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chapter_8/exercises.md
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@ -4009,3 +4009,270 @@ Omitted.
g. What _are_ you? (Do not answer this on paper; just think about it.) g. What _are_ you? (Do not answer this on paper; just think about it.)
Omitted. Omitted.
---
Page 567
**Exercise Set 8.4**
1.
a. Use the Caesar cipher to encrypt the message WHERE SHALL WE MEET.
b. Use the Caesar cipher to decrypt the message LQ WKH FDIHWHULD.
2.
a. Use the Caesar cipher to encrypt the message AN APPLE A DAY.
b. Use the Caesar cipher to decrypt the message NHHSV WKH GRFWRU DZDB.
3. Let $a = 25, $b = 19$, and $n = 3$.
a. Verify that $3 | (25 - 19)$.
b. Explain why $25 \equiv 19 (\mod 3)$.
c. What value of $k$ has the property that $25 = 19 + 3k$?
d. What is the (nonnegative) remainder obtained when $25$ is divided by $3$?
When $19$ is divided by $3$?
e. Explain why $25 \mod 3 = 19 \mod 3$.
4. Let $a = 68$, $b = 33$, and $n = 7$.
a. Verify that $7 | (68 - 33)$.
b. Explain why $68 \equiv 33(\mod 7)$.
c. What value of $k$ has the property that $68 = 33 + 7k$?
d. What is the (nonnegative) remainder obtained when $68$ is divided by $7$?
When $33$ is divided by $7$?
e. Explain why $68 \mod 7 = 33 \mod 7$.
5. Prove the transitivity of modular congruence. That is, prove that for all
integers $a$, $b$, $c$, and $n$ with $n > 1$, if $a \equiv b(\mod n)$ and
$b \equiv c(\mod n)$ then $a \equiv c(\mod n)$.
6. Prove that the distinct equivalence classes of the relation of congruence
modulo $n$ are the sets $[0], [1], [2], \dots, [n - 1]$, where for each
$a = 0, 1, 2, \dots, n - 1$,
$$ [a] = \{m \in \mathbb{Z} | m \equiv a (\mod n)\} $$
7. Verify the following statements.
a. $128 \equiv 2(\mod 7)$ and $61 \equiv 5(\mod 7)$
b. $(128 + 61) \equiv (2 + 5)(\mod 7)$
c. $(128 - 61) \equiv (2 - 5)(\mod 7)$
d. $(128 \cdot 61) \equiv (2 \cdot 5)(\mod 7)$
e. $128^2 = 2^2(\mod 7)$
8. Verify the following statements.
a. $45 \equiv 3(\mod 6)$ and $104 \equiv 2(\mod 6)$
b. $(45 + 104) \equiv (3 + 2)(\mod 6)$
c. $(45 - 104) \equiv (3 - 2)(\mod 6)$
d. $(45 \cdot 104) \equiv (3 \cdot 2)(\mod 6)$
e. $45^2 \equiv 3^2(\mod 6)$
In 9-11, prove each of the following statements, assuming that $a$, $b$, $c$,
$d$, and $n$ are integers with $n > 1$ and that $a \equiv c(\mod n)$ and
$b \equiv d(\mod n)$.
9.
a. $(a + b) \equiv (c + d)(\mod n)$
b. $(a - b) \equiv (c - d)(\mod n)$
10. $a^2 \equiv c^2(\mod n$
11. $a^m \equiv c^m(\mod n)$ for every integer $m \geq 1$ (Use mathematical
induction on $m$.)
12.
a. Prove that for every integer $n \geq 0$, $10^n \equiv 1(\mod 9)$.
b. Use part (a) to prove that a positive integer is divisible by $9$ if, and
only if, the sum of its digits is divisible by $9$.
13.
a. Prove that for every integer $n \geq 1$, $10^n \equiv (-1)^n(\mod 11)$ .
b. Use part (a) to prove that a positive integer is divisible by $11$ if, and
only if, the alternating sum of its digits is divisible by $114. (For instance,
the alternating sum of the digits of $82,379$ is $8 - 2 + 3 - 7 + 9 = 11$ and
$82,379 = 11 \cdot 7489$.)
14. Use the technique of Example 8.4.4 to find $14^2 \mod 55$, $14^4 \mod 55$,
$14^8 \mod 55$, and $14^{16} \mod 55$.
