Let $a,b,c$ be integers, no two of which are zero and $d=\gcd(a,b,c)$. Show that $d= \gcd(\gcd(a,b),c)$.

Question

I did like this

Let $a=dx;b=dy;c=dz$
Let $\gcd(a,b)=d_1$
Here $d_1 ~ge d$
Then question reduces to prove that $\gcd(d_1,dz)=d$ I know $d$ divides $d_1$ but how can I show it is the greatest integer?

Answer 1

We let $a = dx$, $b = dy$, $c = dz$ such that $d$ is the largest value possible for this to happen (i.e. $d = \gcd(a,b,c)$). Then, we know that, in this case, $\gcd(a,b) = d \cdot \gcd(x,y)$, and therefore $\gcd(\gcd(a,b),c) = \gcd(d \cdot \gcd(x,y), dz) = d \cdot \gcd(\gcd(x,y),z)$.

Assume, for the sake of contradiction that $\gcd(\gcd(x,y),z) > 1$ and $\gcd(x,y,z) = 1$. Then we know that some prime $p$ divides both $z$ and $\gcd(x,y)$, so $p | z$ and $p | \gcd(x,y)$. However, for that to happen, $p | x$ and $p |y$, so $p | \gcd(x,y,z)$, which is a contradiction of the assumption. THerefore, $d \cdot \gcd(\gcd(x,y),z) = d$.