gcd$(a,b)=1$ and $ax=by\Rightarrow a|y$

Question

Define $gcd(a,d)=1$ as $m|a\wedge m|b\Rightarrow m=\pm 1$.

Of course if we use the unique prime decomposition it is trivial. Is there any other easier proofs avoiding using that theorem?

Answer 1

Sure, there exists $u,v$ such that $au+bv=1$, then $auy+bvy=y=auy+axv=a(uy+xv)$

Answer 2

We know by Bezout's Thm there exists integers x and y such that $ax+by=1$. [I will provide a proof for Bezout's Thm here below that way you will be able to prove this entire thing from scratch.] Thus, $axy+by^2=y$. Since $ax=by$, then $axy=by^2$. Thus, $axy+by^2=y$ becomes $axy+axy=y$ by substitution that being $axy=by^2$. Hence, $a(xy+xy)=y$. So, $a|y$.

There was an excellet previous proof on this site for Bezout's Thm, and it went something like this:

Let m and n be integers that are relatively prime. Then, we know $\exists x,y \in \mathbb{Z}$ such that $mx+ny=1$.

Proof by Contradiction: Suppose $k$ is the smallest positive integer that fits the equation $mx+ny=k$. Now, consider $k>1$ and show a contradiction. We know $k\mid m$ and $k\mid n$ does not occur because of the following reasoning. Suppose that $k\mid m$ and $k\mid n$ and show a contradiction. Since $k>1$, then m and n would not be relatively prime. This is a contradiction, because they are relatively prime. Thus, we know that $k\nmid n$ is true, or $k\nmid m$ is true (possibly both could be true as well). Consider the case that $k\nmid m$ (note that we will consider the other case, $k\nmid n$, for the possibility that it is not this one later). Since $k\nmid m$, we can use division theorem to conclude $kj+r=m$ when $0<r<|k|=k$ for some $j\in \mathbb{Z}$. Thus, \begin{align*} r &= m-kj\\ &=m-(mx+ny)j, \text{ because } mx+ny=k\\ &=m-mjx-njy \\ &=m(1-jx)+n(-jy) \\ &<k \text{ due to division theorem statement.} \end{align*}
Looking at the second to last step we see that $r$ is a linear combination of m and n and that $0<r<k$. Again, we previously defined $k$ as smallest linear combination for m and n but now r is. This is a contradiction. Thus, we have proven the theorem for this case.

Now, if you want to prove Bezout's theorem completely you do this. [I do not know how in depth you are needing to go with this.]

Lemma: Let $a,b\in \mathbb{Z}$ when $a=b=0$ does not occur. The smallest positive integer value of the set $M=\lbrace ax+by| x,y\in \mathbb{Z}\rbrace$ yields the $gcd(a,b)$.

Proof: Let $a,b\in \mathbb{Z}$ when $a=b=0$ does not occur. It should be apparent that the set $\lbrace ax+by| x,y\in \mathbb{Z}\rbrace$ has elements $\geq 1$. Now, we will define the smallest positive integer value of this set to be k, so that $k=ax+by$. We know that we can express $a=qk+r$ when $0\leq r<k$ for some $r\in \mathbb{Z}$ based off of how we can partition the set of integers. Therefore, we know \begin{align*} r &=a-qk \\ & = a-q(ax+by), \text{ because $k=ax+by$} \\ & = a-aqx-bqy \\ & = a(1-qx)+b(-qy) \\ & \in M. \end{align*} Since $0\leq r<k$, we know that if $r\neq 0$, then r would be the smallest positive element in the set M which is a contradiction. Thus, $r=0$. Hence, $a=qk$ and $k|a$. We can repeat this same process for b and conclude that $k|b$ as well. By definition of greatest common divisor, we know that $gcd(a,b)\geq k$. Since $gcd(a,b)|a$ and $gcd(a,b)|b$, then $\exists i_1, i_2 \in \mathbb{Z}$ ST $gcd(a,b)i_1=a$ and $gcd(a,b)i_2=b$. Since $k=ax+by$, we can substitute our values for a and b to get $k=gcd(a,b)i_1x+gcd(a,b)i_2y$. Thus, $k=gcd(a,b)(i_1x+i_2y)$ and $gcd(a,b)|k$. Hence, we know $k=gcd(a,b)i_3$ for some $i_3\in \mathbb{Z}$. Particularly, we know $i_3=1$, and we will show this. This is because if $i_3=0$, then $k=0$ which is a contradiction. Also, we know $i_3$ cannot be negative as k would be negative. Lastly, if $i_3>1$, then $k=gcd(a,b)i_3\leq gcd(a,b)$ would be a contradiction. Thus, $i_3=1$, and $gcd(a,b)i_3=k$ becomes $gcd(a,b)=k$.