15. Use the result of exercise 14 and the technique of Example 8.4.5 to find
$14^{27} \mod 55$.
In 16-18, use the techniques of Example 8.4.4 and Example 8.4.5 to find the
given numbers.
16. $675^{307} \mod 713$
17. $89^{307} \mod 713$
18. $48^{307} \mod 713$
In 19-24, use the RSA cipher from Examples 8.4.9 and 8.4.10. In 19-21, translate
the message into its numeric equivalent and encrypt it. In 22-24, decrypt the
cipher-text and translate the result into letters of the alphabet to discover
the message.
19. HELLO
20. WELCOME
21. EXCELLENT
22. 13 20 20 09
23. 08 05 15
24. 51 14 49 15
25. Use Theorem 5.2.2 to prove that if $a$ and $n$ are positive integers and
$a^{n - 1}$ is prime, then $a = 2$ and $n$ is prime.
In 26 and 27, use the extended Euclidean algorithm to find the greatest common
divisor of the given numbers and express it as a linear combination of the two
numbers.
26. 6664 and 765
27. 4158 and 1568
Exercises 28 and 29 refer to the following formal version of the extended
Euclidean algorithm.
**Algorithm 8.4.1 Extended Euclidean Algorithm**
_[Given integers $A$ and $B$ with $A > B > 0$, this algorithm computes
$\text{gcd}(A, B) and finds integers $s$ and $t$ such that
$sA + tB = \text{gcd}(A, B)$.]_
**Input:** $A$, $B$ _[integers with $A > B > 0$]_
**Algorithm Body:**
$a := A, b := B, s := 1, t := 0, u := 0, v := 1\\ \textit{[pre-codndition: } a =
sA + tB \textit{ and } b = uA + vB,\\ \text{gcd}(a, b) = \text{gcd}(A, B)
\textit{]}\\ \textbf{while} (b \neq 0) \\ \ \ \textit{[loop invariant: } a =
sA + tB \textit{ and } b = uA + vB,\\ \ \ \text{gcd}(a, b) = \text{gcd}(A, B)\\ \ \ r:= a \mod b, q := a \text{ div } b\\ \ \ a := b, b := r\\ \ \ \textit{newu } := s - uq, \textit{newv } := t - vq\\ \ \ s := u, t := v\\ \ \ u:= \textit{newu}, v := \textit{newv}\\ \textbf{end while}\\ gcd := a\\ \textit{[post condition: } \text{gcd}(A, B) = a = sA + tB \textit{]}$
**Output:** $\text{gcd}\textit{[a positive integer]}, s, t \textit{[integers]}$
In 28 and 29, for the given values of $A$ and $B$, make a table showing the
values of $s$, $t$, and $sA + tB$ before the start of the while loop and after
each iteration of the loop
28. $A = 330$, $B = 156$
29. $A = 284$, $B = 168$
30. Finis the proof of Theorem 8.4.5 by proving that if $a$, $b$, and $c$ are as
in the proof, then $c | b$.
31.
a. Find an inverse for $210$ modulo $13$.
b. Find a positive inverse for $210$ modulo $13$.
c. Find a positive solution for the congruence $210x \equiv 8 (\mod 13)$.
32.
a. Find an inverse for $41$ modulo $660$.
b. Find the least positive solution for the following congruence:
$41x \equiv 125(\mod 660)$.
33. Use Theorem 8.4.5 to prove that for all integers $a$, $b$, and $c$, if
$\text{gcd}(a, b) = 1$ and $a | c$ and $b | c$, then $ab | c$.
34. Give a counterexample to show that the statement of exercise 33 is false if
the hypothesis that $\text{gcd}(a, b) = 1$ is removed.
35. Corollary 8.4.7 guarantees the existence of an inverse modulo $n$ for an
integer $a$ when $a$ and $n$ are relatively prime. Use Euclid's lemma to
prove that the inverse is unique modulo $n$. In other words, show that if
$s$ and $t$ are any two integers whose product with $a$ is congruent to $1$
modulo $n$, then $s$ and $t$ are congruent to each other modulo $n$.
In 36, 37, 39, and 40, use the RSA cipher with public key
$n = 713 = 23 \cdot 31$ and $e = 43$. In 36 and 37, encode the messages into
their numeric equivalents and encrypt them. In 39 and 40, decrypt the given
ciphertext and find the original messages.
