problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
2. Decipher the ciphertext message LFDPH LVDZL FRQTX HUHG that has been enciphered using the Caesar cipher. | 2. I CAME I SAW I CONQUERED | I CAME I SAW I CONQUERED | Logic and Puzzles | math-word-problem | Yes | Yes | number_theory | false | 738,954 |
3. Encipher the message SURRENDER IMMEDIATELY using the affine transformation $C \equiv 11 P+18(\bmod 26)$. | 3. IEXXK FZKXC UUKZC STKJW | IEXXK FZKXC UUKZC STKJW | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,955 |
4. Decipher the message RTOLK TOIK, which was enciphered using the affine transformation $C \equiv 3 P+24(\bmod 26)$. | 4. PHONE HOME | PHONE HOME | Logic and Puzzles | math-word-problem | Yes | Yes | number_theory | false | 738,956 |
7. Given two ciphers, plaintext may be enciphered by using one of the ciphers, and by then using the other cipher. This procedure produces a product cipher.
a) Find the product cipher obtained by using the transformation $C \equiv 5 P+13$ $(\bmod 26)$ followed by the transformation $C \equiv 17 P+3(\bmod 26)$.
b) Find ... | 7. a) $C \equiv 7 P+16(\bmod 26)$
b) $C \equiv a c P+b c+d(\bmod 26)$ | C \equiv 7 P+16(\bmod 26) | Algebra | math-word-problem | Yes | Yes | number_theory | false | 738,959 |
8. A Vignère cipher operates in the following way. A sequence of letters $\ell_{1}, \ell_{2}, \ldots, \ell_{n}$, with numerical equivalents $k_{1}, k_{2}, \ldots, k_{n}$, serves as the key. Plaintext messages are split into blocks of length $n$. To encipher a plaintext block of letters with numerical equivalents $p_{1}... | 8. a) VSPFXH HIPKLB KIPMIE GTG
b) EXPLOSIVES INSIDE | not found | Other | math-word-problem | Yes | Yes | number_theory | false | 738,960 |
1. Using the digraphic cipher that sends the plaintext block $P_{1} P_{2}$ to the ciphertext block $C_{1} C_{2}$ with
$$\begin{array}{l}
C_{1} \equiv 3 P_{1}+10 P_{2} \quad(\bmod 26) \\
C_{2} \equiv 9 P_{1}+7 P_{2} \quad(\bmod 26)
\end{array}$$
encipher the message BEWARE OF THE MESSENGER. | 1. RL OQ NZ OF XM CQ KE QI VD AZ | RL OQ NZ OF XM CQ KE QI VD AZ | Algebra | math-word-problem | Yes | Yes | number_theory | false | 738,961 |
3. A cryptanalyst has determined that the two most common digraphs in a ciphertext message are RH and NI and guesses that these ciphertext digraphs correspond to the two most common diagraphs in English text, TH and HE. If the plaintext was enciphered using a Hill digraphic cipher described by
$$\begin{array}{l}
C_{1} ... | 3. $\left(\begin{array}{cc}3 & 24 \\ 24 & 25\end{array}\right)$ | \left(\begin{array}{cc}3 & 24 \\ 24 & 25\end{array}\right) | Algebra | math-word-problem | Yes | Yes | number_theory | false | 738,963 |
4. How many pairs of letters remain unchanged when encryption is performed using the following digraphic ciphers
a)
$$\begin{array}{l}
C_{1} \equiv 4 P_{1}+5 P_{2}(\bmod 26) \\
C_{2} \equiv 3 P_{1}+P_{2}(\bmod 26)
\end{array}$$
b)
$$\begin{array}{l}
C_{1} \equiv 7 P_{1}+17 P_{2}(\bmod 26) \\
C_{2} \equiv P_{1}+6 P_{2} ... | 4. a) 1 b) 13 c) 26 | 4. a) 1 b) 13 c) 26 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,965 |
10. Find the $6 \times 6$ enciphering matrix corresponding to the product cipher obtained by first using the Hill cipher with enciphering matrix $\left(\begin{array}{ll}3 & 1 \\ 2 & 1\end{array}\right)$, followed by using the Hill cipher with enciphering matrix $\left(\begin{array}{lll}1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 &... | $10 .\left(\begin{array}{llllll}5 & 2 & 0 & 0 & 0 & 0 \\ 3 & 1 & 3 & 1 & 0 & 0 \\ 2 & 1 & 3 & 1 & 0 & 0 \\ 0 & 0 & 2 & 1 & 3 & 1 \\ 0 & 0 & 2 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 & 5 & 2\end{array}\right)$ | \left(\begin{array}{llllll}5 & 2 & 0 & 0 & 0 & 0 \\ 3 & 1 & 3 & 1 & 0 & 0 \\ 2 & 1 & 3 & 1 & 0 & 0 \\ 0 & 0 & 2 & 1 & 3 & 1 \\ 0 & 0 & 2 & 1 & 2 & | Algebra | math-word-problem | Yes | Yes | number_theory | false | 738,971 |
2. Suppose a cryptanalyst discovers a message $P$ that is not relatively prime to the enciphering modulus $n=p q$ used in a RSA cipher.
a) Show that the cryptanalyst can factor $n . \quad(P, n)=p$ or $q$
b) Show that it is extremely unlikely that such a message can be discovered by demonstrating that the probability th... | None | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,982 |
3. What is the ciphertext that is produced when the RSA cipher with key $(e, n)=(3,2669)$ is used to encipher the message BEST WISHES? | 3. 12151224147100230116 | 12151224147100230116 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,983 |
5. Harold and Audrey have as their RSA keys $(3,23 \cdot 47)$ and $(7,31 \cdot 59)$, respectively.
a) Using the method in the text, what is the signed ciphertext sent by Harold to Audrey, when the plaintext message is CHEERS HAROLD?
