🚧 Setup for 8.4
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@ -4009,3 +4009,270 @@ Omitted.
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g. What _are_ you? (Do not answer this on paper; just think about it.)
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Omitted.
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---
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Page 567
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**Exercise Set 8.4**
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1.
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a. Use the Caesar cipher to encrypt the message WHERE SHALL WE MEET.
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b. Use the Caesar cipher to decrypt the message LQ WKH FDIHWHULD.
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2.
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a. Use the Caesar cipher to encrypt the message AN APPLE A DAY.
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b. Use the Caesar cipher to decrypt the message NHHSV WKH GRFWRU DZDB.
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3. Let $a = 25, $b = 19$, and $n = 3$.
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a. Verify that $3 | (25 - 19)$.
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b. Explain why $25 \equiv 19 (\mod 3)$.
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c. What value of $k$ has the property that $25 = 19 + 3k$?
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d. What is the (nonnegative) remainder obtained when $25$ is divided by $3$?
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When $19$ is divided by $3$?
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e. Explain why $25 \mod 3 = 19 \mod 3$.
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4. Let $a = 68$, $b = 33$, and $n = 7$.
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a. Verify that $7 | (68 - 33)$.
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b. Explain why $68 \equiv 33(\mod 7)$.
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c. What value of $k$ has the property that $68 = 33 + 7k$?
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d. What is the (nonnegative) remainder obtained when $68$ is divided by $7$?
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When $33$ is divided by $7$?
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e. Explain why $68 \mod 7 = 33 \mod 7$.
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5. Prove the transitivity of modular congruence. That is, prove that for all
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integers $a$, $b$, $c$, and $n$ with $n > 1$, if $a \equiv b(\mod n)$ and
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$b \equiv c(\mod n)$ then $a \equiv c(\mod n)$.
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6. Prove that the distinct equivalence classes of the relation of congruence
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modulo $n$ are the sets $[0], [1], [2], \dots, [n - 1]$, where for each
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$a = 0, 1, 2, \dots, n - 1$,
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$$ [a] = \{m \in \mathbb{Z} | m \equiv a (\mod n)\} $$
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7. Verify the following statements.
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a. $128 \equiv 2(\mod 7)$ and $61 \equiv 5(\mod 7)$
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b. $(128 + 61) \equiv (2 + 5)(\mod 7)$
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c. $(128 - 61) \equiv (2 - 5)(\mod 7)$
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d. $(128 \cdot 61) \equiv (2 \cdot 5)(\mod 7)$
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e. $128^2 = 2^2(\mod 7)$
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8. Verify the following statements.
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a. $45 \equiv 3(\mod 6)$ and $104 \equiv 2(\mod 6)$
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b. $(45 + 104) \equiv (3 + 2)(\mod 6)$
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c. $(45 - 104) \equiv (3 - 2)(\mod 6)$
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d. $(45 \cdot 104) \equiv (3 \cdot 2)(\mod 6)$
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e. $45^2 \equiv 3^2(\mod 6)$
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In 9-11, prove each of the following statements, assuming that $a$, $b$, $c$,
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$d$, and $n$ are integers with $n > 1$ and that $a \equiv c(\mod n)$ and
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$b \equiv d(\mod n)$.
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9.
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a. $(a + b) \equiv (c + d)(\mod n)$
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b. $(a - b) \equiv (c - d)(\mod n)$
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10. $a^2 \equiv c^2(\mod n$
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11. $a^m \equiv c^m(\mod n)$ for every integer $m \geq 1$ (Use mathematical
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induction on $m$.)
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12.
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a. Prove that for every integer $n \geq 0$, $10^n \equiv 1(\mod 9)$.
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b. Use part (a) to prove that a positive integer is divisible by $9$ if, and
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only if, the sum of its digits is divisible by $9$.
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13.
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a. Prove that for every integer $n \geq 1$, $10^n \equiv (-1)^n(\mod 11)$ .
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b. Use part (a) to prove that a positive integer is divisible by $11$ if, and
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only if, the alternating sum of its digits is divisible by $114. (For instance,
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the alternating sum of the digits of $82,379$ is $8 - 2 + 3 - 7 + 9 = 11$ and
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$82,379 = 11 \cdot 7489$.)
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14. Use the technique of Example 8.4.4 to find $14^2 \mod 55$, $14^4 \mod 55$,
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$14^8 \mod 55$, and $14^{16} \mod 55$.
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15. Use the result of exercise 14 and the technique of Example 8.4.5 to find
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$14^{27} \mod 55$.
