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luku21.tex
163
luku21.tex
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@ -9,9 +9,9 @@ because many questions involving integers
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are very difficult to solve even if they
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seem simple at first glance.
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As an example, let's consider the following equation:
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As an example, let us consider the following equation:
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\[x^3 + y^3 + z^3 = 33\]
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It's easy to find three real numbers $x$, $y$ and $z$
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It is easy to find three real numbers $x$, $y$ and $z$
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that satisfy the equation.
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For example, we can choose
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\[
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@ -28,9 +28,8 @@ is an open problem in number theory.
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In this chapter, we will focus on basic concepts
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and algorithms in number theory.
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We will start by discussing divisibility of numbers
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and important algorithms for primality testing
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and factorization.
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Throughout the chapter, we will assume that all numbers
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are integers, if not otherwise stated.
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\section{Primes and factors}
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@ -38,9 +37,8 @@ and factorization.
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\index{factor}
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\index{divisor}
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A number $a$ is a \key{factor} or \key{divisor} of a number $b$
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if $b$ is divisible by $a$.
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if $a$ divides $b$.
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If $a$ is a factor of $b$,
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we write $a \mid b$, and otherwise we write $a \nmid b$.
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For example, the factors of the number 24 are
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@ -52,8 +50,8 @@ For example, the factors of the number 24 are
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A number $n>1$ is a \key{prime}
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if its only positive factors are 1 and $n$.
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For example, the numbers 7, 19 and 41 are primes.
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The number 35 is not a prime because it can be
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divided into factors $5 \cdot 7 = 35$.
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The number 35 is not a prime, because it can be
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divided into the factors $5 \cdot 7 = 35$.
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For each number $n>1$, there is a unique
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\key{prime factorization}
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\[ n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k},\]
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@ -75,7 +73,7 @@ The factors are
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The \key{sum of factors} of $n$ is
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\[\sigma(n)=\prod_{i=1}^k (1+p_i+\ldots+p_i^{\alpha_i}) = \prod_{i=1}^k \frac{p_i^{a_i+1}-1}{p_i-1},\]
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where the latter form is based on the geometric sum formula.
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where the latter formula is based on the geometric sum formula.
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For example, the sum of factors of the number 84 is
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\[\sigma(84)=\frac{2^3-1}{2-1} \cdot \frac{3^2-1}{3-1} \cdot \frac{7^2-1}{7-1} = 7 \cdot 4 \cdot 8 = 224.\]
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@ -91,23 +89,23 @@ and the product of the factors is $\mu(84)=84^6=351298031616$.
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\index{perfect number}
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A number $n$ is \key{perfect} if $n=\sigma(n)-n$,
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i.e., the number equals the sum of its divisors
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between $1 \ldots n-1$.
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For example, the number 28 is perfect because
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it equals the sum $1+2+4+7+14$.
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i.e., $n$ equals the sum of its divisors
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between $1$ and $n-1$.
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For example, the number 28 is perfect,
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because $28=1+2+4+7+14$.
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\subsubsection{Number of primes}
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It is easy to show that there is an infinite number
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of primes.
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If the number would be finite,
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If the number of primes would be finite,
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we could construct a set $P=\{p_1,p_2,\ldots,p_n\}$
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that contains all the primes.
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that would contain all the primes.
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For example, $p_1=2$, $p_2=3$, $p_3=5$, and so on.
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However, using this set, we could form a new prime
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However, using $P$, we could form a new prime
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\[p_1 p_2 \cdots p_n+1\]
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that is larger than all elements in $P$.
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This is a contradiction, and the number of the primes
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This is a contradiction, and the number of primes
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has to be infinite.
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\subsubsection{Density of primes}
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@ -115,14 +113,14 @@ has to be infinite.
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The density of primes means how often there are primes
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among the numbers.
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Let $\pi(n)$ denote the number of primes between
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$1 \ldots n$. For example, $\pi(10)=4$ because
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there are 4 primes between $1 \ldots 10$: 2, 3, 5 and 7.
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$1$ and $n$. For example, $\pi(10)=4$, because
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there are 4 primes between $1$ and $10$: 2, 3, 5 and 7.
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It's possible to show that
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It is possible to show that
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\[\pi(n) \approx \frac{n}{\ln n},\]
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which means that primes appear quite often.
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which means that primes are quite frequent.
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For example, the number of primes between
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$1 \ldots 10^6$ is $\pi(10^6)=78498$,
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$1$ and $10^6$ is $\pi(10^6)=78498$,
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and $10^6 / \ln 10^6 \approx 72382$.
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\subsubsection{Conjectures}
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@ -138,7 +136,7 @@ For example, the following conjectures are famous:
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Each even integer $n>2$ can be represented as a
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sum $n=a+b$ so that both $a$ and $b$ are primes.
