand, take a common divisor of a and r, call it e. Then a = et and r = es forintegers r and s. If e is a divisor of b, then it is a common divisor of b and a, so we’d be done.Toward this, notice that b = aq+ r = (et)q+ es = e(tq+ s), where tq+ s is an integer, so e | b, andwe are finished.

[Chapter 3, Problem 40] For example, consider a = 4, b = 6, and c = 10, or a = 6, b = 16, andc = 21. We notice that a cannot be prime, but must have two factors, dividing b and c, respectively.

[Chapter 3, Problem 44] Since (a,m) = 1,by Bezout’s theorem, there exist x, y ∈ Z for which


ax+my = 1. Similarly, there exist w, z ∈ Z for which bw +mz = 1 since (b,m) = 1. Then

1 · 1 = (ax+my)(bw +mz)

1 = abxw + axmz + bwmy +m2yz

1 = ab(xw) +m(axz + bwy +myz).

Since X = xw, Y = axz+ bwy+myz ∈ Z, we have integers X,Y for which (ab)X+mY = 1, whichwe know exist if and only if (ab,m) = 1.

[Chapter 3, Problem 45] Suppose that am + bn = e, for some integer e. Let d = (a, b). Thend is a common divisor of a and b, so d | a and d | b. In other words, a = dk and b = dj for someintegers j, k. Then

e = am+ bn = (dk)m+ (dj)n = d(km+ jn),

and km+ jn ∈ Z, so d | e.

[Chapter 3, Problem 46] We know that d | a and d | b, so a = dk and b = dj for some integersj, k. Then

d = ra+ sb = r(dk) + s(dj) = d(rk + sj),

so that 1 = rk + sj, which means that (r, s) = 1.

[Chapter 3, Problem 49] By Bezout’s theorem, there exist integers x, y, x′, y′ for which ax+my = dand (ab)x′ +my′ = 1. Multiplying the right- and left-hand sides of these equations, we see that

a2bxx′ + axmy′ +myabx′ +m2yy′ = d,

or alternatively,ab(axx′) +m(axy′ + abx′y +myy′) = d.

This means that (ab,m) | d, and (ab,m) is at most d.However, d is a common divisor of ab and m: since (a,m) = d, d | a and d | m. Since d | a,

a = dk for some k ∈ Z, so ab = (dk)b = d(kb), so d | ab. Therefore, d | ab and d | m, so that thegreatest common divisor is at least d. Hence (ab,m) = d.

[Chapter 3, Problem 60] To check your work, we give some relevant numbers:

(i) The Euclidean Algorithm gives (242, 1870) = 22, and reverse substitution or the 2× 2 matrixmethod gives 242 · 31 + 1870 · −4 = 22. Then 242 · (31 · 3) + 1870 · (−4 · 3) = 66, so thatX0 = 31 · 3 = 93 and Y0 = −4 · 3 = −12 is one solution; all solutions are of the formX = 93 + 1870

22 k = 93 + 85k and Y = −12− 24222 k = −12− 11k for some integer k.

(ii) The Euclidean Algorithm gives (870, 327) = 3, and reverse substitution or the 2 × 2 matrixmethod gives 327 · −141 + 870 · 53 = 3. Then 327 · (−141 · 22) + 870 · (53 · 22) = 66, so thatX0 = −141 · 22 = −3102 and Y0 = 53 · 22 = 1166 is one solution; all solutions are of the formX = −3102 + 870

3 k = −3102k + 290k and Y = 1166− 3273 k = 1166− 109k.

(iii) The Euclidean Algorithm gives (870, 327) = 3. Since 3 - 56, Proposition 10 gives that thereare no solutions.


[Chapter 3, Problem 62] We first note that we can find that (133, 203) = 7 using the EuclideanAlgorithm. To check your work, we record the answers:

(i) There are no solutions by Proposition 10 since 7 - 38.

(ii) There are no solutions by Proposition 10 since 7 - 40.

(ii) Using reverse substitution or the 2×2 matrix method, we find that 133·−3+203·2 = 7. Then133 · −18 + 203 · 12 = 42, and all solutions are of the form x = −18 + 203

7 k = −18 + 29k andy = 12− 133

7 k = 12− 19k, where k is any integer. We can find that the smallest non-negativex occurs when k = 1, so that x = 11 and y = −7.

(iv) There are no solutions by Proposition 10 since 7 - 44.

[Chapter 3, Problem 65] Now, (13, 16) = 1, and using reverse substitution in the EuclideanAlgorithm, or the 2 × 2 matrix method, we obtain the formula 13 · 5 + 16 · −4 = 1, so that13 · 25 + 16 · −20 = 5. Every solution to the linear diophantine equation 13X + 16Y = 5 is of theform X = 25 + 16k and Y = −20 − 13k, where k is an integer. In particular, if k = −2, we havethe solution X = −7, Y = 6.

Essentially, we need to fill up the 16-quart jug 6 times, bringing it back, and then take backenough water to fill the 13 quart jug 7 times. We can do this in several ways.

[Chapter 4, Problem 1] Write a = pe11 pe22 · · · penn , b = pd11 p

d22 · · · pdnn , and c = p`11 p

`22 · · · p`nn . Here,

p1, . . . , pn are distinct primes, and the exponents ei, di, ì are nonnegative for all 1 ≤ i ≤ n.Then bc = pd1+`11 pd2+`22 · · · pdn+`nn , so that a | bc means that each ei ≤ di + ì, 1 ≤ i ≤ n.

Moreover, (a, b) = 1 means that min{ei, di} = 0 for all i ≤ i ≤ n.Now, it remains to show that a | c, or ei ≤ ì for all 1 ≤ i ≤ n. However, if ei ≤ di, so

that ei0, then certainly ei ≤ ì always. On the other hand, if di ≤ ei, so that di = 0, then eachei ≤ di + ì = 0 + ì = ì as well.

[Chapter 4, Problem 2] We prove the statement by induction on n.Basis case. Suppose that n = 1, so that p | a1; there is nothing to prove.Inductive hypothesis. Assume that for some n ≥ 1, that whenever p | a1a2 · · · an, then p | aj forsome j satisfying 1 ≤ j ≤ n.

Suppose now that p | b1b2 · · · bn+1, where the bi are integers. Then by the Property of primes,since p | b1 · (b2 · · · bn+1), either p | b1 · · · bn, or p | bn+1. In the first case, the inductive hypothesisimplies that b | bj for some j, 1 ≤ j ≤ n. Thus, b | bj for some j, 1 ≤ j ≤ n+ 1.

[Chapter 4, Problem 3] First suppose that (a, b) = 1. By way of contradiction, suppose that p isa prime number and p | a and p | b. Then p is a common divisor of a and b, so that p ≤ (a, b) = 1,a contradiction, since a prime must be at least 2.

On the other hand, suppose that no prime number divides both a and b. Let d = (a, b) andsuppose, by way of contradiction, that d > 1. Then write d = p1 · · · p` into a product of primes,where ` ≥ 1 since d 6= 1. In particular, p1 | d, so p1 is a common divisor of both a and b, acontradiction.

Alternatively, we can prove this by using the characterization of greatest common divisors interms of the exponents appearing in the products of prime of a and b.

[Chapter 4, Problem 4] Write n = p1 · · · p` and q = q1 · · · qt as products of primes pj and qj .


First suppose that (n, qr) = 1. Now, qr = (q1 · · · qt)r = qr1 · · · qrt , and the primes appearing inthe unique prime factorization of qr are still q1, . . . , qt. Then by Ch 3 #4, no prime pi can equalany prime qj . Then (n, q) = 1 as well, by Ch 4 #3.

Similarly, if (n, q) = 1, we have that pi 6= qj for all 1 ≤ i ≤ ` and 1 ≤ j ≤ t. Then looking atthe prime factorization of qr above, we see that (n, qr) = 1 again by Ch 4 #3.

[Chapter 4, Problem 5] If n is not prime, we can write n = ab, where 1 < a, b < n. If botha >

√n and b >

√n, then n = ab > (

√n)2 = n, a contradiction. Thus, n has a divisor a ≥ 2.

Since by the Fundamental Theorem of Arithmetic, a can be written as a product of primes, it hasa prime divisor p. Now, since p | a and a | n, we have that p | n, and n has a prime divisor.

[Chapter 4, Problem 10] Suppose, by way of contradiction, that for some integers a, b > 0 that√p = a

b , so that p = a2

b2, and a2 = pb2.

Write a = pepe11 · · · pe`` and b = pdpd11 · · · p

d`` , where each pi is unique and does not equal p, and

e ≥ 0, d ≥ 0, each ei ≥ 0 and each di ≥ 0. Then as p = a2

b2, and a2 = pb2, we have that

p2ep2e11 · · · p2e`` = p · p2dp2d11 · · · p2d``

= p2d+1p2d11 · · · p2d`` .

In particular, 2e = 2d+ 1; since one side is even and one is odd, this is a contradiction, and√p is

irrational.

[Chapter 4, Problem 13] Suppose, by way of contradiction, that (100)1/5 = ab for some integers

a, b > 0. Then 100 = a5

b5, and a5 = 100b5.

Write a = 2e1 · 5e2 · pe33 · · · pe`` and b = 2d1 · 5d2 · pd33 · · · p

d`` , where each pi is unique and does not

equal 2 or 5, and each ei ≥ 0 and each di ≥ 0. Then as a5 = 100b5 and 100 = 22 · 52, we have that

25e1 · 55e2 · p5e33 · · · p5e`` = 100 · 25d1 · 55d2 · p5d11 · · · p5d``

= 25d1+2 · 55d2+2 · p5d33 · · · p5d`` .

The exponents in these factorizations are equal by the Fundamental Theorem of Arithmetic. Inparticular, 5e1 = 5d1 + 2, so that 2 = 5(e1 − d1), and two is a multiple of five, a contradiction. Weconclude that (100)1/5 is irrational.

[Chapter 4, Problem 17] Since the prime factorizations are 217 ·37 ·56 and 217 ·312 ·53, respectively,their greatest common divisor equals

217 · 37 · 53.

[Chapter 4, Problem 18] . We have that a = p3 · qe11 qe22 · · · qenn and b = p3 · qd11 q

d22 · · · qdnn , where the

primes p, q1, . . . , qn are distinct, and the exponents di, ei are nonnegative for all 1 ≤ i ≤ n. Since(a, b) = p3, whenever some di > 0, then ei = 0.

Then a2 = p9 · q3e11 q3e22 · · · q3enn and b2 = p9 · q3d11 q3d22 · · · q3dnn . Whenever some 3di > 0, we havethat di > 0, forcing 3ei = ei = 0. Therefore, (a2, b2) = p9.

