To prove a property of greatest common divisor Suppose integer $d$ is the greatest common divisor of integer $a$ and $b$,
how to prove, there exist whole number $r$ and $s$, so that 
$$d = r \cdot a + s \cdot b $$
?
i know a proof in abstract algebra, hope to find a number theory proof?
for abstract algebra proof, it's in Michael Artin's book "Algebra".

An approach through elementary number-theory:
It suffices to prove this for relatively prime $a$ and $b$, so suppose this is so. Denote the set of integers $0\le k\le b$ which is relatively prime to $b$ by $\mathfrak B$. Then $a$ lies in the residue class of one of elements in $\mathfrak B$.
Define a map $\pi$ from $\mathfrak B$ into itself by sending $k\in \mathfrak B$ to the residue class of $ka$. If $k_1a\equiv k_2a\pmod b$, then, as $\gcd (a,b)=1$, $b\mid (k_1-k_2)$, so that $k_1=k_2$ (Here $k_1$ and $k_2$ are positive integers less than $b$.). Hence this map is injective. Since the set $\mathfrak B$ is finite, it follows that $\pi$ is also surjectie.
So there is some $k$ such that $ka\equiv 1\pmod b$. This means that there is some $l$ with $ka-1=lb$, i.e. $ka-lb=1$.
Barring mistakes. Thanks and regards then.  

P.S. The reduction step is: Given $a, b$ with $\gcd(a,b)=d$, we know that $\gcd(\frac{a}{d},\frac{b}{d})=1$. So, if the relatively prime case has been settled, then there are $m$ and $n$ such that $m\frac{a}{d}+n\frac{b}{d}=1$, and hence $ma+nb=d$.

