Let $a,b,x \in \mathbb{N}$ and $u,v\in\mathbb{Z}$ so that $x=ua+vb$. If $x\mid a$ and $x\mid b$ then $x=\gcd(a,b)$.
My idea was to perform a proof by contradiction. Assume $x$ is not the $\gcd$, then let $y=\gcd(a,b)$ and clearly since $x$ still divides $a$ and $b$, $y>x$. Then writing $y$ as $y=sa+tb=skx+tlx=x(sk+tl)>x$ but I can't seem to find a contradiction from this point on. MaybeI should do this proof in another way.
Any help would be appreciated!
My answer relies on the following theorem:
If $a$ and $b$ are co-prime natural numbers, then there exist integers $u$ and $v$ such that $ua+vb=1$. The converse is also true.
Proof:Since $x|a$ and $x|b$, let $a=kx, b=lx,$ where $k,x\in\mathbb{N}$. Given, $x=ua+vb\implies x=ukx+vbx\implies 1=uk+vb$. Now, since we know that $u,v\in\mathbb{Z}$, due to the converse of the above theorem, we conclude that $k$ and $l$ are co-prime. This means $x$ is the greatest common factor of $a$ and $b$.(Other common factor will be factors of $x$).