36. HELP
37. COME
38. Find the least positive inverse for $43$ modulo $660$.
39. 675 089 089 048
40. 028 018 675 129
41.
a. Use mathematical induction and Euclid's lemma to prove that for every
positive integer $s$, if $p$ and $q_1, q_2, \dots, q_s$ are 0rime numbers and
$p | q_1q_2 \cdots q_s$, then $p = q_i$ for some $i$ with $1 \leq i \leq s$.
b. The uniqueness part of the unique factorization theorem for the integers says
that given any integer $n$, if
$$ n = p_1p_2 \cdots p_r = q_1q_2 \cdots q_s $$
for some positive integers $r$ and $s$ and prime numbers
$p_1 \leq p_2 \leq \cdots \leq p_r$ and $q_1 \leq q_2 \leq \cdots \leq q_s$,
then $r = s$ and $p_i = q_i$ for every integer $i$ with $1 \leq i \leq r$.
Use the result of part (a) to fill in the details of the following sketch of a
proof:
Suppose that $n$ is an integer with two different prime factorizations:
$n = p_1p_2 \cdots p_t = q_1q_2 \cdots q_u$. All the prime factors that appear
on both sides can be cancelled (as many times as they appear on both sides) to
arrive at the situation where $p_1p_2 \cdots p_r = q_1q_2 \cdots q_s$,
$p_1 \leq p_2 \leq \cdots \leq p_r$, $q_1 \leq q_2 \leq \cdots \leq q_s$, and
$p_i \neq q_j$ for any integers $i$ and $j$. Then use part (a) to deduce a
contradiction, and conclude that the prime factorization of $n$ is unique
except, possibly, for the order in which the prime factors are written.
42. According to Fermat's little theorem, if $p$ is a prime number and $a$ and
$p$ are relatively prime, then $a^{p - 1} \equiv 1 (\mod p)$. Verify that
this theorem gives correct results for the following:
a. $a = 15$ and $p = 7$
b. $a = 8$ and $p = 11$
43. Fermat's little theorem can be used to show that a number is not prime by
finding a number $a$ relatively prime to $p$ with the property that
$a^{p - 1} \cancel{\equiv} 1(\mod p)$. However, it cannot be used to show
that a number _is_ prime. Find an example to illustrate this fact. That is,
find integers $a$ and $p$ such that $a$ and $p$ are relatively prime and
$a^{p - 1} \equiv 1(\mod p)$ but $p$ is not prime.

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chapter_8/notes.md
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@ -459,3 +459,469 @@ $A$,
$$ (a, b) R (c, d) \Leftrightarrow ad = bc $$ $$ (a, b) R (c, d) \Leftrightarrow ad = bc $$
The fact is that $R$ is an equivalence relation. The fact is that $R$ is an equivalence relation.
---
Page 549
**Theorem 8.4.1 Modular Equivalences**
Let $a$, $b$, and $n$ be any integers and suppose $n > 1$. The following
statements are all equivalent:
1. $n | (a - b)$
2. $a \equiv b (\mod n)$
3. $a = b + kn$ for some integer $k$
4. $a$ and $b$ have the same (nonnegative) remainder when divided by $n$
5. $a \mod n = b \mod n$
**Proof:**
We will show that
$(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (4) \Rightarrow (5) \Rightarrow (1)$.
It will follow by the transitivity of if-then that all five statements are
equivalent.
So let $a$, $b$, and $n$ be any integers with $n > 1$.
_Proof that $(1) \Rightarrow (2)$:_
Suppose that $n | (a - b)$. By definition of congruence module $n$, we can
immediately conclude that $a \equiv b (\mod n)$.
_Proof that $(2) \Rightarrow (3)$:_
Suppose $a \equiv b (\mod n)$. By definition of congruence modulo $n$,
$n | (a - b)$. Thus, by definition of divisibility, $a - b = kn$, for some
integer $k$. Adding $b$ to both sides gives that $a = b + kn$.
