how do we prove $p|q\cdot r \rightarrow p=q$ or $p=r$ (all primes)? I know this is one of the most fundamental basis of arithmetic but I can't find the result by myself.
how do we prove $p|q\cdot r\rightarrow p=q$ or $p=r$? ($p, q, r$ being prime numbers) 
 A: If $p\mid q$, we are done.
Suppose that $p\not\mid q$. Since $p$ is only divisible by itself and $1$, $\gcd(p,q)=1$. By Bezout's Identity, there are $a$ and $b$ so that
$$
ap+bq=1\tag{1}
$$
Multiply $(1)$ by $r$ to get
$$
(ar)p+b(qr)=r\tag{2}
$$
Since $p\mid qr$, there is a $k$ so that $kp=qr$. Using this in $(2)$ gives
$$
r=p(ar+bk)\tag{3}
$$
which says that $p\mid r$.
QED
A: The fully rigorous chain of proofs actually follows the following order: 
You can prove the division algorithm using base axioms of the natural numbers (commutativity, distributivity, associativity, multiplicative identity, well-ordering principle), which is the claim that, given $a,b\in\Bbb N$, there exist $q,r\in\Bbb N$ such that $a=bq+r$ and $0\leq r<|b|$.  This easily generalizes to $\Bbb Z$.
From there, you can prove that Linear Diophantine Equations always have a solution: given $a,b\in\Bbb Z$, there exist $x,y\in\Bbb Z$ such that $ax+by=\gcd(a,b)$.
After you've proven this, what you're looking for is easy to prove.  Consider $a,b,c\in\Bbb N$ such that $a|bc$ and $\gcd(a,b)=1$.  Then there exist $x,y\in \Bbb Z$ such that $ax+by=1$, or $acx+bcy=c$.  Note that, since $a|bc$, $\exists k\in \Bbb Z$ such that $ak=bc$, so we have $acx+aky=a(cx+ky)=c$.  Since $cx+ky$ is an integer by closure under addition and multiplication, $a|c$.
Note: unique prime factorization, as assumed by other answers, is not an axiom of the integers. Sure, it's perfectly okay to use it most of the time, but it is in fact a consequence of this chain of reasoning, and it's always best to be fully rigorous.  Luckily, RobJohn did it the right way.
(Ask if you want proofs of the first two; that would take a bit longer).
A: Hint : what are prime divisors of $q \cdot r$ ?
A: This is actually almost axiomatic, but you just have to consider the prime factors of $qr$, which are $q$ and $r$. Therefore - if $p$ divides $qr$ one of them have to be equal to $p$, since unique factorisation holds in $\Bbb Z$.
