This turned out to be very obvious based on the definition of a prime number, so obvious I felt like not bothering, but I still want to see if there is a problem with the proof so far, seeings I am in general just terrible at proofs.
I observed that the greatest common divisor of the largest divisor of $n \cdot p_n$ and the largest divisor of $n\cdot\bigl\lfloor \frac{p_n}{n} \bigr\rfloor$ is always equal to $n$.
I then assumed this must be because $\bigl\lfloor \frac{p_n}{n} \bigr\rfloor$ and $p_n$ share no common divisor greater than 1, i.e they are coprime.
So the question became: Prove that $p_n$ and $\bigl\lfloor \frac{p_n}{n} \bigr\rfloor$ must be coprime.
And this is what have done so far:
since $\lnot (n | p_n)$ is true $\forall n \in \mathbb N$,
$\frac{p_n}{n}-\bigl\lfloor \frac{p_n}{n} \bigr\rfloor \ne 0\,\, \forall n \in \mathbb N$
and since we know :
$\gcd(k,p_n)=1$ $\forall k \leq p_n -1$
and
$\bigl\lfloor \frac{p_n}{n} \bigr\rfloor \lt p_n -1$ $\forall n \gt 1 \in \mathbb N$
we can substitute $k=\bigl\lfloor \frac{p_n}{n} \bigr\rfloor$
therefore
$\gcd(\bigl\lfloor \frac{p_n}{n} \bigr\rfloor,p_n)=1$
Hence showing that $\bigl\lfloor \frac{p_n}{n} \bigr\rfloor$ and $p_n$ are coprime.