_Proof that $(3) \Rightarrow (4)$:_
Suppose that $a = b + kn$, for some integer $k$. Use the quotient-remainder
theorem to divide $a$ by $n$ to obtain
$$ a = qn + r \text{ where } q \text{ and } r \text{ are integers and } 0 \leq r < n $$
So $r$ is the remainder obtained when $a$ is divided by $n$. Substituting
$b + kn$ for $a$ in the equation $a = qn + r$ gives that
$$ b + kn = qn + r $$
and subtracting $kn$ from both sides and factoring out $n$ yields
$$ b = (q - k)n + r $$
Now since $0 \leq r < n$, the uniqueness property of the quotient-remainder
theorem guarantees that $r$ is also the remainder obtained when $b$ is divided
by $n$. Thus $a$ and $b$ have the same remainder when divided by $n$.
_Proof that $(4) \Rightarrow (5)$:_
Suppose that $a$ and $b$ have the same remainder when divided by $n$. It follows
immediately from the definition of the $\mod$ function that
$a \mod n = b \mod n$.
_Proof that $(5) \Rightarrow (1)$:_
Suppose that $a \mod n = b \mod n$. By definition of the $\mod$ function, $a$
and $b$ have the same remainder when divided by $n$. Thus, by the
quotient-remainder theorem, we can write
$$ a = q_1n + r \text{ and } b = q_2n + r \text{ where } q_1, q_2 \text{ and } r \text{ are integers and } 0 \leq r < n $$
It follows that
$$ a - b = (q_1n + r) - (q_2n + r) = (q_1 - q_2)n $$
Therefore, since $q_1 - q_2$ is an integer, $n | (a - b)$.
---
Page 550
**Definition**
Given integers $a$ and $n$ with $n > 1$, **the residue of $a$ modulo $n$** is
$a \mod n$, the nonnegative remainder obtained when $a$ is divided by $n$. The
numbers $0, 1, 2, \dots, n - 1$ are called a **complete set of residues modulo
$n$**. To **reduce a number modulo $n$** means to set it equal to its residue
modulo $n$. If a modulus $n > 1$ is fixed throughout a discussion and an integer
$a$ is given, the words "modulo $n$" are often dropped and we simply speak of
**the residue of $a$**.
---
Page 550
**Theorem 8.4.2 Congruence Modulo $n$ Is an Equivalence Relation**
If $n$ is any integer with $n > 1$, congruence modulo $n$ is an equivalence
relation on the set of all integers. The distinct equivalence classes of the
relation are the sets $[0], [1], [2], \dots, [n - 1]$, where for each
$a = 0, 1, 2, \dots, n - 1$,
$$ [a] = \{m \in \mathbb{Z} | m \equiv a (\mod n)\} $$
or, equivalently,
$$ [a] = \{m \in \mathbb{Z} | m = a + kn \text{ for some integer } k\} $$
**Proof:**
Suppose $n$ is any integer with $n > 1$. We must show that congruence modulo $n$
is reflexive, symmetric, and transitive.
_Proof of reflexivity:_
Suppose $a$ is any integer. To show that $a \equiv a (\mod n)$, we must show
that $n | (a - a)$. Now $a - a = 0$, and $n | 0$ because $0 = n \cdot 0$.
Therefore $a \equiv a (\mod n)$.
_Proof of symmetry:_
Suppose $a$ and $b$ are any integers such that $a \equiv b(\mod n)$. We must
show that $b \equiv a (\mod n)$. Now since $a \equiv b(\mod n)$, then
$n | (a - b)$. Thus, by definition of divisibility, $a - b = nk$, for some
integer $k$. Multiplying both sides of this equation by $-1$ to obtain
$$ -(a - b) = -nk $$
or, equivalently,
$$ b - a = n(-k) $$
Thus, by definition of divisibility $n | (b - a)$, and so, by definition of
congruence modulo $n$, $b \equiv a (\mod n)$.
_Proof of transitivity:_
This is left as exercise 5 at the end of the section.
_Proof that the distinct equivalence classes are
$[0], [1], [2], \dots, [n - 1]$:_
This is left as exercise 6 at the end of the section.
---
Page 551
**Theorem 8.4.3 Modular Arithmetic**
Let $a$, $b$, $c$, $d$, and $n$ be integers with $n > 1$, and suppose
$$ a \equiv c (\mod n) \text{ and } b \equiv d(\mod n) $$
Then
1. $(a + b) \equiv (c + d)(\mod n)$
2. $(a - b) \equiv (c - d)(\mod n)$
3. $ab \equiv cd(\mod n)$
4. $a^m \equiv c^m(\mod n)$ for every positive integer $m$
**Proof:**
Because we will make greatest use of part 3 of this theorem, we prove it here
and leave the proofs of the remaining parts of the theorem to exercises 9-11 at
the end of the section.
