713 lines
20 KiB
TeX
713 lines
20 KiB
TeX
\chapter{Number theory}
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\index{number theory}
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\key{Number theory} is a branch of mathematics
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that studies integers.
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Number theory is a fascinating field,
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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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\[x^3 + y^3 + z^3 = 33\]
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It's easy to find three real numbers $x$, $y$ and $z$
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that satisfy the equation.
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For example, we can choose
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\[
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\begin{array}{lcl}
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x = 3, \\
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y = \sqrt[3]{3}, \\
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z = \sqrt[3]{3}.\\
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\end{array}
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\]
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However, nobody knows if there are any three
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\emph{integers} $x$, $y$ and $z$
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that would satisfy the equation, but this
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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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\section{Primes and factors}
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\index{divisibility}
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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 $a$ 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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1, 2, 3, 4, 6, 8, 12 and 24.
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\index{prime}
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\index{prime decomposition}
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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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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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where $p_1,p_2,\ldots,p_k$ are primes and
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$\alpha_1,\alpha_2,\ldots,\alpha_k$ are positive numbers.
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For example, the prime factorization for the number 84 is
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\[84 = 2^2 \cdot 3^1 \cdot 7^1.\]
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The \key{number of factors} of a number $n$ is
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\[\tau(n)=\prod_{i=1}^k (\alpha_i+1),\]
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because for each prime $p_i$, there are
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$\alpha_i+1$ ways to choose how many times
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it appears in the factor.
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For example, the number of factors
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of the number 84 is
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$\tau(84)=3 \cdot 2 \cdot 2 = 12$.
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The factors are
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1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84.
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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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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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The \key{product of factors} of $n$ is
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\[\mu(n)=n^{\tau(n)/2},\]
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because we can form $\tau(n)/2$ pairs from the factors,
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each with product $n$.
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For example, the factors of the number 84
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produce the pairs
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$1 \cdot 84$, $2 \cdot 42$, $3 \cdot 28$, etc.,
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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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\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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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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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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\[p_1 p_2 \cdots p_n+1\]
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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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has to be infinite.
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\subsubsection{Density of primes}
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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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It's 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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For example, the number of primes between
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$1 \ldots 10^6$ 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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There are many \emph{conjectures} involving primes.
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Most people think that the conjectures are true,
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but nobody has been able to prove them.
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For example, the following conjectures are famous:
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\begin{itemize}
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\index{Goldbach's conjecture}
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\item \key{Goldbach's conjecture}:
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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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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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\index{Legendre's conjecture}
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\item \key{Legendre's conjecture}:
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There is always a prime between numbers
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$n^2$ and $(n+1)^2$, where $n$ is any positive integer.
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\end{itemize}
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\subsubsection{Basic algorithms}
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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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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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and if none of them divides $n$, then $n$ is prime.
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\begin{lstlisting}
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bool prime(int n) {
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if (n < 2) return false;
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for (int x = 2; x*x <= n; x++) {
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if (n%x == 0) return false;
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}
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return true;
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}
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\end{lstlisting}
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\noindent
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The following function \texttt{factors}
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constructs a vector that contains the prime
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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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If $n>1$, it is prime and the last factor.
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\begin{lstlisting}
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vector<int> factors(int n) {
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vector<int> f;
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for (int x = 2; x*x <= n; x++) {
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while (n%x == 0) {
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f.push_back(x);
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n /= x;
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}
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}
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if (n > 1) f.push_back(n);
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return f;
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}
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\end{lstlisting}
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Note that each prime factor appears in the vector
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as many times as it divides the number.
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For example, $24=2^3 \cdot 3$,
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so the result of the function is $[2,2,2,3]$.
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\subsubsection{Sieve of Eratosthenes}
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\index{sieve of Eratosthenes}
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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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The algorithm builds an array $\texttt{a}$
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where indices $2,3,\ldots,n$ 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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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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because the number $x$ divides them.
