Please explain this step in proving the square root of 3 is irrational Assume that

$$3 = \frac{p^2}{q^2}$$

So, 

$$ 3 q^2 = p^2$$

So $p^2$ is divisible by $3$. How we can conclude this?
Since $p$ is divisible by $3$, it is $3$ times some number, let's say $m$. Then we substitute, and we get
$$
3q^2 = (3m)^2 = 9m^2
$$
$$
q^2 = 3m^2
$$
Now $q$ has to be divisible by 3, so by the same argument, for some $n$,
$$
(3n)^2 = 3m^2
$$
$$
3n^2 = m^2
$$
And then you can keep dividing by $3$ indefinetely, and still end up with integers, but this is a contradiction, so there can be no such $p$ and $q$ in the first place.
Every number can be written uniquely as the product of primes, since $3q^2 = p^2$ their representation as the product of the primes must be the same, and since 3 is in the representation of the left hand side, it must also be in the representation  of the right hand side, $p^2$ and so it divides $p^2$