TITLE: How to prove $\gcd(dm,dn)=d\cdot\gcd(m,n)$
QUESTION [5 upvotes]: I want to prove the following equation :
$$
(dm,dn) = d\cdot(m,n) 
$$
where 
$$
(m,n) = \gcd(m,n) \\
(dm,dn) = \gcd(dm,dn)
$$
I tried this :
$$
(dm,dn) \rightarrow \exists g_1 \in Z : g_1|dm, g_1|dn \rightarrow g_1|(dm\cdot x+dn\cdot y) \rightarrow g_1|d\cdot (mx+ny) \\
\rightarrow g_1=\frac{d\cdot (mx+ny)}{t}
$$
And the same for $ (m,n)$ :
$$
g_2=\frac{mx+ny}{t}
$$
If i insert $g_1$ and $g_2$ i get :
$$
d\cdot\frac{mx+ny}{t}=d\cdot \frac{mx+ny}{t}
$$
Is this right?

REPLY [5 votes]: You can also deduce it by Bézout's Lemma:
Put $g:=(m,n)$ and $G:=(dm,dn)$. There exist integers $a,b$ such that $$an+bm=g.$$
Therefore $a(dn)+b(dm)=dg$, what implies that $G|dg$.
On the other hand, since $g|m$ and $g|n$ we get $dg|dm$ and $dg|dn$. Hence $dg|G$ and thus $G=dg$.