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In 16-18, use the techniques of Example 8.4.4 and Example 8.4.5 to find the
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given numbers.
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16. $675^{307} \mod 713$
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17. $89^{307} \mod 713$
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18. $48^{307} \mod 713$
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In 19-24, use the RSA cipher from Examples 8.4.9 and 8.4.10. In 19-21, translate
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the message into its numeric equivalent and encrypt it. In 22-24, decrypt the
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cipher-text and translate the result into letters of the alphabet to discover
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the message.
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19. HELLO
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20. WELCOME
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21. EXCELLENT
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22. 13 20 20 09
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23. 08 05 15
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24. 51 14 49 15
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25. Use Theorem 5.2.2 to prove that if $a$ and $n$ are positive integers and
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$a^{n - 1}$ is prime, then $a = 2$ and $n$ is prime.
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In 26 and 27, use the extended Euclidean algorithm to find the greatest common
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divisor of the given numbers and express it as a linear combination of the two
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numbers.
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26. 6664 and 765
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27. 4158 and 1568
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Exercises 28 and 29 refer to the following formal version of the extended
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Euclidean algorithm.
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**Algorithm 8.4.1 Extended Euclidean Algorithm**
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_[Given integers $A$ and $B$ with $A > B > 0$, this algorithm computes
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$\text{gcd}(A, B) and finds integers $s$ and $t$ such that
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$sA + tB = \text{gcd}(A, B)$.]_
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**Input:** $A$, $B$ _[integers with $A > B > 0$]_
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**Algorithm Body:**
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$a := A, b := B, s := 1, t := 0, u := 0, v := 1\\ \textit{[pre-codndition: } a =
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sA + tB \textit{ and } b = uA + vB,\\ \text{gcd}(a, b) = \text{gcd}(A, B)
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\textit{]}\\ \textbf{while} (b \neq 0) \\ \ \ \textit{[loop invariant: } a =
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sA + tB \textit{ and } b = uA + vB,\\ \ \ \text{gcd}(a, b) = \text{gcd}(A, B)\\ \ \ r:= a \mod b, q := a \text{ div } b\\ \ \ a := b, b := r\\ \ \ \textit{newu } := s - uq, \textit{newv } := t - vq\\ \ \ s := u, t := v\\ \ \ u:= \textit{newu}, v := \textit{newv}\\ \textbf{end while}\\ gcd := a\\ \textit{[post condition: } \text{gcd}(A, B) = a = sA + tB \textit{]}$
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**Output:** $\text{gcd}\textit{[a positive integer]}, s, t \textit{[integers]}$
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In 28 and 29, for the given values of $A$ and $B$, make a table showing the
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values of $s$, $t$, and $sA + tB$ before the start of the while loop and after
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each iteration of the loop
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28. $A = 330$, $B = 156$
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29. $A = 284$, $B = 168$
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30. Finis the proof of Theorem 8.4.5 by proving that if $a$, $b$, and $c$ are as
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in the proof, then $c | b$.
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31.
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a. Find an inverse for $210$ modulo $13$.
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b. Find a positive inverse for $210$ modulo $13$.
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c. Find a positive solution for the congruence $210x \equiv 8 (\mod 13)$.
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32.
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a. Find an inverse for $41$ modulo $660$.
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b. Find the least positive solution for the following congruence:
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$41x \equiv 125(\mod 660)$.
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33. Use Theorem 8.4.5 to prove that for all integers $a$, $b$, and $c$, if
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$\text{gcd}(a, b) = 1$ and $a | c$ and $b | c$, then $ab | c$.
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34. Give a counterexample to show that the statement of exercise 33 is false if
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the hypothesis that $\text{gcd}(a, b) = 1$ is removed.
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35. Corollary 8.4.7 guarantees the existence of an inverse modulo $n$ for an
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integer $a$ when $a$ and $n$ are relatively prime. Use Euclid's lemma to
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prove that the inverse is unique modulo $n$. In other words, show that if
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$s$ and $t$ are any two integers whose product with $a$ is congruent to $1$
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modulo $n$, then $s$ and $t$ are congruent to each other modulo $n$.
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In 36, 37, 39, and 40, use the RSA cipher with public key
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$n = 713 = 23 \cdot 31$ and $e = 43$. In 36 and 37, encode the messages into
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their numeric equivalents and encrypt them. In 39 and 40, decrypt the given
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ciphertext and find the original messages.
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36. HELP
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37. COME
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38. Find the least positive inverse for $43$ modulo $660$.
