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If $a|(b+c)$ and $\gcd(b,c)=1$, prove that $\gcd(a,b)=1$ and $\gcd(a,c)=1$.

I started with:

Suppose $a|(b+c)$ and $\gcd(b,c)=1$. This means that $ak=b+c$, for some integer $k$. And $1|b$ and $1|c$.

I know I can solve this using the theorem that if $\gcd(a,b)=1$ then there exists integers $u$ and $v$ such that $au+bv=1$.

But I was wondering how else I can solve this, without using this theorem.

Mohsen Shahriari
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Sammy
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    Hint: Let $p$ be a prime dividing $\gcd(a,b)$ then show that the assumptions imply that $p,|,c$ as well. – lulu Oct 06 '20 at 22:00

3 Answers3

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The fact that $\gcd(b,c)=1$ implies there are $k,l \in \Bbb Z$ such that $kb + lc= 1$ (Bézout identity). $a|(b+c)$ tells us $ma = b+c$ for some $m \in \Bbb Z$.

Can you find a Bezout identity for the pairs $(a,b)$ and $(a,c)$ from that?

Henno Brandsma
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Hint $\ \,a\mid b\!+\!c\,\Rightarrow (a,b)\mid \color{#c00}c,\,$ but $\,(a,b)\mid \color{#c00}b\,$ too so $\,(a,b)=1\,$ by $\,\color{#c00}{(b,c)=1}$

Bill Dubuque
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  • $\begin{align} {\bf Or:}\ \ a\mid b!+!c&\Rightarrow (a,b) = (a,b,b!+!c) = (a,\color{#c00}{b,c}) = \color{#c00}1\ {\bf Or:}\ \ a\mid b!+!c&\Rightarrow (a,bc)= (a,\color{#c00}{bc,b+c}) = \color{#c00}1\ \ {\rm by\ Post1} \end{align}\ \ $ for Post1 – Bill Dubuque Oct 06 '20 at 23:01
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One thing I like to do is figure if $k$ is a common (positive) divisor of $a$ and $b$, and $a|b+c$ then $k|b+c$. But we also have $k|b$ so $k| (b+c)-b = c$. So $k$ is a common (positive) divisor of $b,c$. But $\gcd(b,c) = 1$ so the only common (positive) divisor of $b,c$ is one so $k=1$. So the only common divisor of $a$ and $b$ is $1$. So $\gcd(a,b)=1$.

And doing similar to prove $\gcd(a,c) =1$ is nearly entirely the same. If $d$ is a common divisor of $a$ and $c$ then $d|b+c$ and $(b+c)-c=b$ and so $d$ is a common divisor of $b$ and $c$ which are relatively prime.

....

Or maybe even easier: $a|b+c$. So $\gcd(a,b)|a$ so $\gcd(a,b)|b+c$ but $\gcd(a,b)|b$ so $\gcd(a,b)|(b+c)-b=c$. So $\gcd(a,b)|b$ and $\gcd(a,b)|c$ so $\gcd(a,b)|\gcd(b,c)=1$. So $\gcd(a,b) = 1$.

fleablood
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