[Chapter 4, Problem 20] Suppose first that (a, b) = 1. Then a and b have no divisor larger than1, so therefore no prime divisor. On the other hand, assume that a and b have no common prime


divisor. If d = (a, b), then we know that p | d for some prime p, and since d | a and d | b, we havethat p | a and p | b, a contradiction.

[Chapter 4, Problem 21] Throughout, we will use the characterization of being relatively primeproved in Ch 4, #3.

Suppose, by way of contradiction, that there exists a prime p such that p | ab and p | c. By theProperty of Primes, p | a or p | b. If p | a, then since p | c as well, (a, b) > 1. Likewise, if p | b, then(a, b) > 1. In either case, we reach a contradiction, so we must have that (ab, c) = 1.

[Chapter 4, Problem 23] Throughout, we will use the characterization of being relatively primeproved in Ch 4, #3.

Say, by way of contradiction, that there exists a prime p such that p | c and p | b. Since c | a,a = ck for an integer k, and p | a as well. This means that (a, b) > 1, a contradiction.

[Chapter 4, Problem 27] We can prove this statement using Bezout’s theorem, of the FTA:Bezout method. Fix integers x and y so that ax + by = c, which exist by Bezout’s Theorem.Then as a | bc, ak = bc for some integer k, and acx+ bcy = c2, so that a(cx+ky) = acx+aky = c2,and a | c2.FTA method. Write

a = pe11 · · · penn , b = pd11 · · · p

dnn , and c = p`11 · · · p

`nn ,

where p1, . . . , pn are unique primes, and all exponents are nonnegative. Since a | bc and

bc = pd1+`11 · · · pdn+`nn ,

we have that ei ≤ di + ì for every 1 ≤ i ≤ n. Moreover, since

(a, b) = pmin{e1,d1}1 · · · pmin{en,dn}

n

and (a, b) | c, we have that min{ei, di} ≤ ì for all 1 ≤ i ≤ n. Since

c2 = (p`11 · · · p`nn )2 = p2`11 · · · p

2`nn ,

in order to verify that a | c2, it is enough to check that each ei ≤ 2ì for 1 ≤ i ≤ n. Indeed,

If ei ≤ di, then ei = min{ei, di} ≤ ì ≤ 2ì, and

If di ≤ ei, then ei ≤ di + ì = min{ei, di}+ ì ≤ ì + ì = 2ì,

so we are finished.

[Chapter 4, Problem 43] Suppose that p1, . . . , pr are all the primes of the form 4n − 1 for anyinteger n. Every integer can be written in the form 4k + r for some k ≥ 0, 0 ≤ r < 4. If a primep = 4k + r, then r cannot equal 0 or 2 unless p = 2; else p is even, and not prime. Thus, everyprime not equal to 2 is of the form 4k+ 1 or 4k+ 3. We notice that 4n− 1 = 4(n− 1) + 3, so thatour primes fall into the second category.

Since the number N = 4p1 · · · pr − 1 is bigger than any of our primes pi, all the primes in itsunique prime factorization are of the form 4ki + 1, and possibly 2 as well. However, 4p1 · · · pr iseven, so that 4p1 · · · pr − 1 is odd, and 2 cannot appear. Thus, we have that

N = 4p1 · · · pr − 1 = (4k1 + 1)(4k2 + 1) · · · (4kt + 1).


By induction, we can show that the right hand side, (4k1 + 1)(4k2 + 1) · · · (4kt + 1), is of the form4M + 1 for some integer M . Thus, 4p1 · · · pr − 1 = 4M + 1, which means that 4p1 · · · pr − 4M = 2,so that 4 | 2, a contradiction.

Thus, some prime in the factorization of N has the form 4m + 1 = 4(m − 13, so is one of thepi. Renumber the primes so that i = 1, and p1 | N . Then p1 | N , and N = 4p1 · · · pr − 1, sop1k = 4p1 · · · pr − 1, and 1 = p1(4p2 · · · pr − k), and p1 | 1, a contradiction.

Therefore, there are infinitely many primes of the form 4n− 1.

[Chapter 5, Problem 4] Since 365 = 7 · 52 + 1, 365 ≡ 1 mod 7. Since 2006 is not a leap year,this means that the day of the week shifted (forward) by one, so that New Years Day landed on aMonday in 2007.

[Chapter 5, Problem 5] To check your work, the answers are: (i) 1912, 1925, 1938, 1951, 1964, 1977, 1990;(ii) (noting that 1776 ≡ 1 mod 25), 1901, 1926, 1951, 1976; (iii) (noting that 1914 ≡ 24 mod 27)1914, 1941, 1968, 1995

[Chapter 5, Problem 6] For example, we could choose a = 45 is solution. (In fact, all solutionsare integer a for which a ≡ 45 mod 56, which we will show later in the course.)

[Chapter 5, Problem 9] We know that that a − b = mj for some integer j, so that ka − kb =k(a− b) = m(kj), and since kj is still an integer, ka ≡ kb mod m.

[Chapter 5, Problem 11] We proceed by induction on e ≥ 1. The basis case, when e = 1, is justthe assumption.Inductive hypothesis. Suppose that for some e ≥ 1 that ae ≡ be mod m. Then using Proposition4 (ii-m),

ae+1 ≡ ae · a ≡ be · b ≡ be+1 mod m.

[Chapter 5, Problem 12] No, for example, consider a = 21. Then

2 6≡ 6 mod 12, but 2 · 21 ≡ 0 ≡ 6 · 21 mod 12.

In general, we can find examples such that c is not relatively prime to 12; that is, 2 or 3 divides c.

[Chapter 5, Problem 13] The integer a cannot be a multiple of five.The reasoning is as follows: 15a ≡ ca mod 25 if and only if a(15 − c) is a multiple of 25. By

writing out a, 15 − c, and 25 into products of primes, we see that the power of 5 appearing in aplus the power of 5 appearing in 15 − c must be at least 2. However, if a is a multiple of 5, thenthe power of 5 in 15− c need not be 2 in order to ensure that 15− c is a multiple of 5.

[Chapter 5, Problem 14] We have that 4 ≡ 4 mod 9, 42 ≡ 16 ≡ 7 mod 9, and 43 ≡ 4 ·7 ≡ 1 mod 9.Now, one way to proceed for the second part is the following: let e ≥ 1 and use the Division

Algorithm to write e = 3q + r, where 0 ≤ r < 3. Then

4e ≡ (43)q · 4r ≡ 1q · 4r ≡ 4r mod 9,


so that

4e ≡

1 if e ≡ 0 mod 3

4 if e ≡ 1 mod 3

7 if e ≡ 2 mod 3

mod 9. (1)

Indeed, it is not hard to show this; for example, if e ≡ 1 mod 3, then e = 3k + 1 for some integerk, and

4e ≡ 43k+1 ≡ (43)k · 4 ≡ 1k ≡ 4 mod 9.

We can then proceed by induction on n ≥ 1.Basis case. For n = 1, we see that 6 · 4 ≡ 24 ≡ 6 mod 9 since 24− 6 = 18 = 2 · 9.Inductive hypothesis. Suppose that for some n ≥ 1, 6 · 4n ≡ 6 mod 24.

Then

6 · 4n+1 ≡ (6 · 4n) · 4≡ 6 · 4 by the Inductive hypothesis

≡ 24 ≡ 6 mod 9,

and we conclude that the statement holds for all n ≥ 1.Alternatively, we could check for all cases (n ≡ 0, 1, or 2 mod 3) that this statement holds using

(1).

[Chapter 5, Problem 15] 72 ≡ 49 ≡ 15 mod 17 since 49− 15 = 34 = 17 · 2, so that 74 = (72)2 ≡152 ≡ (−2)2 ≡ 4 mod 17, and 78 ≡ 16 mod 17, so that 716 ≡ 162 ≡ (−1)2 ≡ 1 mod 17.

Thus, as 546 = 16 · 34 + 2, 7546 = (716)34 · 72 ≡ 1 · 15 mod 17.

[Chapter 5, Problem 16] To check your work, here are the final answers: (i) 4, (ii) 6, (iii) 4, (iv)7

[Chapter 5, Problem 17]

16 ≡ 1 mod 7.

26 ≡ (23)2 ≡ 82 ≡ 12 ≡ 1 mod 7.

36 ≡ (32)3 ≡ 93 ≡ 23 ≡ 8 ≡ 1 mod 7.

46 ≡ (42)3 ≡ 163 ≡ 23 ≡ 8 ≡ 1 mod 7.

55 ≡ (−2)6 ≡ 26 ≡ 1 mod 7.

66 ≡ (−1)6 ≡ 1 mod 7.


(i) Since (a + b)2 − (a2 + b2) = (a2 + 2ab + b2) − (a2 + b2) = 2ab, a multiple of 2, the statementholds.

(ii) We proceed by induction on n.


Basis case. For n = 1, a21 ≡ a21 mod m by the Reflexive property of congruences.Inductive hypothesis. Suppose that for some n ≥ 1, (a1 + · · ·+ an)2 ≡ a21 + · · ·+ a2n mod m.

Then modulo 2,

(a1 + · · ·+ an+1)2 ≡ ((a1 + · · ·+ an) + an+1)

2

≡ (a1 + · · ·+ an)2 + 2 · (a1 + · · ·+ an) · an+1 + a2n+1

≡ a21 + · · ·+ a2n + a2n+1

since 2 · (a1 + · · · + an) · an+1 ≡ 0 mod 2 since the left hand side is a multiple of 2, and by theinductive hypothesis.

[Chapter 5, Problem 20] First notice that an integer a is even if and only if 2 | a, which is trueif and only if a ≡ 0 mod 2. Likewise, a is odd if and only if a = 2k+ 1 for some integer k, which isequivalent to the condition that 2 | (a− 1), or a ≡ 1 mod 2.

If a is even, then a2 is even, so a2 ≡ 0 ≡ a mod 2. If a is odd, then a2 is odd, so a2 ≡ 1 ≡ a mod 2.

[Chapter 5, Problem 22] First, we can check by hand that a6 ≡ 1 mod 7 for all a = 0, 1, . . . , 6.Then since 67 = 6 · 11 + 1, we have that

a67 ≡ a6·11+1 ≡ (a6)11 · a ≡ 111 · a ≡ a mod 7.

Alternatively, we can check each congruence for a = 0, 1, . . . , 6 by hand by doing some arithmetic.

