Let $a, b, c$ be integers, no two of which are zero, and $d=\gcd(a, b, c)$. Show that
$d=\gcd(\gcd(a,b),c)=\gcd(a,\gcd(b,c))=\gcd(\gcd(a,c),b)$.
Here is what I have tried, but I'm unsure if the part that I prove $d=\gcd(a,b)$ is correct:
Proof:
$d=\gcd(a,b,c)\implies d\mid a$, $d\mid b$, and $d\mid c$.
Let $d=\gcd(a,b)$.
So, $d=ax+by$, some $x,y\in \mathbb{Z}$, by Theorem 2.3.
Let $n\mid a$ and $n\mid b$, $n\in \mathbb{N}$.
Then, $n\mid (ax+by)$ by Theorem 2.2(g) $\implies n\mid d$.
So, $d=\gcd(a,b)$ by Theorem 2.6.
Similarly, we can show $d=\gcd(a,c)=\gcd(b,c)$.
Thus, we have $\gcd(\gcd(a,b),c)=\gcd(d,c)$.
Since $d\mid c$, we have $dk=c$, some $k\in \mathbb{Z}$.
So, we have $\gcd(d,c)=d\cdot \gcd(1,k)=d\cdot 1=d$ by Theorem 2.7.
Similarly, we can prove that $\gcd(a,\gcd(b,c))=\gcd(\gcd(a,c),b)=d.\blacksquare$
Theorems
2.2(g): If $a\mid b$ and $a\mid c$, then $a\mid (bx+cy)$ for arbitrary integers $x$ and $y$.
2.3: Given integers $a$ and $b$, not both of which are zero, there exist integers $x$ and $y$ such that $\gcd(a,b)=ax+by$.
2.6: Let $a,b$ be integers, not both zero. For a positive integer $d$, $d=\gcd(a,b)$ if and only if
(a) $d\mid a$ and $d\mid b$.
(b) Whenever $c\mid a$ and $c\mid b$, then $c\mid d$.
2.7: If $k>0$, then $\gcd(ka,kb)=k\cdot \gcd(a,b)$.