b) Using the method in the text, what is the signed ciphertext sent by Audrey to Harold... | 5. a) 037103540858085800871359035400000087154317970535 b) 00190977 075603700343064702740872082100730845074000000008014808030415
045802740740 | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,985 |
6. Let $H$ be a fixed integer. Let each individual have two pairs of enciphering keys: $k=(e, n)$ and $k^{*}=\left(e, n^{*}\right)$ with $n<H<n^{*}$, where $n$ and $n^{*}$ are both the product of two primes. Using the RSA cipher system, individual $i$ can send a signed message $P$ to individual $j$ by sending $E_{k_{j}... | 6. c) 00420056048104810763000000510000029402620995049505430972 00000734015206470972 | 00420056048104810763000000510000029402620995049505430972 00000734015206470972 | Number Theory | proof | Yes | Yes | number_theory | false | 738,987 |
7. a) Show that if individuals $i$ and $j$ have enciphering keys $k_{i}=\left(e_{i}, n_{i}\right)$ and $k_{j}=\left(e_{j}, n_{j}\right)$, respectively, where both $n_{i}$ and $n_{j}$ are products of two distinct primes, then individual $i$ can send a signed message $P$ to individual $j$ without needing to change the si... | 7. d) 13831812035200001383013010801351138318120130097212080956 000009721515093712971208227315150000 | 13831812035200001383013010801351138318120130097212080956 000009721515093712971208227315150000 | Number Theory | proof | Yes | Yes | number_theory | false | 738,988 |
4. Find all subsets of the integers $2,3,4,7,11,13,16$ that have 18 as their sum. | 4. $18=2+16=2+3+13=3+4+11=7+11$ | 18=2+16=2+3+13=3+4+11=7+11 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 738,993 |
5. Find the sequence obtained from the super-increasing sequence $(1,3,5,10,20,41,80)$ when modular multiplication is applied with multiplier $w=17$ and modulus $m=162$. | 5. $(17,51,85,8,16,49,64)$ | (17,51,85,8,16,49,64) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,994 |
8. Find the sequence obtained by applying successively the modular multiplications with multipliers and moduli $(7,92),(11,95)$, and $(6,101)$, respectively, on the super-increasing sequence $(3,4,8,17,33,67)$. | 8. $(44,37,74,72,50,24)$ | (44,37,74,72,50,24) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,998 |
10. A multiplicative knapsack problem is a problem of the following type: Given positive integers $a_{1}, a_{2}, \ldots, a_{n}$ and a positive integer $P$, find the subset, or subsets, of these integers with product $P$, or equivalently, find all solutions of
$$P=a_{1}^{x_{1}} a_{2}^{x_{1}} \cdots a_{n}^{x}$$
where $x_... | 10. a) $60=2 \cdot 3 \cdot 10=2 \cdot 5 \cdot 6=6 \cdot 10$
b) $15960=8 \cdot 21 \cdot 95$ | 15960=8 \cdot 21 \cdot 95 | Number Theory | proof | Yes | Yes | number_theory | false | 739,000 |
Theorem 8.1. If $a$ and $n$ are relatively prime integers with $n>0$, then the positive integer $x$ is a solution of the congruence $a^{x} \equiv 1(\bmod n)$ if and only if $\operatorname{ord}_{n} a \mid x$ | Proof. If $\operatorname{ord}_{n} a \mid x$, then $x=k \cdot \operatorname{ord}_{n} a$ where $k$ is a positive integer. Hence,
$$a^{x}=a^{k \cdot \text { ord } a}=\left(a^{\text {ord } a}\right)^{k} \equiv 1(\bmod n)$$
Conversely, if $a^{x} \equiv 1(\bmod n)$, we first use the division algorithm to write
$$x=q \cdot \... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,005 |
Corollary 8.1. If $a$ and $n$ are relatively prime integers with $n>0$, then $\operatorname{ord}_{n} a \mid \phi(n)$ | Proof. Since $(a, n)=1$, Euler's theorem tells us that
$$a^{\phi(n)} \equiv 1(\bmod n)$$
Using Theorem 8.1, we conclude that $\operatorname{ord}_{n} a \mid \phi(n)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,006 |
Theorem 8.2. If $a$ and $n$ are relatively prime integers with $n>0$, then $a^{i} \equiv a^{j},(\bmod n)$ where $i$ and $j$ are nonnegative integers, if and only if $i \equiv j\left(\bmod \operatorname{ord}_{n} a\right)$ | Proof. Suppose that $i \equiv j\left(\bmod _{\operatorname{ord}_{n}} a\right)$, and $0 \leqslant j \leqslant i$. Then, we have $i=j+k \cdot \operatorname{ord}_{n} a$, where $k$ is a positive integer. Hence,
$$a^{i}=a^{j+k \cdot \text { ord } a}=a^{j}\left(a^{\text {ord } a}\right)^{k} \equiv a^{j}(\bmod n)$$
since $a^{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,007 |
Theorem 8.3. If $r$ and $n$ are relatively prime positive integers with $n>0$ and if $r$ is a primitive root modulo $n$, then the integers
$$r^{1}, r^{2}, \ldots, r^{\phi(n)}$$
form a reduced residue set modulo $n$. | Proof. To demonstrate that the first $\phi(n)$ powers of the primitive root $r$ form a reduced residue set modulo $n$, we only need to show that they are all relatively prime to $n$, and that no two are congruent modulo $n$.
Since $(r, n)=1$, it follows from problem 8 of Section 2.1 that $\left(r^{k}, n\right)=1$ for ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,009 |
Theorem 8.4. If $\operatorname{ord}_{m} a=t$ and if $u$ is a positive integer, then $\operatorname{ord}_{m}\left(a^{u}\right)=t /(t, u)$. | Proof. Let $s=\operatorname{ord}_{m}\left(a^{u}\right), \quad v=(t, u), \quad t=t_{1} v, \quad$ and $u=u_{1} v$. From Proposition 2.1 , we know that $\left(t_{1}, u_{1}\right)=1$.
Note that
$$\left(a^{u}\right)^{t_{1}}=\left(a^{u_{1} v}\right)^{(t / v)}=\left(a^{t}\right)^{u_{1}} \equiv 1(\bmod m)$$
since $\operatorna... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,010 |
Corollary 8.2. Let $r$ be a primitive root modulo $m$ where $m$ is an integer, $m>1$. Then $r^{u}$ is a primitive root modulo $m$ if and only if $(u, \phi(m))=1$. | Proof. From Theorem 8.4 , we know that
$$\begin{aligned}
\operatorname{ord}_{m} r^{u} & =\operatorname{ord}_{m} r /\left(u, \operatorname{ord}_{m} r\right) \\
& =\phi(m) /(u, \phi(m))
\end{aligned}$$
Consequently, $\operatorname{ord}_{m} r^{u}=\phi(m)$, and $r^{u}$ is a primitive root modulo $m$, if and only if $(u, \... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,011 |
Theorem 8.5. If the positive integer $m$ has a primitive root, then it has a total of $\phi(\phi(m))$ incongruent primitive roots. | Proof. Let $r$ be a primitive root modulo $m$. Then Theorem 8.3 tells us that the integers $r, r^{2}, \ldots, r^{\phi(m)}$ form a reduced residue system modulo $m$. From Corollary 8.2 , we know that $r^{u}$ is a primitive root modulo $m$ if and only if $(u, \phi(m))=1$. Since there are exactly $\phi(\phi(m))$ such inte... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,012 |
2. Find a primitive root modulo
a) 4
d) 13
b) 5
e) 14
c) 10
f) 18 . | 2. a) 3
b) 2,3
c) 3,7
d) $2,6,7,11$
e) 3,5
f) 5,11 | a) 3, b) 2,3, c) 3,7, d) 2,6,7,11, e) 3,5, f) 5,11 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,014 |
14. Let $p$ be a prime divisor of the Fermat number $F_{n}=2^{2^{\circ}}+1$.
a) Show that $\operatorname{ord}_{p} 2=2^{n+1}$.
b) From part (a), conclude that $2^{n+1} \mid(p-1)$, so that $p$ must be of the form $2^{n+1} k+1$. | None | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,027 |
3. Find $n$ ! for $n$ equal to each of the first ten positive integers. | 3. $1,2,6,24,120,720,5040,40320,362880,3628800$ | 3628800 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 739,030 |
Lagrange's Theorem.
