If $d$ is a common divisor of $m$ and $n$, then so it is of $n$ and $m-n$ I am having trouble proving the following statement: 

Prove that for all integers $m$ and $n$, if $d$ is a common divisor of $m$ and $n$ (but $d$ is not necessarily the GCD) then $d$ is a common divisor of $n$ and $m - n$.

I've noticed that for any integers $m,n,d$ that $m - n = kd$ (where $k$ is an integer as well). Any help or hints on how to prove this statement would be greatly appreciated.
 A: For integers $d,m$, we say that $d$ divides $m$ if $\frac md$ is an integer. That is to say, $m = m'd$ for some integer $m' = \frac md$.
So we know that $m = m'd, n = n'd$. Hence $m-n = \ldots$
A: If $d$ divides $m$, then $m=m_1d$ for some integer $m_1$. Similarly, if $d$ divides $n$, then $n=n_1d$ for some integer $n_1$. And hence
$$
m-n=d(m_1-n_1).
$$
A: Here is an answer that is a bit more formal than the others.  "$\;d\;$ divides $\;n\;$", often written as $\;d | n\;$, is defined like this:
$$
(0) \;\;\; d | n \;\equiv\; \langle \exists q \in \mathbb Z :: q \cdot d = n \rangle
$$
You are asked to prove
$$
d | m \:\land\: d | n \;\Rightarrow\; d | n \:\land\: d | (m-n)
$$
for all $\;d,n,m\;$, or logically equivalently
$$
(1) \;\;\; d | m \:\land\: d | n \;\Rightarrow\; d | (m-n)
$$
Starting with the consequent --since that is the most complex part-- we calculate as follows:
\begin{align}
& d | (m-n) \\
\equiv & \;\;\;\;\;\text{"definition $(0)$"} \\
& \langle \exists q \in \mathbb Z :: q \cdot d = m - n \rangle \\
\equiv & \;\;\;\;\;\text{"use assumptions $\;d | m\;$ and $\;d | n\;$ with $(0)$ to choose $\;q_1,q_2 \in \mathbb Z\;$"} \\
& \langle \exists q \in \mathbb Z :: q \cdot d = q_1 \cdot d - q_2 \cdot d \rangle \\
\equiv & \;\;\;\;\;\text{"divide by $\;d\;$; move $\;d = 0\;$ out of $\;\exists q\;$"} \\
& d = 0 \;\lor\; \langle \exists q \in \mathbb Z :: q = q_1 - q_2 \rangle \\
\equiv & \;\;\;\;\;\text{"one-point rule"} \\
& d = 0 \;\lor\; (q_1 - q_2) \in \mathbb Z \\
\equiv & \;\;\;\;\;\text{"right hand part is true since $\;q_1,q_2 \in \mathbb Z\;$"} \\
& \text{true} \\
\end{align}
This proves $(1)$.
