I was studying non-abelian groups of order $pq$ where $p$ and $q$ are primes such that $q$ divides $p - 1$. I compute two quantities $A$ and $B$ for this group which are given by
$A = (p -1)(q - 1) + \sqrt{(p -1)^2(q - 1)^2 + 4p(p -1)(q - 1)}$
and
$B = \frac{2p^2(p -1)(q - 1) + 2p(q - 1)^2}{pq - 1}$.
We observe that $A < B$ if $q^2 < p + q + 1$. But the condition $q^2 < p + q + 1$ is not true in general. There are many pairs of primes $p, q$ such that $q$ divides $p - 1$ satisfying as well as not satisfying $q^2 < p + q + 1$. For example, (a) $p = 43, q = 7$; (b) $p = 53, q = 13$; (c) $p = 67, q = 11$; (d) $p = 89, q = 11$ are examples of primes satisfying $q\mid p - 1$ but not $q^2 < p + q + 1$. Examples of pair of primes satisfying $q\mid p - 1$ and $q^2 < p + q + 1$ are easy to find. Now, I have the following questions:
How to find the pairs of primes $p, q$ such that $q\mid p - 1$ and $q^2 < p + q + 1$. Is there any characterization of such primes?
How to find the pairs of primes $p, q$ such that $q\mid p - 1$ and $q^2 > p + q + 1$. Is there any characterization of such primes?