[Chapter 5, Problem 44] To illustrate the method, we give the steps to solve (i): Using theEuclidean Algorithm, we find that 12 · −12 + 29 · 5 = 1, so that 12 · −12 ≡ 1 mod 29; that is,12 · 17 mod 1 mod 29. Multiplying the equation 12x ≡ 5 mod 29 through by 17, we see that

x ≡ (17 · 12)x ≡ 17 · (12x) ≡ 17 · 5 ≡ 85 ≡ 27 mod 29.

(i) All integers x of the form x = 27+29k for some integer k; in other words, all integers x satisfyingx ≡ 27 mod 29.

(ii) No solution exists since (12, 38) 6= 1.

(iii) All integers x for which x ≡ 20 mod 47.

(iv) No solution exists since (12, 56) 6= 1.

(v) All integers x for which x ≡ 60 mod 65.

[Chapter 5, Problem 45] Here are the final answers:

(i) All integers of the form x = 7k for an integer k; in other words, all integers x for whichx ≡ 0 mod 7.

(ii) All integers of the form x = 9 + 11k for an integer k, or all integers x for which x ≡ 9 mod 11.

(iii) All integers of the form x = 15+23k for an integer k, or all integers x for which x ≡ 15 mod 23.

(iv) No solutions exist since (12, 24) = 12, and 12 - 42.


(v) All integers of the form x = 16+25k for an integer k, or all integers x for which x ≡ 16 mod 25.

[Chapter 5, Problem 48] Using the Euclidean Algorithm, we find that 313 · −71 + 463 · 48 = 1,so that 313 · −71 ≡ 1 mod 463, and x ≡ −71 ≡ 392 mod 463 is the solution.

[Chapter 5, Problem 49] Using the Euclidean Algorithm, we find that 7 · −31 + 218 · 1 = 1, sothat 7 · −31 ≡ 1 mod 218, and x ≡ −31 ≡ 187 mod 218 is the solution.

[Chapter 5, Problem 50] Using the previous problem, multiplying the equation 7x ≡ 13 mod 218through by 187, we see that

x ≡ (187 · 7)x ≡ 187(7x) ≡ 187 · 13 ≡ 2431 ≡ 33 mod 218,

and x ≡ 33 mod 218 is the solution.

[Chapter 6, Problem 5] The tables are as follows:

+ [0]4 [1]4 [2]4 [3]4[0]4 [0]4 [1]4 [2]4 [3]4[1]4 [1]4 [2]4 [3]4 [0]4[2]4 [2]4 [3]4 [0]4 [1]4[3]4 [3]4 [0]4 [1]4 [2]4

· [0]4 [1]4 [2]4 [3]4[0]4 [0]4 [0]4 [0]4 [0]4[1]4 [0]4 [1]4 [2]4 [3]4[2]4 [0]4 [2]4 [0]4 [2]4[3]4 [0]4 [3]4 [2]4 [1]4

[Chapter 6, Problem 6] The tables are as follows:

+ [0]5 [1]5 [2]5 [3]5 [4]5[0]5 [0]5 [1]5 [2]5 [3]5 [4]5[1]5 [1]5 [2]5 [3]5 [4]5 [0]5[2]5 [2]5 [3]5 [4]5 [0]5 [1]5[3]5 [3]5 [4]5 [0]5 [1]5 [2]5[4]5 [4]5 [0]5 [1]5 [2]5 [3]5

· [0]5 [1]5 [2]5 [3]5 [4]5[0]5 [0]5 [0]5 [0]5 [0]5 [0]5[1]5 [0]5 [1]5 [2]5 [3]5 [4]5[2]5 [0]5 [2]5 [4]5 [1]5 [3]5[3]5 [0]5 [3]5 [1]5 [4]5 [2]5[4]5 [0]5 [4]5 [3]5 [2]5 [1]5

[Chapter 6, Problem 7] We give the addition table; the multiplication table can be found usingthe method of Ch 6 #6.

+ [0]7 [1]7 [2]7 [3]7 [4]7 [5]7 [6]7[0]7 [0]7 [1]7 [2]7 [3]7 [4]7 [5]7 [6]7[1]7 [1]7 [2]7 [3]7 [4]7 [5]7 [6]7 [0]7[2]7 [2]7 [3]7 [4]7 [5]7 [6]7 [0]7 [1]7[3]7 [3]7 [4]7 [5]7 [6]7 [0]7 [1]7 [2]7[4]7 [4]7 [5]7 [6]7 [0]7 [1]7 [2]7 [3]7[5]7 [5]7 [6]7 [0]7 [1]7 [2]7 [3]7 [4]7[6]7 [6]7 [0]7 [1]7 [2]7 [3]7 [4]7 [5]7


· [0]7 [1]7 [2]7 [3]7 [4]7 [5]7 [6]7[0]7 [0]7 [0]7 [0]7 [0]7 [0]7 [0]7 [0]7[1]7 [0]7 [1]7 [2]7 [3]7 [4]7 [5]7 [6]7[2]7 [0]7 [1]7 [2]7 [3]7 [4]7 [5]7 [6]7[3]7 [0]7 [1]7 [2]7 [3]7 [4]7 [5]7 [6]7[4]7 [0]7 [1]7 [2]7 [3]7 [4]7 [5]7 [6]7[5]7 [0]7 [1]7 [2]7 [3]7 [4]7 [5]7 [6]7[6]7 [0]7 [1]7 [2]7 [3]7 [4]7 [5]7 [6]7

[Chapter 6, Problem 8] We will show (i):

[6]10[x]10 = [4]10 if and only if [6x]10 = [4]10

if and only if 6x ≡ 4 mod 10.

As (6, 10) = 2 > 1, 6 has no multiplicative inverse modulo 10, so we can’t use the method we’velearned to finish this.

We can turn 6x ≡ 4 mod 10 into a linear Diophantine equation involving integers and solve, orwe can simply make a table of all classes to find all the solutions:

[x]10 [0]10 [1]10 [2]10 [3]10 [4]10 [5]10 [6]10 [7]10 [8]10 [9]10[6]10 · [x]10 = [6x]10 [0]10 [6]10 [2]10 [8]10 [4]10 [0]10 [6]10 [2]10 [8]10 [4]10

(i) [x]10 = [4]10 or [x]10 = [9]10, meaning x = 10k + 4 or x = 10k + 9 for some integer k.

(ii) [x]7 = [2]7.

(iii) [x]7 = [4]7.

(iv) [x]17 = [9]17.

(v) [x]19 = [10]19.

[Chapter 6, Problem 9] By making a table of all values, [x]11 = 4 or [x]11 = 7 is a solution; inother words, [x]11 is a solution precisely when x is an integer of the form 11k+ 4 or 11k+ 7, k ∈ Z.

[Chapter 6, Problem 10] By making a table of all values, we find that x for which [x]13 = 4 or[x]13 = 9 are solutions.

[Chapter 6, Problem 18] For example, {2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016} (wesometimes did not use the set notation).

[Chapter 6, Problem 22] We see that, along with 0, we obtain all least positive residues0, 1, 2, · · · , 12. Below, all congruences are modulo 13:

21 = 2 25 ≡ 3 · 2 ≡ 6 29 ≡ 9 · 2 ≡ 522 = 4 26 ≡ 6 · 2 ≡ 12 210 ≡ 5 · 2 ≡ 1023 = 8 27 ≡ 12 · 2 ≡ 11 211 ≡ 10 · 2 ≡ 724 ≡ 8 · 2 ≡ 3 28 ≡ 3 · 3 ≡ 9 212 ≡ 7 · 2 ≡ 1


[Chapter 6, Problem 23] To demonstrate the method, we will show that 3 is a primitive root,and that 4 is not. In the first table, we see that, after we add 0, all of 0, 1, 2, 3, 4, 5, 6 are obtained.In the second, we see that this is not the case. All congruences below are modulo 7:

31 = 3 34 ≡ 2 · 2 ≡ 4 41 = 4 44 ≡ 2 · 2 ≡ 432 = 9 ≡ 2 35 = 4 · 3 ≡ 5 42 ≡ 2 45 ≡ 4 · 4 ≡ 233 ≡ 2 · 3 ≡ 6 36 ≡ 6 · 6 ≡ 1 43 ≡ 4 · 2 ≡ 1 46 ≡ 1 · 1 ≡ 1

[Chapter 6, Problem 25] To check your work, note that the primitive roots modulo 17 that areless than 17 are 3, 5, 6, 7, 10, 11, 12, and 14. (To get all primitive roots, we can add multiplies of 17to each of these.)

[Chapter 6, Problem 33] First, we note that since b is a primitive root modulo p, one ofb, b2, . . . , bp−1 must be congruent to 1 modulo p. Thus, the minimum e for which be ≡ 1 mod p isless than or equal to p− 1.

Suppose, by way of contradiction, that be ≡ 1 mod p and e < p− 1. Then

be+1 ≡ be · b ≡ 1 · b ≡ b mod p.

But then 0, b, b2, . . . , be ≡ 1, b is part of a complete set of representatives. This can’t happen, sinceb ≡ b mod m; that is, they represent the same class! We know that each must represent a differentclass in Z/mZ, and this is a contradiction. We conclude that the smallest such e is p− 1.

[Chapter 6, Problem 34] Take any m consecutive integers, n, n + 1, · · · , n + (m − 1). Suppose,by way of contradictions, that two are congruent modulo m: For some 0 ≤ k < j ≤ m − 1 andn+k ≡ n+j mod m. Then, adding −n to both sides, we see that k ≡ j mod m. This can’t happensince k and j are not equal and are less than m, a contradiction.

Alternatively, we could find the remainder r in the division algorithm when dividing by n, andthen carefully identify what must happen to the remainders of n+ 1, n+ 2, . . . as well.


(i) By inspection, or by using E. A. method, [4]−1 = [10], [5]−1 = [8], and [7]−1 = [2].

(ii) For (a), If X ∈ Z/13Z such that [4]X = [7], then if we multiply this element, that is [7], on theleft by [4]−1, we find [1]X = [4]−1[7] = [10][7] = [70] = [5] by using associativity of multiplicationin Z/13Z. The other parts can be solved in the same way: (b) [5], (c) [3].


(a) Suppose that for some 1 ≤ i < j ≤ m that bai ≡ baj mod m. Then [bai] = [baj ] in Z/mZ(subscripts m omitted), i.e., [b][ai] = [b][aj ]. Since (b,m) = 1, [b] has a multiplicative inverse inZ/mZ, so that

[b]−1([b][ai]) = [b]−1([b][aj ])

([b]−1[b])[ai] = ([b]−1[b])[aj ]

[1][ai] = [1][aj ]

[ai] = [aj ],

meaning that ai ≡ aj mod m, which forces i = j since a1, a2, . . . , am is a CSR modulo m.