Let $f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$ be a polynomial of degree $n$ with integer coefficients and with leading coefficient $a_{n}$ not divisible by $p$. Then $f(x)$ has at most $n$ incongruent roots modulo $p$. | Proof. To prove the theorem, we use mathematical induction. When $n=1$, we have $f(x)=a_{1} x+a_{0}$ with $p \backslash a_{1}$. A root of $f(x)$ modulo $p$ is a solution of the linear congruence $a_{1} x \equiv-a_{0}(\bmod p)$. By Theorem 3.7, since $\left(a_{1}, p\right)=1$, this linear congruence has exactly one solu... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,034 |
Theorem 8.6. Let $p$ be prime and let $d$ be a divisor of $p-1$. Then the polynomial $x^{d}-1$ has exactly $d$ incongruent roots modulo $p$. | Proof. Let $p-1=d e$. Then
$$\begin{aligned}
x^{p-1}-1 & =\left(x^{d}-1\right)\left(x^{d(e-1)}+x^{d(e-2)}+\cdots+x^{e}+1\right) \\
& =\left(x^{d}-1\right) g(x)
\end{aligned}$$
From Fermat's little theorem, we see that $x^{p-1}-1$ has $p-1$ incongruent roots modulo $p$. Furthermore, from Corollary 2.2, we know that any... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,035 |
Theorem 8.7. Let $p$ be a prime and let $d$ be a positive divisor of $p-1$. Then the number of incongruent integers of order $d$ modulo $p$ is equal to $\phi(d)$. | Proof. For each positive integer $d$ dividing $p-1$, let $F(d)$ denote the number of positive integers of order $d$ modulo $p$ that are less than $p$. Since the order modulo $p$ of an integer not divisible by $p$ divides $p-1$, it follows that
$$p-1=\sum_{d \mid p-1} F(d)$$
From Theorem 6.6, we know that
$$p-1=\sum_{d... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,036 |
1. Find the number of primitive roots of the following primes:
a) 7
d) 19
b) 13
e) 29
c) 17
f) 47 . | 1. a) 2
b) 4
c) 8
d) 6
e) 12
f) 22 | 2, 4, 8, 6, 12, 22 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,038 |
5. a) Use Lagrange's theorem to show that if $p$ is a prime and $f(x)$ is a polynomial of degree $n$ with integer coefficients and more than $n$ roots modulo $p$, then $p$ divides every coefficient of $f(x)$.
b) Let $p$ be prime. Using part (a), show that every coefficient of the polynomial $f(x)=(x-1)(x-2) \cdots(x-p+... | None | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,043 |
12. In this problem, we develop a threshold scheme for protection of master keys in a computer system, different than the scheme discussed in Section 7.6. Let $f(x)$ be a randomly chosen polynomial of degree $r-1$, with the condition that $K$, the master key, is the constant term of the polynomial. Let $p$ be a prime, ... | 12. c) $22,37,8,6,8,38,26$ | 22,37,8,6,8,38,26 | Algebra | proof | Yes | Yes | number_theory | false | 739,050 |
Theorem 8.8. If $p$ is an odd prime with primitive root $r$, then either $r$ or $r+p$ is a primitive root modulo $p^{2}$. | Proof. Since $r$ is a primitive root modulo $p$, we know that
$$\operatorname{ord}_{p} r=\phi(p)=p-1$$
Let $n=\operatorname{ord}_{p} r$, so that
$$r^{n} \equiv 1\left(\bmod p^{2}\right)$$
Since a congruence modulo $p^{2}$ obviously holds modulo $p$, we have
$$r^{n} \equiv 1(\bmod p)$$
From Theorem 8.1, it follows th... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,052 |
Theorem 1.3. Let $b$ be a positive integer with $b>1$. Then every positive integer $n$ can be written uniquely in the form
$$n=a_{k} b^{k}+a_{k-1} b^{k-1}+\cdots+a_{1} b+a_{0}$$
where $a_{j}$ is an integer with $0 \leqslant a_{j} \leqslant b-1$ for $j=0,1, \ldots, k$ and the initial coefficient $a_{k} \neq 0$ | Proof. We obtain an expression of the desired type by successively applying the division algorithm in the following way. We first divide $n$ by $b$ to obtain
$$n=b q_{0}+a_{0}, \quad 0 \leqslant a_{0} \leqslant b-1$$
Then we divide $q_{0}$ by $b$ to find that
$$q_{0}=b q_{1}+a_{1}, \quad 0 \leqslant a_{1} \leqslant b-... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,053 |
Theorem 8.9. Let $p$ be an odd prime, then $p^{k}$ has a primitive root for all positive integers $k$. Moreover, if $r$ is a primitive root modulo $p^{2}$, then $r$ is a primitive root modulo $p^{k}$, for all positive integers $k$. | Proof. From Theorem 8.8, we know that $p$ has a primitive root $r$ that is also a primitive root modulo $p^{2}$, so that
$$r^{p-1} \not \equiv 1\left(\bmod p^{2}\right)$$
Using mathematical induction, we will prove that for this primitive root r ,
$$r^{p^{k-2}(p-1)} \not \equiv 1\left(\bmod p^{k}\right)$$
for all posi... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,054 |
Theorem 8.10. If $a$ is an odd integer, and if $k$ is an integer, $k>3$, then $a^{\phi\left(2^{k}\right) / 2}=a^{2^{k-2}} \equiv 1\left(\bmod 2^{k}\right)$. | Proof. We prove this result using mathematical induction. If $a$ is an odd integer, then $a=2 b+1$, where $b$ is an integer. Hence,
$$a^{2}=(2 b+1)^{2}=4 b^{2}+4 b+1=4 b(b+1)+1$$
Since either $b$ or $b+1$ is even, we see that $8 \mid 4 b(b+1)$, so that
$$a^{2} \equiv 1(\bmod 8)$$
This is the congruence of interest wh... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,055 |
Theorem 8.11. Let $k \geqslant 3$ be an integer. Then $\operatorname{ord}_{2^{\star}} 5=\phi\left(2^{k}\right) / 2=2^{k-2}$ | Proof. Theorem 8.10 tells us that
$$5^{2^{k-1}} \equiv 1\left(\bmod 2^{k}\right)$$
for $k \geqslant 3$. From Theorem 8.1 , we see that $\operatorname{ord}_{2^{k}} 5 \mid 2^{k-2}$. Therefore, if we show that $\operatorname{ord}_{2^{\star}} 5 \backslash 2^{k-3}$, we can conclude that
$$\operatorname{ord}_{2} 5=2^{k-2}$$
... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,056 |
Theorem 8.12. If $n$ is a positive integer that is not a prime power or twice a prime power, then $n$ does not have a primitive root. | Proof. Let $n$ be a positive integer with prime-power factorization
$$n=p_{1}^{t_{1}} p_{2}^{t_{2}} \cdots p_{m}^{t_{m}}$$
Let us assume that the integer $n$ has a primitive root $r$. This means that $(r, n)=1$ and $\operatorname{ord}_{n} r=\phi(n)$. Since $(r, n)=1$, we know that $\left(r, p^{t}\right)=1$, whenever $... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,057 |
Theorem 8.13. If $p$ is an odd prime and $t$ is a positive integer, then $2 p^{t}$ possesses a primitive root. In fact, if $r$ is a primitive root modulo $p^{t}$, then if $r$ is odd it is also a primitive root modulo $2 p^{t}$, while if $r$ is even, $r+p^{t}$ is a primitive root modulo $2 p^{t}$. | Proof. If $r$ is a primitive root modulo $p^{t}$, then
$$r^{\phi\left(p^{t}\right)} \equiv 1\left(\bmod p^{t}\right)$$
and no positive exponent smaller than $\phi\left(p^{t}\right)$ has this property. From Theorem 6.4 , we note that $\phi\left(2 p^{t}\right)=\phi(2) \phi\left(p^{t}\right)=\phi\left(p^{t}\right)$, so th... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,058 |
Corollary 1.1. Every positive integer may be represented as the sum of distinct powers of two. | Proof. Let n be a positive integer. From Theorem 1.3 with $b=2$, we know that $n=a_{k} 2^{k}+a_{k-1} 2^{k-1}+\cdots+a_{1} 2+a_{0}$ where each $a_{j}$ is either 0 or 1. Hence, every positive integer is the sum of distinct powers of 2 . | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,064 |
Theorem 8.15. Let $m$ be a positive integer with primitive root $r$, and let $a$ and $b$ be integers relatively prime to $m$. Then
(i) ind $_{r} 1 \equiv 0(\bmod \phi(m))$.
