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@ -9,9 +9,9 @@ because many questions involving integers
are very difficult to solve even if they are very difficult to solve even if they
seem simple at first glance. seem simple at first glance.
As an example, let's consider the following equation: As an example, let us consider the following equation:
\[x^3 + y^3 + z^3 = 33\] \[x^3 + y^3 + z^3 = 33\]
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. that satisfy the equation.
For example, we can choose 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 In this chapter, we will focus on basic concepts
and algorithms in number theory. and algorithms in number theory.
We will start by discussing divisibility of numbers Throughout the chapter, we will assume that all numbers
and important algorithms for primality testing are integers, if not otherwise stated.
and factorization.
\section{Primes and factors} \section{Primes and factors}
@ -38,9 +37,8 @@ and factorization.
\index{factor} \index{factor}
\index{divisor} \index{divisor}
A number $a$ is a \key{factor} or \key{divisor} of a number $b$ 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 $b$, If $a$ is a factor of $b$,
we write $a \mid b$, and otherwise we write $a \nmid b$. we write $a \mid b$, and otherwise we write $a \nmid b$.
For example, the factors of the number 24 are 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} A number $n>1$ is a \key{prime}
if its only positive factors are 1 and $n$. if its only positive factors are 1 and $n$.
For example, the numbers 7, 19 and 41 are primes. For example, the numbers 7, 19 and 41 are primes.
The number 35 is not a prime because it can be The number 35 is not a prime, because it can be
divided into factors $5 \cdot 7 = 35$. divided into the factors $5 \cdot 7 = 35$.
For each number $n>1$, there is a unique For each number $n>1$, there is a unique
\key{prime factorization} \key{prime factorization}
\[ n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k},\] \[ n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k},\]
@ -75,7 +73,7 @@ The factors are
The \key{sum of factors} of $n$ is The \key{sum of factors} of $n$ is
\[\sigma(n)=\prod_{i=1}^k (1+p_i+\ldots+p_i^{\alpha_i}) = \prod_{i=1}^k \frac{p_i^{a_i+1}-1}{p_i-1},\] \[\sigma(n)=\prod_{i=1}^k (1+p_i+\ldots+p_i^{\alpha_i}) = \prod_{i=1}^k \frac{p_i^{a_i+1}-1}{p_i-1},\]
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 For example, the sum of factors of the number 84 is
\[\sigma(84)=\frac{2^3-1}{2-1} \cdot \frac{3^2-1}{3-1} \cdot \frac{7^2-1}{7-1} = 7 \cdot 4 \cdot 8 = 224.\] \[\sigma(84)=\frac{2^3-1}{2-1} \cdot \frac{3^2-1}{3-1} \cdot \frac{7^2-1}{7-1} = 7 \cdot 4 \cdot 8 = 224.\]
@ -91,23 +89,23 @@ and the product of the factors is $\mu(84)=84^6=351298031616$.
\index{perfect number} \index{perfect number}
A number $n$ is \key{perfect} if $n=\sigma(n)-n$, A number $n$ is \key{perfect} if $n=\sigma(n)-n$,
i.e., the number equals the sum of its divisors i.e., $n$ equals the sum of its divisors
between $1 \ldots n-1$. between $1$ and $n-1$.
For example, the number 28 is perfect because For example, the number 28 is perfect,
it equals the sum $1+2+4+7+14$. because $28=1+2+4+7+14$.
\subsubsection{Number of primes} \subsubsection{Number of primes}
It is easy to show that there is an infinite number It is easy to show that there is an infinite number
of primes. of primes.
If the number would be finite, If the number of primes would be finite,
we could construct a set $P=\{p_1,p_2,\ldots,p_n\}$ 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. 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
\[p_1 p_2 \cdots p_n+1\] \[p_1 p_2 \cdots p_n+1\]
that is larger than all elements in $P$. 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. has to be infinite.
\subsubsection{Density of primes} \subsubsection{Density of primes}
@ -115,14 +113,14 @@ has to be infinite.
