Proof if $\gcd(a,b)=\gcd(b,c)=\gcd(a,c)=d$ then $\gcd(a,b,c)=d$

Question

How can we prove for $a$, $b$ and $c$ positive integers that if

$$\gcd(a,b)=\gcd(b,c)=\gcd(a,c)=d$$ then $$\gcd(a,b,c)=d$$

For $a$ and $b$ co-prime numbers, $\gcd(a,b)=1$, means that the pairs $(b,c)$ and $(a,c)$ are also co-prime numbers, then from the $\gcd$ commutativity and associativity:

$$\gcd(a,b,c)=\gcd(\gcd(a,b),c)=\gcd(\gcd(a,c),b)=\gcd(a,gcd(b,c))$$ $$\gcd(a,b,c)=\gcd(1,c)=\gcd(1,b)=\gcd(a,1)=1$$ This is correct because if $a$, $b$ and $c$ are mutually prime numbers their $\gcd$ is $1$, is this a special case only and we can found counterexamples for the above claim?

Answer 1

Let us merely assume that $\gcd(a,b)=d$ and that $d$ divides $c.$ Clearly, $d$ is a common divisor of $a,b,$ and $c.$ Let $e$ be any other common divisor of $a,b,$ and $c.$ Since $e$ is a common divisor of $a$ and $b,$ it follows that $e$ divides $\gcd(a,b)=d.$ Hence $d$ is the greatest common divisor of $a,b,$ and $c.$

Answer 2

If $gcd(a,b)=d\Rightarrow a=md,b=nd$ with $ m,n\in\mathbb{N}$ and so $gcd(m,n)=1$. If $gcd(m,n)=1$, then $gcd(n,m,k)=1$ for any $k\in\mathbb{N}$. Can you work out from here?

Answer 3

Assume that $d = gcd(,)$ and $ \mid $. Let $e = gcd(a,b,c)$. Since $d$ divides $a$, $b$, and $c$, we must have $e \geq d$. But if we assume $e > d$ this contradicts our assumption that $d = gcd(a,b)$. Thus, $e = d$.