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For example, if $n=20$, the array becomes:
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\draw (0,0) grid (19,1);
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\node at (0.5,0.5) {$0$};
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\node at (1.5,0.5) {$0$};
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\node at (2.5,0.5) {$2$};
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\node at (3.5,0.5) {$0$};
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\node at (4.5,0.5) {$3$};
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\node at (5.5,0.5) {$0$};
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\node at (6.5,0.5) {$2$};
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\node at (7.5,0.5) {$3$};
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\node at (8.5,0.5) {$5$};
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\node at (9.5,0.5) {$0$};
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\node at (10.5,0.5) {$3$};
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\node at (11.5,0.5) {$0$};
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\node at (12.5,0.5) {$7$};
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\node at (13.5,0.5) {$5$};
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\node at (14.5,0.5) {$2$};
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\node at (15.5,0.5) {$0$};
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\node at (16.5,0.5) {$3$};
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\node at (17.5,0.5) {$0$};
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\node at (18.5,0.5) {$5$};
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\footnotesize
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\node at (0.5,1.5) {$2$};
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\node at (1.5,1.5) {$3$};
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\node at (2.5,1.5) {$4$};
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\node at (3.5,1.5) {$5$};
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\node at (4.5,1.5) {$6$};
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\node at (5.5,1.5) {$7$};
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\node at (6.5,1.5) {$8$};
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\node at (7.5,1.5) {$9$};
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\node at (8.5,1.5) {$10$};
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\node at (9.5,1.5) {$11$};
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\node at (10.5,1.5) {$12$};
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\node at (11.5,1.5) {$13$};
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\node at (12.5,1.5) {$14$};
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\node at (13.5,1.5) {$15$};
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\node at (14.5,1.5) {$16$};
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\node at (15.5,1.5) {$17$};
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\node at (16.5,1.5) {$18$};
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\node at (17.5,1.5) {$19$};
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\node at (18.5,1.5) {$20$};
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\end{tikzpicture}
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\end{center}
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The following code implements the sieve of
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Eratosthenes.
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The code assumes that each element in
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\texttt{a} is initially zero.
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\begin{lstlisting}
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for (int x = 2; x <= n; x++) {
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if (a[x]) continue;
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for (int u = 2*x; u <= n; u += x) {
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a[u] = x;
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}
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}
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\end{lstlisting}
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The inner loop of the algorithm will be executed
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$n/x$ times for any $x$.
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Thus, an upper bound for the running time
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of the algorithm is the harmonic sum
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\index{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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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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\subsubsection{Euclid's algorithm}
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\index{greatest common divisor}
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\index{least common multiple}
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\index{Euclid's algorithm}
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The \key{greatest common divisor} of
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numbers $a$ and $b$, $\gcd(a,b)$,
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is the greatest number that divides both $a$ and $b$,
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and the \key{least common multiple} of
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$a$ and $b$, $\textrm{lcm}(a,b)$,
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is the smallest number that is divisible by
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both $a$ and $b$.
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For example,
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$\gcd(24,36)=12$ and
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$\textrm{lcm}(24,36)=72$.
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The greatest common divisor and the least common multiple
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are connected as follows:
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\[\textrm{lcm}(a,b)=\frac{ab}{\textrm{gcd}(a,b)}\]
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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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\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}(b,a \bmod b) & b \neq 0\\
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\end{cases}
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\end{equation*}
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For example,
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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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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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\subsubsection{Euler's totient function}
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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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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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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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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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\[ \varphi(n) = \prod_{i=1}^k p_i^{\alpha_i-1}(p_i-1). \]
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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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\section{Modular arithmetic}
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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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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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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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In particular, the following formulas can be used:
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\[
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\begin{array}{rcl}
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(x+y) \bmod m & = & (x \bmod m + y \bmod m) \bmod m \\
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(x-y) \bmod m & = & (x \bmod m - y \bmod m) \bmod m \\
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(x \cdot y) \bmod m & = & (x \bmod m \cdot y \bmod m) \bmod m \\
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(x^k) \bmod m & = & (x \bmod m)^k \bmod m \\
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\end{array}
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\]
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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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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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x^n = \begin{cases}
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1 & n = 0\\
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x^{n/2} \cdot x^{n/2} & \text{$n$ is even}\\
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x^{n-1} \cdot x & \text{$n$ is odd}
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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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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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when it is even.
