The application of Bezout's identity on $\gcd(a, b) = 1$ and $\gcd(a, c) = 1$, then $\gcd(a, bc) = 1.$

Question

If $\gcd(a, b) = 1$ and $\gcd(a, c) = 1$, then $\gcd(a, bc) = 1$.

A proof using Bezout's Identity was written as follows:

Using Bezout's Identity:

Since there are $x,y,u,v$ so that $\color{#C00000}{ax+by=1}$ and $\color{#00A000}{au+cv=1}$, we have $$ \begin{align} \color{#C00000}{by}\color{#00A000}{cv} &=\color{#C00000}{(1-ax)}\color{#00A000}{(1-au)}\\ &=1-a(x+u-axu)\\ \color{#0000FF}{a}(x+u-axu)+\color{#0000FF}{bc}vy &=1 \end{align} $$ Therefore, $(\color{#0000FF}{a},\color{#0000FF}{bc})=1$

However, as far as I understand, Bezout's Identity does not work in the converse. In other words, $\color{#FF00FF}{a(k) + bc(j) = 1}$ does not imply $(\color{#0000FF}{a},\color{#0000FF}{bc})=1$.

Do I not understand Bezout's Identity, or does this proof work for some other reason?

Does this answer your question? If $\gcd(a,b)=1$ and $\gcd(a,c)=1$, then $\gcd(a,bc)=1$ — Andrew V, Sep 07 '20 at 02:31
It does work in the converse. Well, the statement of Bezout's theorem is that the gcd is the smallest positive number that is an integer linear combination of the given numbers, and there's no smaller positive number than $1$. — Sarvesh Ravichandran Iyer, Sep 07 '20 at 02:32
@AndréMuchon The link you provided contains the exact proof I was questioning. — Zanilen, Sep 07 '20 at 03:02
@TeresaLisbon Thank you! That clarified Bezout's Identity and answered my question. — Zanilen, Sep 07 '20 at 03:03
Every common divisor $d$ of $,a,bc,$ also divides $,a,k+bc,j = 1,,$ including their greatest common divisor. — Bill Dubuque, Sep 07 '20 at 03:46

score 1 · Answer 1 · edited Sep 07 '20 at 02:53

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Any time $ax+by=1$, we have $(x,y)=1$. For if $c\mid x\land c\mid y$, then $c \mid (ax+by)=1$.

That's actually the easy direction. Bezout's is the harder direction.

edited Sep 07 '20 at 02:53

Community

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answered Sep 07 '20 at 02:32

score 1 · Accepted Answer · answered Sep 07 '20 at 02:32

This very well does prove what we wanted. If you take a look at the proof of Bezout's theorem, you will notice that the smallest positive value of the expression $\ ax+by$ where all of them are integers is indeed $\ (a, b)$. Hence, if the value of the expression can be equal to the smallest positive integer, then $\ (a, b)=1$. I hope it answers your question.

The application of Bezout's identity on $\gcd(a, b) = 1$ and $\gcd(a, c) = 1$, then $\gcd(a, bc) = 1.$

2 Answers2