For any integers \(a,b\) with \(a \not= 0\text{,}\) there exists unique integers \(q\) and \(r\) for which
\begin{equation*} b = aq + r, \quad 0 \le r \lt \abs{a}. \end{equation*}The intger \(b\) is called the dividend, \(a\) the divisor, \(q\) the quotient, and \(r\) the remainder.
Saved for a later chapter.
Let \(a,b,c \in \mathbb{Z}\) be nonzero. We say \(c\) is a common divisor of \(a\) and \(b\) if and only if \(c\) divides \(a\) and \(c\) divides \(b\text{.}\) The greatest common divisor of \(a,b\) is denoted \(\gcd (a,b)\text{,}\) and is the common divisor of \(a,b\) which is greater than every other common divisor.
For any \(a,b,x,y \in \mathbb{Z}\text{,}\) any divisor of \(a,b\) also divides \(ax+by\text{.}\) \(ax+by\) is called a linear combination of \(a,b\text{.}\)
Let \(a,b,x,y \in \mathbb{Z}\) be arbitrary. Let \(c\) be any divisor of \(a\) and \(b\) so there exist integers \(r,s\) so that
\begin{equation*} a = rc, \quad b = sc. \end{equation*}Thus
\begin{equation*} ax + by = (rc)x + (sc)y = c (rx+sy). \end{equation*}Since \(rx+sy \in \mathbb{Z}\text{,}\) we have shown \(c\) divides \(ax+by\text{.}\)
The following lemma is the first proof we encounter where the key idea to get started is not obvious, even after a bit of playing around. At this point in the course, you would not be expected to come up with a proof like this, although later in the course you would.
For any nonzero \(a,b \in \mathbb{Z}\text{,}\) the smallest positive linear combination of \(a,b\) is a common divisor.
Let \(a,b\) be arbitrary nonzero integers. Assume \(x,y \in \mathbb{Z}\) so that \(s := ax+by\) is positive, but as small as possible (i.e., it is the smallest positive linear combination of \(a,b\)).
By Division Algorithm, we may write \(a = sq + r\) for some \(q,r \in \mathbb{Z}\) with \(0 \le r \lt \abs{s} = s\text{.}\) Now
\begin{equation*} r = a - sq = a - (ax + by)q = a(1-xq) + b(yq) \end{equation*}is another (nonnegative) linear combination of \(a,b\text{.}\) But we know \(r\) cannot be positive, otherwise it would violate the minimality of \(s\text{.}\) Therefore, \(r = 0\) and hence \(a = sq\text{.}\) Thus \(s\) divides \(a\text{.}\) A nearly identical argument shows \(s\) divides \(b\text{,}\) and so \(s\) is a common divisor.
For any nonzero integers \(a,b\text{,}\) their greatest common divisor is their smallest linear combination.
Let \(a,b\) be arbitrary nonzero integers and let \(s\) denote their smallest positive linear combination. By Lemma 1.7.4 \(s\) is a common divisor of \(a,b\text{.}\) Let \(x\) be any other positive common divisor of \(a,b\text{.}\) By Theorem 1.7.3, we know \(t\) divides \(s\text{,}\) and so there is some (positive) integer \(k\) for which \(s = tk\text{.}\) Since \(k \in \mathbb{N}\text{,}\) we have \(k \ge 1\) and so \(s = tk \ge t\text{.}\) Therefore \(s\) is greater than any other common divisor and hence \(s = \gcd(a,b)\text{.}\)
We say nonzero integers \(a,b\) are relatively prime, or coprime, if \(\gcd(a,b) = 1\text{.}\)
For any relatively prime \(a,b \in \mathbb{Z}\) (necessarily nonzero) and for any \(c \in \mathbb{Z}\) there exist \(x,y \in \mathbb{Z}\) so that \(ax+by = c\text{.}\)
Suppose \(a,b \in \mathbb{Z}\) are relatively prime. By Theorem 1.7.5 there exist \(r,s\) so that \(ar+bs = 1\text{.}\) Then
\begin{equation*} a(rc)+b(sc) = (ar+bs)c = c. \end{equation*}Setting \(x = rc, y = sc \in \mathbb{Z}\) proves the result.
Let \(a,b,p \in \mathbb{Z}\) with \(p\) prime. If \(p\) divides \(ab\text{,}\) then either \(p\) divides \(a\) or \(p\) divides \(b\text{.}\)
Let \(a,b,p \in \mathbb{Z}\) with \(p\) prime and suppose \(p\) divides \(ab\text{.}\) Thus \(ab = pk\) for some integer \(k\text{.}\) Note that \(\gcd(a,p)\) can be only either \(1\) or \(p\) since \(p\) has no other divisors.
Then \(p\) divides \(a\) and we are finished.
By Theorem 1.7.5 there exist \(x,y \in \mathbb{Z}\) so that \(ax + py = 1\text{.}\) Multiplying both sides by \(b\text{,}\) we obtain
\begin{equation*} b = (ax+py)b = abx+pyb = pkx+pyb = p(kx+yb). \end{equation*}Therefore \(p\) divides \(b\text{.}\)