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Antti H S Laaksonen 2017-02-08 23:43:26 +02:00
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@ -9,9 +9,9 @@ because many questions involving integers
are very difficult to solve even if they
seem simple at first glance.
As an example, let's consider the following equation:
As an example, let us consider the following equation:
\[x^3 + y^3 + z^3 = 33\]
It's easy to find three real numbers $x$, $y$ and $z$
It is easy to find three real numbers $x$, $y$ and $z$
that satisfy the equation.
For example, we can choose
\[
@ -28,9 +28,8 @@ is an open problem in number theory.
In this chapter, we will focus on basic concepts
and algorithms in number theory.
We will start by discussing divisibility of numbers
and important algorithms for primality testing
and factorization.
Throughout the chapter, we will assume that all numbers
are integers, if not otherwise stated.
\section{Primes and factors}
@ -38,9 +37,8 @@ and factorization.
\index{factor}
\index{divisor}
A number $a$ is a \key{factor} or \key{divisor} of a number $b$
if $b$ is divisible by $a$.
if $a$ divides $b$.
If $a$ is a factor of $b$,
we write $a \mid b$, and otherwise we write $a \nmid b$.
For example, the factors of the number 24 are
@ -52,8 +50,8 @@ For example, the factors of the number 24 are
A number $n>1$ is a \key{prime}
if its only positive factors are 1 and $n$.
For example, the numbers 7, 19 and 41 are primes.
The number 35 is not a prime because it can be
divided into factors $5 \cdot 7 = 35$.
The number 35 is not a prime, because it can be
divided into the factors $5 \cdot 7 = 35$.
For each number $n>1$, there is a unique
\key{prime factorization}
\[ n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k},\]
@ -75,7 +73,7 @@ The factors are
The \key{sum of factors} of $n$ is
\[\sigma(n)=\prod_{i=1}^k (1+p_i+\ldots+p_i^{\alpha_i}) = \prod_{i=1}^k \frac{p_i^{a_i+1}-1}{p_i-1},\]
where the latter form is based on the geometric sum formula.
where the latter formula is based on the geometric sum formula.
For example, the sum of factors of the number 84 is
\[\sigma(84)=\frac{2^3-1}{2-1} \cdot \frac{3^2-1}{3-1} \cdot \frac{7^2-1}{7-1} = 7 \cdot 4 \cdot 8 = 224.\]
@ -91,23 +89,23 @@ and the product of the factors is $\mu(84)=84^6=351298031616$.
\index{perfect number}
A number $n$ is \key{perfect} if $n=\sigma(n)-n$,
i.e., the number equals the sum of its divisors
between $1 \ldots n-1$.
For example, the number 28 is perfect because
it equals the sum $1+2+4+7+14$.
i.e., $n$ equals the sum of its divisors
between $1$ and $n-1$.
For example, the number 28 is perfect,
because $28=1+2+4+7+14$.
\subsubsection{Number of primes}
It is easy to show that there is an infinite number
of primes.
If the number would be finite,
If the number of primes would be finite,
we could construct a set $P=\{p_1,p_2,\ldots,p_n\}$
that contains all the primes.
that would contain all the primes.
For example, $p_1=2$, $p_2=3$, $p_3=5$, and so on.
However, using this set, we could form a new prime
However, using $P$, we could form a new prime
\[p_1 p_2 \cdots p_n+1\]
that is larger than all elements in $P$.
This is a contradiction, and the number of the primes
This is a contradiction, and the number of primes
has to be infinite.
\subsubsection{Density of primes}
@ -115,14 +113,14 @@ has to be infinite.
The density of primes means how often there are primes
among the numbers.
Let $\pi(n)$ denote the number of primes between
$1 \ldots n$. For example, $\pi(10)=4$ because
there are 4 primes between $1 \ldots 10$: 2, 3, 5 and 7.
$1$ and $n$. For example, $\pi(10)=4$, because
there are 4 primes between $1$ and $10$: 2, 3, 5 and 7.
It's possible to show that
It is possible to show that
\[\pi(n) \approx \frac{n}{\ln n},\]
which means that primes appear quite often.
which means that primes are quite frequent.