(ii) $\operatorname{ind}_{r}(a b) \equiv \operatorname{ind}_{r} a+\operatorname{ind}_{r} b(\bmod \phi(m))$
(iii) ind $_{r} a^{k} \equiv k \cdot \op... | Proof of $(i)$. From Euler's theorem, we know that $r^{\phi(m)} \equiv 1(\bmod m)$ Since $r$ is a primitive root modulo $m$, no smaller positive power of $r$ is congruent to 1 modulo $m$. Hence, ind ${ }_{r} 1=\phi(m) \equiv 0(\bmod \phi(m))$.
Proof of (ii). To prove this congruence, note that from the definition of i... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,070 |
Theorem 8.16. Let $m$ be a positive integer with a primitive root. If $k$ is a positive integer and $a$ is an integer relatively prime to $m$, then the congruence $x^{k} \equiv a(\bmod m)$ has a solution if and only if
$$a^{\phi(m) / d} \equiv 1(\bmod m)$$
where $d=(k, \phi(m))$. Furthermore, if there are solutions of ... | Proof. Let $r$ be a primitive root modulo the positive integer $m$. We note that the congruence
$$x^{k} \equiv a(\bmod m)$$
holds if and only if
$$k \cdot \operatorname{ind}_{r} x \equiv \operatorname{ind}_{r} a(\bmod \phi(m))$$
Now let $d=(k, \phi(m))$ and $y=\operatorname{ind}_{r} x$, so that $x \equiv r^{y}(\bmod m... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,071 |
Lemma 8.1. Let $p$ be an odd prime and let $e$ and $q$ be positive integers. Then the number of incongruent solutions of the congruence $x^{q-1} \equiv 1\left(\bmod p^{e}\right)$ is $\left(q, p^{e-1}(p-1)\right)$ | Proof. Let $r$ be a primitive root of $p^{e}$. By taking indices with respect to $r$, we see that $x^{q} \equiv 1\left(\bmod p^{e}\right)$ if and only if $q y \equiv 0\left(\bmod \phi\left(p^{e}\right)\right)$ where $y=\operatorname{ind}_{r} x$. Using Theorem 3.7, we see that there are exactly $\left(q, \phi\left(p^{e}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,073 |
1. Write out a table of indices modulo 23 with respect to the primitive root 5 . | 1. ind $_{5} 1=22, \operatorname{ind}_{5} 2=2, \operatorname{ind}_{5} 3=16, \operatorname{ind}_{5} 4=4, \operatorname{ind}_{5} 5=1, \operatorname{ind}_{5} 6=18, \operatorname{ind}_{5} 7=19$,
ind $_{5} 8=6$, ind $_{5} 9=10$, ind $_{5} 10=3$, ind $_{5} 11=9$, ind $_{5} 12=20$, ind $_{5} 13=14$, ind $_{5} 14=21$,
$\operat... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,074 |
1. Convert $(1999)_{10}$ from decimal to base 7 notation. Convert $(6105)_{7}$ from base 7 to decimal notation. | 1. $(5554)_{7},(2112)_{10}$ | 5554_7, 2112_{10} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,075 |
3. Find all the solutions of the congruences
a) $3^{x} \equiv 2(\bmod 23)$
b) $13^{x} \equiv 5(\bmod 23)$ | 3. a) $x \equiv 7,18 \quad(\bmod 22)$
b) no solution | no solution | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,077 |
4. For which positive integers $a$ is the congruence $a x^{4} \equiv 2(\bmod 13)$ solvable? | 4. $a \equiv 2,5$, or $6(\bmod 13)$ | a \equiv 2,5,6(\bmod 13) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,078 |
5. For which positive integers $b$ is the congruence $8 x^{7} \equiv b(\bmod 29)$ solvable? | 5. $b \equiv 8,9,20$, or $21(\bmod 29)$ | b \equiv 8,9,20,21(\bmod 29) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,079 |
6. Find the solutions of $2^{x} \equiv x(\bmod 13)$, using indices to the base 2 modulo 13 . | 6. $x \equiv 10,16,57,59,90,99,115,134,144,145,149$, or $152(\bmod 156)$ | x \equiv 10,16,57,59,90,99,115,134,144,145,149,152(\bmod 156) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,080 |
2. Convert $(101001000)_{2}$ from binary to decimal notation and $(1984)_{10}$ from decimal to binary notation. | 2. $(328)_{10},(11111000000)_{2}$ | 328, 11111000000 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,086 |
12. Let $n=2^{t} \cdot p_{1}^{t} p_{2}^{t} \cdots p_{m}^{t}$ be the prime-power factorization of $n$. Let $a$ be an integer relatively prime to $n$. Let $r_{1}, r_{2}, \ldots, r_{m}$ be primitive roots of $p_{1}^{t_{1}}, p_{2}^{t_{2}}, \ldots, p_{m}^{t}$, respectively, and let $\gamma_{1}=\operatorname{ind}_{r_{1}} a\l... | 12. b) $(0,0,1,1),(0,0,1,4)$
d) $x \equiv 17(\bmod 60)$ | 17 | Number Theory | proof | Yes | Yes | number_theory | false | 739,087 |
