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?
 A: It depends on what definition of greatest common divisor you use. You probably use the second one.
Definition 1

Given natural numbers $a$ and $b$, the natural number $d$ is their greatest common divisor if
  
  
*
  
*$d\mid a$ and $d\mid b$
  
*for all $c$, if $c\mid a$ and $c\mid b$, then $c\mid d$
  

Theorem. The greatest common divisor exists and is unique.
Proof. Euclidean algorithm.
Definition 2

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 $\gcd(a,b)=\max(\operatorname{Div}(a)\cap\operatorname{Div}(b))$
Extension to $\mathbb{Z}$, for both definitions
If $a,b\in\mathbb{Z}$, then $\gcd(a,b)=\gcd(|a|,|b|)$, adding $\gcd(0,0)=0$ for definition 2.
Proof of the statement using definition 1
With this definition, the statement is obvious.
Proof of the statement using definition 2
Let $p\mid a$ and $p\mid b$. We need to show that $p\mid\gcd(a,b)$. It is not restrictive to assume $p,a,b>0$.
It is true that $\operatorname{Div}(p)\subseteq\operatorname{Div}(a)\cap\operatorname{Div}(b)$, but this just implies that $p\le\gcd(a,b)$, not that it is a divisor thereof.
The proof can be accomplished by using the fact that $\gcd(a,b)=ax+by$ for some integers $x$ and $y$ (Bézout's theorem). With this it is easy: $a=pr$, $b=ps$, so
$$
\gcd(a,b)=ax+by=prx+psy=p(rx+sy)
$$
How to prove Bézout's theorem is beyond the scope of this answer.
A: One alternative is:
By Bezout's Lemma, we know that for all $(a, b)$, $\exists x, y \in \mathbb{Z}. ax + by = \gcd(a, b)$.
So we have to prove that:
$$p|ax+by$$
Since, $p|a \implies a = pk_1$ and $p|b \implies b = pk_2$ for $k_1, k_2 \in \mathbb{Z}^+$.
$$\implies p|pk_1x + pk_2y$$
$$\implies p|p(k_1x + k_2y)$$
which is true.
A: 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. 