_Proof of Part 3:_Proof Suppose $a$, $b$, $c$, $d$, and $n$ are integers with
$n > 1$, and suppose $a \equiv b(\mod n)$ and $c \equiv d(\mod n)$. By Theorem
8.4.1, there exists integers $s$ and $t$ such that
$$ a = c + sn \text{ and } b = d + tn $$
Then
$$ ab = (c + sn)(d + tn) $$
$$ = cd + ctn + snd + sntn $$
$$ = cd + n(ct + sd + stn) $$
Let $k = ct + sd + stn$. Then $k$ is an integer because it is a sum of products
of integers, and $ab = cd + nk$. Thus by Theorem 8.4.1, $ab \equiv cd(\mod n)$.
---
Page 552
**Corollary 8.4.4**
Let $a$, $b$, and $n$ be integers with $n > 1$. Then
$$ ab \equiv [(a \mod n)(b \mod n)](\mod n) $$
or, equivalently,
$$ ab \mod n = [(a \mod n)(b \mod n)]\mod n $$
In particular, if $m$ is a positive integer, then
$$ a^m \equiv [(a \mod n)^m](\mod n) $$
---
Page 555
**Definition**
An integer $d$ is said to be a **linear combination of integers** $a$ and $b$
if, and only if, there exist integers $s$ and $t$ such that $as + bt = d$.
---
Page 555
**Theorem 8.4.5 Writing a Greatest Common Divisor as a Linear Combination**
For all integers $a$ and $b$, not both zero, if $d = \text{gcd}(a, b)$, then
there exist integers $s$ and $t$ such that $as + bt = d$.
**Proof:**
Given integers $a$ and $b$, not both zero, and given $d = \text{gcd}(a, b)$, let
$$ S = \{x | x \text{ is a positive integer and } x = as + bt \text{ for some integers } s \text{ and } t\} $$
Note that $S$ is a nonempty set because (1) if $a > 0$ then
$1 \cdot a + 0 \cdot b \in S$, (2) if $a < 0$ then
$(-1) \cdot a + 0 \cdot b \in S$, and (3) if $a = 0$ then, by assumption,
$b \neq 0$, and hence $0 \cdot a + 1 \cdot b \in S$ or
$0 \cdot a + (-1) \cdot b \in S$. Thus, because $S$ is a nonempty subset of
positive integers, by the well-ordering principle for the integers there is a
least element $c$ in $S$. By definition of $S$,
$$ c = as + bt \text{ for some integers } s \text{ and } t $$
We will show that (1) $c \geq d$, and (2) $c \leq d$, and we will therefore be
able to conclude that $c = d = \text{gcd}(a, b)$.
_(1) Proof that $c \geq d$:_
_[In this part of the proof, we show that $d$ is a divisor of $c$ and thus that
$d \leq c$.]_ Because $d = \text{gcd}(a, b)$, by definition of greatest common
divisor, $d | a$ and $d | b$. Hence $a = dx$ and $b = dy$ for some integers $x$
and $y$. Then
$$ c = as + bt $$
$$ = (dx)s + (dy)t $$
$$ = d(xs + y) $$
Now $xs + yt$ is an integer because it is a sum of products of integers. Thus,
by definition of divisibility, $d | c$. Both $c$ and $d$ are positive, and
hence, by Theorem 4.4.1, $c \geq d$.
_(2) Proof that $c \leq d$:_
_[In this part of the proof, we show that $c$ is a divisor of both $a$ and $b$
and therefore that $c$ is less than or equal to the greatest common divisor of
$a$ and $b$, which is $d$.]_ Apply the quotient-remainder theorem to the
division of $a$ by $c$ to obtain
$$ a = cq + r \text{ for some integers } q \text{ and } r \text{ with } 0 \leq r < c $$
Thus for some integers $q$ and $r$ with $0 \leq r < c$,
$$ r = q - cq $$
Now $c = as + bt$. Therefore, for some integers $q$ and $r$ with $0 \leq r < c$,
$$ r = a - (as + bt)q $$
$$ = a(1 - sq) - btq $$
Thus $r$ is a linear combination of $a$ and $b$. If $r > 0$, then $r$ would be
in $S$, and so $r$ would be a smaller element of $S$ than $c$, which would
contradict the fact that $c$ is the least element of $S$. Hence $r = 0$. By
substitution into (8.4.4),
$$ a = cq $$
and therefore $c | a$.