The density of primes means how often there are primes The density of primes means how often there are primes
among the numbers. among the numbers.
Let $\pi(n)$ denote the number of primes between Let $\pi(n)$ denote the number of primes between
$1 \ldots n$. For example, $\pi(10)=4$ because $1$ and $n$. For example, $\pi(10)=4$, because
there are 4 primes between $1 \ldots 10$: 2, 3, 5 and 7. 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
\[\pi(n) \approx \frac{n}{\ln n},\] \[\pi(n) \approx \frac{n}{\ln n},\]
which means that primes appear quite often. which means that primes are quite frequent.
For example, the number of primes between 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 $10^6 / \ln 10^6 \approx 72382$. and $10^6 / \ln 10^6 \approx 72382$.
\subsubsection{Conjectures} \subsubsection{Conjectures}
@ -138,7 +136,7 @@ For example, the following conjectures are famous:
Each even integer $n>2$ can be represented as a Each even integer $n>2$ can be represented as a
sum $n=a+b$ so that both $a$ and $b$ are primes. sum $n=a+b$ so that both $a$ and $b$ are primes.
\index{twin prime} \index{twin prime}
\item \key{twin prime}: \item \key{Twin prime conjecture}:
There is an infinite number of pairs There is an infinite number of pairs
of the form $\{p,p+2\}$, of the form $\{p,p+2\}$,
where both $p$ and $p+2$ are primes. 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, If a number $n$ is not prime,
it can be represented as a product $a \cdot b$, it can be represented as a product $a \cdot b$,
where $a \le \sqrt n$ or $b \le \sqrt n$, 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 Using this observation, we can both test
if a number is prime and find the prime factorization if a number is prime and find the prime factorization
of a number in $O(\sqrt n)$ time. of a number in $O(\sqrt n)$ time.
The following function \texttt{prime} checks The following function \texttt{prime} checks
if the given number $n$ is prime. if the given number $n$ is prime.
The function tries to divide the number by The function attempts to divide $n$ by
all numbers between $2 \ldots \sqrt n$, all numbers between $2$ and $\lfloor \sqrt n \rfloor$,
and if none of them divides $n$, then $n$ is prime. and if none of them divides $n$, then $n$ is prime.
\begin{lstlisting} \begin{lstlisting}
@ -181,7 +179,7 @@ factorization of $n$.
The function divides $n$ by its prime factors, The function divides $n$ by its prime factors,
and adds them to the vector. and adds them to the vector.
The process ends when the remaining number $n$ 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. If $n>1$, it is prime and the last factor.
\begin{lstlisting} \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 The \key{sieve of Eratosthenes} is a preprocessing
algorithm that builds an array using which we algorithm that builds an array using which we
can efficiently check if a given number between $2 \ldots n$ 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 $\texttt{a}$ 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 The value $\texttt{a}[k]=0$ means
that $k$ is prime, that $k$ is prime,
and the value $\texttt{a}[k] \neq 0$ 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 $\texttt{a}[k]$. of its prime factors is $\texttt{a}[k]$.
The algorithm iterates through the numbers The algorithm iterates through the numbers
$2 \ldots n$ one by one. $2 \ldots n$ one by one.
Always when a new prime $x$ is found, Always when a new prime $x$ is found,
the algorithm records that the multiples 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. 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{center}
\begin{tikzpicture}[scale=0.7] \begin{tikzpicture}[scale=0.7]
@ -301,12 +299,12 @@ of the algorithm is the harmonic sum
\[\sum_{x=2}^n n/x = n/2 + n/3 + n/4 + \cdots + n/n = O(n \log n).\] \[\sum_{x=2}^n n/x = n/2 + n/3 + n/4 + \cdots + n/n = O(n \log n).\]
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 because the inner loop will be executed only if
the number $x$ is prime. the number $x$ is prime.
It can be shown that the time complexity of the It can be shown that the time complexity of the
algorithm is only $O(n \log \log n)$ algorithm is only $O(n \log \log n)$,
that is very near to $O(n)$. a complexity very near to $O(n)$.
