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The problem is to prove: Let a,b belong to the set of integers, and a,b not equal to 0. Then for d belonging to the natural numbers, d=gcd(a,b) if and only if (1) d divides a and d divides b (2) if c is a common divisor of a and b then c divides d.

Bezout's lemma: For a,b belonging to the set of natural numbers, n belonging to the integers, we have gcd(a,b) divides n if and only if we can find x,y belonging to the integers such that n=ax+by. I.e. The set {ax+by:x,y belong to the integers} is precisely the set of integer multiples of gcd(a,b).

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  • $\begingroup$ What are your thoughts on this problem? $\endgroup$ – paw88789 Nov 13 '14 at 11:52
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Definition: If $a$ and $b$ are integers with one of them not zero, then the greatest common divisor, $gcd(a,b)$, of $a$ and $b$, is the positive integer $d$ satisfying:

(1) $d \mid a$ and $d \mid b$.

(2) If $c \mid a$ and $c \mid b$, then $c \leq d$.

Theorem: Suppose $a$, $b$ are integers not both $0$. Then, for an integer $d>0$, $d=gcd(a,b)$ if and only if

(a) $d \mid a$ and $d \mid b$.

(b) If $c \mid a$ and $c \mid b$, then $c \mid d$.

Proof: Suppose $d=gcd(a,b)$. Then (a) is (1) in the definition. Now, taking $n=d$ in Bezout's Lemma gives $d=ax+by$ for certain integers $x$ and $y$; so, if $c \mid a$ and $c \mid b$, we have $a=cm$ and $b=cn$ for certain integers $m$ and $n$, so that $$d=(cm)x+(cn)y=c(mx+ny).$$ Then (b) follows. Conversely, suppose $d>0$ is an integer and that (a) and (b) hold. Then (1) is (a) from the statement. Now, if $c \mid a$ and $c \mid b$, then, by (b), $c \mid d$, so that $c \leq d$. This is (2) from the defintion. Therefore $d=gcd(a,b)$.

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