面临数论的学习挑战吗？别担心！我们的number-theory-guide团队专业为您解决整数理论、素数理论、同余理论、费马定理等方面的问题。我们拥有深厚的专业背景和丰富的经验，能帮您完成高水平的作业和论文，让您的学习之路一路顺风！

以下是一些我们可以帮助您解决的问题：

数论基础：素数定理、质数分布、同余理论等。

整数和素数理论：整数的性质、质因数分解、素数的分布等。

代数数论：代数整数、域的理论、伽罗华理论等。

解析数论和概率数论：黎曼zeta函数、素数定理、费马定理等。

其他相关主题，如：数论在密码学、计算机科学、组合数学中的应用、数论算法、整数序列等。

Use this method (called the Euclidean algorithm) to find the following greatest common divisors. We shall study this algorithm in much more detail later in the term.

$$

\operatorname{gcd}(64,81), \operatorname{gcd}(6,121), \operatorname{gcd}(169,273), \operatorname{gcd}(51,187), \operatorname{gcd}(999,2187) \text {. }

$$

The following is maybe the most important elementary fact about greatest common divisors and relatively prime integers.

The Euclidean algorithm is a method for computing the greatest common divisor (gcd) of two integers. The key idea is to repeatedly apply the algorithm to pairs of integers where the larger integer is replaced by the difference between the two. We stop when one of the integers is 0, and the other one will be the gcd.

Here’s how you would compute the gcd for the given pairs:

$\text{gcd}(64, 81)$: 81 and 64 have no common factor other than 1 since one is odd and the other is even. Hence, $\text{gcd}(64, 81) = 1$.

$\text{gcd}(6, 121)$: 121 is odd and 6 is even. They have no common factor other than 1. Hence, $\text{gcd}(6, 121) = 1$.

$\text{gcd}(169, 273)$: Using the Euclidean algorithm, 273 = 169*1 + 104, 169 = 104*1 + 65, 104 = 65*1 + 39, 65 = 39*1 + 26, 39 = 26*1 + 13, 26 = 13*2, So, $\text{gcd}(169, 273) = 13$.

$\text{gcd}(51, 187)$: The Euclidean algorithm yields: 187 = 51*3 + 34, 51 = 34*1 + 17, 34 = 17*2, So, $\text{gcd}(51, 187) = 17$.

$\text{gcd}(999, 2187)$: The Euclidean algorithm gives: 2187 = 999*2 + 189, 999 = 189*5 + 84, 189 = 84*2 + 21, 84 = 21*4, So, $\text{gcd}(999, 2187) = 21$.

As for the elementary fact about greatest common divisors and relatively prime integers, it might be the statement that two integers are relatively prime if and only if their gcd is 1.

Prove this. Hint: This is one of many times in the near future we will be using the assumption $\operatorname{gcd}(a, b)=1$ by using that there are integers $x$ and $y$ with $a x+b x=1$. As we have $a \mid b c$ there is a integer $k$ such $b c=a k$. Multiply $a x+b x=1$ by $c$ to get $a c x+b c x=c$. Now replace $b c$ by $a k$ and the result is $a c x+a k x=c$. Now it should be easy to show $a \mid c$.

The following is really a corollary to the last result, but is important enough to get promoted to a Proposition.

The goal is to prove that if $\text{gcd}(a, b) = 1$ and $a | bc$, then $a | c$.

We start with the assumptions that $\text{gcd}(a, b) = 1$ and $a | bc$. This means there exist integers $x$ and $y$ such that $ax + by = 1$, and there exists an integer $k$ such that $bc = ak$.

Multiplying the equation $ax + by = 1$ by $c$ gives us $acx + bcy = c$. Then, we substitute $ak$ for $bc$ in this equation to get $acx + aky = c$.

Now, factoring out the $a$ from the left side of the equation gives $a(cx + ky) = c$. Since $cx + ky$ is an integer (let’s call it $n$), we have shown that $c = an$ for some integer $n$.

Therefore, we have shown that $a | c$ given the original assumptions, which completes the proof.

The proposition that follows from this result might be something like: If $\text{gcd}(a, b) = 1$ and $a | bc$, then $\text{gcd}(a, c) = 1$. This is because if $a | c$, as we’ve shown, then $a$ is a common divisor of $a$ and $c$, and hence the greatest common divisor of $a$ and $c$ is not greater than $a$ itself. So the gcd cannot be any number other than 1.

**E-mail:** help-assignment@gmail.com **微信:**shuxuejun

**help-assignment™是一个服务全球中国留学生的专业代写公司专注提供稳定可靠的北美、澳洲、英国代写服务专注于数学，统计，金融，经济，计算机科学，物理的作业代写服务**