\subsubsection{Euclid's algorithm} \subsubsection{Euclid's algorithm}
@ -331,7 +329,7 @@ are connected as follows:
\key{Euclid's algorithm} provides an efficient way \key{Euclid's algorithm} provides an efficient way
to find the greatest common divisor of two numbers. 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*} \begin{equation*}
\textrm{gcd}(a,b) = \begin{cases} \textrm{gcd}(a,b) = \begin{cases}
a & b = 0\\ a & b = 0\\
@ -342,11 +340,9 @@ For example,
\[\textrm{gcd}(24,36) = \textrm{gcd}(36,24) \[\textrm{gcd}(24,36) = \textrm{gcd}(36,24)
= \textrm{gcd}(24,12) = \textrm{gcd}(12,0)=12.\] = \textrm{gcd}(24,12) = \textrm{gcd}(12,0)=12.\]
The time complexity of Euclid's algorithm The time complexity of Euclid's algorithm
is $O(\log n)$ where $n=\min(a,b)$. is $O(\log n)$, where $n=\min(a,b)$.
The worst case is when The worst case for the algorithm is
$a$ and $b$ are successive Fibonacci numbers. the case when $a$ and $b$ are consecutive Fibonacci numbers.
In this case, the algorithm goes through
all smaller Fibonacci numbers.
For example, For example,
\[\textrm{gcd}(13,8)=\textrm{gcd}(8,5) \[\textrm{gcd}(13,8)=\textrm{gcd}(8,5)
=\textrm{gcd}(5,3)=\textrm{gcd}(3,2)=\textrm{gcd}(2,1)=\textrm{gcd}(1,0)=1.\] =\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{coprime}
\index{Euler's totient function} \index{Euler's totient function}
Numbers $a$ and $b$ are coprime Numbers $a$ and $b$ are \key{coprime}
if $\textrm{gcd}(a,b)=1$. if $\textrm{gcd}(a,b)=1$.
\key{Euler's totient function} $\varphi(n)$ \key{Euler's totient function} $\varphi(n)$
returns the number of coprime numbers to $n$ returns the number of coprime numbers to $n$
between $1 \ldots n$. between $1$ and $n$.
For example, $\varphi(12)=4$, For example, $\varphi(12)=4$,
because the numbers 1, 5, 7 and 11 because the 1, 5, 7 and 11
are coprime to the number 12. are coprime to 12.
The value of $\varphi(n)$ can be calculated The value of $\varphi(n)$ can be calculated
using the prime factorization of $n$ from the prime factorization of $n$
by the formula using the formula
\[ \varphi(n) = \prod_{i=1}^k p_i^{\alpha_i-1}(p_i-1). \] \[ \varphi(n) = \prod_{i=1}^k p_i^{\alpha_i-1}(p_i-1). \]
For example, $\varphi(12)=2^1 \cdot (2-1) \cdot 3^0 \cdot (3-1)=4$. 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. 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} \index{modular arithmetic}
In \key{modular arithmetic}, In \key{modular arithmetic},
the set of available numbers is restricted so the set of available numbers is limited so
that only numbers $0,1,2,\ldots,m-1$ can be used that only numbers $0,1,2,\ldots,m-1$ may be used,
where $m$ is a constant. where $m$ is a constant.
Each number $x$ is Each number $x$ is
represented by the number $x \bmod m$: represented by the number $x \bmod m$:
@ -386,23 +382,22 @@ the remainder after dividing $x$ by $m$.
For example, if $m=17$, then $75$ For example, if $m=17$, then $75$
is represented by $75 \bmod 17 = 7$. is represented by $75 \bmod 17 = 7$.
Often we can take the remainder before doing a Often we can take the remainder before doing
calculation. calculations.