16. a) Show that the probability that $n$ is a strong pseudoprime for a base $b$ randomly chosen with $1 \leqslant b \leqslant n-1$ is near $(n-1) / 4$ only when $n$ has a prime factorization of the form $n=p_{1} p_{2}$ where $p_{1}=1+2 q_{1}$ and $p_{2}=1+4 q_{2}$ with $q_{1}$ and $q_{2}$ prime or $n=p_{1} p_{2} p_{3}... | 16. b) $(49938.99876) /(4 \cdot 49939 \cdot 99877)=.24999249 \ldots$ | 0.24999249 | Number Theory | proof | Yes | Yes | number_theory | false | 739,091 |
Theorem 8.17. If $n$ is a positive integer and if an integer $x$ exists such that
$$x^{n-1} \equiv 1(\bmod n)$$
and
$$x^{(n-1) / q} \not \equiv 1(\bmod n)$$
for all prime divisors $q$ of $n-1$, then $n$ is prime. | Proof. Since $x^{n-1} \equiv 1(\bmod n)$, Theorem 8.1 tells us that $\operatorname{ord}_{n} x \mid(n-1)$ We will show that $\operatorname{ord}_{n} x=n-1$. Suppose that $\operatorname{ord}_{n} x \neq n-1$. Since $\operatorname{ord}_{n} x \mid(n-1)$, there is an integer $k$ with $n-1=k \cdot \operatorname{ord}_{n} x$ and... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,092 |
Corollary 8.4. If $n$ is an odd positive integer and if $x$ is a positive integer such that
$$x^{(n-1) / 2} \equiv-1(\bmod n)$$
and
$$x^{(n-1) / q} \not \equiv 1(\bmod n)$$
for all odd prime divisors $q$ of $n-1$, then $n$ is prime. | Proof. Since $x^{(n-1) / 2} \equiv-1(\bmod n)$, we see that
$$x^{n-1}=\left(x^{(n-1) / 2}\right)^{2} \equiv(-1)^{2} \equiv 1(\bmod n)$$
Since the hypotheses of Theorem 8.17 are met, we know that $n$ is prime. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,093 |
3. Convert $(100011110101)_{2}$ and $(11101001110)_{2}$ from binary to hexadecimal. | 3. $(8 F 5)_{16},(74 E)_{16}$ | (8F5)_{16},(74E)_{16} | Other | math-word-problem | Yes | Yes | number_theory | false | 739,097 |
Theorem 8.20. Let $n$ be a positive integer with prime-power factorization
$$n=2^{t_{0}} p_{1}^{t_{1}} p_{2}^{t_{2}} \cdots p_{m}^{t_{m}}$$
Then $\lambda(n)$, the minimal universal exponent of $n$, is given by
$$\lambda(n)=\left[\lambda\left(2^{t_{0}}\right), \phi\left(p_{1}^{t_{1}}\right), \ldots, \phi\left(p_{m}^{t_... | Proof. Let $a$ be an integer with $(a, n)=1$. For convenience, let
$$M=\left[\lambda\left(2^{t_{0}}\right), \phi\left(p_{1}^{t_{1}}\right), \phi\left(p_{2}^{t_{2}}\right), \ldots, \phi\left(p_{m}^{t_{m}}\right)\right]$$
Since $M$ is divisible by all of the integers $\lambda\left(2^{t_{0}}\right), \phi\left(p_{1}^{t_{1... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,102 |
Theorem 8.21. If $n>2$ is a Carmichael number, then $n=q_{1} q_{2} \cdots q_{k}$, where the $q_{j}$ 's are distinct primes such that $\left(q_{j}-1\right) \mid(n-1)$ for $j=1,2, \ldots, k$ | Proof. If $n$ is a Carmichael number, then
$$b^{n-1} \equiv 1(\bmod n)$$
for all positive integers $b$ with $(b, n)=1$. Theorem 8.20 tells us that there is an integer $a$ with $\operatorname{ord}_{n} a=\lambda(n)$, where $\lambda(n)$ is the minimal universal exponent, and since $a^{n-1} \equiv 1(\bmod n)$, Theorem 8.1 ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,103 |
Theorem 8.22. A Carmichael number must have at least three different odd prime factors. | Proof. Let $n$ be a Carmichael number. Then $n$ cannot have just one prime factor, since it is composite, and is the product of distinct primes. So assume that $n=p q$, where $p$ and $q$ are odd primes with $p>q$. Then
$$n-1=p q-1=(p-1) q+(q-1) \equiv q-1 \not \equiv 0(\bmod p-1)$$
which shows that $(p-1) \backslash(n-... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,104 |
1. Find $\lambda(n)$, the minimal universal exponent of $n$, for the following values of $n$
a) 100
e) $2^{4} \cdot 3^{3} \cdot 5^{2} \cdot 7$
b) 144
f) $2^{5} \cdot 3^{2} \cdot 5^{2} \cdot 7^{3} \cdot 11^{2} \cdot 13 \cdot 17 \cdot 19$
c) 222
g) 10 !
d) 884
h) 20 !. | a) 20
b) 12
c) 36
d) 48
e) 180
f) 388080
g) 8640
h) 125411328000 | 48 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,105 |
2. Find all positive integers $n$ such that $\lambda(n)$ is equal to
a) 1
d) 4
b) 2
e) 5
c) 3
f) 6 . | 2. a) 1,2
b) $3,4,6,8,12,24$
c) no solution
d) $5,10,15,16,20,30,40,48,60$
$80,120,240$
e) no solution
f) $7,9,14,18,21,28,36,42,56,63,72,84,126$.