For example, the number of primes between
$1 \ldots 10^6$ is $\pi(10^6)=78498$,
$1$ and $10^6$ is $\pi(10^6)=78498$,
and $10^6 / \ln 10^6 \approx 72382$.
\subsubsection{Conjectures}
@ -138,7 +136,7 @@ For example, the following conjectures are famous:
Each even integer $n>2$ can be represented as a
sum $n=a+b$ so that both $a$ and $b$ are primes.
\index{twin prime}
\item \key{twin prime}:
\item \key{Twin prime conjecture}:
There is an infinite number of pairs
of the form $\{p,p+2\}$,
where both $p$ and $p+2$ are primes.
@ -153,15 +151,15 @@ $n^2$ and $(n+1)^2$, where $n$ is any positive integer.
If a number $n$ is not prime,
it can be represented as a product $a \cdot b$,
where $a \le \sqrt n$ or $b \le \sqrt n$,
so it certainly has a factor between $2 \ldots \sqrt n$.
so it certainly has a factor between $2$ and $\lfloor \sqrt n \rfloor$.
Using this observation, we can both test
if a number is prime and find the prime factorization
of a number in $O(\sqrt n)$ time.
The following function \texttt{prime} checks
if the given number $n$ is prime.
The function tries to divide the number by
all numbers between $2 \ldots \sqrt n$,
The function attempts to divide $n$ by
all numbers between $2$ and $\lfloor \sqrt n \rfloor$,
and if none of them divides $n$, then $n$ is prime.
\begin{lstlisting}
@ -181,7 +179,7 @@ factorization of $n$.
The function divides $n$ by its prime factors,
and adds them to the vector.
The process ends when the remaining number $n$
has no factors between $2 \ldots \sqrt n$.
has no factors between $2$ and $\lfloor \sqrt n \rfloor$.
If $n>1$, it is prime and the last factor.
\begin{lstlisting}
@ -210,24 +208,24 @@ so the result of the function is $[2,2,2,3]$.
The \key{sieve of Eratosthenes} is a preprocessing
algorithm that builds an array using which we
can efficiently check if a given number between $2 \ldots n$
is prime and find one prime factor of the number.
is prime and, if it is not, find one prime factor of the number.
The algorithm builds an array $\texttt{a}$
where indices $2,3,\ldots,n$ are used.
whose positions $2,3,\ldots,n$ are used.
The value $\texttt{a}[k]=0$ means
that $k$ is prime,
and the value $\texttt{a}[k] \neq 0$
means that $k$ is not a prime but one
means that $k$ is not a prime and one
of its prime factors is $\texttt{a}[k]$.
The algorithm iterates through the numbers
$2 \ldots n$ one by one.
Always when a new prime $x$ is found,
the algorithm records that the multiples
of $x$ ($2x,3x,4x,\ldots$) are not primes
of $x$ ($2x,3x,4x,\ldots$) are not primes,
because the number $x$ divides them.
For example, if $n=20$, the array becomes:
For example, if $n=20$, the array is as follows:
\begin{center}
\begin{tikzpicture}[scale=0.7]
@ -301,12 +299,12 @@ of the algorithm is the harmonic sum
\[\sum_{x=2}^n n/x = n/2 + n/3 + n/4 + \cdots + n/n = O(n \log n).\]
In fact, the algorithm is even more efficient
In fact, the algorithm is even more efficient,
because the inner loop will be executed only if
the number $x$ is prime.
It can be shown that the time complexity of the
algorithm is only $O(n \log \log n)$
that is very near to $O(n)$.
algorithm is only $O(n \log \log n)$,
a complexity very near to $O(n)$.
\subsubsection{Euclid's algorithm}
@ -331,7 +329,7 @@ are connected as follows:
\key{Euclid's algorithm} provides an efficient way
to find the greatest common divisor of two numbers.
The algorithm is based on the formula
The algorithm is based on the following formula:
\begin{equation*}
\textrm{gcd}(a,b) = \begin{cases}
a & b = 0\\
@ -342,11 +340,9 @@ For example,
\[\textrm{gcd}(24,36) = \textrm{gcd}(36,24)
= \textrm{gcd}(24,12) = \textrm{gcd}(12,0)=12.\]
The time complexity of Euclid's algorithm
is $O(\log n)$ where $n=\min(a,b)$.