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The following function calculates the number
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$x^n \bmod m$:
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\begin{lstlisting}
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int modpow(int x, int n, int m) {
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if (n == 0) return 1%m;
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int u = modpow(x,n/2,m);
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u = (u*u)%m;
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if (n%2 == 1) u = (u*x)%m;
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return u;
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}
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\end{lstlisting}
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\subsubsection{Fermat's theorem and Euler's theorem}
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\index{Fermat's theorem}
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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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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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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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\subsubsection{Modular inverse}
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\index{modular inverse}
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The modular inverse of $x$ modulo $m$
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is a number $x^{-1}$ such that
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\[ x x^{-1} \bmod m = 1. \]
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For example, if $x=6$ and $m=17$,
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then $x^{-1}=3$, because $6\cdot3 \bmod 17=1$.
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Using modular inverses, we can do divisions
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for remainders, because division by $x$
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corresponds to multiplication by $x^{-1}$.
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For example, to evaluate the value of $36/6 \bmod 17$,
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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.
|
|
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.
|
|
It turns out that the number $x^{-1} \bmod m$ exists
|
|
exactly when $x$ and $m$ are coprime.
|
|
|
|
If a modular inverse exists, it can be
|
|
calculated using the formula
|
|
\[
|
|
x^{-1} = x^{\varphi(m)-1}.
|
|
\]
|
|
If $m$ is prime, the formula becomes
|
|
\[
|
|
x^{-1} = x^{m-2}.
|
|
\]
|
|
For example, if $x=6$ and $m=17$, then
|
|
\[x^{-1}=6^{17-2} \bmod 17 = 3.\]
|
|
Using this formula, we can calculate the modular inverse
|
|
efficiently using the modular exponentation algorithm.
|
|
|
|
The above formula can be derived using Euler's theorem.
|
|
First, the modular inverse should satisfy the following equation:
|
|
\[
|
|
x x^{-1} \bmod m = 1.
|
|
\]
|
|
On the other hand, according to Euler's theorem,
|
|
\[
|
|
x^{\varphi(m)} \bmod m = xx^{\varphi(m)-1} \bmod m = 1,
|
|
\]
|
|
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.
|
|
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}
|
|
variable whose value is $123456789$.
|
|
After this, the value will be multiplied by itself,
|
|
and the result is
|
|
$123456789^2 \bmod 2^{32} = 2537071545$.
|
|
|
|
\begin{lstlisting}
|
|
unsigned int x = 123456789;
|
|
cout << x*x << "\n"; // 2537071545
|
|
\end{lstlisting}
|
|
|
|
\section{Yhtälönratkaisu}
|
|
|
|
\index{Diofantoksen yhtxlz@Diofantoksen yhtälö}
|
|
|
|
\key{Diofantoksen yhtälö} on muotoa
|
|
\[ ax + by = c, \]
|
|
missä $a$, $b$ ja $c$ ovat vakioita
|
|
ja tehtävänä on ratkaista muuttujat $x$ ja $y$.
|
|
Jokaisen yhtälössä esiintyvän luvun tulee
|
|
olla kokonaisluku.
|
|
Esimerkiksi jos yhtälö on $5x+2y=11$, yksi ratkaisu
|
|
on valita $x=3$ ja $y=-2$.
|
|
|
|
\index{Eukleideen algoritmi@Eukleideen algoritmi}
|
|
|
|
Diofantoksen yhtälön voi ratkaista
|
|
tehokkaasti Eukleideen algoritmin avulla,
|
|
koska Eukleideen algoritmia laajentamalla
|
|
pystyy löytämään luvun $\textrm{syt}(a,b)$
|
|
lisäksi luvut $x$ ja $y$,
|
|
jotka toteuttavat yhtälön
|
|
\[
|
|
ax + by = \textrm{syt}(a,b).
|
|
\]
|
|
|
|
Diofantoksen yhtälön ratkaisu on olemassa, jos $c$ on
|
|
jaollinen $\textrm{syt}(a,b)$:llä,
|
|
ja muussa tapauksessa yhtälöllä ei ole ratkaisua.
|
|
|
|
\index{laajennettu Eukleideen algoritmi@laajennettu Eukleideen algoritmi}
|
|
|
|
\subsubsection*{Laajennettu Eukleideen algoritmi}
|
|
|
|
Etsitään esimerkkinä luvut $x$ ja $y$,
|
|
jotka toteuttavat yhtälön
|
|
\[
|
|
39x + 15y = 12.
|
|
\]
|
|
Yhtälöllä on ratkaisu, koska $\textrm{syt}(39,15)=3$
|
|
ja $3 \mid 12$.