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\index{twin prime}
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\item \key{twin prime}:
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\item \key{Twin prime conjecture}:
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There is an infinite number of pairs
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of the form $\{p,p+2\}$,
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where both $p$ and $p+2$ are primes.
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@ -153,15 +151,15 @@ $n^2$ and $(n+1)^2$, where $n$ is any positive integer.
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If a number $n$ is not prime,
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it can be represented as a product $a \cdot b$,
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where $a \le \sqrt n$ or $b \le \sqrt n$,
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so it certainly has a factor between $2 \ldots \sqrt n$.
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so it certainly has a factor between $2$ and $\lfloor \sqrt n \rfloor$.
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Using this observation, we can both test
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if a number is prime and find the prime factorization
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of a number in $O(\sqrt n)$ time.
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The following function \texttt{prime} checks
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if the given number $n$ is prime.
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The function tries to divide the number by
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all numbers between $2 \ldots \sqrt n$,
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The function attempts to divide $n$ by
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all numbers between $2$ and $\lfloor \sqrt n \rfloor$,
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and if none of them divides $n$, then $n$ is prime.
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\begin{lstlisting}
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@ -181,7 +179,7 @@ factorization of $n$.
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The function divides $n$ by its prime factors,
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and adds them to the vector.
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The process ends when the remaining number $n$
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has no factors between $2 \ldots \sqrt n$.
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has no factors between $2$ and $\lfloor \sqrt n \rfloor$.
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If $n>1$, it is prime and the last factor.
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\begin{lstlisting}
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@ -210,24 +208,24 @@ so the result of the function is $[2,2,2,3]$.
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The \key{sieve of Eratosthenes} is a preprocessing
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algorithm that builds an array using which we
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can efficiently check if a given number between $2 \ldots n$
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is prime and find one prime factor of the number.
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is prime and, if it is not, find one prime factor of the number.
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The algorithm builds an array $\texttt{a}$
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where indices $2,3,\ldots,n$ are used.
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whose positions $2,3,\ldots,n$ are used.
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The value $\texttt{a}[k]=0$ means
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that $k$ is prime,
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and the value $\texttt{a}[k] \neq 0$
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means that $k$ is not a prime but one
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means that $k$ is not a prime and one
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of its prime factors is $\texttt{a}[k]$.
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The algorithm iterates through the numbers
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$2 \ldots n$ one by one.
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Always when a new prime $x$ is found,
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the algorithm records that the multiples
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of $x$ ($2x,3x,4x,\ldots$) are not primes
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of $x$ ($2x,3x,4x,\ldots$) are not primes,
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because the number $x$ divides them.
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For example, if $n=20$, the array becomes:
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For example, if $n=20$, the array is as follows:
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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@ -301,12 +299,12 @@ of the algorithm is the harmonic sum
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\[\sum_{x=2}^n n/x = n/2 + n/3 + n/4 + \cdots + n/n = O(n \log n).\]
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In fact, the algorithm is even more efficient
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In fact, the algorithm is even more efficient,
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because the inner loop will be executed only if
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the number $x$ is prime.
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It can be shown that the time complexity of the
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algorithm is only $O(n \log \log n)$
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that is very near to $O(n)$.
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algorithm is only $O(n \log \log n)$,
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a complexity very near to $O(n)$.
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\subsubsection{Euclid's algorithm}
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@ -331,7 +329,7 @@ are connected as follows:
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\key{Euclid's algorithm} provides an efficient way
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to find the greatest common divisor of two numbers.
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The algorithm is based on the formula
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The algorithm is based on the following formula:
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\begin{equation*}
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\textrm{gcd}(a,b) = \begin{cases}
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a & b = 0\\
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\[\textrm{gcd}(24,36) = \textrm{gcd}(36,24)
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= \textrm{gcd}(24,12) = \textrm{gcd}(12,0)=12.\]
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The time complexity of Euclid's algorithm
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is $O(\log n)$ where $n=\min(a,b)$.
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The worst case is when
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$a$ and $b$ are successive Fibonacci numbers.
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In this case, the algorithm goes through
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all smaller Fibonacci numbers.
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is $O(\log n)$, where $n=\min(a,b)$.
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The worst case for the algorithm is
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the case when $a$ and $b$ are consecutive Fibonacci numbers.
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For example,
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\[\textrm{gcd}(13,8)=\textrm{gcd}(8,5)
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=\textrm{gcd}(5,3)=\textrm{gcd}(3,2)=\textrm{gcd}(2,1)=\textrm{gcd}(1,0)=1.\]
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@ -356,18 +352,18 @@ For example,
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\index{coprime}
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\index{Euler's totient function}
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Numbers $a$ and $b$ are coprime
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Numbers $a$ and $b$ are \key{coprime}
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if $\textrm{gcd}(a,b)=1$.