The worst case is when
$a$ and $b$ are successive Fibonacci numbers.
In this case, the algorithm goes through
all smaller Fibonacci numbers.
is $O(\log n)$, where $n=\min(a,b)$.
The worst case for the algorithm is
the case when $a$ and $b$ are consecutive Fibonacci numbers.
For example,
\[\textrm{gcd}(13,8)=\textrm{gcd}(8,5)
=\textrm{gcd}(5,3)=\textrm{gcd}(3,2)=\textrm{gcd}(2,1)=\textrm{gcd}(1,0)=1.\]
@ -356,18 +352,18 @@ For example,
\index{coprime}
\index{Euler's totient function}
Numbers $a$ and $b$ are coprime
Numbers $a$ and $b$ are \key{coprime}
if $\textrm{gcd}(a,b)=1$.
\key{Euler's totient function} $\varphi(n)$
returns the number of coprime numbers to $n$
between $1 \ldots n$.
between $1$ and $n$.
For example, $\varphi(12)=4$,
because the numbers 1, 5, 7 and 11
are coprime to the number 12.
because the 1, 5, 7 and 11
are coprime to 12.
The value of $\varphi(n)$ can be calculated
using the prime factorization of $n$
by the formula
from the prime factorization of $n$
using the formula
\[ \varphi(n) = \prod_{i=1}^k p_i^{\alpha_i-1}(p_i-1). \]
For example, $\varphi(12)=2^1 \cdot (2-1) \cdot 3^0 \cdot (3-1)=4$.
Note that $\varphi(n)=n-1$ if $n$ is prime.
@ -377,8 +373,8 @@ Note that $\varphi(n)=n-1$ if $n$ is prime.
\index{modular arithmetic}
In \key{modular arithmetic},
the set of available numbers is restricted so
that only numbers $0,1,2,\ldots,m-1$ can be used
the set of available numbers is limited so
that only numbers $0,1,2,\ldots,m-1$ may be used,
where $m$ is a constant.
Each number $x$ is
represented by the number $x \bmod m$:
@ -386,23 +382,22 @@ the remainder after dividing $x$ by $m$.
For example, if $m=17$, then $75$
is represented by $75 \bmod 17 = 7$.
Often we can take the remainder before doing a
calculation.
Often we can take the remainder before doing
calculations.
In particular, the following formulas can be used:
\[
\begin{array}{rcl}
(x+y) \bmod m & = & (x \bmod m + y \bmod m) \bmod m \\
(x-y) \bmod m & = & (x \bmod m - y \bmod m) \bmod m \\
(x \cdot y) \bmod m & = & (x \bmod m \cdot y \bmod m) \bmod m \\
(x^k) \bmod m & = & (x \bmod m)^k \bmod m \\
x^n \bmod m & = & (x \bmod m)^n \bmod m \\
\end{array}
\]
\subsubsection{Modular exponentiation}
Often there is need to efficiently calculate
the remainder of $x^n$.
There is often need to efficiently calculate
the value of $x^n \bmod m$.
This can be done in $O(\log n)$ time
using the following recursion:
\begin{equation*}
@ -413,13 +408,13 @@ using the following recursion:
\end{cases}
\end{equation*}
It's important that in the case of an even $n$,
the number $x^{n/2}$ is calculated only once.
It is important that in the case of an even $n$,
the value of $x^{n/2}$ is calculated only once.
This guarantees that the time complexity of the
algorithm is $O(\log n)$ because $n$ is always halved
algorithm is $O(\log n)$, because $n$ is always halved
when it is even.
The following function calculates the number
The following function calculates the value of
$x^n \bmod m$:
\begin{lstlisting}
@ -438,12 +433,12 @@ int modpow(int x, int n, int m) {
\index{Euler's theorem}
\key{Fermat's theorem} states that
\[x^{m-1} \bmod m = 1,\]
\[x^{m-1} \bmod m = 1\]
when $m$ is prime and $x$ and $m$ are coprime.
