# 数学代写|数论作业代写number theory代考|MATH201

## 数学代写|数论作业代写number theory代考|The Sum and Number of Divisors

Problem 6.3.1. Prove that there are infinitely many pairs of integers $m$ and $n$ with $\sigma\left(m^2\right)=\sigma\left(n^2\right)$.

Solution 6.3.1. There are infinitely many integers $k$ such that $\operatorname{gcd}(k, 10)=1$. Let us consider $m=5 k, n=4 k$. This implies there exist infinitely many such $m, n$. Suppose $k$ is prime with $k \neq 2,5$. Now $m^2=5^2 k^2$ and $n^2=4^2 k^2=2^4 k^2$. Theorem 6.2.2 yields
$$\begin{gathered} \sigma\left(m^2\right)=\frac{5^3-1}{5-1} \cdot \frac{k^3-1}{k-1}=31\left(\frac{k^3-1}{k-1}\right) \ \sigma\left(n^2\right)=\frac{2^5-1}{2-1} \cdot \frac{k^3-1}{k-1}=31\left(\frac{k^3-1}{k-1}\right) \end{gathered}$$
Thus there are infinitely many pairs of integers $m$ and $n$ with $\sigma\left(m^2\right)=\sigma\left(n^2\right)$.
Problem 6.3.2. If $n$ is a square-free integer, prove that $\tau(n)=2^s$, where $s$ is the number of prime divisors of $n$.

Solution 6.3.2. Since $n$ is square-free, therefore $n=p_1 p_2 \cdots p_r$ where each $p_i \neq p_j \neq$ for $i \neq j$. From Theorem 6.2.2, we obtain
$\tau(n)=\left(k_1+1\right)\left(k_2+1\right) \cdots\left(k_s+1\right)$, with $k_i=1$ for all $i$
Thus $\tau(n)=(1+1)(1+1) \cdots(1+1)=2 \cdot 2 \cdots 2=2^s$ as there are $s$ terms
Problem 6.3.3. Prove that the following statements are equivalent:

1. $\tau(n)$ is an odd integer.

## 数学代写|数论作业代写number theory代考|Greatest Integer Function

In this section we are going to discuss a special type of arithmetic function called greatest integer function. The domain of definition of this function is the set of real numbers and the range set is the set of integers. This function is very much useful for calculating continued fractions. The definition of the function as follows.

Definition 6.6.1. For an arbitrary real number $x$, the largest integer less than or equal to $x$ and denoted by $[x]$ is called the greatest integer function.

For an example we have $[2.2]=2$ and $[-2.2]=-3$. Here for every real number $x$, there is a unique real number $\theta$ such that $x=[x]+\theta, 0 \leq \theta<1$, where $\theta$ is the fractional part of $x$. This $\theta$ sometimes denoted as ${x}$ such that $x=[x]+{x}, \forall x \in \mathbb{R}$. Actually the greatest integer function for any real number $x$ follows the inequality $x-1<[x] \leq x$. In our next proposition we have shown division algorithm using this inequality.

Proposition 6.6.1. For any $x \in \mathbb{R}$, prove division algorithm by the inequality $x-1<[x] \leq x$

Proof. Let $q=\left[\frac{m}{n}\right]$ and $r=m-n\left[\frac{m}{n}\right]$, clearly $m=n q+r$ and we will show that the remainder satisfies the above inequality. As $\frac{m}{n} \in \mathbb{R}$ then $\left(\frac{m}{n}\right)-1<\left[\frac{m}{n}\right] \leq \frac{m}{n}$. Now multiplying by $-n$ the above inequality and changing the order of inequality we have, $-m \leq-n\left[\frac{m}{n}\right]<n-m$. Adding $m$ we get, $0 \leq m-n\left[\frac{m}{n}\right]<n \Rightarrow$ $0 \leq r<n$. We are to show this $q$ and $r$ are unique. Let us assume that they are not unique then $m=n q_1+r_1=n q_2+r_2$ for $q_1, q_2$ are quotients and $0 \leq r_1, r_2<n$ where $r_1, r_2$ are remainders. Now subtracting these two equations we have, $0=n\left(q_1-q_2\right)+\left(r_1-r_2\right)$ thus $\left(r_2-r_1\right)=n\left(\left(q_1-q_2\right)\right)$ which implies $n \mid\left(r_1-r_2\right)$ but this is possible only if $r_1-r_2=0$. Therefore $r_1=r_2$ and $q_1-q_2$, which shows that $q$ is unique quotient and $r$ is unique remainder.

Now we will discuss few properties related to this greatest integer function.

# 数论代考

## 数学代写|数论作业代写number theory代考|The Sum and Number of Divisors

$$\sigma\left(m^2\right)=\frac{5^3-1}{5-1} \cdot \frac{k^3-1}{k-1}=31\left(\frac{k^3-1}{k-1}\right) \sigma\left(n^2\right)=\frac{2^5-1}{2-1} \cdot \frac{k^3-1}{k-1}=31\left(\frac{k^3-1}{k-1}\right)$$

$$\tau(n)=\left(k_1+1\right)\left(k_2+1\right) \cdots\left(k_s+1\right) \text { ，和 } k_i=1 \text { 对所有人 } i$$

1. $\tau(n)$ 是一个奇数。

## 数学代写|数论作业代写number theory代考|Greatest Integer Function

$-m \leq-n\left[\frac{m}{n}\right]<n-m$. 添加 $m$ 我们得到， $0 \leq m-n\left[\frac{m}{n}\right]<n \Rightarrow 0 \leq r<n$. 我们要展示这个 $q$ 和 $r$ 是独一无二的。让我们假设它们不是唯一的 $m=n q_1+r_1=n q_2+r_2$ 为了 $q_1, q_2$ 是商和
$0 \leq r_1, r_2<n$ 在哪里 $r_1, r_2$ 是余数。现在减去这两个方程，我们有， $0=n\left(q_1-q_2\right)+\left(r_1-r_2\right)$ 因 此 $\left(r_2-r_1\right)=n\left(\left(q_1-q_2\right)\right)$ 这意味着 $n \mid\left(r_1-r_2\right)$ 但这只有在 $r_1-r_2=0$. 所以 $r_1=r_2$ 和 $q_1-q_2$ ， 这表明 $q$ 是唯一的商并且 $r$ 是唯一余数。

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