In particular, the following formulas can be used: In particular, the following formulas can be used:
\[ \[
\begin{array}{rcl} \begin{array}{rcl}
(x+y) \bmod m & = & (x \bmod m + y \bmod m) \bmod m \\ (x+y) \bmod m & = & (x \bmod m + y \bmod m) \bmod m \\
(x-y) \bmod m & = & (x \bmod m - y \bmod m) \bmod m \\ (x-y) \bmod m & = & (x \bmod m - y \bmod m) \bmod m \\
(x \cdot y) \bmod m & = & (x \bmod m \cdot y \bmod m) \bmod m \\ (x \cdot y) \bmod m & = & (x \bmod m \cdot y \bmod m) \bmod m \\
(x^k) \bmod m & = & (x \bmod m)^k \bmod m \\ x^n \bmod m & = & (x \bmod m)^n \bmod m \\
\end{array} \end{array}
\] \]
\subsubsection{Modular exponentiation} \subsubsection{Modular exponentiation}
There is often need to efficiently calculate
Often there is need to efficiently calculate the value of $x^n \bmod m$.
the remainder of $x^n$.
This can be done in $O(\log n)$ time This can be done in $O(\log n)$ time
using the following recursion: using the following recursion:
\begin{equation*} \begin{equation*}
@ -413,13 +408,13 @@ using the following recursion:
\end{cases} \end{cases}
\end{equation*} \end{equation*}
It's important that in the case of an even $n$, It is important that in the case of an even $n$,
the number $x^{n/2}$ is calculated only once. the value of $x^{n/2}$ is calculated only once.
This guarantees that the time complexity of the 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. when it is even.
The following function calculates the number The following function calculates the value of
$x^n \bmod m$: $x^n \bmod m$:
\begin{lstlisting} \begin{lstlisting}
@ -438,12 +433,12 @@ int modpow(int x, int n, int m) {
\index{Euler's theorem} \index{Euler's theorem}
\key{Fermat's theorem} states that \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. when $m$ is prime and $x$ and $m$ are coprime.
This also yields This also yields
\[x^k \bmod m = x^{k \bmod (m-1)} \bmod m.\] \[x^k \bmod m = x^{k \bmod (m-1)} \bmod m.\]
More generally, \key{Euler's theorem} states that 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. when $x$ and $m$ are coprime.
Fermat's theorem follows from Euler's theorem, Fermat's theorem follows from Euler's theorem,
because if $m$ is a prime, then $\varphi(m)=m-1$. 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 $2 \cdot 3 \bmod 17$, we can use the formula $2 \cdot 3 \bmod 17$,
because $36 \bmod 17 = 2$ and $6^{-1} \bmod 17 = 3$. 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 For example, if $x=2$ and $m=4$, the equation
\[ x x^{-1} \bmod m = 1. \] \[ x x^{-1} \bmod m = 1 \]
can't be solved, because all multiples of the number 2 cannot be solved, because all multiples of the number 2
are even, and the remainder can never be 1 when $m=4$. are even and the remainder can never be 1 when $m=4$.
It turns out that the number $x^{-1} \bmod m$ exists It turns out that the value of $x^{-1} \bmod m$
exactly when $x$ and $m$ are coprime. can be calculated exactly when $x$ and $m$ are coprime.
If a modular inverse exists, it can be If a modular inverse exists, it can be
calculated using the formula calculated using the formula
@ -500,14 +495,14 @@ so the numbers $x^{-1}$ and $x^{\varphi(m)-1}$ are equal.
\subsubsection{Computer arithmetic} \subsubsection{Computer arithmetic}
In a computers, unsigned integers are represented modulo $2^k$ In programming, unsigned integers are represented modulo $2^k$,
where $k$ is the number of bits. where $k$ is the number of bits of the data type.
A usual consequence of this is that a number wraps around A usual consequence of this is that a number wraps around
if it becomes too large. if it becomes too large.
For example, in C++, numbers of type \texttt{unsigned int} For example, in C++, numbers of type \texttt{unsigned int}
are represented modulo $2^{32}$. 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 $123456789$. variable whose value is $123456789$.