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\key{Euler's totient function} $\varphi(n)$
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returns the number of coprime numbers to $n$
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between $1 \ldots n$.
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between $1$ and $n$.
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For example, $\varphi(12)=4$,
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because the numbers 1, 5, 7 and 11
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are coprime to the number 12.
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because the 1, 5, 7 and 11
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are coprime to 12.
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The value of $\varphi(n)$ can be calculated
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using the prime factorization of $n$
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by the formula
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from the prime factorization of $n$
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using the formula
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\[ \varphi(n) = \prod_{i=1}^k p_i^{\alpha_i-1}(p_i-1). \]
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For example, $\varphi(12)=2^1 \cdot (2-1) \cdot 3^0 \cdot (3-1)=4$.
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Note that $\varphi(n)=n-1$ if $n$ is prime.
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@ -377,8 +373,8 @@ Note that $\varphi(n)=n-1$ if $n$ is prime.
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\index{modular arithmetic}
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In \key{modular arithmetic},
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the set of available numbers is restricted so
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that only numbers $0,1,2,\ldots,m-1$ can be used
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the set of available numbers is limited so
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that only numbers $0,1,2,\ldots,m-1$ may be used,
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where $m$ is a constant.
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Each number $x$ is
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represented by the number $x \bmod m$:
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@ -386,23 +382,22 @@ the remainder after dividing $x$ by $m$.
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For example, if $m=17$, then $75$
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is represented by $75 \bmod 17 = 7$.
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Often we can take the remainder before doing a
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calculation.
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Often we can take the remainder before doing
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calculations.
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In particular, the following formulas can be used:
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\[
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\begin{array}{rcl}
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(x+y) \bmod m & = & (x \bmod m + y \bmod m) \bmod m \\
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(x-y) \bmod m & = & (x \bmod m - y \bmod m) \bmod m \\
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(x \cdot y) \bmod m & = & (x \bmod m \cdot y \bmod m) \bmod m \\
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(x^k) \bmod m & = & (x \bmod m)^k \bmod m \\
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x^n \bmod m & = & (x \bmod m)^n \bmod m \\
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\end{array}
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\]
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\subsubsection{Modular exponentiation}
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Often there is need to efficiently calculate
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the remainder of $x^n$.
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There is often need to efficiently calculate
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the value of $x^n \bmod m$.
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This can be done in $O(\log n)$ time
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using the following recursion:
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\begin{equation*}
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@ -413,13 +408,13 @@ using the following recursion:
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\end{cases}
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\end{equation*}
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It's important that in the case of an even $n$,
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the number $x^{n/2}$ is calculated only once.
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It is important that in the case of an even $n$,
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the value of $x^{n/2}$ is calculated only once.
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This guarantees that the time complexity of the
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algorithm is $O(\log n)$ because $n$ is always halved
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algorithm is $O(\log n)$, because $n$ is always halved
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when it is even.
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The following function calculates the number
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The following function calculates the value of
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$x^n \bmod m$:
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\begin{lstlisting}
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\index{Euler's theorem}
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\key{Fermat's theorem} states that
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\[x^{m-1} \bmod m = 1,\]
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\[x^{m-1} \bmod m = 1\]
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when $m$ is prime and $x$ and $m$ are coprime.
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This also yields
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\[x^k \bmod m = x^{k \bmod (m-1)} \bmod m.\]
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More generally, \key{Euler's theorem} states that
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\[x^{\varphi(m)} \bmod m = 1,\]
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\[x^{\varphi(m)} \bmod m = 1\]
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when $x$ and $m$ are coprime.
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Fermat's theorem follows from Euler's theorem,
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because if $m$ is a prime, then $\varphi(m)=m-1$.
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we can use the formula $2 \cdot 3 \bmod 17$,
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because $36 \bmod 17 = 2$ and $6^{-1} \bmod 17 = 3$.
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However, a modular inverse doesn't always exist.
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However, a modular inverse does not always exist.
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For example, if $x=2$ and $m=4$, the equation
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\[ x x^{-1} \bmod m = 1. \]
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can't be solved, because all multiples of the number 2
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are even, and the remainder can never be 1 when $m=4$.
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It turns out that the number $x^{-1} \bmod m$ exists
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exactly when $x$ and $m$ are coprime.
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\[ x x^{-1} \bmod m = 1 \]
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cannot be solved, because all multiples of the number 2
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are even and the remainder can never be 1 when $m=4$.
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It turns out that the value of $x^{-1} \bmod m$
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can be calculated exactly when $x$ and $m$ are coprime.