An almost identical argument establishes that $c | b$ and is left as exercise 30
at the end of the section.
Because $c | a$ and $c | b$, $c$ is a common divisor of $a$ and $b$. Hence $c$
is less than or equal to the greatest common divisor of $a$ and $b$. In other
words, $c \leq d$.
From (1) and (2), we conclude that $c = d$. It follows that $d$, the greatest
common divisor of $a$ and $b$, is equal to $as + bt$.
---
Page 557
**Definition**
Given any integer $a$ and any positive integer $n$, if there exists an integer
$s$ such that $as \equiv 1(\mod n)$, then $s$ is called **an inverse for $a$
modulo $n$.**
---
Page 557
**Definition**
Integers $a$ and $b$ are **relatively prime** if, and only if,
$\text{gcd}(a, b) = 1$. Integers $a_1, a_2, a_3, \dots, a_n$ are **pairwise
relatively prime** if, and only if, $\text{gcd}(a_i, a_j) = 1$ for all integers
$i$ and $j$ with $1 \leq i$, $j \leq n$, and $i \neq j$.
---
Page 557
**Corollary 8.4.6**
If $a$ and $b$ are relatively prime integers, then there exist integers $s$ and
$t$ such that $as + bt = 1$.
---
Page 558
**Corollary 8.4.7 Existence of Inverses Modulo $n$**
For all integers $a$ and $n$, if $\text{gcd}(a, n) = 1$, then there exists an
integer $s$ such that $as \equiv 1(\mod n)$, and so $s$ is an inverse for $a$
modulo $n$.
**Proof:**
Suppose $a$ and $n$ are integers and $\text{gcd}(a, n) = 1$. By Corollary 8.4.6,
there exist integers $s$ and $t$ such that
$$ as + nt = 1 $$
Subtracting $nt$ from both sides gives that
$$ as = 1 - nt = 1 + (-t)n $$
Thus, by definition of congruence modulo $n$,
$$ as \equiv 1(\mod n) $$
---
Page 562
**Theorem 8.4.8 Euclid's Lemma**
For all integers $a$, $b$, and $c$, if $\text{gcd}(a, c) = 1$ and $a | bc$, then
$a | b$.
**Proof:**
Suppose $a$, $b$, and $c$ are integers, $\text{gcd}(a, c) = 1$, and $a | bc$.
_[We must show that $a | b$.]_ By Theorem 8.4.5, there exist integers $s$ and
$t$ so that
$$ as + ct = 1 $$
Multiply both sides of this equation by $b$ to obtain
$$ bas + bct = b $$
Since $a | bc$, by definition of divisibility there exists an integer $k$ such
that
$$ bc = ak $$
Substituting (8.4.8) into (8.4.7), rewriting, and factoring out an $a$ gives
that
$$ b = bas + (ak)t = a(bs + kt) $$
Let $r = bs + kt$. Then $r$ is an integer (because $b$, $s$, $k$, and $t$ are
all integers), and $b = ar$. Thus $a | b$ by definition of divisibility.
Page 562
**Theorem 8.4.9 Cancellation Theorem for Modular Congruence**
For all integers $a$, $b$, and $c$, and $n$ with $n > 1$, if
$\text{gcd}(c, n) = 1$ and $ac = bc(\mod n)$, then $a \equiv b(\mod n)$.
**Proof:**
Suppose $a$, $b$, $c$, and $n$ are integers, $\text{gcd}(c, n) = 1$, and
$ac \equiv bc(\mod n)$. _[We must show that $a \equiv b(\mod n)$.]_ By
definition of congruence modulo $n$,
$$ n | (ac - bc) $$
and so, since
$$ ac - bc = (a - b)c $$
then
$$ n | (a - b)c $$
Because $\text{gcd}(c, n) = 1$, we may apply Euclid's lemma to obtain
$$ n | (a - b) $$
and so, by definition of congruence modulo $n$,
$$ a \equiv b(\mod n) $$
---
Page 563
**Theorem 8.4.10 Fermat's Little Theorem**
If $p$ is any prime number and $a$ is any integer such that $p \cancel{|} a$,
then $a^{p - 1} \equiv 1(\mod p)$.