(b) Suppose that (b,m) = d > 1. Then since d | m and d | b, m = dk and b = dj for some integersk and j satisfying 1 < k < m and 1 < j < b. Since a1, . . . , am is a CSR modulo m, there existsdistinct integers i and j for which 1 ≤ i, j ≤ m and

ai ≡ k mod m and aj ≡ 0 mod m.

Then we certainly have that baj ≡ b · 0 ≡ 0 mod m. On the other hand, we have that

bai ≡ bk ≡ (dj)k ≡ j(dk) ≡ jm ≡ 0 mod m,

so that ba1, . . . , bam cannot be a CSR modulo m.

[Chapter 6, Problem 43] To check your work, the final answers are: (i) [17], [16], and [12],respectively; (ii) (a) X = [12], (b) X = [19], (c) X = [11]

[Chapter 6, Problem 44] We know that [a] ∈ Z/12Z is invertible if and only if (a, 12) = 1(and a 6= 0). By unique factorization in Z, since 12 = 22 · 3, the integers relatively prime to 12are exactly those that do not have 2 or 3 as divisors; thus, the invertible elements of Z/12Z are{[1], [5], [7], [11]}. By inspection, or by using the E. A. method, [1]−1 = [1], [5]−1 = [5], [7]−1 = [7],and [11]−1 = [11].

[Chapter 6, Problem 47] Since [a]m = [a′]m, we have that a ≡ a′ mod m, and a = a′ + mk forsome integer k. Then since (a,m) = 1, by Bezout’s theorem, there exist integers x, y for which

ax+my = 1.

Then1 = (a′ +mk)x+my = a′x+mkx+my = a′x+m(kx+ y),

and this identity means that (a′,m) = 1 as well.

[Chapter 6, Problem 48] First suppose that [a]mn is a unit in Z/mnZ. This happens if and onlyif (a,mn) = 1. Write out the prime factorizations m = pe11 · · · p

e`` , n = pd11 · · · p

d`` where the pi are

distinct primes, and each eidj ≥ 0. Since (a,mn) = 1, we can write a = qt11 · · · qtkk , where qj are

primes not equal to any pi. Thus, a and m have no prime factors in common, and (a,m) = 1, and,similarly, a and n have no prime factors in common, so that (a, n) = 1.

The other direction is similar.


(i) By inspection, it is clear X = [4] is a solution.

(ii) (12, 20) = 4 (by uniquely factoring each of 12 and 20), so we know by the Proposition from class(by Proposition 8 in the book) that the solutions are exactly

[204 k]

= [5k] with k = 0, 1, 2, 3, or[0], [5], [10], and [15].

(iii) By the Proposition from class (Proposition 7 in the book), we can add the solutions from (i)and (ii) to get [4], [9], [14], and [19].



(i) Note that [14] = [2] and [18] = [6], so [3] is clearly a solution.

(ii) (14, 12) = 2 (by uniquely factoring each of 14 and 12), so by the same method as Ch 6 #56,the solutions are exactly [0], and [6].

(iii) By the same method as Ch 6 #56, the solutions are exactly [3], and [9].

[Chapter 6, Problem 58] We are asked to show that [36]X = [6] has no solutions in Z/45Z. Bymeans of contradiction, suppose such a solution exists. Since we are working in Z/45Z, we havethat X is of the form X = [x] for some [x] ∈ Z/45Z. Then, the equation [36]X = [6] becomes[36x] = [6]. However, we know this occurs if and only if 36x ≡ 6 mod 45, which itself occurs if andonly if 36x = 6 + 45k for some k ∈ Z. Rearranging, we see that 36x − 45k = 6. However, this isimpossible. To see why, note that 9 divides both 36 and 45 (in fact, 9 = (36, 45)). Thus, dividingthe equation 36x − 45k = 6 by 9 shows that 4x − 5k = 2

3 , which is impossible since the left handside must be an integer.


(i) If [a] is a unit, it has a multiplicative inverse [a]−1. Then multiplying both sides of [a]X = b by[a]−1, we see that

X = ([a] · [a]−1)X = [a]−1 · ([a]X) = [a]−1 · [b]

is a solution.

(ii) If [a]X = [0] then multiplying both sides by [a]−1,

X = ([a]−1[a])X = [a]−1([a]X) = [a]−1 · [0] = [0].

(iii) Suppose that X and Y are both solutions; that is, [a]X = [b] and [a]Y = [b]. Then [a]X = [a]Y ,and

X = ([a] · [a]−1)X = [a]−1 · ([a]X) = [a]−1 · ([a]Y ) = ([a] · [a]−1)Y = Y,

and X = Y .

[Chapter 6, Problem 60] To check your work, the final answers are: (i) X = [6], (ii) (i) X = [8],(iii) (i) X = [4]

[Chapter 6, Problem 61] To check your work, the final answers are: (i) X = [3], [8], [11], (ii) (i)X = [1], [6], [10], (iii) (i) No solutions

[Chapter 7, Problem 1] We know that (−1) + 1 = 0 by the property of additive inverses, so that

0 = 0 · (−1) = ((−1) + 1) · (−1) = (−1) · (−1) + 1 · (−1) = (−1) · (−1) + (−1)

by Ch 7, # 7, the Distributive Law, and property of multiplicative identity. Then (−1) · (−1) = 1since the equation above defines the additive inverse of −1, which is 1.


[Chapter 7, Problem 2] Call b = −(−a). Then b = −(−a) satisfies the property b+ (−a) = 0 =(−a) + b. We aim to show that b = a. We have that

(b+ (−a)) + a = 0 + a

b+ ((−a) + a) = a by associativity and property of additive inverses

b+ 0 = a by property of additive inverses

b = a. by property of additive identity

[Chapter 7, Problem 3] Say a+ b = d and a+ c = d. Then a+ b = a+ c, and adding −a to bothsides:

(−a) + (a+ b) = (−a) + (a+ c)

(−a+ a) + b = (−a+ a) + c by associativity

0 + b = 0 + c by property of additive inverses

b = c. by property of additive identity

[Chapter 7, Problem 4] We have that a · b = a · c. Then

b · (a · b) = b · (a · c)(b · a) · b = (b · a) · c by associativity

1 · b = 1 · c by property of multiplicative inverses

b = c by property of multiplicative identity

[Chapter 7, Problem 5] Say ax = d and ay = d for x, y ∈ R. Then ax = ay, and

a−1 · (a · x) = a−1 · (a · y)

(a−1 · a) · x = (a−1 · a) · y by associativity

1 · x = 1 · y by property of multiplicative inverses

x = y, by property of multiplicative identity

and there is a unique solution.

[Chapter 7, Problem 6] We will show that (−a) · b = −(ab); the other equality is similar. In orderto show this, we want to show that (−a) · b satisfies the property of the additive identity of ab; thatis, we must have that (−a) · b+ a · b = 0 (and adding in the other order, ab+ (−a) · b = 0, but weare in an abelian group, so this will follow).

We know that (−a) + a = 0, so multiplying both sides by b, we obtain that

((−a) + a) · b = 0 · b(−a) · b+ a · b = 0 by Distributive Law and argument of Ch 7,#7.


[Chapter 7, Problem 7] We know that 1 + 0 = 1 since 0 is the additive identity, so that

b · (1 + 0) = b · 1b · 1 + b · 0 = b · 1 by Distributive Law

b+ b · 0 = b by property of multiplicative identity

−b+ (b+ b · 0) = −b+ b

(−b+ b) + b · 0 = −b+ b by associativity

0 + b · 0 = 0 by property of additive inverses

b · 0 = 0 by property of additive identity.

[Chapter 7, Problem 8] Suppose that a ∗ b = a ∗ c. If d is the inverse of a, then

d ∗ (a ∗ b) = d ∗ (a ∗ c), and

(d ∗ a) ∗ b = (d ∗ a) ∗ c by associativity, so that

e ∗ b = e ∗ c by the property of inverses, and

b = c by the property of the identity e.

[Chapter 7, Problem 9] We claim that if d is the inverse of a, then x = d ∗ b is a solution. Indeed,

a ∗ (d ∗ b) = (a ∗ d) ∗ b by associativity

= e ∗ b since a and d are inverses

= b since e is the identity.

[Chapter 7, Problem 16] The inverse is[m+12

]m

since if m is odd, m+12 ∈ Z, and

[2]m ·[m+ 1

2

]m

=

[2 · m+ 1

2

]m

= [m+ 1]m = [1]m

by the definition of multiplication in Z/mZ.

[Chapter 7, Problem 21] In each case, the number of zero divisors is m − ϕ(m) − 1, where ϕdenotes the Euler phi function. To check your work, the final answers are as follows: m = 6: 3,m = 7: 0 (since 7 is prime), m = 8: 3, m = 9: 2, m = 10: 5, m = 11: 0 (since 11 is prime), m = 12:7, m = 13: 0 (since 13 is prime), m = 14: 7

[Chapter 7, Problem 22] Say, by way of contradiction, that R has no zero divisors, but that s ∈ Sis a zero divisor. Then there is s′ ∈ S satisfying s · s′ = 0, where s, s′ are nonzero elements of S.Then s, s′ are also nonzero elements of R, and s is a zero divisor of R, a contradiction.

[Chapter 7, Problem 27] Let a ∈ Z/mZ such that a 6= 0 and a has an inverse in Z/mZ. Supposeb ∈ Z/mZ such that a · b = 0. Then

0 = a−1 · 0 = a−1 · (a · b) = (a−1 · a) · b = 1 · b = b

So since b = 0 for all b such that a · b = 0, a is not a zero diviser. So if a ∈ Z/mZ is a zero diviser,a cannot have an inverse.


[Chapter 7, Problem 28] To check your work, we list all the zero divisors, and their complementaryzero divisors:

Zero divisor Complementary zero divisors

[3] [5], [10][5] [3], [6], [9], [12][6] [5], [10][9] [5], [10][10] [3], [6], [9], [12][12] [5], [10]

[Chapter 7, Problem 32] To check your work, here are the inverses: [53]−1 = [62], [73] is notinvertible since 73 | 365, [93]−1 = [157], and [113]−1 = [42].

[Chapter 7, Problem 33] Let z ∈ R satisfy z2 = 1. Then, by distributivity,

(z + 1)(z − 1) = (z + 1) · z + (z + 1) · −1 = (z2 + z) + (−z − 1) = z2 − 1 = 0

As R has no zero divisers, (z + 1) = 0 or (z − 1) = 0. So z ∈ {1,−1}. Note that the assumptionthat R be commutative is not needed.