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If a modular inverse exists, it can be
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calculated using the formula
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\subsubsection{Computer arithmetic}
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In a computers, unsigned integers are represented modulo $2^k$
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where $k$ is the number of bits.
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In programming, unsigned integers are represented modulo $2^k$,
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where $k$ is the number of bits of the data type.
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A usual consequence of this is that a number wraps around
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if it becomes too large.
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For example, in C++, numbers of type \texttt{unsigned int}
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are represented modulo $2^{32}$.
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The following code defines an \texttt{unsigned int}
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The following code declares an \texttt{unsigned int}
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variable whose value is $123456789$.
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After this, the value will be multiplied by itself,
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and the result is
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@ -525,7 +520,7 @@ cout << x*x << "\n"; // 2537071545
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A \key{Diophantine equation} is of the form
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\[ ax + by = c, \]
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where $a$, $b$ and $c$ are constants,
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and our tasks is to solve variables $x$ and $y$.
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and we should find the values of $x$ and $y$.
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Each number in the equation has to be an integer.
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For example, one solution for the equation
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$5x+2y=11$ is $x=3$ and $y=-2$.
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@ -538,25 +533,25 @@ It turns out that we can extend Euclid's algorithm
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so that it will find numbers $x$ and $y$
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that satisfy the following equation:
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\[
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ax + by = \textrm{syt}(a,b)
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ax + by = \textrm{gcd}(a,b)
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\]
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A Diophantine equation can be solved if
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$c$ is divisible by
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$\textrm{gcd}(a,b)$,
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and otherwise it can't be solved.
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and otherwise the equation cannot be solved.
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\index{extended Euclid's algorithm}
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\subsubsection*{Extended Euclid's algorithm}
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As an example, let's find numbers $x$ and $y$
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As an example, let us find numbers $x$ and $y$
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that satisfy the following equation:
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\[
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39x + 15y = 12
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\]
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The equation can be solved, because
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$\textrm{syt}(39,15)=3$ and $3 \mid 12$.
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$\textrm{gcd}(39,15)=3$ and $3 \mid 12$.
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When Euclid's algorithm calculates the
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greatest common divisor of 39 and 15,
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it produces the following sequence of function calls:
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@ -580,15 +575,15 @@ and by multiplying this by 4, the result is
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\[
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39 \cdot 8 + 15 \cdot (-20) = 12,
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\]
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so a solution for the original equation is
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so a solution for the equation is
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$x=8$ and $y=-20$.
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A solution for a Diophantine equation is not unique,
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but we can form an infinite number of solutions
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if we know one solution.
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If the pair $(x,y)$ is a solution, then also the pair
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If a pair $(x,y)$ is a solution, then also all pairs
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\[(x+\frac{kb}{\textrm{gcd}(a,b)},y-\frac{ka}{\textrm{gcd}(a,b)})\]
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is a solution where $k$ is any integer.
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are solutions, where $k$ is any integer.
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\subsubsection{Chinese remainder theorem}
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|
@ -656,7 +651,7 @@ as the sum $8^2+5^2+5^2+3^2$.
|
|||
\key{Zeckendorf's theorem} states that every
|
||||
positive integer has a unique representation
|
||||
as a sum of Fibonacci numbers such that
|
||||
no two numbers are the same of successive
|
||||
no two numbers are equal or consecutive
|
||||
Fibonacci numbers.
|
||||
For example, the number 74 can be represented
|
||||
as the sum $55+13+5+1$.
|
||||
|
@ -685,7 +680,7 @@ all primitive Pythagorean triples.
|
|||
Each such triple is of the form
|
||||
\[(n^2-m^2,2nm,n^2+m^2),\]
|
||||
where $0<m<n$, $n$ and $m$ are coprime
|
||||
and at least one of the numbers $n$ and $m$ is even.
|
||||
and at least one of $n$ and $m$ is even.
|
||||
For example, when $m=1$ and $n=2$, the formula
|
||||
produces the smallest Pythagorean triple
|
||||
\[(2^2-1^2,2\cdot2\cdot1,2^2+1^2)=(3,4,5).\]
|
||||
|
@ -703,8 +698,8 @@ and the number 12 is not prime, because
|
|||
\[11! \bmod 12 = 0 \neq 11.\]
|
||||
|
||||
Hence, Wilson's theorem tells us whether a number
|
||||
is prime. However, in practice, the formula can't be
|
||||
used for large values of $n$, because it's difficult
|
||||
to calculate the number $(n-1)!$ if $n$ is large.
|
||||
is prime. However, in practice, the theorem cannot be
|
||||
applied to large values of $n$, because it is difficult
|
||||
to calculate the value of $(n-1)!$ when $n$ is large.
|
||||
|
||||
|
||||
|
|
Loading…
Reference in New Issue