|
|
Kun Eukleideen algoritmi laskee lukujen
|
|
39 ja 15 suurimman
|
|
yhteisen tekijän, syntyy ketju
|
|
\[
|
|
\textrm{syt}(39,15) = \textrm{syt}(15,9)
|
|
= \textrm{syt}(9,6) = \textrm{syt}(6,3)
|
|
= \textrm{syt}(3,0) = 3. \]
|
|
Algoritmin aikana muodostuvat jakoyhtälöt ovat:
|
|
\[
|
|
\begin{array}{lcl}
|
|
39 - 2 \cdot 15 & = & 9 \\
|
|
15 - 1 \cdot 9 & = & 6 \\
|
|
9 - 1 \cdot 6 & = & 3 \\
|
|
\end{array}
|
|
\]
|
|
Näiden yhtälöiden avulla saadaan
|
|
\[
|
|
39 \cdot 2 + 15 \cdot (-5) = 3
|
|
\]
|
|
ja kertomalla yhtälö 4:lla tuloksena on
|
|
\[
|
|
39 \cdot 8 + 15 \cdot (-20) = 12,
|
|
\]
|
|
joten alkuperäisen yhtälön ratkaisu on $x=8$ ja $y=-20$.
|
|
|
|
Diofantoksen yhtälön ratkaisu ei ole yksikäsitteinen,
|
|
vaan yhdestä ratkaisusta on mahdollista muodostaa
|
|
äärettömästi muita ratkaisuja.
|
|
Kun yhtälön ratkaisu on $(x,y)$,
|
|
niin myös
|
|
\[(x+\frac{kb}{\textrm{syt}(a,b)},y-\frac{ka}{\textrm{syt}(a,b)})\]
|
|
on ratkaisu, missä $k$ on mikä tahansa kokonaisluku.
|
|
|
|
\subsubsection{Kiinalainen jäännöslause}
|
|
|
|
\index{kiinalainen jxxnnzslause@kiinalainen jäännöslause}
|
|
|
|
\key{Kiinalainen jäännöslause} ratkaisee yhtälöryhmän muotoa
|
|
\[
|
|
\begin{array}{lcl}
|
|
x & = & a_1 \bmod m_1 \\
|
|
x & = & a_2 \bmod m_2 \\
|
|
\cdots \\
|
|
x & = & a_n \bmod m_n \\
|
|
\end{array}
|
|
\]
|
|
missä kaikki parit luvuista $m_1,m_2,\ldots,m_n$
|
|
ovat suhteellisia alkulukuja.
|
|
|
|
Olkoon $x^{-1}_m$ luvun $x$ käänteisluku
|
|
modulo $m$ ja
|
|
\[ X_k = \frac{m_1 m_2 \cdots m_n}{m_k}.\]
|
|
Näitä merkintöjä käyttäen yhtälöryhmän ratkaisu on
|
|
\[x = a_1 X_1 {X_1}^{-1}_{m_1} + a_2 X_2 {X_2}^{-1}_{m_2} + \cdots + a_n X_n {X_n}^{-1}_{m_n}.\]
|
|
Tässä ratkaisussa jokaiselle luvulle $k=1,2,\ldots,n$
|
|
pätee, että
|
|
\[a_k X_k {X_k}^{-1}_{m_k} \bmod m_k = a_k,\]
|
|
sillä
|
|
\[X_k {X_k}^{-1}_{m_k} \bmod m_k = 1.\]
|
|
Koska kaikki muut summan osat ovat jaollisia luvulla
|
|
$m_k$, ne eivät vaikuta jakojäännökseen ja
|
|
koko summan jakojäännös $m_k$:lla on $a_k$.
|
|
|
|
Esimerkiksi yhtälöryhmän
|
|
\[
|
|
\begin{array}{lcl}
|
|
x & = & 3 \bmod 5 \\
|
|
x & = & 4 \bmod 7 \\
|
|
x & = & 2 \bmod 3 \\
|
|
\end{array}
|
|
\]
|
|
ratkaisu on
|
|
\[ 3 \cdot 21 \cdot 1 + 4 \cdot 15 \cdot 1 + 2 \cdot 35 \cdot 2 = 263.\]
|
|
|
|
Kun yksi ratkaisu $x$ on löytynyt,
|
|
sen avulla voi muodostaa äärettömästi
|
|
erilaisia ratkaisuja, koska kaikki luvut muotoa
|
|
\[x+m_1 m_2 \cdots m_n\]
|
|
ovat ratkaisuja.