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39. 675 089 089 048
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40. 028 018 675 129
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41.
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a. Use mathematical induction and Euclid's lemma to prove that for every
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positive integer $s$, if $p$ and $q_1, q_2, \dots, q_s$ are 0rime numbers and
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$p | q_1q_2 \cdots q_s$, then $p = q_i$ for some $i$ with $1 \leq i \leq s$.
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b. The uniqueness part of the unique factorization theorem for the integers says
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that given any integer $n$, if
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$$ n = p_1p_2 \cdots p_r = q_1q_2 \cdots q_s $$
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for some positive integers $r$ and $s$ and prime numbers
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$p_1 \leq p_2 \leq \cdots \leq p_r$ and $q_1 \leq q_2 \leq \cdots \leq q_s$,
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then $r = s$ and $p_i = q_i$ for every integer $i$ with $1 \leq i \leq r$.
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Use the result of part (a) to fill in the details of the following sketch of a
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proof:
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Suppose that $n$ is an integer with two different prime factorizations:
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$n = p_1p_2 \cdots p_t = q_1q_2 \cdots q_u$. All the prime factors that appear
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on both sides can be cancelled (as many times as they appear on both sides) to
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arrive at the situation where $p_1p_2 \cdots p_r = q_1q_2 \cdots q_s$,
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$p_1 \leq p_2 \leq \cdots \leq p_r$, $q_1 \leq q_2 \leq \cdots \leq q_s$, and
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$p_i \neq q_j$ for any integers $i$ and $j$. Then use part (a) to deduce a
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contradiction, and conclude that the prime factorization of $n$ is unique
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except, possibly, for the order in which the prime factors are written.
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42. According to Fermat's little theorem, if $p$ is a prime number and $a$ and
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$p$ are relatively prime, then $a^{p - 1} \equiv 1 (\mod p)$. Verify that
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this theorem gives correct results for the following:
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a. $a = 15$ and $p = 7$
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b. $a = 8$ and $p = 11$
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43. Fermat's little theorem can be used to show that a number is not prime by
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finding a number $a$ relatively prime to $p$ with the property that
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$a^{p - 1} \cancel{\equiv} 1(\mod p)$. However, it cannot be used to show
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that a number _is_ prime. Find an example to illustrate this fact. That is,
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find integers $a$ and $p$ such that $a$ and $p$ are relatively prime and
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$a^{p - 1} \equiv 1(\mod p)$ but $p$ is not prime.
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@ -459,3 +459,469 @@ $A$,
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$$ (a, b) R (c, d) \Leftrightarrow ad = bc $$
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The fact is that $R$ is an equivalence relation.
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---
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Page 549
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**Theorem 8.4.1 Modular Equivalences**
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Let $a$, $b$, and $n$ be any integers and suppose $n > 1$. The following
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statements are all equivalent:
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1. $n | (a - b)$
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2. $a \equiv b (\mod n)$
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3. $a = b + kn$ for some integer $k$
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4. $a$ and $b$ have the same (nonnegative) remainder when divided by $n$
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5. $a \mod n = b \mod n$
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**Proof:**
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We will show that
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$(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (4) \Rightarrow (5) \Rightarrow (1)$.
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It will follow by the transitivity of if-then that all five statements are
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equivalent.
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So let $a$, $b$, and $n$ be any integers with $n > 1$.
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_Proof that $(1) \Rightarrow (2)$:_
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Suppose that $n | (a - b)$. By definition of congruence module $n$, we can
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immediately conclude that $a \equiv b (\mod n)$.
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_Proof that $(2) \Rightarrow (3)$:_
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Suppose $a \equiv b (\mod n)$. By definition of congruence modulo $n$,
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$n | (a - b)$. Thus, by definition of divisibility, $a - b = kn$, for some
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integer $k$. Adding $b$ to both sides gives that $a = b + kn$.
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_Proof that $(3) \Rightarrow (4)$:_
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Suppose that $a = b + kn$, for some integer $k$. Use the quotient-remainder
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theorem to divide $a$ by $n$ to obtain
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$$ a = qn + r \text{ where } q \text{ and } r \text{ are integers and } 0 \leq r < n $$
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So $r$ is the remainder obtained when $a$ is divided by $n$. Substituting
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$b + kn$ for $a$ in the equation $a = qn + r$ gives that
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$$ b + kn = qn + r $$
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and subtracting $kn$ from both sides and factoring out $n$ yields
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$$ b = (q - k)n + r $$
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Now since $0 \leq r < n$, the uniqueness property of the quotient-remainder
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theorem guarantees that $r$ is also the remainder obtained when $b$ is divided
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by $n$. Thus $a$ and $b$ have the same remainder when divided by $n$.