After this, the value will be multiplied by itself, After this, the value will be multiplied by itself,
and the result is and the result is
@ -525,7 +520,7 @@ cout << x*x << "\n"; // 2537071545
A \key{Diophantine equation} is of the form A \key{Diophantine equation} is of the form
\[ ax + by = c, \] \[ ax + by = c, \]
where $a$, $b$ and $c$ are constants, 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. Each number in the equation has to be an integer.
For example, one solution for the equation For example, one solution for the equation
$5x+2y=11$ is $x=3$ and $y=-2$. $5x+2y=11$ 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 $y$ so that it will find numbers $x$ and $y$
that satisfy the following equation: that satisfy the following equation:
\[ \[
ax + by = \textrm{syt}(a,b) ax + by = \textrm{gcd}(a,b)
\] \]
A Diophantine equation can be solved if A Diophantine equation can be solved if
$c$ is divisible by $c$ is divisible by
$\textrm{gcd}(a,b)$, $\textrm{gcd}(a,b)$,
and otherwise it can't be solved. and otherwise the equation cannot be solved.
\index{extended Euclid's algorithm} \index{extended Euclid's algorithm}
\subsubsection*{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: that satisfy the following equation:
\[ \[
39x + 15y = 12 39x + 15y = 12
\] \]
The equation can be solved, because 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 When Euclid's algorithm calculates the
greatest common divisor of 39 and 15, greatest common divisor of 39 and 15,
it produces the following sequence of function calls: it produces the following sequence of function calls:
@ -580,15 +575,15 @@ and by multiplying this by 4, the result is
\[ \[
39 \cdot 8 + 15 \cdot (-20) = 12, 39 \cdot 8 + 15 \cdot (-20) = 12,
\] \]
so a solution for the original equation is so a solution for the equation is
$x=8$ and $y=-20$. $x=8$ and $y=-20$.
A solution for a Diophantine equation is not unique, A solution for a Diophantine equation is not unique,
but we can form an infinite number of solutions but we can form an infinite number of solutions
if we know one solution. 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
\[(x+\frac{kb}{\textrm{gcd}(a,b)},y-\frac{ka}{\textrm{gcd}(a,b)})\] \[(x+\frac{kb}{\textrm{gcd}(a,b)},y-\frac{ka}{\textrm{gcd}(a,b)})\]
is a solution where $k$ is any integer. are solutions, where $k$ is any integer.
\subsubsection{Chinese remainder theorem} \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 \key{Zeckendorf's theorem} states that every
positive integer has a unique representation positive integer has a unique representation
as a sum of Fibonacci numbers such that 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. Fibonacci numbers.
For example, the number 74 can be represented For example, the number 74 can be represented
as the sum $55+13+5+1$. as the sum $55+13+5+1$.
@ -685,7 +680,7 @@ all primitive Pythagorean triples.
Each such triple is of the form Each such triple is of the form
\[(n^2-m^2,2nm,n^2+m^2),\] \[(n^2-m^2,2nm,n^2+m^2),\]
where $0<m<n$, $n$ and $m$ are coprime 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 For example, when $m=1$ and $n=2$, the formula
produces the smallest Pythagorean triple produces the smallest Pythagorean triple
\[(2^2-1^2,2\cdot2\cdot1,2^2+1^2)=(3,4,5).\] \[(2^2-1^2,2\cdot2\cdot1,2^2+1^2)=(3,4,5).\]
@ -703,8 +698,8 @@ and the number 12 is not prime, because
\[11! \bmod 12 = 0 \neq 11.\] \[11! \bmod 12 = 0 \neq 11.\]
Hence, Wilson's theorem tells us whether a number Hence, Wilson's theorem tells us whether a number
is prime. However, in practice, the formula can't be is prime. However, in practice, the theorem cannot be
used for large values of $n$, because it's difficult applied to large values of $n$, because it is difficult
to calculate the number $(n-1)!$ if $n$ is large. to calculate the value of $(n-1)!$ when $n$ is large.