A prime number problem...........

[Md Ashraful Kader]

$p$ and $q$ are two prime number. Again, $(p^2-q)$ and $(p-q^2)$ are also prime. If you divide $(p^2-
q)$ by a composite number $n$ where $n<p$ you’ll get a remainder of $14$. If you divide $(p-q^2+
14)$ by the same number what will you get as remainder this time?

[Phlembac Adib Hasan]

Clearly, $p-q^2>0$ implies $p>q$. Now note that $q$ has to be $2$. Otherwise both $p^2-q$ and $p-q^2$ become even which is impossible. (since they are primes)
Now assume $p-4>\dfrac p 2$. (You can check out other small cases one by one)
The condition $p^2-q\equiv 14\pmod n$ turns into
\[(p+4)(p-4)\equiv 0\pmod n\]
$p-4=p-q^2$ is a prime $>\dfrac p 2$. Hence $\gcd (p-4,n)=1$ because $n<p$.
Thus $n|p+4$. Therefore,
\[p-q^2+14= p+10\equiv 6\pmod n\]
So the reminder is $\boxed 6.$

[Phlembac Adib Hasan]

In my opinion, this is a secondary level question. So, I'm moving it to the appropriate forum.