# 数学代写|数论作业代写number theory代考|MATH230

## 数学代写|数论作业代写number theory代考|Euler’s Theorem

The first published proof of Fermat’s little theorem(stated in chapter 5 of this book) was given by Euler in 1736 , where he had taken a prime $p$ and an integer $a$. But later in the year, 1760 he succeeded in generalizing the result from prime $p$ to an arbitrary integer $n$. This generalization is known as Euler’s generalization of Fermat’s theorem. The present section deals with the proof and related ideas associated with this remarkable theorem.

Now, as a precursor to launch the proof of Euler’s generalization of Fermat’s theorem, we need the following lemma:

Lemma 7.4.1. Let $n>1$ and $\operatorname{gcd}(a, n)=1$. If $k_1, k_2, \cdots, k_{\dot{\phi}(n)}$ are the positive integers less than and prime to $n$, then $a k_1, a k_2, \cdots, a k_{\varphi(n)}$ are congruent modulo $n$ to $k_1, k_2, \cdots, k_{\phi(n)}$ in some order.

Proof. Here we are going to show that no two of the integers $a k_1, a k_2, \cdots, a k_{\phi(n)}$ are congruent modulo $n$. For if, $a k_i \equiv a k_j(\bmod n)$ holds with $1 \leq i<j \leq \phi(n)$ then $k_i \equiv k_j(\bmod n)$, which is a contradiction since this two integers are less than $n$. Since, $\operatorname{gcd}\left(k_i, n\right)=1 \forall i$ and $\operatorname{gcd}(a, n)=1$ then from the worked out Problem 2.6.1) $\operatorname{gcd}\left(a k_i, n\right)=1 \forall i$. Let us fix $a k_j$ for some integer $j$, there exists unique integer $b$ where $0 \leq b<n$ for which $a k_j \equiv b(\bmod n)$. Since, $\operatorname{gcd}(b, n)=\operatorname{gcd}\left(a k_j, n\right)=1$, so $b$ must be one of the integers $k_1, k_2, \cdots, k_{\phi(n)}$.

This is true for all $j$. This proves that the numbers $a k_1, a k_2, \cdots, a k_{\phi(n)}$ and the numbers $k_1, k_2, \cdots, k_{\phi(n)}$ are identical with respect to modulo $n$ in a certain order.

We now represent an example to make a lucid understanding of this lemma. For that let us take $n=9$ and the set ${1,2,4,5,7,8}$ is a reduce system modulo 9. Since $\operatorname{ged}(2,9)=1$ then we have, $2 \cdot 1=2,2 \cdot 2=4,2 \cdot 4=8,2 \cdot 5=10,2 \cdot 7=$ $14,2 \cdot 8=16$ is also a reduce system modulo 9 .

## 数学代写|数论作业代写number theory代考|Properties of -function

Present section deals with some curious properties of Euler’s phi function related with some arithmetic functions. Discussion of this chapter commence with an important property of totient $(\phi)$ function, where the sum of values of $\phi(d)$ where $d$ is the divisor of any positive integer $n$ is always equal to $n$ itself. Famous German mathematician Carl Friedrich Gauss was the first person to notice that.
Theorem 7.6.1. For each positive integer $n \geq 1, n=\sum_{d \mid n} \phi(d)$ where $d$ is positive divisor of $n$ :

Proof. Let us choose $n=1$ then, $\sum_{d \mid 1} \phi(d)=\phi(1)=1=n$. Thus the equality is true in this case. Now we are only to prove the result for any positive integer $n>1$. Let us choose a set $S_n={1,2,3, \cdots, n}$ and $\left|S_n\right|$ be the number of elements in $S_n$, then clearly $\left|S_n\right|=n$. For each divisor $d$ of $n$ we denote $S_d$ be the set of all integers not exceeding $n$ and $\operatorname{gcd}(m, n)=d$ for each $m \in S_d$. Now from the proposition (2.4.2) we have $\operatorname{gcd}(m, n)=d$ if and only if $\operatorname{gcd}\left(\frac{m}{d}, \frac{n}{d}\right)=1$. We now have to show that each $S_d$ has $\phi\left(\frac{n}{d}\right)$ number of elements. Here for a particular $d$ all the elements of $S_d$ are multiples of $d$ and less than or equal to n. Thus the elements of $S_d$ are $d, 2 d, 3 d, \cdots,\left(\frac{n}{d}\right) d$. Now, let $a d \in S_d$ be any element where $\operatorname{gcd}\left(a, \frac{n}{d}\right)=e$. Then clearly $\operatorname{gcd}(a d, n)=e d$. Here $e d=d$ if and only if $e=1$ imply that only $a d$ in $S_d$ are those whose $\operatorname{gcd}\left(a, \frac{n}{d}\right)=1$ that is the number $\phi\left(\frac{n}{d}\right)$. Since each integers of the set ${1,2,3, \cdots, n}$ lies in exactly one class $S_d$, we have the formula $n=\sum_{d \mid n} \phi\left(\frac{n}{d}\right)$. But $d$ runs through all positive divisors of $n$ so does $\frac{n}{d}$. Thus finally we have, $n=\sum_{d \mid n} \phi\left(\frac{n}{d}\right)=\sum_{d \mid n} \phi(d)$.

# 数论代考

## 数学代写|数论作业代写number theory代考|Euler’s Theorem

$\operatorname{gcd}\left(k_i, n\right)=1 \forall i$ 和 $\operatorname{gcd}(a, n)=1$ 然后从制定出的问题2.6.1) $\operatorname{gcd}\left(a k_i, n\right)=1 \forall i$. 让我们修复 $a k_j$ 对 于某个整数 $j$, 存在唯一整数 $b$ 在哪里 $0 \leq b<n$ 为了哪个 $a k_j \equiv b(\bmod n)$. 自从，
$\operatorname{gcd}(b, n)=\operatorname{gcd}\left(a k_j, n\right)=1$ ，所以b必须是整数之一 $k_1, k_2, \cdots, k_{\phi(n)}$.

$2 \cdot 1=2,2 \cdot 2=4,2 \cdot 4=8,2 \cdot 5=10,2 \cdot 7=14,2 \cdot 8=16$ 也是一个减少系统模 9 。

## 数学代写|数论作业代写number theory代考|Properties of -function

$\operatorname{gcd}\left(k_i, n\right)=1 \forall i$ 和 $\operatorname{gcd}(a, n)=1$ 然后从制定出的问题2.6.1) $\operatorname{gcd}\left(a k_i, n\right)=1 \forall i$. 让我们修复 $a k_j$ 对 于某个整数 $j$, 存在唯一整数 $b$ 在哪里 $0 \leq b<n$ 为了哪个 $a k_j \equiv b(\bmod n)$. 自从，
$\operatorname{gcd}(b, n)=\operatorname{gcd}\left(a k_j, n\right)=1$ ，所以b必须是整数之一 $k_1, k_2, \cdots, k_{\phi(n)}$.

$2 \cdot 1=2,2 \cdot 2=4,2 \cdot 4=8,2 \cdot 5=10,2 \cdot 7=14,2 \cdot 8=16$ 也是一个减少系统模 9 。

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