[Chapter 7, Problem 34] The order is six since [3]2 = [9] = [2] 6= [1], [3]3 = [3]2 · [3] = [2] · [3] =[6] 6= [1], [3]4 = ([3]2)2 = [2]2 = [4] 6= [1], [3]5 = [3]4 · [3] = [4] · [3] = [12] = [5] 6= [1], and[3]6 = ([3]3)2 = [6]2 = [36] = [1].


(i) The elements are [0] = [0] + [0]i, [1] = [1] + [0]i, [2] = [2] + [0]i, [1]i = [0] + [1]i, [1] + [1]i, [2] +[1]i, [2]i = [0] + [2]i, [1] + [2]i, [2] + [2]i.

(ii) To check your work, we list the multiplicative inverses here:

Element [1] [2] [1]i [1] + [1]i [2] + [1]i [2]i [1] + [2]i [2] + [2]i

Multiplicative inverse [1] [2] [2]i [2] + [1]i [1] + [1]i [1]i [2] + [2]i [1] + [2]i

(iii) The order is 8. We can confirm this by calculating (1 + i)n for 1 ≤ n ≤ 8 .

[Chapter 7, Problem 37] Since Q[i] is a subset of the set of all complex numbers C, we can checkthat it is a subring (a subset that is also a ring), if it is closed under addition, multiplication, takingnegatives, and contains the additive and multiplicative identities.

First we check that Q[i] is closed under addition, multiplication, and taking negatives: Ifa, b, c, d ∈ Q, then

(a+ bi) + (c+ di) = (a+ c) + (b+ d)i, and

(a+ bi) · (c+ di) = (ac− bd) + (ad+ bc)i, and

− (a+ bi) = −a+ (−b)i since (a+ bi) + (−a+ (−b)i) = 0,

and a+ c, b+ d, ac− bd, ad+ bc,−a,−b ∈ Q.The additive and multiplicative inverses of C are the typical “0” and “1,” which are also in Q[i]

since 0 = 0 + 0i and 1 = 1 + 0i. Thus, Q[i] is a subring of C, so it is also a ring.


Now, Q[i] is commutative, since C is. Thus, as 0 6= 1 (so that Q[i] has at least two elements),in order to show that Q[i] is a field, it suffices to show that every nonzero element a + bi has amultiplicative inverse in Q[i]. We know that its inverse in C is 1

a+bi , so it suffices to show that1

a+bi ∈ Q[i] (that is, 1a+bi = c+ di for some c, d ∈ Q). In fact,

1

a+ bi=a− bia− bi

· 1

a+ bi=

a− bia2 + b2

=a

a2 + b2+

(−b

a2 + b2

)i.

[Chapter 9, Problem 2] To check your work, the orders of [1], [2], [4], [5], [7], [8] are, respectively,1, 6, 3, 6, 4, 2.

[Chapter 9, Problem 5] To check your work, the orders of [1], [2], · · · , [12] are, respectively,1, 12, 3, 6, 4, 12, 12, 4, 3, 6, 12, 2.

[Chapter 9, Problem 13] Since 3 has order 162, we have that, by Proposition 3, the order of 326

is 162/(26, 162). Using the Euclidean Algorithm, we can find that (26, 162) = 2, so that the orderis 162/2 = 81. Likewise, since (162, 27) = 27, the order of 327 is 162/27 = 6.

[Chapter 9, Problem 14] For each, we should use the Euclidean Algorithm to find the appropriategreatest common divisor. For each, we find the order of [2] in Z/mZ by checking powers.. Thenwe apply Proposition 3. (i) The order of [2] in Z/11Z is 10, so the order of [224] in Z/11Z is10/(10, 24) = 10/2 = 5. (ii) The order of [2] in Z/17Z is 8, so the order of [224] in Z/11Zis 8/(8, 24) = 8/8 = 1. (iii) The order of [2] in Z/31Z is 5, so the order of [224] in Z/11Z is5/(5, 24) = 5/1 = 5.

[Chapter 9, Problem 15] We know that

282 ≡ (241)2 ≡ 822 ≡ (−1)2 ≡ 1 mod 83.

This means that the order of 2 modulo 83 is a divisor of 82. Since 82 = 2 · 41, 22 ≡ 4 6≡ 1 mod 83,and 241 ≡ 82 6≡ 1 mod 83, the only possibility for the order is 82 itself.


(i) Since 32 ≡ 9 mod 14, 33 ≡ 27 ≡ 13 mod 14, 34 ≡ 92 ≡ 81 ≡ 11 mod 14, 35 ≡ 11 · 3 ≡ 33 ≡5 mod 14, and 36 ≡ (32)3 ≡ (−1)3 ≡ 1 mod 14, the order of 6.

(ii) We have that 59 ≡ 3 mod 14, so by part (i), 593 ≡ 1 mod 14. Then since 110 = 6 · 18 + 2, wehave that

59110 ≡ (596)18 · 592 ≡ 118 · 32 ≡ 9 mod 14.

[Chapter 9, Problem 17] First, we calculate that φ(11) = 10 since 11 is prime. Since [5]2 = [3],we have that [5]10 = ([5]2)5 = [3]5 = [243] = 1 since 244 = 11 · 22 + 1.

[Chapter 9, Problem 21] We can check that 167 is prime by checking that no primes less than√167 ≈ 12.92 are not factors. By FLT, we have that the order of 7 modulo 167 is a divisor of

166 = 2 · 83. The only positive factors of 166 are then 1, 2, 83, and itself. Since 7 6≡ 1 mod 167, and72 = 49 6≡ 1 mod 167, the order of 7 is at least 83, so at least 80.


[Chapter 9, Problem 23] First, we can check that 97 is prime by checking that it has no primefactors less than

√97 ≈ 9.84. Then by FLT, even element that is not a multiple of 97 has an order

dividing 96 = 25 · 3. Since 5 is not a divisor of 96, there is no element of order 5.

[Chapter 9, Problem 25] By FLT, we have that since (2, 23) = 1, then 222 ≡ 1 mod 23. Then247 ≡ (222)2 · 23 ≡ 12 · 8 ≡ 8 mod 23.

[Chapter 9, Problem 27] By FLT version 2, for every integer n, we have that n5 ≡ n mod 5 andn3 ≡ n mod 3. Thus, n5 = n+ 5k and n3 = n+ 5j for some integers k and j. Thus,

n5/5 + n3/3 + 7n/15 = (n+ 5k)/5 + (n+ 5j)/3 + 7n/15

= n/5 + k + n/3 + j + 7n/15

= k + j + 3n/15 + 5n/15 + 7n/15

= k + j + 15n/15 = k + j + n,

an integer.

[Chapter 9, Problem 31] We know that by FLT, n10 ≡ 1 mod 11 if 11 mod n. Thus, for theseintegers,

n111 ≡ (n10)11 · n1 ≡ 111 · n ≡ n mod 11.

On the other hand, if 11 - n, then n ≡ 0 mod 11, so n111 ≡ 0111 ≡ 0 ≡ n mod 11.Alternatively, the alternate characterization of FLT states that n11 ≡ n mod 11 for all integers

n. Thus,n111 = (n11)10 · n ≡ n10 · n ≡ n11 ≡ n mod 11.

[Chapter 9, Problem 34] We are asked to find the least non-negative residue of 3255 mod 29. ByFermat’s Little Theorem, 328 ≡ 1 mod 29, and since 225 = 8 · 28 + 1, we have that

3225 ≡ 328·8+1 ≡ (328)8 · 3 ≡ 3 mod 29.

[Chapter 9, Problem 40] First, we calculate that Φ(10) = 4 since the positive integers less that10 that are relatively prime to 10 are 1, 3, 7, and 9. Moreover, [1]4 = [1], [3]4 = ([3]2)2 = [9]2 =[−1]2 = [1], [7]4 = ([7]2)2 = [49]2 = [−1]2 = [1], and [9]4 = [−1]4 = [1].

[Chapter 9, Problem 43] We can find that Φ(25) = 25−5 = 20 since the integers between 1 and 25that have a factor in common with 25 are the multiples of five: 5, 10, 15, 20, 25. Thus, we have thata20 ≡ 1 mod 25 for all integers a relatively prime to 25. Note that since 48 ≡ −2 mod 25, we canreplace 48 by −2 if we wish, to make calculations easier. Since (−2, 25) = 1, (−2)20 ≡ 1 mod 25.

(−2)322 ≡ ((−2)25)12 · (−2)22 ≡ 112 · (−2)22 ≡ (−2)22 mod 25.

Since (−2)2 = 4 mod 25, so (−2)4 = 42 ≡ 16 mod 25, (−2)8 ≡ 162 = 256 ≡ 6 mod 25, and(−2)16 ≡ 62 = 36 ≡ 11 mod 25. Thus, (−2)22 = (−2)16 ·(−2)4 ·(−2)2 ≡ 11 ·16 ·4 = 704 ≡ 4 mod 25.

[Chapter 9, Problem 45] Note that Φ(33) = Φ(11 ·3) = Φ(11) ·Φ(3) = 10 ·2 = 20. Then by Euler’stheorem, if (n, 33) = 1, we have that n10 ≡ 1 mod 33. This means that n100 = (n10)10 ≡ 110 ≡1 mod 33, and multiplying through by n, we have that n101 ≡ n mod 33; that is, 33 | (n101 − n).


On the other hand, for each 1 ≤ n ≤ 33 for which (n, 33) 6= 1, then we can check thatn101 ≡ n mod 33 by hand.

[Chapter 9, Problem 49] By Euler’s theorem, we have that since (5, 26) = 1 and ϕ(26) =ϕ(2)ϕ(13) = 1 · 12 = 12, 512 ≡ 1 mod 26. Then 5 · 511 ≡ 1 mod 26, so 511 is an inverse of 5 modulo26.

[Chapter 9, Problem 51] We see that (a) holds since all of 1, 2, . . . , p− 1 are relatively prime top since they are less than p. For (b), consider the integers 1, 2, . . . , pn. The numbers that are notrelatively prime to p are multiples of p since it is prime; these are p, 2p, 3p, . . . , pn−1 · p. There arepn−1 of these. Thus, Φ(pn) = pn − pn−1.

[Chapter 10, Problem 5] Notice that m = 3337 = 47 · 71, so that ϕ(m) = 46 · 70 = 3220. Wecheck that e and d are consistent encryption and decryption exponents by verifying that

ed ≡ 11 · 1171 ≡ 12881 ≡ 1 mod ϕ(m) = 3220.

To encode “NO,” we first translate the message into 14 : 15. Next, we find the least nonnegativeresidue of each piece to the exponent e = 11 modulo m = 3337:

1411 ≡ 2570 mod 3337, 1511 ≡ 2637 mod 3337.