This also yields
\[x^k \bmod m = x^{k \bmod (m-1)} \bmod m.\]
More generally, \key{Euler's theorem} states that
\[x^{\varphi(m)} \bmod m = 1,\]
\[x^{\varphi(m)} \bmod m = 1\]
when $x$ and $m$ are coprime.
Fermat's theorem follows from Euler's theorem,
because if $m$ is a prime, then $\varphi(m)=m-1$.
@ -465,13 +460,13 @@ For example, to evaluate the value of $36/6 \bmod 17$,
we can use the formula $2 \cdot 3 \bmod 17$,
because $36 \bmod 17 = 2$ and $6^{-1} \bmod 17 = 3$.
However, a modular inverse doesn't always exist.
However, a modular inverse does not always exist.
For example, if $x=2$ and $m=4$, the equation
\[ x x^{-1} \bmod m = 1. \]
can't be solved, because all multiples of the number 2
are even, and the remainder can never be 1 when $m=4$.
It turns out that the number $x^{-1} \bmod m$ exists
exactly when $x$ and $m$ are coprime.
\[ x x^{-1} \bmod m = 1 \]
cannot be solved, because all multiples of the number 2
are even and the remainder can never be 1 when $m=4$.
It turns out that the value of $x^{-1} \bmod m$
can be calculated exactly when $x$ and $m$ are coprime.
If a modular inverse exists, it can be
calculated using the formula
@ -500,14 +495,14 @@ so the numbers $x^{-1}$ and $x^{\varphi(m)-1}$ are equal.
\subsubsection{Computer arithmetic}
In a computers, unsigned integers are represented modulo $2^k$
where $k$ is the number of bits.
In programming, unsigned integers are represented modulo $2^k$,
where $k$ is the number of bits of the data type.
A usual consequence of this is that a number wraps around
if it becomes too large.
For example, in C++, numbers of type \texttt{unsigned int}
are represented modulo $2^{32}$.
The following code defines an \texttt{unsigned int}
The following code declares an \texttt{unsigned int}
variable whose value is $123456789$.
After this, the value will be multiplied by itself,
and the result is
@ -525,7 +520,7 @@ cout << x*x << "\n"; // 2537071545
A \key{Diophantine equation} is of the form
\[ ax + by = c, \]
where $a$, $b$ and $c$ are constants,
and our tasks is to solve variables $x$ and $y$.
and we should find the values of $x$ and $y$.
Each number in the equation has to be an integer.
For example, one solution for the equation
$5x+2y=11$ is $x=3$ and $y=-2$.
@ -538,25 +533,25 @@ It turns out that we can extend Euclid's algorithm
so that it will find numbers $x$ and $y$
that satisfy the following equation:
\[
ax + by = \textrm{syt}(a,b)
ax + by = \textrm{gcd}(a,b)
\]
A Diophantine equation can be solved if
$c$ is divisible by
$\textrm{gcd}(a,b)$,
and otherwise it can't be solved.
and otherwise the equation cannot be solved.
\index{extended Euclid's algorithm}
\subsubsection*{Extended Euclid's algorithm}
As an example, let's find numbers $x$ and $y$
As an example, let us find numbers $x$ and $y$
that satisfy the following equation:
\[
39x + 15y = 12
\]
The equation can be solved, because
$\textrm{syt}(39,15)=3$ and $3 \mid 12$.
$\textrm{gcd}(39,15)=3$ and $3 \mid 12$.
When Euclid's algorithm calculates the
greatest common divisor of 39 and 15,
it produces the following sequence of function calls:
@ -580,15 +575,15 @@ and by multiplying this by 4, the result is
\[
39 \cdot 8 + 15 \cdot (-20) = 12,
\]
so a solution for the original equation is
so a solution for the equation is
$x=8$ and $y=-20$.
A solution for a Diophantine equation is not unique,
but we can form an infinite number of solutions
if we know one solution.
If the pair $(x,y)$ is a solution, then also the pair
If a pair $(x,y)$ is a solution, then also all pairs
\[(x+\frac{kb}{\textrm{gcd}(a,b)},y-\frac{ka}{\textrm{gcd}(a,b)})\]
is a solution where $k$ is any integer.
are solutions, where $k$ is any integer.
\subsubsection{Chinese remainder theorem}
@ -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.