$168,252,504$ | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,106 |
1. Find the sequence of two-digit pseudo-random numbers generated using the middle-square method, taking 69 as the seed. | 1. $69,76,77,92,46,11,12,14,19,36,29,84,5,25,62,84,5,25,62, \ldots$ | 62 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,126 |
2. Find the first ten terms of the sequence of pseudo-random numbers generated by the linear congruential method with $x_{0}=6$ and $x_{n+1} \equiv 5 x_{n}+2(\bmod 19)$. What is the period length of this generator? | 2. $6.13,10,14,15,1,7,18,16,6,13, \ldots$ period length is 9 | 9 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,127 |
6. a) Show that if $b$ is a negative integer less than -1 , then every integer $n$ can be uniquely written in the form
$$n=a_{k} b^{k}+a_{k-1} b^{k-1}+\cdots+a_{1} b+a_{0}$$
where $\quad a_{k} \neq 0 \quad$ and $0 \leqslant a_{j}<|b| \quad$ for $j=0,1,2, \ldots, k$. We write $n=\left(a_{k} a_{k-1} \ldots a_{1} a_{0}\ri... | 6. b) $-39,26$ c) $(1001)_{-2},(110011)_{-2},(1001101)_{-2}$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,130 |
9. Another way to generate pseudo-random numbers is to use the Fibonacci generator. Let $m$ be a positive integer. Two initial integers $x_{0}$ and $x_{1}$ less than $m$ are specified and the rest of the sequence is generated recursively by the congruence $x_{n+1} \equiv x_{n}+x_{n-1}(\bmod m), \quad 0 \leqslant x_{n+1... | 9. $1,24,25,18,12,30,11,10$ | 1,24,25,18,12,30,11,10 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,135 |
Theorem 8.26. If $m$ is a positive integer, $m>2$, with a primitive root, then the maximal $\pm 1-$ exponent $\lambda_{0}(m)$ equals $\phi(m) / 2=\lambda(m) / 2$ | Proof. We first note that if $m$ has a primitive root, then $\lambda(m)=\phi(m)$. From problem 5 of Section 6.1 , we know that $\phi(m)$ is even, so that $\phi(m) / 2$ is an integer, if $m>2$. Euler's Theorem tells us that
$$a^{\phi(m)}=\left(a^{\phi(m) / 2}\right)^{2} \equiv 1(\bmod m)$$
for all integers $a$ with $(a,... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,140 |
Lemma 9.1. Let $p$ be an odd prime and $a$ an integer not divisible by $p$. Then, the congruence
$$x^{2} \equiv a(\bmod p)$$
has either no solutions or exactly two incongruent solutions modulo $p$. | Proof. If $x^{2} \equiv a(\bmod p)$ has a solution, say $x=x_{0}$, then we can easily demonstrate that $x=-x_{0}$ is a second incongruent solution. Since $\left(-x_{0}\right)^{2}=x_{0}^{2} \equiv a(\bmod p)$, we see that $-x_{0}$ is a solution. We note that $x_{0} \not \equiv-x_{0}(\bmod p)$, for if $x_{0} \equiv-x_{0}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,149 |
Theorem 9.1. If $p$ is an odd prime, then there are exactly $(p-1) / 2$ quadratic residues of $p$ and $(p-1) / 2$ quadratic nonresidues of $p$ among the integers $1,2, \ldots, p-1$ | Proof. To find all the quadratic residues of $p$ among the integers $1,2, \ldots, p-1$ we compute the least positive residues modulo $p$ of the squares of the integers $1,2, \ldots, p-1$. Since there are $p-1$ squares to consider and since each congruence $x^{2} \equiv a(\bmod p)$ has either zero or two solutions, ther... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,150 |
Theorem 9.2. Let $p$ be an odd prime and $a$ and $b$ integers not divisible by $p$. Then
(i) if $a \equiv b(\bmod p)$, then $\left(\frac{a}{p}\right)=\left(\frac{b}{p}\right)$.
(ii) $\left(\frac{a}{p}\right)\left(\frac{b}{p}\right)=\left(\frac{a b}{p}\right)$.
(iii) $\left(\frac{a^{2}}{p}\right)=1$. | Proof of $(i)$. If $a \equiv b(\bmod p)$, then $x^{2} \equiv a(\bmod p)$ has a solution if and only if $x^{2} \equiv b(\bmod p)$ has a solution. Hence, $\left(\frac{a}{p}\right)=\left(\frac{b}{p}\right)$
Proof of (ii). By Euler's criterion, we know that
$$\left(\frac{a}{p}\right) \equiv a^{(p-1) / 2}(\bmod p),\left(\fr... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,151 |
Theorem 9.3. If $p$ is an odd prime, then
$$\left(\frac{-1}{p}\right)=\left\{\begin{aligned}
1 & \text { if } p \equiv 1(\bmod 4) \\
-1 & \text { if } p \equiv-1(\bmod 4)
\end{aligned}\right.$$ | Proof. By Euler's criterion, we know that
$$\left(\frac{-1}{p}\right) \equiv(-1)^{(p-1) / 2}(\bmod p)$$
If $p \equiv 1(\bmod 4)$, then $p=4 k+1$ for some integer $k$. Thus,
$$(-1)^{(p-1) / 2}=(-1)^{2 k}=1$$
so that $\left(\frac{-1}{p}\right)=1$. If $p \equiv 3(\bmod 4)$, then $p=4 k+3$ for some integer $k$. Thus,
$$(-... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,152 |
Gauss' Lemma. Let $p$ be an odd prime and $a$ an integer with $(a, p)=1$. If $s$ is the number of least positive residues modulo $p$ of the integers $a, 2 a, 3 a, \ldots,((p-1) / 2) a$ that are greater than $p / 2$, then the Legendre symbol $\left(\frac{a}{p}\right)=(-1)^{s}$ | Proof. Let $u_{1}, u_{2}, \ldots, u_{s}$ represent the least positive residues of the integers $a, 2 a, 3 a, \ldots,((p-1) / 2) a$ that are greater than $p / 2$, and let $v_{1}, v_{2}, \ldots, v_{t}$ be the least positive residues of these integers that are less than $p / 2$. Since $(j a, p)=1$ for all $j$ with $1 \leq... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,154 |
Theorem 9.4. If $p$ is an odd prime, then
$$\left(\frac{2}{p}\right)=(-1)^{\left(p^{2}-1\right) / 8}$$
Hence, 2 is a quadratic residue of all primes $p \equiv \pm 1(\bmod 8)$ and a quadratic nonresidue of all primes $p \equiv \pm 3(\bmod 8)$. | Proof. From Gauss' lemma, we know that if $s$ is the number of least positive residues of the integers
$$1 \cdot 2,2 \cdot 2,3 \cdot 2, \ldots,\left(\frac{p-1}{2}\right) \cdot 2$$
that are greater than $p / 2$, then $\left(\frac{2}{p}\right)=(-1)^{s}$. Since all these integers are less than $p$, we only need to count t... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,155 |
1. Find all the quadratic residues of
a) 3
c) 13
b) 5
d) 19 . | 1. a) 1
b) 1,4
c) $1,3,4,9,10,12$
d) $1,4,5,6,7,9,11,16,17$ | 1,3,4,9,10,12 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,156 |
