Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $.


How do I prove it? I don't even intuitively get this... after so long time of trying to prove it.
For the intuition: take the primes $p_1=5$ and $p_2=11$. Then $p_1\cdot p_2=55$ and indeed the only divisors (among the natural numbers) of the product are $1,5,11$, and $55$. Now take not both numbers prime, say $p_1=24$ and $p_2=11$. Then $p_1\cdot p_2=264$ and the divisors of the product this time certainly include $1,24,11$, and $264$, but also $2,3,6,12$.

For the proof: suppose $p_1,p_2$ are prime and that $n\mid p_1p_2$, where $n\in \mathbb N$. If $n\ne 1$ then it has a prime divisor. Let $q$ be an arbitrary prime divisor of $q$. Then in particular, $q$ divides $p_1p_2$. Now, according to Euclid's Lemma, if a prime divides a product, then it divides one of the factors. Thus, either $q\mid p_1$ or $q\mid p_2$. Assume, without loss of generality, that $q\mid p_2$. But since $p_2$ is prime, it follows that $q$ is either $1$ or $q=p$. Since $q$ is prime, the possibility of $q=1$ is (explicitly in the definition of prime number) ruled out. We conclude that $q=p$. 

Now consider $m=n/q=n/p_1$. It follows that $m\mid p_2$. Thus, either $m=1$ or $m=p$. In the case $m=1$, it follows that $n=p_1$. If $m=p_2$ it follows that $n=p_1p_2$. 

To conclude, we just showed that the only possible prime divisor of the product $p_1p_2$ are $1,p_1,p_2$, and $p_1p_2$.
Let there be a divisor $q$ of $p_1p_2$ which is not in the list.
This means, that $q$ should be either divisor of $p_1$ or $p_2$ which meanse, that $p_1$ or $p_2$ is not a prime.