The encoded message is 2570 : 2637.Then we can decrypt this message by finding the least nonnegative residue of each term to the

exponent d = 1171 modulo m = 3337:

25701171 ≡ 14 mod 3337, 26371171 ≡ 15 mod 3337,

and we recover the original message, “14 : 15,” or “NO.”

[Chapter 10, Problem 6] First, we find (using a computer) that m = 1399 · 2503, where bothfactors are prime. We check that e and d are inverses modulo ϕ(m) = 1398 · 2502 = 3497796:

ed ≡ 17 · 1440269 ≡ 24484573 ≡ 1 mod 3497796.

Then we encrypt the message by finding least nonnegative residues modulo m = 3501697:

555217 ≡ 947771 mod 3501697

30717 ≡ 889719 mod 3501697

456217 ≡ 3245212 mod 3501697

158717 ≡ 2608977 mod 3501697

and the encoded message is:

947771 : 889719 : 3245212 : 2608977

Now try decoding yourself!

[Chapter 12, Problem 2] There is no solution: the first equation says that x must be even, andthe second says that x must be odd.


[Chapter 12, Problem 3] To check your work, the answer is 325. (All solutions are congruent to325 modulo 420.)

[Chapter 12, Problem 6] To check your answers, all solutions are precisely all integers x for whichx ≡ 47 mod 120.

[Chapter 12, Problem 8] To check your answers, all solutions are precisely all integers x for whichx ≡ 103 mod 504.

[Chapter 12, Problem 10] Suppose that m and nq are coprime. Then the system of equations

x ≡ 0 mod n

x ≡ 0 mod q

has the solution x = 0 by inspection. Moreover, we can check that any x ∈ Z satisfying thecongruence x ≡ 0 mod nq is also a solution to the above system, since this says that x is a multipleof nq, and so a multiple of n and of q. Therefore, any x ∈ ZZ satisfying the following system is asolution.

x ≡ 1 mod m

x ≡ 0 mod nq

Moreover, the system has a solution by the CRT since (m,nq) = 1.Now suppose that there is a solution x0 ∈ Z to the system given. Then x0 = 1 + mk and

x0 = nqj for some integers j, k. Then 1 + mk = nqj, so that 1 = m(−k) + nq(j), which forces(m,nq) = 1.

[Chapter 12, Problem 11] All solutions are integer x for which x ≡ 5600 mod 8800, and thesolution closest to zero is x0 = −3200.

[Chapter 12, Problem 12] We are trying to solve the system of equations

x ≡ 1 mod 3

x ≡ 6 mod 10

x ≡ 3 mod 7.

To check your answer, all solutions are precisely integers x for which x ≡ 136 mod 210, so there are136 unwounded survivors, the only positive solution not exceeding 208.

[Chapter 12, Problem 16] The first two congruences are equivalent to the third by the CRT. Ifx ≡ 5 mod 12, we have that x = 5 + 12k for some integer k, so x = 5 + 6(2k), and x ≡ 5 mod 6.So, all solutions are values of x for which x ≡ 5 mod 12; the smallest positive solution is x = 5.

[Chapter 12, Problem 26] By multiplying the first equation by a multiplicative inverse of 7 modulo17 (5), and the second by an inverse of 13 modulo 22 (17), we can translate this problem into thesystem

x ≡ 5 · 2 ≡ 10 mod 17

x ≡ 17 · 21 ≡ 5 mod 22.


Since (17, 22) = 1, we can apply our method to find a solution via CRT. To check your answer, allsolutions are integers x of the form x = 27 + 374k, k ∈ Z.

[Chapter 12, Problem 34] To check your work, the answers are recorded:

(i) e1 ≡ 400 mod 525, e2 ≡ 126 mod 525

(ii) x = 192

(iii) x = 17

[Chapter 12, Problem 35] To check your work, the answers are recorded:

(i) e1 ≡ 1365, e2 ≡ 495, e3 ≡ 286 mod 11 · 13 · 15

(ii) x = 278

(iii) All x ≡ 548 mod 2145

[Chapter 13, Problem 1] For example, f(x) = xp + [1] and g(x) = x+ [1].

[Chapter 13, Problem 2] Since φ(6) = 2, by Euler’s Theorem, [x]2 = [1] for all units [x] ∈ Z/6Z,we could choose, for example, q(x) = x.

[Chapter 13, Problem 3] Let R be a commutative ring which has no zero divisers. Let f, g ∈R[x] \ {0}. Then we can write f = anx

n + an−1xn−1 . . .+ a0 where n ∈ N, ai ∈ R when 0 ≤ i ≤ n

and an 6= 0 and g = bmxm + bm−1x

m−1 . . .+ b0 where m ∈ N, bi ∈ R when 0 ≤ i ≤ m and bm 6= 0.So the coefficient of xm+n in f · g is an · bm 6= 0. So f · g is not the zero polynomial. So R[x] hasno zero divisors.

[Chapter 13, Problem 4] There are lots of different ways to approach this problem. Here isone: Let R be a commutative ring, and suppose that a and b are non-zero elements such thatab = 0. Then, the polynomials f(x) = ax and g(x) = bx each have degree 1, but the productf(x)g(x) = abx2 = 0 does not have degree deg f(x) + deg g(x). Another example is the following:If a and b are as above and now f(x) = ax and h(x) = bx + 1, then f(x) and g(x) again havedegree 1 but the product f(x)g(x) = (ax)(bx + 1) = abx2 + ax = ax has degree 1 (which is againnot equal to deg f(x) + deg g(x).

[Chapter 13, Problem 5] Since units cannot be zero divisors, the proposition holds. Say first thatp(x)q(x) = 1. Then by the proposition,

deg p+ deg q = deg(pq) = deg 1 = 0.

Thus, deg p = 0 (and deg q = 0). On the other hand, if deg p = 0, then p = a0, a constant in F , soalso a unit since F is a field. Thus, we can take q(x) = a−10 .

[Chapter 13, Problem 6] For part (i), we need to show that 1 + 2x is a unit in Z/4Z[x]. In fact,it is its own inverse, as

(1 + 2x)(1 + 2x) = 1 + 2x+ 2x+ 4x2 = 1 + 4x+ 4x2 = 1.


Next, start with (iii), and show by induction that the coefficients are multiples of [2] (that is,either [0] or [2]. Then for (ii), show that if p(x) is a unit with constant term a0, then p(x) − a0must be a zero divisor.

For part (iv), we need to find an element of Z/4Z[x] that is neither a unit nor a zero-divisor.It is easy to check that x is such an element: Note that x cannot have a unit, because no matterwhat f(x) is, xf(x) has no constant term, and so could never equal 1. From this, we see that xcan have no inverse.

To see why x is a non-zero divisor, let 0 6= f(x) = a0 + a1x+ · · · anxn with an 6= 0. Then it isclear that the highest term of xf(x) is anx

n+1, and so is non-zero (as an 6= 0). This shows that xcan never multiply a non-zero element into zero.

[Chapter 14, Problem 1] Here, we write the quotient and remainder to check your answers:

(i) x2 + 2x− 3,−7

(ii) x2 − 3x+ 8,−27x+ 7

(iii) 2x2 − 3x, 1

(iv) 12x

2 + 12x+ 1

2 , 0

(v) 0, 3x2 − x− 1

[Chapter 14, Problem 2] No, we cannot always cancel the coefficients of g of xk when k < deg g.For example, try g(x) = 4x2 + x+ 1 and f(x) = 2x− 1.

[Chapter 14, Problem 3] For example, in Z/8Z[x], [6]x = ([2]x + [1]) · [3] + [5] and [6]x =([2]x+ [1]) · [7] + [1].

[Chapter 14, Problem 4] Suppose that r, s ∈ R are complementary zero divisors. Then, forexample, if f = rx and g = sx, deg f = deg g = 1, but

deg(fg) = deg(rsx2) = deg(0x2) = deg 0 = −∞ < 2 = deg f + deg g.

[Chapter 14, Problem 7] As f(x) = (x2 − 3)(x+ 1)q(x) + (x2 + 2x+ 5), then

f(x) = (x2 − 3) ((x+ 1)q(x) + 1)− (x2 − 3) + (x2 + 2x+ 5) = (x2 − 3) ((x+ 1)q(x) + 1) + (2x+ 8)

and, as deg(2x+ 8) < deg(x2 − 3), the remainder is 2x+ 8.

[Chapter 14, Problem 8] We can use the Root Theorem since Q is a field, so that we needto find k such that f(k) = 0, if f(x) = x3 − kx2 − 2x + (k + 3). Calculating, we get thatf(k) = k3 − k3 − 2k + (k + 3) = 3− k, so k = 3 is the only solution.

[Chapter 14, Problem 10] By plugging in values, we obtain (i) [0], [2], [5], and [12]; and (ii)[0], [2], [12], and [20].


Aternatively, using the Chinese Remainder Theorem, we see that a ∈ Z satisfies thecongruence a2 − 2a ≡ 0 mod 15 if and only if it satisfies both equations

a2 − 2a ≡ 0 mod 3 and

a2 − 2a ≡ 0 mod 5.

Since a2 − 2a = a(a − 2) and Z/3Z and Z/5Z are fields so that these polynomials have at mosttwo roots, a = 0 and a = 2 are solutions. Thus, we have four systems whose solutions solve thecongruence desired:

a ≡ 0 mod 3 a ≡ 0 mod 3 a ≡ 2 mod 3 a ≡ 2 mod 3

a ≡ 0 mod 5 a ≡ 2 mod 5 a ≡ 0 mod 5 a ≡ 2 mod 5.

The solutions are a ≡ 0, 12, 5 and 2 mod 15, respectively, so [0], [12], [5], and [2] are all the roots.For the second part, we can proceed both ways: We can turn the congruence a2 − 2a mod 30

into the two congruences:

a2 − 2a ≡ 0 mod 2 and

a2 − 2a ≡ 0 mod 15,

and proceed as above, using the four solutions to the second equation. On the other hand, we coulddirectly turn the equation into the system of three equations

a2 − 2a ≡ 0 mod 2,

a2 − 2a ≡ 0 mod 3, and

a2 − 2a ≡ 0 mod 5.

Each will have two solutions (0 and 2). However, modulo 2, both solutions are equal, so we willobtain 1 · 2 · 2 = 4 total solutions.

[Chapter 14, Problem 15] To check your answer, the greatest common divisor obtained using theEuclidean Algorithm is [2]x+ [2].

[Chapter 14, Problem 22] To check your answer, we obtain r(s) = x+ [1], s(x) = [1].

[Chapter 14, Problem 25] To check your work, we give the final answer:

(i) [1] is a greatest common divisor.