2. Find the value of the Legendre symbols $\left(\frac{j}{7}\right)$, for $j=1,2,3,4,5$, and 6 . | 2. $1,1,-1,1,-1,-1$ | 1,1,-1,1,-1,-1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,157 |
15. Find all solutions of the congruence $x^{2} \equiv 1(\bmod 15)$. | 15. $x \equiv 1,4,11,14(\bmod 15)$ | x \equiv 1,4,11,14(\bmod 15) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,171 |
Lemma 9.2. If $p$ is an odd prime and $a$ is an odd integer not divisible by $p$, then
$$\left(\frac{a}{p}\right)=(-1)^{T(a, p)}$$
where
$$T(a, p)=\sum_{j=1}^{(p-1) / 2}[j a / p]$$ | Proof. Consider the least positive residues of the integers $a, 2 a, \ldots,((p-1) / 2) a$; let $u_{1}, u_{2}, \ldots, u_{s}$ be those greater than $p / 2$ and let $v_{1}, v_{2}, \ldots, v_{t}$ be those less than $p / 2$. The division algorithm tells us that
$$j a=p[j a / p]+\text { remainder }$$
where the remainder is... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,199 |
4. Find a congruence describing all primes for which 5 is a quadratic residue. | 4. $p \equiv \pm 1(\bmod 5)$ | p \equiv \pm 1(\bmod 5) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,204 |
5. Find a congruence describing all primes for which 7 is a quadratic residue. | 5. $p \equiv \pm 1, \pm 3, \pm 9(\bmod 28)$ | p \equiv \pm 1, \pm 3, \pm 9(\bmod 28) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,205 |
Theorem 9.5. Let $n$ be an odd positive integer and let $a$ and $b$ be integers relatively prime to $n$. Then
(i) if $a \equiv b(\bmod n)$, then $\left(\frac{a}{n}\right)=\left(\frac{b}{n}\right)$,
(ii) $\left(\frac{a b}{n}\right)=\left(\frac{a}{n}\right)\left(\frac{b}{n}\right)$,
(iii) $\left(\frac{-1}{n}\right)=(-1)^... | Proof. In the proof of all four parts of this theorem we use the prime factorization $n=p_{1}^{t_{1}} p_{2}^{t_{2}} \cdots p_{m}^{t_{m}}$
Proof of $(i)$. We know that if $p$ is a prime dividing $n$, then $a \equiv b(\bmod p)$. Hence, from Theorem 9.2 (i), we have $\left(\frac{a}{b}\right)=\left(\frac{b}{p}\right)$. Co... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,211 |
Theorem 9.6. Let $n$ and $m$ be relatively prime odd positive integers. Then
$$\left(\frac{n}{m}\right)\left(\frac{m}{n}\right)=(-1)^{\frac{m-1}{2} \frac{n-1}{2}}$$ | Proof. Let the prime factorizations of $m$ and $n$ be $m=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{s}^{a}$ and $n=q_{1}^{b_{1}} q_{2}^{b_{2}} \cdots q_{r}^{b}$. We see that
$$\left(\frac{m}{n}\right)=\prod_{i=1}^{r}\left(\frac{m}{q_{i}}\right)^{b_{i}}=\prod_{i=1}^{r} \prod_{j=1}^{s}\left(\frac{p_{j}}{q_{j}}\right)^{b_{i} a... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,212 |
2. For which positive integers $n$ that are relatively prime to 15 does the Jacobi symbol $\left(\frac{15}{n}\right)$ equal 1 ? | 2. $n \equiv 1,7,11,17,43,49,53$, or $59(\bmod 60)$ | n \equiv 1,7,11,17,43,49,53,59(\bmod 60) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,216 |
14. A Cantor expansion of a positive integer $n$ is a sum
$$n=a_{m} m!+a_{m-1}(m-1)!+\cdots+a_{2} 2!+a_{1} 1!$$
where each $a_{j}$ is an integer with $0 \leqslant a_{j} \leqslant j$.
a) Find Cantor expansions of 14,56 , and 384 .
b) Show that every positive integer has a unique Cantor expansion. | 14. а) $14=2 \cdot 3!+1 \cdot 2!, 56=2 \cdot 4!+1 \cdot 3!+1 \cdot 2!, 384=3 \cdot 5!+1 \cdot 4$ ! | 14=2 \cdot 3!+1 \cdot 2!, 56=2 \cdot 4!+1 \cdot 3!+1 \cdot 2!, 384=3 \cdot 5!+1 \cdot 4! | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,219 |
Proposition 9.1. If $n$ is an Euler pseudoprime to the base $b$, then $n$ is a pseudoprime to the base $b$. | Proof. If $n$ is an Euler pseudoprime to the base $b$, then
$$b^{(n-1) / 2} \equiv\left(\frac{b}{n}\right)(\bmod n)$$
Hence, by squaring both sides of this congruence, we find that
$$\left(b^{(n-1) / 2}\right)^{2} \equiv\left(\frac{b}{n}\right)^{2}(\bmod n)$$
Since $\left(\frac{b}{n}\right)= \pm 1$, we see that $b^{n... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,232 |
Theorem 9.8. If $n$ is a strong pseudoprime to the base $b$, then $n$ is an Euler pseudoprime to this base . | Proof. Let $n$ be a strong pseudoprime to the base $b$. Then if $n-1=2^{s} t$, where $t$ is odd, either $b^{t} \equiv 1(\bmod n)$ or $b^{2^{\prime} t} \equiv-1(\bmod n)$ where $0 \leqslant r \leqslant s-1$. Let $n=\prod_{i=1} p_{i}^{a_{i}}$ be the prime-power factorization of $n$.
First, consider the case where $b^{t} ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,233 |
Theorem 9.9. If $n \equiv 3(\bmod 4)$ and $n$ is an Euler pseudoprime to the base $b$, then $n$ is a strong pseudoprime to the base $b$. | Proof. From the congruence $n \equiv 3(\bmod 4)$, we know that $n-1=2^{2} \cdot t$ where $t=(n-1) / 2$ is odd. Since $n$ is an Euler pseudoprime to the base $b$, it follows that
$$b^{t}=b^{(n-1) / 2} \equiv\left(\frac{b}{n}\right)(\bmod n)$$
Since $\quad\left(\frac{b}{n}\right)= \pm 1, \quad$ we know that either $\qua... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,234 |
Theorem 9.10. If $n$ is an Euler pseudoprime to the base $b$ and $\left(\frac{b}{n}\right)=-1$, then $n$ is a strong pseudoprime to the base $b$. | Proof. We write $n-1=2^{s} t$, where $t$ is odd and $s$ is a positive integer. Since $n$ is an Euler pseudoprime to the base $b$, we have
$$b^{2^{2-1} t}=b^{(n-1) / 2} \equiv\left(\frac{b}{n}\right)(\bmod n)$$
But since $\left(\frac{b}{n}\right)=-1$, we see that
$$b^{t 2^{i-1}} \equiv-1(\bmod n)$$
This is one of the ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,235 |
Lemma 9.3. If $n$ is an odd positive integer that is not a perfect square, then there is at least one integer $b$ with $1<b<n,(b, n)=1$, and $\left(\frac{b}{n}\right)=-1$, where $\left(\frac{b}{n}\right)$ is the Jacobi symbol. | Proof. If $n$ is prime, the existence of such an integer $b$ is guaranteed by Theorem 9.1. If $n$ is composite, since $n$ is not a perfect square, we can write $n=r s$ where $(r, s)=1$ and $r=p^{e}$, with $p$ an odd prime and $e$ an odd positive integer.