**Proof:**
Suppose $p$ is any prime number and $a$ is any integer such that
$p \cancel{|} a$. Note that $a \neq 0$ because otherwise $p$ would divide $a$.
Consider the set of integers
$$ S = \{a, 2a, 3a, \dots, (p - 1)a\} $$
We claim that no two elements of $S$ are congruent modulo $p$. For suppose
$sa \equiv ra(\mod p)$ for some integers $s$ and $r$ with
$1 \leq r < s \leq p - 1$. Then, by definition of congruence modulo $p$,
$$ p | (sa - ra) \text{ or, equivalently, } p | (s - r)a $$
Now $p \cancel{|} a$ by hypothesis, and because $p$ is prime,
$\text{gcd}(a, p) = 1$. Thus, by Euclid's lemma, $p | (s - r)$, But this is
impossible because $0 < s - r < p$.
Consider the function $F$ from $S$ to the set $T = \{1, 2, 3, \dots, (p - 1)\}$
that sends each element of $S$ to its residue modulo $p$. Then $F$ is one-to-one
because no two elements of $S$ are congruent modulo $p$. In Section 9.4 we prove
that if a function from one finite set to another is one-to-one, then it is also
onto. Hence $F$ is onto, and so $p - 1$ residues of the $p - 1$ elements of $S$
are exactly the numbers $1, 2, 3 \dots, (p - 1)$.
It follows by Theorem 8.4.3(3) that
$$ a \cdot 2a \cdot 3a \cdots (p - 1)a \equiv [1 \cdot 2 \cdot 3 \cdots (p - 1)](\mod p) $$
or, equivalently,
$$ a^{p - 1}(p - 1)! \equiv (p - 1)!(\mod p) $$
Now because $p$ is prime, $p$ and $(p - 1)!$ are relatively prime. Thus, by the
cancellation theorem for modular congruence (Theorem 8.4.9),
$$ a^{p - 1} \equiv 1(\mod p) $$

42
chapter_8/test_yourself.md
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@ -113,3 +113,45 @@ a partition of $A$
equivalence class of $R$ for each ____. equivalence class of $R$ for each ____.
rational number rational number
---
Page 566
**Test Yourself**
1. When letters of the alphabet are encrypted using the Caesar cipher, the
encrypted version of the letter is ____.
2. If $a$, $b$, and $n$ are integers with $n > 1$, all of the following are
different ways to express the fact that $n | (a - b)$: ____, ____, ____,
____.
3. If $a$, $b$, $c$, $d$, $m$, and $n$ are integers with $n > 1$ and if
$a \equiv c(\mod n)$ and $b \equiv d(\mod n)$, then $a + b \equiv$ ____,
$a - b \equiv$ ____, $ab \equiv$ ____, and $a^m \equiv$ ____
4. If $a$, $n$, and $k$ are positive integers with $n > 1$, an efficient way to
compute $a^k(\mod n)$ is to write $k$ as a ____ and use the facts about
computing products and powers modulo $n$.
5. To express a greatest common divisor of two integers as a linear combination
of the integers, use the extended version of the ____ algorithm.
6. To find an inverse for a positive integer $a$ modulo an integer $n$ with
$n > 1$, you express the number $1$ as ____.
7. To encrypt a message $M$ using RSA cryptography with public key $pq$ and $e$,
you use the formula ____, and to decrypt a message $C$, you use the formula
____, where ____.
8. Euclid's lemma says that for all integers $a$, $b$, and $c$ if
$\text{gcd}(a, c) = 1$ and $a | bc$, then ____.
9. Fermat's little theorem says that if $p$ is any prime number and $a$ is any
integer such that $p | a$, then ____.
10. The crux of the proof that the RSA cipher works is that if (1) $p$ and $q$
are distinct large prime numbers, (2) $M < pq$, (3) $M$ is relatively prime
to $pq$, (4) $e$ is relatively prime to $(p - 1)(q - 1)$, and (5) $d$ is a
positive inverse for $e$ modulo $(p - 1)(q - 1)$, then $M =$ ____.