Does Bezout's theorem work both ways?

[Heisenberg7]

Homework Statement

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Relevant Equations

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Let's assume that for integers $m$ and $n$ (and integers $x_1$ and $y_1$) the following is satisfied: $$mx_1 + ny_1 = 1$$ Then by this theorem (for some integer $k$), if $k|a$ and $k|b$ then $k|mx+ny$ for all integers $x$ and $y$. So it must be, $$k | mx_1 + ny_1$$ for all common factors of $m$ and $n$. In other words, $$k|1$$ for all common factors of $m$ and $n$. But this implies that $k=1$ since $k$ is an integer. Thus $(m, n) = 1$. Since I've never seen this be mentioned before, I'll be happy to see where I went wrong.

Thanks in advance

 

[Heisenberg7]

Made a little correction. Numbers $x_1$ and $y_1$ are integers as well.

 

[julian]

By Bezout's theorem, if $(m,n)=1$ there exists $x$ and $y$ such that

\begin{align*}
mx+ny=1 .
\end{align*}

Are you asking if we can find integers $x_1$ and $y_1$ such that

\begin{align*}
mx_1+ny_1=1
\end{align*}

then is $(m,n)=1$? In that case, $(m,n) | mx_1+ny_1 \Rightarrow (m,n) | 1 \Rightarrow (m,n)=1$.

 

[Heisenberg7]

By Bezout's theorem, if $(m,n)=1$ there exists $x$ and $y$ such that

\begin{align*}
mx+ny=1 .
\end{align*}

Are you asking if we can find integers $x_1$ and $y_1$ such that

\begin{align*}
mx_1+ny_1=1
\end{align*}

then is $(m,n)=1$? In that case, $(m,n) | mx_1+ny_1 \Rightarrow (m,n) | 1 \Rightarrow (m,n)=1$.

Yes. That's what I'm asking for. Seems like a useful property for problem solving. So, I am right?

 

[julian]

It is correct. It follows trivially. As $(m,n)$ divides $m$ and $n$ by definition, and given that we have found $x_1$ and $x_2$ such that $mx_1+ny_1=1$, we have $(m,n) | 1$. It may be useful in proving two integers are coprime.