Now let $t$ be a quadratic nonresidue of the prime $p$; such a $... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,236 |
Lemma 9.4. Let $n$ be an odd composite integer. Then there is at least one integer $b$ with $1<b<n,(b, n)=1$, and
$$b^{(n-1) / 2} \not \equiv\left(\frac{b}{n}\right)(\bmod n)$$ | Proof. Assume that for all positive integers not exceeding $n$ and relatively prime to $n$, that
$$b^{(n-1) / 2} \equiv\left(\frac{b}{n}\right)(\bmod n)$$
Squaring both sides of this congruence tells us that
$$b^{n-1} \equiv\left(\frac{b}{n}\right)^{2} \equiv( \pm 1)^{2}=1(\bmod n)$$
if $(b, n)=1$. Hence, $n$ must be ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,237 |
Theorem 9.11. Let $n$ be an odd composite integer. Then, the number of positive integers less then $n$, relatively prime to $n$, that are bases to which $n$ is an Euler pseudoprime, is less than $\phi(n) / 2$. | Proof. From Lemma 9.4 , we know that there is an integer $b$ with $1<b<n,(b, n)=1$, and
$$b^{(n-1) / 2} \not \equiv\left(\frac{b}{n}\right)(\bmod n)$$
Now, let $a_{1}, a_{2}, \ldots, a_{m}$ denote the positive integers less than $n$ satisfying $1 \leqslant a_{j} \leqslant n,\left(a_{j}, n\right)=1$, and
$$a_{j}^{(n-1)... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,238 |
Theorem 10.1. Let $\alpha$ and $\beta$ be rational numbers. Then $\alpha+\beta, \alpha-\beta, \alpha \beta$, and $\alpha / \beta$ (when $\beta \neq 0$ ) are rational. | Proof. Since $\alpha$ and $\beta$ are rational, it follows that $\alpha=a / b$ and $\beta=c / d$, where $a, b, c$, and $d$ are integers with $b \neq 0$ and $d \neq 0$. Then, each of the numbers
$$\begin{aligned}
\alpha+\beta & =a / b+c / d=(a d+b c) / b d \\
\alpha-\beta & =a / b-c / d=(a d-b c) / b d \\
\alpha \beta &... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,248 |
Proposition 10.1. The number $\sqrt{2}$ is irrational. | Proof. Suppose that $\sqrt{2}=a / b$, where $a$ and $b$ are relatively prime integers with $b \neq 0$. Then, we have
$$2=a^{2} / b^{2}$$
so that
$$2 b^{2}=a^{2}$$
Since $2 \mid a^{2}$, problem 31 of Section 2.3 tells us that $2 \mid a$. Let $a=2 c$, so that
$$b^{2}=2 c^{2}$$
Hence, $2 \mid b^{2}$, and by problem 31 o... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,249 |
Theorem 10.2. Let $\alpha$ be a root of the polynomial $x^{n}+c_{n-1} x^{n-1}+$ $\cdots+c_{1} x+c_{0}$ where the coefficients $c_{0}, c_{1}, \ldots, c_{n-1}$, are integers with $c_{0} \neq 0$. Then $\alpha$ is either an integer or an irrational number. | Proof. Suppose that $\alpha$ is rational. Then we can write $\alpha=a / b$ where $a$ and $b$ are relatively prime integers with $b \neq 0$. Since $\alpha$ is a root of $x^{n}+c_{n-1} x^{n-1}+\cdots+c_{1} x+c_{0}$, we have
$$(a / b)^{n}+c_{n-1}(a / b)^{n-1}+\cdots+c_{1}(a / b)+c_{0}=0$$
Multiplying by $b^{n}$, we find ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,250 |
Theorem 10.5. The real number $\alpha, 0 \leqslant \alpha<1$, has a terminating base $b$ expansion if and only if $\alpha$ is rational and $\alpha=r / s$, where $0 \leqslant r<s$ and every prime factor of $s$ also divides $b$. | Proof. First, suppose that $\alpha$ has a terminating base $b$ expansion,
$$\alpha=\left(. c_{1} c_{2} \ldots c_{n}\right)_{b}$$
Then
$$\begin{aligned}
\alpha & =\frac{c_{1}}{b}+\frac{c_{2}}{b^{2}}+\cdots+\frac{c_{n}}{b^{n}} \\
& =\frac{c_{1} b^{n-1}+c_{2} b^{n-2}+\cdots+c_{n}}{b^{n}}
\end{aligned}$$
so that $\alpha$ ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,255 |
7. Find the base 8 expansions of the following numbers
a) $1 / 3$
d) $1 / 6$
b) $1 / 4$
e) $1 / 12$
c) $1 / 5$
f) $1 / 22$. | 7. a) $(\overline{.25})_{8}$ b) $(.2)_{8}$ c) $(\overline{1463})_{8}$ d) $(.1 \overline{25})_{8}$ e) $(.05 \overline{52})_{8}$ f) $(.0 \overline{2721350564})_{8}$ | a) (\overline{.25})_{8} \quad b) (.2)_{8} \quad c) (\overline{1463})_{8} \quad d) (.1 \overline{25})_{8} \quad e) (.05 \overline{52})_{8} \quad f) (.0 \overline{2721350564})_{8} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,263 |
Proposition 1.6. If $f$ is $O(g)$ and $c$ is a positive constant, then $c f$ is $O(g)$. | Proof. If $f$ is $O(g)$, then there is a constant $K$ such that $f(x)<K g(x)$ for all $x$ under consideration. Hence $c f(x)<(c K) g(x)$. Therefore, $c f$ is $O(g)$. | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,264 |
8. Find the fraction, in lowest terms, represented by the following expansions
a) .12
b) $.1 \overline{2}$
c) $\overline{12}$. | 8. a) $\frac{3}{25}$
b) $\frac{11}{90}$
c) $\frac{4}{33}$ | \frac{11}{90} | Other | math-word-problem | Yes | Yes | number_theory | false | 739,265 |
9. Find the fraction, in lowest terms, represented by the following expansions
a) $(.123)_{7}$
c) $(. \overline{17})_{11}$
b) $(.0 \overline{13})_{6}$
d) $(. \overline{A B C})_{16}$. | 9. a) $\frac{66}{343}$
b) $\frac{3}{70}$
c) $\frac{3}{20}$
d) $\frac{916}{1365}$ | \frac{3}{20} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,266 |
12. Find the pre-period and period lengths of the base 12 expansions of the following rational numbers
a) $1 / 4$
d) $5 / 24$
b) $1 / 8$
c) $7 / 10$
e) $17 / 132$
f) $7 / 360$. | 12. a) 1,0
b) 2,0
c) 1,4
d) 2,1
e) 1,1
f) 2,4 | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,269 |
Proposition 1.7. If $f_{1}$ is $O\left(g_{1}\right)$ and $f_{2}$ is $O\left(g_{2}\right)$, then $f_{1}+f_{2}$ is $O\left(g_{1}+g_{2}\right)$ and $f_{1} f_{2}$ is $O\left(g_{1} g_{2}\right)$. | Proof. If $f$ is $O\left(g_{1}\right)$ and $f_{2}$ is $O\left(g_{2}\right)$, then there are constants $K_{1}$ and $K_{2}$ such that $f_{1}(x)<K_{1} g_{1}(x)$ and $f_{2}(x)<K_{2} g_{2}(x)$ for all $x$ under consideration. Hence
$$\begin{aligned}
f_{1}(x)+f_{2}(x) & \leqslant K_{1} g_{1}(x)+K_{2} g_{2}(x) \\
& \leqslant ... | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,275 |
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