## Section6.1The Integers

Before we consider a theory of polynomials, we will consider some basic properties of a very familiar ring: the integers. We have seen that $\Z$ is an integral domain. The rings of polynomials we will consider will also be integral domains. A theme for this chapter is that many of the properties of the integers can be generalized to polynomials. But first, we need to consider what those properties are.

We begin with the question of division. Since $\Z$ is not a field, not every element has a multiplicative inverse (in fact, only $1$ and $-1$ do). But sometimes, we can still divide. Can we always? In a sense, yes!

###### Proof.

This is a perfect example of the existence-and-uniqueness type of proof. We must first prove that the numbers $q$ and $r$ actually exist. Then we must show that if $q'$ and $r'$ are two other such numbers, then $q = q'$ and $r = r'\text{.}$

Existence of $q$ and $r\text{.}$ Let

\begin{equation*} S = \{ a - bk : k \in {\mathbb Z} \text{ and } a - bk \geq 0 \}\text{.} \end{equation*}

If $0 \in S\text{,}$ then $b$ divides $a\text{,}$ and we can let $q = a/b$ and $r = 0\text{.}$ If $0 \notin S\text{,}$ we can use the Well-Ordering Principle. We must first show that $S$ is nonempty. If $a \gt 0\text{,}$ then $a - b \cdot 0 \in S\text{.}$ If $a \lt 0\text{,}$ then $a - b(2a) = a(1 - 2b) \in S\text{.}$ In either case $S \neq \emptyset\text{.}$ By the Well-Ordering Principle, $S$ must have a smallest member, say $r = a - bq\text{.}$ Therefore, $a = bq + r\text{,}$ $r \geq 0\text{.}$ We now show that $r \lt b\text{.}$ Suppose that $r \gt b\text{.}$ Then

\begin{equation*} a - b(q + 1)= a - bq - b = r - b \gt 0\text{.} \end{equation*}

In this case we would have $a - b(q + 1)$ in the set $S\text{.}$ But then $a - b(q + 1) \lt a - bq\text{,}$ which would contradict the fact that $r = a - bq$ is the smallest member of $S\text{.}$ So $r \leq b\text{.}$ Since $0 \notin S\text{,}$ $r \neq b$ and so $r \lt b\text{.}$

Uniqueness of $q$ and $r\text{.}$ Suppose there exist integers $r\text{,}$ $r'\text{,}$ $q\text{,}$ and $q'$ such that

\begin{equation*} a = bq + r, 0 \leq r \lt b \quad \text{and}\quad a = bq' + r', 0 \leq r' \lt b\text{.} \end{equation*}

Then $bq + r = bq' + r'\text{.}$ Assume that $r' \geq r\text{.}$ From the last equation we have $b(q - q') = r' - r\text{;}$ therefore, $b$ must divide $r' - r$ and $0 \leq r'- r \leq r' \lt b\text{.}$ This is possible only if $r' - r = 0\text{.}$ Hence, $r = r'$ and $q = q'\text{.}$

Let $a$ and $b$ be integers. If $b = ak$ for some integer $k\text{,}$ we write $a \mid b\text{.}$ An integer $d$ is called a common divisor of $a$ and $b$ if $d \mid a$ and $d \mid b\text{.}$ The greatest common divisor of integers $a$ and $b$ is a positive integer $d$ such that $d$ is a common divisor of $a$ and $b$ and if $d'$ is any other common divisor of $a$ and $b\text{,}$ then $d' \mid d\text{.}$ We write $d = \gcd(a, b)\text{;}$ for example, $\gcd( 24, 36) = 12$ and $\gcd(120, 102) = 6\text{.}$ We say that two integers $a$ and $b$ are relatively prime if $\gcd( a, b ) = 1\text{.}$

###### Proof.

Let

\begin{equation*} S = \{ am + bn : m, n \in {\mathbb Z} \text{ and } am + bn \gt 0 \}\text{.} \end{equation*}

Clearly, the set $S$ is nonempty; hence, by the Well-Ordering Principle $S$ must have a smallest member, say $d = ar + bs\text{.}$ We claim that $d = \gcd( a, b)\text{.}$ Write $a = dq + r'$ where $0 \leq r' \lt d\text{.}$ If $r' \gt 0\text{,}$ then

\begin{align*} r'& = a - dq\\ & = a - (ar + bs)q\\ & = a - arq - bsq\\ & = a( 1 - rq ) + b( -sq )\text{,} \end{align*}

which is in $S\text{.}$ But this would contradict the fact that $d$ is the smallest member of $S\text{.}$ Hence, $r' = 0$ and $d$ divides $a\text{.}$ A similar argument shows that $d$ divides $b\text{.}$ Therefore, $d$ is a common divisor of $a$ and $b\text{.}$

