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Here is what I got so far:

Let $\gcd(ka,km) = d$

Then by the Euclidean Algorithm, we have integers $s,t$ such that:

$$s(ka) + t(km) = d \implies k|d $$

Let $$d/k = g = sa + tm. \tag{1}$$

So now I need to show that $g$ is indeed = $\gcd(a,m)$. Let $\gcd(a,m) = h$

Now since $k|d \implies d|a , d|m \space $ (since $$\gcd(ka,km) = d \implies g| \gcd(a,m) = h \\\implies h = dj ,$$ for some integer $j$.

Putting this into (1):

$$ h/jk = g = sa + tm $$

since $$h = dj \implies dj/j = kg \\ \implies d = kg. $$

So $g$ is the $\gcd(a,m)$?

Is this a correct proof? Or have I gone in some sort of circle? Many thanks for any hints or direction!

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    $\begingroup$ Looking at your proof it seems you have the logical steps correct in your mind but while typing here you have made lot of mistakes. For example $k\mid d$ implies that $g = d/k$ divides $a, m$ and not $d \mid a, d \mid m$. If you correct this part then you should get $h = gj$. Next you should try to establish $g = hi$ for some integer $i$. This will give $d/k = g = h$. $\endgroup$
    – Paramanand Singh
    Aug 28, 2013 at 9:04
  • $\begingroup$ You can get $g = hi$ in following way. Clearly $h$ divides $a, m$ so that $a = hl, m = hn$ so that $ g = sa + tm = shl + thn = h(sl + tn) = hi$. $\endgroup$
    – Paramanand Singh
    Aug 28, 2013 at 9:10

2 Answers 2

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You seem to be on the right track, although I'm not so sure about your last step. Also, I would avoid using fractions when possible. Here's my approach.

Let $g = \gcd(a,m)$ and let $d = \gcd(ka,km)$. We want to show that $d = kg$.

Since $d = \gcd(ka,km)$, we know by the Extended Euclidean Algorithm that $d = (ka)s + (km)t$ for some $s,t \in \Bbb Z$. But then observe that $d = k(as + mt)$. Hence, it suffices to prove that $g = as + mt$.

Since $g = \gcd(a,m)$, we know by the Extended Euclidean Algorithm that $g$ is the smallest possible integer that can be expressed in the form $ax + my$, where $x,y \in \Bbb Z$. But since $as + mt$ can also be expressed in this form, we know that $\boxed{g \leq as + mt}$.

Since $g=ax+my$, we may scale this equation by $k$ to obtain $kg = (ka)x + (km)y$. But recall that $d$ is the smallest possible integer that can be expressed in the form $(ka)s + (km)t$, where $s,t \in \Bbb Z$. Hence, we have $k(as + mt)=d \leq kg$ so that cancelling the $k$ from both sides yields the inequality $\boxed{as + mt \leq g}$ (we assume, without loss of generality, that $k\geq0$).

Finally, since $g \leq as + mt$ and $as + mt \leq g$, we have shown that $g=as+mt$, as desired.

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This is proven using Bezout's Identity in this answer.

Bezout says that $$ \gcd(a,b)=\inf\{ax+by\gt0:x,y\in\mathbb{Z}\} $$ Then we simply note that $$ \inf\{cax+cby\gt0:x,y\in\mathbb{Z}\} $$ is both $c\gcd(a,b)$ and $\gcd(ac,bc)$.

I believe this is very similar to your proof, but I think you need to exploit the fact that $\gcd(a,b)$ is the smallest positive $ax+by$ for any $x,y\in\mathbb{Z}$.

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