For every natural number $n$, $\gcd(an,bn)=n\gcd(a,b).$ For every natural number I am trying to show that $n$, $\gcd(an,bn)=n\gcd(a,b).$
Here is my attempt.   Put $d = \gcd(a,b)$; we can write  $d=aT+bJ$, where $T$
adn $J$ are integers. Then as $d\mid a$ and $d\mid b$, we can write the equation as $dgT + dhJ$ and if we multiple it by $n$ then it becomes $dngT + dnhJ$ and the common part is $dn$ hence dn is the $\gcd$. 
 A: What is $d$? is it an arbitrary integer?
Usually $d$ is the $\gcd(a,b)$.
If $d = \gcd(a,b) $, then it is clear how to prove the relation, isn't it?
finding the gcd of $an$ and $bn$ is not difficult if you know the gcd of $a$ and $b$.
A: $$
\begin{align}
\gcd(an,bn) = dn \Rightarrow anx+bny = dn \Rightarrow ax+by = d \Rightarrow \gcd(a,b) = d
\end{align}
$$
A: Use the Euclide algorithm. Assume $a>b$.
$$
a = bq + r, 0\le r < b\implies \gcd(a,b )= \gcd(b,r)
$$
now multiply by $n$:$$
na = nbq + nr, 0\le nr < nb\implies \gcd(na,nb )= \gcd(nb,nr)
$$
So at every step of the Euclide algorithm, everything will be multiplied by $n$. Hence so will be the end result:
$$
\gcd(na,nb) = n\gcd(a,b)
$$
A: $d=\gcd(a,b)$ iff $a=da'$ and $'b=db'$ with $\gcd(a',b')=1$.
Then $na=nda'$ and $nb=ndb'$ implies $nd=\gcd(na,nb)$
A: It's not sufficient to show that you can write $nd$ as a linear combination of $na$ and $nb$. This just shows that $nd$ is a multiple of $\gcd(na,nb)$.
However, if $e=\gcd(na,nb)$ and $nd=ke$, you have, for some integers $X$ and $Y$,
$$
e=naX+nbY
$$
and so
$$
ke=nd=nkaX+nkbY
$$
which implies
$$
d=k(aX+bY)
$$
Since $aX+bY=hd$ for some $h$, you get $hk=1$ so $k=1$.
Note: the greatest common divisor is always supposed to be positive.