(ii) The product of the two polynomials is a least common multiple.

[Chapter 14, Problem 26] To check your work, d(x) = 2116 , r(x) = 1

4x+ 516 , and s(x) = −1

4x+ 1116

or any multiple by an element of Q is a solution; for example, d(x) = 1, r(s) = 421x + 5

21 , ands(x) = − 4

21x+ 1121 is also a solution.

[Chapter 14, Problem 28] To check your work, we record the steps of the Euclidean Algorithm:

x4 + x2 + r2 = (x2 − x+ r)(x2 + x+ (2− r)) + ((2− 2r)x+ 2r(r − 1))

x2 − x+ r = ((2− 2r)x+ 2r(r − 1))

(1

2− 2rx+

r − 1

2− 2r

)+ r2

(2− 2r)x+ 2r(r − 1) = r2 ·(

2− 2r

r2x+

2r − 2

r

)+ 0.


Since the last nonzero remainder, r2, is a unit since r 6= 0, the two are relatively prime.

[Chapter 14, Problem 37] If f and g are associates, then f = cg for a nonzero constant polynomialc, but in F2, the only nonzero element is [1], so f = [1]g = g.

[Chapter 14, Problem 41] In this case (compare to Ch 14, #42), both polynomials are writtenas products of irreducible polynomials in Q[x]. We can check this by using the quadratic formulaon the quadratics–they will only have (purely) complex roots; we also know all linear polynomialsare irreducible. Thus, a GCD can be found by finding the lowest power of each irreducible in thepair, and taking the product:

(x2 + 3x+ 6)(x+ 1)3.

[Chapter 14, Problem 42] The only tricky part here is that factorization presented are notirreducible factorizations, since the polynomial x2 − 3x − 4 can be factored as (x − 4)(x + 1). Inlight of this, the two polynomials become

(x− 4)3(x+ 1)3(x− 3)2 and (x− 4)5(x+ 1)2.

Given this factorization, it follows that a GCD is (x− 4)3(x+ 1)2.

[Chapter 14, Problem 43] By Fermat’s Little Theorem, we know that for all integers a, a5 ≡a mod 5. Thus, if p(x) = x5 − x, p(a) = 0 for a = [0], [1], [2], [3], and [4]. Thus, x− a divides p(x)for each of these a. As p(x) is monic and each of the distinct factors x− [a] is relatively prime, weobtain that p(x) = (x− [0])(x− [1])(x− [2])(x− [3])(x− [4]) = x(x+ [4])(x+ [3])(x+ [2])(x+ [1]),where each x− [a] is irreducible since it is linear (which we showed in class).

[Chapter 14, Problem 44] We are asked to find an irreducible (i.e., prime) factorization of thepolynomial f(x) = xp − x ∈ Fp[x]. To do this, we recall the Root Theorem, which states that(x−α) is an irreducible factor of f(x) if and only if f(α) = 0. Combining this with Fermat’s LittleTheorem, which states that f(α) = αp − α = 0 for every α ∈ Fp, we see that every irreduciblepolynomial of the form fα(x) = x − α for some α is an irreducible factor of f(x); moreover, eachpair is relatively prime. Since there are p distinct polynomials fα and f has degree p, it followsthat f is the product of the fα. That is, we have that

f = x(x− [1])(x− [2]) · · · (x− [p]).

Compare this with Chapter 14, Problem 43.

[Chapter 14, Problem 46] We notice by the Degree Proposition that if p(x) is a polynomial thatis not a unit and has degree at most three, if p is not irreducible, then p must have a linear divisor:If p = fg, then deg(p) = deg(f) + deg(g). Since 1 ≤ deg(p) ≤ 3, one of deg(f) and deg(g) must beone, which is the same things as saying that f or g has a root by the Root Theorem.

Using this, we proceed as follows: We showed in class that all linear polynomials (over a field)are irreducible, so that x, x+ [1] are the irreducible polynomials of degree one.

Then to obtain those of degree two, we must find those with no roots: All polynomials of degreetwo are x2, x2 + [1], x2 +x, and x2 +x+ [1]. The first and third have root [0], so that x is a divisor,and the second has root [1], so that x− [1] is a divisor. Since the fourth has no root, it is irreducible.


Note that in general, if a polynomial has no constant term, [0] is a root, so we can skip over it,so the constant term of an irreducible polynomial must be [1]. Moreover, if the polynomial has aneven number of terms, then [1] is a root, so we can focus on those with an odd number of terms.

We can apply the Root Theorem similarly to polynomials of degree three, to obtain the followingirreducible polynomials: x3 + x+ [1] and x3 + x2 + [1].

Finally, we cannot do the same things for polynomials of degree four. If a polynomial of degreefour, p(x), is reducible, then p = fg, where either one of f and g is degree one (so that we canuse the root test), or both are irreducible of degree two. However, since x2 + x + 1 is the onlyirreducible of degree two, the only one in this setting is

(x2 + x+ [1])2 = x4 + x2 + [1].

Ruling these out, we obtain the following irreducible polynomials of degree four:

x4 + x3 + x2 + x+ [1], x4 + x3 + [1], x4 + x+ [1].

[Chapter 15, Problem 8] Recall that if α ∈ C is a non-zero complex number, then its inverseis given by the formula α−1 = 1

α·α · α (it is easy to check that αα−1 = 1). To solve the equationαx = β for x, all we have to do is multiply both sides by α−1 to see that

x = βα−1.

For example, in (ii), we have that α−1 = 113 · (3− 2i), and so

x = (1− i) · α−1 =1

13− 5

13i.


(i) Since 1 is a root, x−1 is a factor. Using long division of polynomials, x3−1 = (x−1)(x2+x+1).The quadratic formula gives roots x = −1+3i

2 , x = −1−3i2 of the equation x2 + x + 1. Thus,

these roots, with x = 1, are all the roots.

(ii) We see that

x8 − 1 = (x4 − 1)(x4 + 1) = (x2 − 1)(x2 + 1)(x2 − i)(x2 + i)

so the roots are ±1, ±√−1 = ±i, ±

√i = ±

√22 +

√22 i, and ±

√−i =

√22 ±

√22 i.

(iii) We see that

x12 − 1 = (x6 − 1)(x6 + 1) = (x3 − 1)(x3 + 1)(x3 − i)(x3 + i)

and by (i) and (iv), the roots coming from the first two cubics are x = 1, x = −1+3i2 , x = −1−3i

2

and x = −1, x = 1+3i2 , x = 1−3i

2 . We can see that x = 3√i = −i is one root of x3 − i, and

long division yields x3 − i = (x + i)(x2 − ix − 1); the quadratic formula then gives roots

x = i2 ±

√32 . On the other hand, x3 + i = 0 has root x = 3

√−i = i, and long division yields

x3 + i = (x− i)(x2 + ix− 1); the quadratic formula then gives roots x = − i2 ±

√32 . Hence we

have found all twelve roots.


(iv) We see that x = −1 is a root, and using long division, we find that x3+1 = (x+1)(x2−x+1).The quadratic equation yields roots x = 1+3i

2 , x = 1−3i2 of x2 − x + 1, so these three are all

the roots.

[Chapter 15, Problem 18] Since 3 + 2i is a root, we know that its complex conjugate 3 − 2i isalso a root, and even more, that

(x− (3 + 2i))(x− (3− 2i)) = x2 − ()x+ (3 + 2i)(3− 2i) = x2 − 6x+ 13

is a factor of f . Using long division of polynomials, we obtain quotient x2 + 2x+ 2, is irreducibleover R since the quadratic formula gives non-real roots. Therefore,

f(x) = (x2 − 6x+ 13)(x2 + 2x+ 2)

and these two quadratic polynomials are irreducible in R[x].

[Chapter 15, Problem 21] By the quadratic formula, the roots of f(x) are 2 − 3i and 1 + i.Therefore, if we include their conjugates as roots, we get a real polynomial:

(x−(2−3i))(x−(2+3i))(x−(1+i))(x−(1−i)) = (x2−4x+13)(x2−2x+2) = x4−6x3+23x2−34x+26.

[Chapter 17, Problem 2] We know the units modulo m(x) are those relatively prime to m(x),and the zero divisors are those polynomials that are not divisible by m(x) such that their greatestcommon divisor with m(x) is at least one (see Ch 17, Proposition 6). Since m(x) = x(x + 1), wehave that the zero divisors are x, x + 1, 2x, and 2x + 2 (some choices of GCDs with m(x) are x,x+ 1, x, and x+ 1 respectively), and the units are 1, 2, x+ 2, and 2x+ 1.

[Chapter 17, Problem 3] The residues of least degree are the last polynomials on each line; allcongruences are modulo m(x) = x4 − 2 (noting that x4 ≡ 2 mod m):

(i) x5 ≡ x · x4 ≡ x(2) ≡ 2x.

(ii) x4 + 2x2 + 1 ≡ 2 + 2x2 + 1 ≡ 2x2 + 5.

(iii) We start calculating:

x4 ≡ 2

x5 ≡ 2x

x6 ≡ x · x5 ≡ 2x2

x7 ≡ x · x6 ≡ 2x3

x8 ≡ (x4)2 ≡ 4

x9 ≡ x · x8 ≡ 4x

x9 ≡ x · x8 ≡ 4x

x10 ≡ x · x9 ≡ 4x2

x11 ≡ x · x10 ≡ 4x3

x12 ≡ x · x11 ≡ 4x4 ≡ 4 · 2 ≡ 8 ≡ 3


x13 ≡ x · x12 ≡ 3x

x14 ≡ x · x13 ≡ x2

x15 ≡ x · x14 ≡ 3x3

x16 ≡ x · x15 ≡ x · 3x3 ≡ 3x4 ≡ 3 · 2 ≡ 1.

Thus, if n ≥ 16 and we write n = 16q + r, 0 ≤ r ≤ 15, using the Division Algorithm, we seethat xn ≡ (x16)q · xr ≡ xr, we can use the table above to find the residue of least degree. (Notethat the least nonnegative residues of 1, x, x2, and x3 are themselves!)

