How to prove $\gcd(dm,dn)=d\cdot\gcd(m,n)$

Question

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?

Answer 1

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 $G=dg$.

Answer 2

I don't think you've shown that the two $t$-s are the same. Try using the formula that $\mbox{lcm}(a,b)\cdot\gcd(a,b)=ab$ to solve for $\gcd(dm,dn)$.

Answer 3

The $\gcd$ is the product of the prime factors of the two numbers, taken with their minimum multiplicities.

Multiplying both by a third number increases the multiplicities identically, and increases the minimum the same way (by $\min(a+c,b+c)=\min(a,b)+c$).