If $a=b+c$ then $(a,b)=(a,c)=(b,c)$

Question

If $a=b+c$ then $(a,b)=(a,c)=(b,c)$

I was thinking of writing the Euclidean algorithm

\begin{align*}a &= b\cdot 1+c\\ b &= c\cdot (-1) + (b+c)\\ c &= a \cdot 1 + (-b)\end{align*} and does it go like the above expression? So we have $(a,b)=(a,c)=(b,c)$

Answer 1

Hint: Use Bezout's theorem, that is:

$(x,y)$ can be written as $ax+by$ for some $a,b \in \mathbb{Z}$ and $(x,y)$ divides any number of the form $nx+by$ with $n,m \in \mathbb{Z}$.

Apply this to $(x,y),(y,z)$ and $(x,z)$ to get that they all divide each other.

Put $d=(x,y),d'=(y,z),d''=(x,z)$.

We have: $d=ax+by$ for some $a,b \in \mathbb{Z}$ hence: $d=a(y+z)+by=(a+b)y+az \ $, since $(a+b),a\in \mathbb{Z}$ we get $d'\vert d$.

We also have $d'=py+qz$ for some $p,q \in \mathbb{Z}$, which gives: $d'=py+q(x-y)=qx+(p-q)y$ hence $d \vert d'$.

So we have $d \vert d'$ and $d' \vert d$, and since both are positive we conclude $d=d'$.

The other equalities follow in the same manner.

Answer 2

If $x+y=z$, then any common factor of $x$ and $y$ is also a factor of $z$. Similarly any common factor of $x$ and $z$ is also a factor of $y$; and any common factor of $y$ and $z$ is also a factor of $x$.

So the set of common factors of any pair (two) of $x,y,z$ is the same as the set of common factors of any other pair. And so all gcf's mentioned in the problem are equal.