GCD : Why does the GCD of two numbers divides their difference? I was working my way through some number theoretic proofs and being a newbie am stuck on this proof :

Why does the gcd of two numbers , say (a,b) - also divides their difference : a-b

My Question :
As I am a novice to induction and proofs , in general , can someone help me out ?
 A: For any common divisor, $d$, of $a$ and $b$, there exist integers $x$ and $y$ such that $a=dx$ and $b=dy$.  Then $a-b=dx-dy=d(x-y)$.  S0 $d$ divides $a-b$.
A: Let $\dfrac aA=\dfrac bB=d=(a,b)\implies a=dA,b=dB$
$ax+by=d(Ax+By)\iff\dfrac{ax+by}d=Ax+By$ which is an integer for any integers $x,y$
Set $x=1,y=-1$
A: That means we have integer numbers $m$, $n$, $c$ with
$$
c = \gcd(a,b)
$$
and 
$$
a = c \cdot m
$$
and 
$$
b = c \cdot n
$$
because $c$ is a common divisor (a divisor of both $a$ and $b$), which means that
$$
a - b = c \cdot m - c \cdot n = c \cdot (m - n)
$$
where $m - n$ is an integer as well. So $c$, the gcd, divides $a-b$ too.
As André Nicolas noted, the property, that it is the greatest of the common divisors is not needed.
A: It's good to start by introducing some notation that allows you to express some of the given information in the form of equations.  For instance, "let $d$ be the gcd of $a$ and $b$" is a good start.  Now, what does gcd mean? for one thing, it means that $d$ is a divisor of both $a$ and $b$, and you can use that to introduce some notation: "let $k$ and $\ell$ be integers such that $kd=a$ and $\ell d = b$".  Now you can do the same thing to the statement that you are trying to prove: "we wish to show that, for some integer $m$, $m d = a-b$".  Now you're almost done! Hint: find a general expression for $m$ in terms of $k$ and $\ell$.  Then "let $m=$" your expression, and show why $md=a-b$, and that completes the proof.
