if $p\mid a$ and $p\mid b$ then $p\mid \gcd(a,b)$ 
I would like to prove the following property :
$$\forall (p,a,b)\in\mathbb{Z}^{3} \quad p\mid a \mbox{ and } p\mid b \implies p\mid \gcd(a,b)$$

Knowing that :
Definition

Given two natural numbers $a$ and $b$, not both zero, their greatest common divisor is the largest divisor of $a$ and $b$.



*

*If $\operatorname{Div}(a)$ denotes the set of divisors of $a$, the greatest common divisor of $a$ and $b$ is $\gcd(a,b)=\max(\operatorname{Div}(a)\cap\operatorname{Div}(b))$

*$$d=\operatorname{gcd}(a,b)\iff \begin{cases}d\in \operatorname{Div}(a)\cap\operatorname{Div}(b) & \\ & \\  \forall x \in \operatorname{Div}(a)\cap\operatorname{Div}(b): x\leq d \end{cases}$$

*$$\forall (a,b) \in \mathbb{N}^{2}\quad a\mid b \iff Div(a) \subset Div(b)$$

*$$\forall  x\in \mathbb{Z}\quad \operatorname{Div}(x)=\operatorname{Div}(-x)  $$

*If $a,b\in\mathbb{Z}$, then $\gcd(a,b)=\gcd(|a|,|b|)$, adding $\gcd(0,0)=0$


Indeed,
Let $(p,a,b)\in\mathbb{Z}^{3} $ such that $p\mid a$ and $p\mid b$ then : 
$p\mid a \iff \operatorname{Div}(p)\subset \operatorname{Div}(a)$ and $p\mid b \iff \operatorname{Div}(p)\subset \operatorname{Div}(b)$ then
$\operatorname{Div}(p)\subset \left( \operatorname{Div}(a)\cap \operatorname{Div}(b)\right) \iff p\mid \gcd(a,b)$
Am I right?

One line proof using ring theory: $(a) \subset (p), (b) \subset (p) \implies (a,b) \subset (p)$. $\mathbb{Z}$ is PID implies $(a,b)$ is principal. We are done. 