Suppose that $d'$ is another common divisor of $a$ and $b\text{,}$ and we want to show that $d' \mid d\text{.}$ If we let $a = d'h$ and $b = d'k\text{,}$ then

\begin{equation*} d = ar + bs = d'hr + d'ks = d'(hr + ks)\text{.} \end{equation*}

So $d'$ must divide $d\text{.}$ Hence, $d$ must be the unique greatest common divisor of $a$ and $b\text{.}$

### Subsection6.1.1The Euclidean Algorithm

Among other things, Theorem 6.2 allows us to compute the greatest common divisor of two integers.

###### Example6.4.

Let us compute the greatest common divisor of $945$ and $2415\text{.}$ First observe that

\begin{align*} 2415 & = 945 \cdot 2 + 525\\ 945 & = 525 \cdot 1 + 420\\ 525 & = 420 \cdot 1 + 105\\ 420 & = 105 \cdot 4 + 0\text{.} \end{align*}

Reversing our steps, $105$ divides $420\text{,}$ $105$ divides $525\text{,}$ $105$ divides $945\text{,}$ and $105$ divides $2415\text{.}$ Hence, $105$ divides both $945$ and $2415\text{.}$ If $d$ were another common divisor of $945$ and $2415\text{,}$ then $d$ would also have to divide $105\text{.}$ Therefore, $\gcd( 945, 2415 ) = 105\text{.}$

If we work backward through the above sequence of equations, we can also obtain numbers $r$ and $s$ such that $945 r + 2415 s = 105\text{.}$ Observe that

\begin{align*} 105 & = 525 + (-1) \cdot 420\\ & = 525 + (-1) \cdot [945 + (-1) \cdot 525]\\ & = 2 \cdot 525 + (-1) \cdot 945\\ & = 2 \cdot [2415 + (-2) \cdot 945] + (-1) \cdot 945\\ & = 2 \cdot 2415 + (-5) \cdot 945\text{.} \end{align*}

So $r = -5$ and $s= 2\text{.}$ Notice that $r$ and $s$ are not unique, since $r = 41$ and $s = -16$ would also work.

To compute $\gcd(a,b) = d\text{,}$ we are using repeated divisions to obtain a decreasing sequence of positive integers $r_1 \gt r_2 \gt \cdots \gt r_n = d\text{;}$ that is,

\begin{align*} b & = a q_1 + r_1\\ a & = r_1 q_2 + r_2\\ r_1 & = r_2 q_3 + r_3\\ & \vdots\\ r_{n - 2} & = r_{n - 1} q_{n} + r_{n}\\ r_{n - 1} & = r_n q_{n + 1}\text{.} \end{align*}

To find $r$ and $s$ such that $ar + bs = d\text{,}$ we begin with this last equation and substitute results obtained from the previous equations:

\begin{align*} d & = r_n\\ & = r_{n - 2} - r_{n - 1} q_n\\ & = r_{n - 2} - q_n( r_{n - 3} - q_{n - 1} r_{n - 2} )\\ & = -q_n r_{n - 3} + ( 1+ q_n q_{n-1} ) r_{n - 2}\\ & \vdots\\ & = ra + sb\text{.} \end{align*}

The algorithm that we have just used to find the greatest common divisor $d$ of two integers $a$ and $b$ and to write $d$ as the linear combination of $a$ and $b$ is known as the Euclidean algorithm.

### Subsection6.1.2Prime Numbers

Let $p$ be an integer such that $p \gt 1\text{.}$ We say that $p$ is a prime number, or simply $p$ is prime, if the only positive numbers that divide $p$ are $1$ and $p$ itself. An integer $n \gt 1$ that is not prime is said to be composite.

