Does Bezout's theorem work both ways?

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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
 
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Made a little correction. Numbers ##x_1## and ##y_1## are integers as well.
 
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##.
 
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julian said:
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?
 
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.
 
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Also, in general ##(m,n)=d## implies ##xm+yn=d## for some integers ##x,y##. The converse is not true, though.
 
nuuskur said:
Also, in general ##(m,n)=d## implies ##xm+yn=d## for some integers ##x,y##. The converse is not true, though.
Well, in case of a prime number ##k## for which ##xm + yn = k## for some integers ##x,y,m,n##, shouldn't it be true? I mean, ##(m,n)|xm+yn \implies (m,n)|k \implies (m,n) = 1 \vee (m,n) = k##. So we only need to verify whether ##(m,n)## equals 1 or ##k##.
 
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