Prove that $gcd(u \times a, u \times b) = |u| \times gcd(a,b)$

Question

The question: "Prove $gcd(u \times a, u \times b) = |u| \times gcd(a,b)$"

We have the Theorem which says that for any intergers $a$ and $b$ both not $0$, have a positive gcd which can be expressed as $gcd(a,b) = x\times a + y \times b$.

My attempts:

First, let $gcd(a,b) = x \times a + y \times b$, but let denote $gcd(a,b)$ as $d$, since $d$ divides $a$ and $b$, it divides $u \times a$ and $u \times b$ also, thus it is a positive common divisor of $u \times a$ and $u \times b$.

Now, let $gcd(u \times a, u \times b) = x' \times (u \times a) + y' \times (u \times b)$, but lets denote $gcd(u \times a, u \times b$) as $p$, since $d$ is a positive common divisor of $u \times a$ and $u \times b$, then $d$ must divide $p$, from which follows that $p = k \times d$ for some integer $k$, but it is the same as: $$gcd(u\times a, u \times b) = k \times gcd(a,b)$$

And since $gcd(u \times a, u \times b) = x'\times (u \times a) + y' \times (u \times b)$, it follows that $gcd(u \times a, u \times b) = u \times (x' \times a + y' \times b)$ then i got: $$gcd(u\times a, u \times b) = k \times (x \times a + y \times b) = u \times (x' \times a + y' \times b)$$

But I got stucked here, then I choosed to try another approach, first trying to prove the equality is valied to all positive integers $u$ using induction.

Let $B = \{u : u \in \mathbb{N} \land [gcd(u \times a, u \times b) = u \times gcd(a,b)]\}$.

First $1 \in B$ since $gcd(1 \times a, 1 \times b) = gcd(a,b) = 1 \times gcd(a,b)$.

Now assume $u \in B$ that is $gcd(u \times a, u \times b) = u \times gcd(a,b)$, now if we take $(u+1) \times gcd(a,b)$ we got $u \times gcd(a,b) + gcd(a,b)$ and by the assumption it is the same as $gcd(u \times a, u \times b) + gcd(a,b)$.

But I got stucked again, I will like to know if its possible to prove $gcd((u+1) \times a, (u+1) \times b) = gcd(u×a,u×b)+gcd(a,b)$.

I really want to know if its possible to continue from some of these two approaches, but other paths will be appreciated too.

Edit: I just noticed the answer from @blackmirror7, add the information to finish my first attempt, I dont think is needed to provide more information on the first approach, but meanwhile I managed to reach the answer following other path as follows:

First assume $a,b$ and $u$ are not $0$ because these are trivial cases, then we can proceed as follows:

$gcd(a,b) = d, \text{such that } 0<d$

Then $a=dm$ and $b=dn$ for some integers $m$ and $n$, and $gcd(m,n)=1$, for if there is some positive common divisor of $m$ and $n$ bigger than $1$, let say $c$ we got:

$m=cr$ and $n=cs$, for some $r,s$, thus $a=dcr$ and $b=dcs$ and $dc$ divides both $a$ and $b$. but as $1<c$ then $d<dc$, which contradicts the fact that $gcd(a,b)=d$.

Now let $gcd(ua, ub)=k$, since $ua=udm$ and $ub=udn$ we know that $ud$ divides both $ua$ and $ub$ then it must divide $k$ too, thus we have $k = pud$ for some $p$.

Since $k$ divides both $ua$ and $ub$, we have $ua = xk$ and $ub = yk$, for some $x,y$, but that is $udm = xpud$ and $udn=ypud$, from which follows $m=xp$ and $n=yp$, but by our previous result that $gdc(m,n)=1$ and the fact $p$ is a common divisor of $m$ and $n$ we have that $p=1$ or $p=-1$.

Finally, since $ua$ and $ub$ are not $0$, we know that $0<k=gcd(ua,ub)$, then $|k| = k$, but $|k| = |pud| = |p|.|u|.|d|$, and since $|p|=1$ and $0<d$ we got $k = |u|.d$, that is $gcd(ua,ub) = |u|.gcd(a,b)$.

I know that the first approach using the Bézout's identity is much more short, but i think its worth to present another approaches too, Hope it help someone.

Answer 1

Let gcd(ua,ub)=k

gcd(ua,ub)= smallest positive integer of the form uax+uby for some integers x,y.

uax+uay=u(ax+by)=k and by the same property of gcd, we have u(ax+by)=u(gcd(a,b))