###### Proof.

Suppose that $p$ does not divide $a\text{.}$ We must show that $p \mid b\text{.}$ Since $\gcd( a, p ) = 1\text{,}$ there exist integers $r$ and $s$ such that $ar + ps = 1\text{.}$ So

\begin{equation*} b = b(ar + ps) = (ab)r + p(bs)\text{.} \end{equation*}

Since $p$ divides both $ab$ and itself, $p$ must divide $b = (ab)r + p(bs)\text{.}$

###### Proof.

We will prove this theorem by contradiction. Suppose that there are only a finite number of primes, say $p_1, p_2, \ldots, p_n\text{.}$ Let $P = p_1 p_2 \cdots p_n + 1\text{.}$ Then $P$ must be divisible by some $p_i$ for $1 \leq i \leq n\text{.}$ In this case, $p_i$ must divide $P - p_1 p_2 \cdots p_n = 1\text{,}$ which is a contradiction. Hence, either $P$ is prime or there exists an additional prime number $p \neq p_i$ that divides $P\text{.}$

###### Proof.

Uniqueness. To show uniqueness we will use induction on $n\text{.}$ The theorem is certainly true for $n = 2$ since in this case $n$ is prime. Now assume that the result holds for all integers $m$ such that $1 \leq m \lt n\text{,}$ and

\begin{equation*} n = p_1 p_2 \cdots p_k = q_1 q_2 \cdots q_l\text{,} \end{equation*}

where $p_1 \leq p_2 \leq \cdots \leq p_k$ and $q_1 \leq q_2 \leq \cdots \leq q_l\text{.}$ By Lemma 6.5 , $p_1 \mid q_i$ for some $i = 1, \ldots, l$ and $q_1 \mid p_j$ for some $j = 1, \ldots, k\text{.}$ Since all of the $p_i$'s and $q_i$'s are prime, $p_1 = q_i$ and $q_1 = p_j\text{.}$ Hence, $p_1 = q_1$ since $p_1 \leq p_j = q_1 \leq q_i = p_1\text{.}$ By the induction hypothesis,

\begin{equation*} n' = p_2 \cdots p_k = q_2 \cdots q_l \end{equation*}

has a unique factorization. Hence, $k = l$ and $q_i = p_i$ for $i = 1, \ldots, k\text{.}$

Existence. To show existence, suppose that there is some integer that cannot be written as the product of primes. Let $S$ be the set of all such numbers. By the Principle of Well-Ordering, $S$ has a smallest number, say $a\text{.}$ If the only positive factors of $a$ are $a$ and $1\text{,}$ then $a$ is prime, which is a contradiction. Hence, $a = a_1 a_2$ where $1 \lt a_1 \lt a$ and $1 \lt a_2 \lt a\text{.}$ Neither $a_1\in S$ nor $a_2 \in S\text{,}$ since $a$ is the smallest element in $S\text{.}$ So

\begin{align*} a_1 & = p_1 \cdots p_r\\ a_2 & = q_1 \cdots q_s\text{.} \end{align*}

Therefore,

\begin{equation*} a = a_1 a_2 = p_1 \cdots p_r q_1 \cdots q_s\text{.} \end{equation*}

So $a \notin S\text{,}$ which is a contradiction.

### Subsection6.1.3Historical Note

Prime numbers were first studied by the ancient Greeks. Two important results from antiquity are Euclid's proof that an infinite number of primes exist and the Sieve of Eratosthenes, a method of computing all of the prime numbers less than a fixed positive integer $n\text{.}$ One problem in number theory is to find a function $f$ such that $f(n)$ is prime for each integer $n\text{.}$ Pierre Fermat (1601?– 1665) conjectured that $2^{2^n} + 1$ was prime for all $n\text{,}$ but later it was shown by Leonhard Euler (1707– 1783) that

\begin{equation*} 2^{2^5} + 1 = 4{,}294{,}967{,}297 \end{equation*}

is a composite number. One of the many unproven conjectures about prime numbers is Goldbach's Conjecture. In a letter to Euler in 1742, Christian Goldbach stated the conjecture that every even integer with the exception of $2$ seemed to be the sum of two primes: $4 = 2 + 2\text{,}$$6 = 3 + 3\text{,}$ $8 =3 + 5\text{,}$$\ldots\text{.}$ Although the conjecture has been verified for the numbers up through $4 \times 10^{18}\text{,}$ it has yet to be proven in general. Since prime numbers play an important role in public key cryptography, there is currently a great deal of interest in determining whether or not a large number is prime.

###### 1.

What is the Division Algorithm? Illustrate your answer with the exampls (a) $a = 4$ and $b = 15$ and (b) $a = -3$ and $b = 5\text{.}$

###### 2.

Use the definition of greatest common divisor given in the section to explain why $\gcd(28, 42) = 14\text{.}$

###### 3.

Is 1 a prime number? Answer this question based on the definition in this section and explain.