[Chapter 17, Problem 5] Here, we omit the [−] notation for brevity and clarity:

· 0 1 x x+ 1 x2 x2 + 1 x2 + x x2 + x+ 1

0 0 0 0 0 0 0 0 01 0 1 x x+ 1 x2 x2 + 1 x2 + x x2 + x+ 1x 0 x x2 x2 + x 1 x+ 1 x2 + 1 x2 + x+ 1x+ 1 0 x+ 1 x2 + x x2 + 1 x2 + 1 x2 + x x+ 1 0x2 0 x2 1 x2 + 1 x x2 + x x+ 1 x2 + x+ 1x2 + 1 0 x2 + 1 x+ 1 x2 + x x2 + x x+ 1 x2 + 1 0x2 + x 0 x2 + x x2 + 1 x3 + 1 x+ 1 x2 + 1 x2 + x 0x2 + x+ 1 0 x2 + x+ 1 x2 + x+ 1 0 x2 + x+ 1 0 0 x2 + x+ 1

[Chapter 17, Problem 7] One complete set of representatives is {0, 1, 2, x, x + 1, x + 2, 2x, 2x +1, 2x+2}. Using Ch 17, Proposition 6, we can find that since x2−1 = (x−1)(x+1) = (x+2)(x+1),by unique factorization, the zero divisors are x+ 1, x+ 2, 2x+ 2,, and 2x+ 1, since they are exactlythose that have a factor in common with x2 − 1. Thus, the units are 1, 2, x, and 2x.

[Chapter 17, Problem 8] Recall that the hint tells us that we can define the order of a unitf(x) ∈ F [x] (if F is a field) as the smallest positive integer k such that xk ≡ 1 mod f(x). Moreover,if N denotes the number of residues of least degree modulo f (so any CSR modulo f has Nelements), if can be shown that xN−1 ≡ 1 mod f . Moreover, it is a fact that the order of x equalsN − 1 if and only if {0, 1, x, x2, . . . , xN−2} is a CSR modulo f .

Since the residues of least degree modulo each of the polynomials given are exactly all poly-nomials of degree at most one (so there are 52 = 25 of them), for this problem, it is enough tocheck whether for each polynomial g(x) given x24 ≡ 1 mod g(x). For (i), we can find that x12 ≡1 mod g(x) so that the order cannot be 24; for (ii), the order is 24 (and since x24 ≡ 1 mod g(x), itcan be shown that the order must divide 24, but no divisor gives 1); and for (iii), x1 ≡ x mod g(x),so it cannot be a CSR. (Note that there are several ways to make conclusions in each part.)

[Chapter 17, Problem 9] We can calculate that x4 ≡ x3+x2+x+1 mod p(x) := x4+x3+x2+x+1,so that

x5 ≡ x ·x4 ≡ x(x3+x2+x+1) ≡ x4+x3+x2+x ≡ (x3+x2+x+1)+(x4+x2+x+1) ≡ 1 mod p(x).

Thus, if n ≥ 5, them using the Division Algorithm to write n = 5q + r, r ≤ 4, shows thatxn ≡ (x5)q · xr ≡ xr mod x4 + x3 + x2 + x + 1. We do not, therefore, obtain a representative forthe polynomial x.


[Chapter 17, Problem 10] We notice that modulo x4 + x + 1, x4 ≡ −x − 1 ≡ x + 1; using this,we obtain the residues of least degree (the last polynomials listed in each part):

(i) x5 ≡ x · x4 ≡ x(x+ 1) ≡ x2 + x.

(ii) x9 ≡ (x4)2 · x ≡ (x+ 1)2 · x ≡ (x2 + 1) · x ≡ x3 + x.

(iii) x13 ≡ x9 · x4 ≡ (x3 + x)(x+ 1) ≡ x4 + x3 + x2 + x ≡ (x+ 1) + x3 + x2 + x ≡ x3 + x2 + 1.

[Chapter 17, Problem 11] To check your work, we list the inverses here:

(i) x3 + x2

(ii) x2 + x

(iii) x2 + x+ 1

[Chapter 17, Problem 13] We must have exactly one polynomial with each residue of least degree;each is of the form ad−1x

d−1 + ad−2xd−2 + . . .+ a1x+ a0, where ai ∈ Fp. Since Fp has p elements,

we have pd such polynomials.


(i) Here, we can find that x3 + x + 1 and x4 + x + 1 are relatively prime using the EuclideanAlgorithm, and back substitution obtains the polynomial f(x) = x2 + 1 (and any polynomialcongruent to this modulo x4 + x+ 1 is a solution.

(ii) Similarly, we can find that 2x+1 and x2+1 are relatively prime using the Euclidean Algorithm;back substitution gives all polynomials f(x) ∈ Q[x] satisfying f(x) ≡ x + 2 mod x2 + 1 assolutions.

(ii) First, to save time, we notice that x9 ≡ (x2)4 · x ≡ (−2)4 · x ≡ 16x mod x2 + 2; we’re reducedto solving 16xf(x) ≡ 1 mod x2 + 1. The Euclidean Algorithm gives that 16x and x2 + 2 arecoprime, and using that x2 + 2 = (16x) ·

(116x)

+ 2 (so that 12(x2 + 2) = (16x)

(132x)

+ 1), weobtain that the inverse of 16x modulo x2 + 2 is − 1

32x. Thus, we can multiply this on both sidesof the congruence to obtain that

f(x) ≡ − 1

32x mod x2 + 2,

and all polynomials f(x) ∈ Q[x] satisfying the congruence are solutions.

[Chapter 17, Problem 19] To check your work, we list the answers (and remember that to useour methods, we need to check that the moduli are relatively prime!):

(i) All solutions are polynomials f(x) ∈ F2[x] such that f(x) ≡ x3 mod (x2 + x)(x2 + x+ 1).

(ii) Polynomials f(x) ∈ F2[x] such that f(x) ≡ x8 + x7 + x5 mod (x4 + x2 + x)(x4 + x3 + 1).

(iii) Polynomials f(x) ∈ F2[x] such that f(x) ≡ x3 + x2 + 1 mod x(x2 + 1).


[Chapter 17, Problem 20] To check your work, the solution is: q(x) = 12x+ 5

2 . Note that any twopoints determine a line, so we could just solve for m and b in the equation for the line q(x) = mx+b,with the points (1, 3) and (−1, 2) on it.

[Chapter 17, Problem 24] To check your work, the answer is f(x) = − 524x

2 + 43x−

18 .

[Chapter 17, Problem 25] To check your work, the answer is f(x) = 32x

2 + 12x.

Additional Problems.

A. We show that −1 ·a = −a. You can show that −a = a ·−1 using a similar argument. Note thatto show that −1 · a is the negative of a, it is enough to show that

(−1 · a) + a = 0.

Indeed,

−1 · a+ a = −1 · a+ 1 · a since 1 is the multiplicative identity

= ((−1) + 1) · a by the Distributive Property

= 0 · a since −1 is the additive inverse of 1

= 0 by Chapter 7, #7.

B.

1. Since the matrix has four entries, and there are three options for each entry, there are 34 = 81elements.

2. We have that

A+B =

([1]3 [2]3[2]3 [0]3

)+

([0]3 [0]3[2]3 [2]3

)=

([1]3 [2]3[4]3 [2]3

)=

([1]3 [2]3[1]3 [2]3

)A ·B =

([1]3 [2]3[2]3 [0]3

)·(

[0]3 [0]3[2]3 [2]3

)=

([1]3 · [0]3 + [2]3 · [2]3 [1]3 · [0]3 + [2]3 · [2]3[0]3 · [2]3 + [0]3 · [2]3 [0]3 · [2]3 + [0]3 · [2]3

)=

([4]3 [4]3[0]3 [0]3

)=

([1]3 [1]3[0]3 [0]3

).

3. Yes, M2(Z/mZ) has the additive identity

([0]3 [0]3[0]3 [0]3

)and the multiplicative identity

([1]3 [0]3[0]3 [1]3

).

It also has additive inverses. Moreover,

−A =

(−[1]3 −[2]3−[2]3 −[0]3

)=

([2]3 [1]3[1]3 [0]3

).

4. For example, we can verify by calculating products that the matrices([1]3 [0]3[0]3 [1]3

),

([2]3 [0]3[0]3 [2]3

),

([1]3 [0]3[0]3 [2]3

),

([2]3 [0]3[0]3 [1]3

),

([0]3 [1]3[1]3 [0]3

),

([0]3 [2]3[2]3 [0]3

)


are each self-inverses. The matrices([0]3 [1]3[2]3 [0]3

)and

([0]3 [2]3[1]3 [0]3

)are multiplicative inverses of one another, as are(

[1]3 [1]3[1]3 [0]3

)and

([0]3 [1]3[1]3 [2]3

)and (

[0]3 [1]3[1]3 [1]3

)and

([1]3 [2]3[2]3 [0]3

).

C. The matrix C =

([0] [−1][1] [0]

)is a unit since we can check that it has inverse

([0] [1]

[−1] [0]

). By

iteratively multiplying the matrix to itself, we find that

C2 =

([−1] [0][0] [−1]

), C3 =

([0] [1]

[−1] [0]

), C4 =

([0] [1][1] [0]

),

and we checked that C4 is it own inverse, so C8 = C4 ·C4 is the multiplicative identity. Therefore,the order is s divisor of 8, but cannot be 1, 2, or 4 by the calculation above, so so the order of C is8.

D.

1. We can check that

f(2− i) = (2− i)3 + (2− i)2 − 15(2− i) + 25

= (2− 11i) + (3− 4i)− 30 + 15i+ 25 = 0.

Thus, by the lemma from class, its conjugate, 2 + i, is also a root. Since x − (2 + i) andx− (2− i) are relatively prime and (x− (2 + i)) (x− (2− i)) = x2 + 4x+ 5, we can use theDivision Algorithm for polynomials to obtain quotient x + 5. Therefore, factorizations intoirreducible polynomials in C[x] and R[x], respectively, are:

f(x) = (x− (2 + i)) (x− (2− i)) (x+ 5) in C[x], and

f(x) = (x2 + 4x+ 5)(x+ 5) in R[x].

2. We can check that

g(1 + 3i) = (28− 96i)− 2(−26− 18i) + 11(−8 + 6i)− 2(1 + 3i) + 10

= 28− 96i+ 52 + 36i− 88 + 66i− 2− 6i+ 10

= 0.

Thus, its complex conjugate, 1 − 3i, is also a root, and since the factors x − (1 + 3i) andx− (1− 3i) are relatively prime, their product, (x− (1 + 3i)) (x− (1− 3i)) = x2 − 2x+ 10,divides g(x). Using the Division Algorithm for polynomials, we the quotient when g(x) isdivided by this polynomial is x2 + 1, which is irreducible over R (since it has no roots), andfactors as (x− i)(x+ i) over C. Therefore, factorizations into irreducible polynomials in C[x]and R[x], respectively, are:

f(x) = (x− (1 + 3i)) (x− (1− 3i)) (x− i)(x+ i) in C[x], and

f(x) = (x2 − 2x+ 10)(x2 + 1) in R[x].

homework solutions introductory modern algebra, spring...

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