|
|
|
|
|
|
\section{Muita tuloksia}
|
|
|
|
\subsubsection{Lagrangen lause}
|
|
|
|
\index{Lagrangen lause@Lagrangen lause}
|
|
|
|
\key{Lagrangen lauseen} mukaan jokainen positiivinen kokonaisluku voidaan
|
|
esittää neljän neliöluvun summana eli muodossa $a^2+b^2+c^2+d^2$.
|
|
Esimerkiksi luku 123 voidaan esittää muodossa $8^2+5^2+5^2+3^2$.
|
|
|
|
\subsubsection{Zeckendorfin lause}
|
|
|
|
\index{Zeckendorfin lause@Zeckendorfin lause}
|
|
\index{Fibonaccin luku@Fibonaccin luku}
|
|
|
|
\key{Zeckendorfin lauseen} mukaan
|
|
jokaiselle positiiviselle kokonaisluvulle
|
|
on olemassa yksikäsitteinen esitys
|
|
Fibonaccin lukujen summana niin, että
|
|
mitkään kaksi lukua eivät ole samat eivätkä peräkkäiset
|
|
Fibonaccin luvut.
|
|
Esimerkiksi luku 74 voidaan esittää muodossa
|
|
$55+13+5+1$.
|
|
|
|
\subsubsection{Pythagoraan kolmikot}
|
|
|
|
\index{Pythagoraan kolmikko@Pythagoraan kolmikko}
|
|
\index{Eukleideen kaava@Eukleideen kaava}
|
|
|
|
\key{Pythagoraan kolmikko} on lukukolmikko $(a,b,c)$,
|
|
joka toteuttaa Pythagoraan lauseen $a^2+b^2=c^2$
|
|
eli $a$, $b$ ja $c$ voivat olla suorakulmaisen
|
|
kolmion sivujen pituudet.
|
|
Esimerkiksi $(3,4,5)$ on Pythagoraan kolmikko.
|
|
|
|
Jos $(a,b,c)$ on Pythagoraan kolmikko,
|
|
niin myös kaikki kolmikot muotoa $(ka,kb,kc)$
|
|
ovat Pythagoraan kolmikoita,
|
|
missä $k>1$.
|
|
Pythagoraan kolmikko on \key{primitiivinen},
|
|
jos $a$, $b$ ja $c$ ovat suhteellisia alkulukuja,
|
|
ja primitiivisistä kolmikoista voi muodostaa
|
|
kaikki muut kolmikot kertoimen $k$ avulla.
|
|
|
|
\key{Eukleideen kaavan} mukaan jokainen primitiivinen
|
|
Pythagoraan kolmikko on muotoa
|
|
\[(n^2-m^2,2nm,n^2+m^2),\]
|
|
missä $0<m<n$, $n$ ja $m$ ovat suhteelliset
|
|
alkuluvut ja ainakin toinen luvuista $n$ ja $m$ on parillinen.
|
|
Esimerkiksi valitsemalla $m=1$ ja $n=2$ syntyy
|
|
pienin mahdollinen Pythagoraan kolmikko
|
|
\[(2^2-1^2,2\cdot2\cdot1,2^2+1^2)=(3,4,5).\]
|
|
|
|
\subsubsection{Wilsonin lause}
|
|
|
|
\index{Wilsonin lause@Wilsonin lause}
|
|
|
|
\key{Wilsonin lauseen} mukaan luku $n$ on alkuluku
|
|
tarkalleen silloin, kun
|
|
\[(n-1)! \bmod n = n-1.\]
|
|
Esimerkiksi luku 11 on alkuluku, koska
|
|
\[10! \bmod 11 = 10,\]
|
|
ja luku 12 ei ole alkuluku, koska
|
|
\[11! \bmod 12 = 0 \neq 11.\]
|
|
|
|
Wilsonin lauseen avulla voi siis tutkia, onko luku alkuluku.
|
|
Tämä ei ole kuitenkaan käytännössä hyvä tapa,
|
|
koska luvun $(n-1)!$ laskeminen on työlästä,
|
|
jos $n$ on suuri luku.
|
|
|
|
|