###### 4.

After reading the section, what questions do you still have? Write at least one well formulated question (even if you think you understand everything).

### Exercises6.1.5Exercises

###### 1.

For each of the following pairs of numbers $a$ and $b\text{,}$ calculate $\gcd(a,b)$ and find integers $r$ and $s$ such that $\gcd(a,b) = ra + sb\text{.}$

1. $14$ and $39$

2. $234$ and $165$

3. $1739$ and $9923$

4. $471$ and $562$

5. $23771$ and $19945$

6. $-4357$ and $3754$

###### 2.

Let $a$ and $b$ be nonzero integers. If there exist integers $r$ and $s$ such that $ar + bs =1\text{,}$ show that $a$ and $b$ are relatively prime.

###### 3.

Let $a$ and $b$ be integers such that $\gcd(a,b) = 1\text{.}$ Let $r$ and $s$ be integers such that $ar + bs = 1\text{.}$ Prove that

\begin{equation*} \gcd(a,s) = \gcd(r,b) = \gcd(r,s) = 1\text{.} \end{equation*}
###### 4.

Let $x, y \in {\mathbb N}$ be relatively prime. If $xy$ is a perfect square, prove that $x$ and $y$ must both be perfect squares.

Hint

Use the Fundamental Theorem of Arithmetic.

###### 5.

Using the division algorithm, show that every perfect square is of the form $4k$ or $4k + 1$ for some nonnegative integer $k\text{.}$

###### 6.

Suppose that $a, b, r, s$ are pairwise relatively prime and that

\begin{align*} a^2 + b^2 & = r^2\\ a^2 - b^2 & = s^2\text{.} \end{align*}

Prove that $a\text{,}$ $r\text{,}$ and $s$ are odd and $b$ is even.

###### 7.

Let $n \in {\mathbb N}\text{.}$ Use the division algorithm to prove that every integer is congruent mod $n$ to precisely one of the integers $0, 1, \ldots, n-1\text{.}$ Conclude that if $r$ is an integer, then there is exactly one $s$ in ${\mathbb Z}$ such that $0 \leq s \lt n$ and $[r] = [s]\text{.}$ Hence, the integers are indeed partitioned by congruence mod $n\text{.}$

###### 8.

Define the least common multiple of two nonzero integers $a$ and $b\text{,}$ denoted by $\lcm(a,b)\text{,}$ to be the nonnegative integer $m$ such that both $a$ and $b$ divide $m\text{,}$ and if $a$ and $b$ divide any other integer $n\text{,}$ then $m$ also divides $n\text{.}$ Prove there exists a unique least common multiple for any two integers $a$ and $b\text{.}$

Hint

Use the Principle of Well-Ordering and the division algorithm.

###### 9.

If $d= \gcd(a, b)$ and $m = \lcm(a, b)\text{,}$ prove that $dm = |ab|\text{.}$

###### 10.

Show that $\lcm(a,b) = ab$ if and only if $\gcd(a,b) = 1\text{.}$

###### 11.

Prove that $\gcd(a,c) = \gcd(b,c) =1$ if and only if $\gcd(ab,c) = 1$ for integers $a\text{,}$ $b\text{,}$ and $c\text{.}$

###### 12.

Let $a, b, c \in {\mathbb Z}\text{.}$ Prove that if $\gcd(a,b) = 1$ and $a \mid bc\text{,}$ then $a \mid c\text{.}$

Hint

Since $\gcd(a,b) = 1\text{,}$ there exist integers $r$ and $s$ such that $ar + bs = 1\text{.}$ Thus, $acr + bcs = c\text{.}$

###### 13.

Let $p \geq 2\text{.}$ Prove that if $2^p - 1$ is prime, then $p$ must also be prime.

###### 14.

Prove that there are an infinite number of primes of the form $6n + 5\text{.}$

Hint

Every prime must be of the form $2\text{,}$ $3\text{,}$ $6n + 1\text{,}$ or $6n + 5\text{.}$ Suppose there are only finitely many primes of the form $6k + 5\text{.}$

###### 15.

Prove that there are an infinite number of primes of the form $4n - 1\text{.}$

###### 16.

Using the fact that $2$ is prime, show that there do not exist integers $p$ and $q$ such that $p^2 = 2 q^2\text{.}$ Demonstrate that therefore $\sqrt{2}$ cannot be a rational number.