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_Proof that $(4) \Rightarrow (5)$:_
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Suppose that $a$ and $b$ have the same remainder when divided by $n$. It follows
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immediately from the definition of the $\mod$ function that
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$a \mod n = b \mod n$.
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_Proof that $(5) \Rightarrow (1)$:_
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Suppose that $a \mod n = b \mod n$. By definition of the $\mod$ function, $a$
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and $b$ have the same remainder when divided by $n$. Thus, by the
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quotient-remainder theorem, we can write
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$$ a = q_1n + r \text{ and } b = q_2n + r \text{ where } q_1, q_2 \text{ and } r \text{ are integers and } 0 \leq r < n $$
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It follows that
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$$ a - b = (q_1n + r) - (q_2n + r) = (q_1 - q_2)n $$
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Therefore, since $q_1 - q_2$ is an integer, $n | (a - b)$.
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---
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Page 550
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**Definition**
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Given integers $a$ and $n$ with $n > 1$, **the residue of $a$ modulo $n$** is
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$a \mod n$, the nonnegative remainder obtained when $a$ is divided by $n$. The
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numbers $0, 1, 2, \dots, n - 1$ are called a **complete set of residues modulo
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$n$**. To **reduce a number modulo $n$** means to set it equal to its residue
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modulo $n$. If a modulus $n > 1$ is fixed throughout a discussion and an integer
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$a$ is given, the words "modulo $n$" are often dropped and we simply speak of
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**the residue of $a$**.
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---
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Page 550
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**Theorem 8.4.2 Congruence Modulo $n$ Is an Equivalence Relation**
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If $n$ is any integer with $n > 1$, congruence modulo $n$ is an equivalence
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relation on the set of all integers. The distinct equivalence classes of the
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relation are the sets $[0], [1], [2], \dots, [n - 1]$, where for each
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$a = 0, 1, 2, \dots, n - 1$,
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$$ [a] = \{m \in \mathbb{Z} | m \equiv a (\mod n)\} $$
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or, equivalently,
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$$ [a] = \{m \in \mathbb{Z} | m = a + kn \text{ for some integer } k\} $$
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**Proof:**
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Suppose $n$ is any integer with $n > 1$. We must show that congruence modulo $n$
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is reflexive, symmetric, and transitive.
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_Proof of reflexivity:_
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Suppose $a$ is any integer. To show that $a \equiv a (\mod n)$, we must show
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that $n | (a - a)$. Now $a - a = 0$, and $n | 0$ because $0 = n \cdot 0$.
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Therefore $a \equiv a (\mod n)$.
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_Proof of symmetry:_
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Suppose $a$ and $b$ are any integers such that $a \equiv b(\mod n)$. We must
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show that $b \equiv a (\mod n)$. Now since $a \equiv b(\mod n)$, then
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$n | (a - b)$. Thus, by definition of divisibility, $a - b = nk$, for some
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integer $k$. Multiplying both sides of this equation by $-1$ to obtain
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$$ -(a - b) = -nk $$
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or, equivalently,
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$$ b - a = n(-k) $$
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Thus, by definition of divisibility $n | (b - a)$, and so, by definition of
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congruence modulo $n$, $b \equiv a (\mod n)$.
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_Proof of transitivity:_
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This is left as exercise 5 at the end of the section.
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_Proof that the distinct equivalence classes are
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$[0], [1], [2], \dots, [n - 1]$:_
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This is left as exercise 6 at the end of the section.
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---
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Page 551
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**Theorem 8.4.3 Modular Arithmetic**
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Let $a$, $b$, $c$, $d$, and $n$ be integers with $n > 1$, and suppose
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$$ a \equiv c (\mod n) \text{ and } b \equiv d(\mod n) $$
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Then
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1. $(a + b) \equiv (c + d)(\mod n)$
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2. $(a - b) \equiv (c - d)(\mod n)$
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3. $ab \equiv cd(\mod n)$
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|
||||
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),
|
||||
|
||||
$$ a^{p - 1} \equiv 1(\mod p) $$
|
||||
|
|
|
|||
|
|
@ -113,3 +113,45 @@ a partition of $A$
|
|||
equivalence class of $R$ for each ____.
|
||||
|
||||
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 $M =$ ____.
|
||||
|
|
|
|||
Loading…
Add table
Add a link
Reference in a new issue