## 数学代写|数论作业代写number theory代考|MATH2088

2022年7月26日

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## 数学代写|数论作业代写number theory代考|Prime Factorization

Fermat’s Little Theorem $2.17$ can be used to show that a given number $q>1$ is composite without knowing any non-trivial divisor. If we find some $a$ coprime to $q$ such that congruence (2.35) does not hold, then $q$ is not a prime number. For example, if we show that $q$ does not divide $2^{q-1}-1$ for $q$ odd, then $q$ is composite.

It should be emphasized that the factorization of a given number $n$ to prime factors is much more difficult than finding whether $n$ is a prime or composite number. According to Gauss, finding the prime number factorization of a given number is a very important and fundamental role of arithmetic [118, Article 329].

The oldest method to factor the number $n$ into prime factors gradually tries to divide $n$ by all primes not exceeding $\sqrt{n}$. For instance, to show that 283 is a prime it is enough to verify that it is not divisible by $2,3,5,7,11$, and 13 . However, it must be emphasized that this method is not very effective. Suppose, for instance, we are able to perform one billion divisions per second. Then a factorization of $n=p q$, where $p$ and $q$ are unknown thirty digits primes, would take more time than the age of the universe $\left(\approx 13.7 \cdot 10^{9}\right.$ years).

To explore this assertion further, let us denote by $\pi(x)$ the number of primes not exceeding $x$ (see Fig. 2.7). Note that Carl Friedrich Gauss when he was a teenager conjectured that the probability that $n$ is a prime number is $1 / \log n$. By $[138,350]$ the value of $\pi(x)$ is approximately equal to $\frac{x}{\log x}$, where the error is less than $15 \%$ for every $x \geq 3000$. In 1896 Jacques Hadamard (1865-1963) and independently also Charles-Jean de la Vallée Poussin (1866-1962) even proved this asymptotic equality for $x \rightarrow \infty$,
$$\left.\pi(x) \approx \frac{x}{\log x} \quad \text { (Prime Number Theorem }\right)$$ Since the integer part of $\sqrt{n}$ has 30 digits, there exist at least $10^{29} /(29 \log 10)$ $\approx 1.5 \cdot 10^{27}$ prime numbers less than $\sqrt{n}$. Because each year has about $3.2 \cdot 10^{7} \mathrm{~s}$, we will perform $3.2 \cdot 10^{16}$ divisions during this time period. But to test all primes not exceeding $\sqrt{n}$, we would need at least $47 \cdot 10^{9}$ years to factor $n$ which is more than three times the age of the universe. Mathematicians compare such an algorithm to an effort to smash an atom by a hammer. By having any additional information about the number to be factored or about its prime factors, the factorization can be done much more efficiently (see e.g. Theorems $4.5$ and 4.17).

## 数学代写|数论作业代写number theory代考|Criteria for Primality

Let $j, m, n$ be positive integers. We say that $m^{j}$ exactly divides $n$ and write $m^{j} | n$, if $m^{j} \mid n$, but $m^{j+1} \nmid n$. For $j=0$ the symbol $m^{0} | n$ means that $m \nmid n$.

Theorem $3.1$ A positive integer $n$ is a prime if and only if $n \mid\left(\begin{array}{l}n \ k\end{array}\right)$ for all $k \in$ ${1, \ldots, n-1}$

Proof $\Rightarrow:$ For $k \in{1, \ldots, n-1}$ the number $n-k$ is also between 1 and $n-1$.
Since $n$ is prime, $n \nmid k !$ and $n \nmid(n-k)$ !. Hence, we have
$$n \mid\left(\begin{array}{l} n \ k \end{array}\right)=\frac{n !}{k !(n-k) !},$$
since $n \mid n !$
$\Leftarrow$ : Conversely, let $n$ be composite and let $p$ be the smallest prime which divides
$n$. Hence, $1<p<n$ and
$$\left(\begin{array}{l} n \ p \end{array}\right)=\frac{n(n-1) \cdots(n-p+1)}{p(p-1) \cdots 2 \cdot 1} .$$
Assume that $p^{j} | n$ for some positive integer $j$. Between $p$ successive numbers $n, n-1, \ldots, n-p+1$ there exists exactly one divisible by $p$. Since $p \mid n$, we obtain
$$p \nmid(n-1)(n-2) \cdots(n-p+1) .$$
Therefore, $p^{j} | n(n=1) \cdots(n=p \neq 1)$ and clearly $p | p(p=1) \cdots 2,1$. Thus by (3.1) we get $p^{j-1} |\left(\begin{array}{l}n \ p\end{array}\right)$. However, $n \nmid\left(\begin{array}{l}n \ p\end{array}\right)$, since $p^{j} | n$.

# 数论作业代写

## 数学代写|数论作业代写number theory代考|Prime Factorization

$$\pi(x) \approx \frac{x}{\log x} \quad \text { (Prime Number Theorem) }$$

## 数学代写|数论作业代写number theory代考|Criteria for Primality

$$n \mid(n k)=\frac{n !}{k !(n-k) !},$$

$n$. 因此， $1<p<n$ 和
$$(n p)=\frac{n(n-1) \cdots(n-p+1)}{p(p-1) \cdots 2 \cdot 1}$$

$$p \nmid(n-1)(n-2) \cdots(n-p+1) .$$

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## MATLAB代写

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