Find all $p, q$ (coprime) such that $\frac{(p-1)(q-1)}{2}$ is a prime number 
Find all $p, q$ (coprime) such that $\frac{(p-1)(q-1)}{2}$ is a prime number.

I found that $(p,q) = (2,7),(7,2),(2,15),(3,8)$ gave prime numbers for $\frac{(p-1)(q-1)}{2}$, but how do we find all?

If $p, q$ are coprime then at most one is even.  Wolog $q$ is odd and $q-1$ is even.
So $(p-1)\frac{q-1}2$ is prime.  So either $p-1 = \pm 1$ or $\frac {q-1}2 = \pm 1$.
Case 1: $p-1 = -1$
If $p-1 = -1$ then $p = 0$ and $0$ is not coprime to anything except $\pm 1$. (Everything divides 0) so $q = \pm 1$.  $\frac {1-1}2 = 0$ is not prime so $q = -1$ and $\frac {q-1}2 = -1$ and ... $ (p-1)\frac{q-1}2=1$ is not prime.
Case 2: $p-1 = 1$
So $p = 2$ and $\frac {q-1}2 $ is prime. So $q = 2k + 1$ for any prime.  i.e. Let $k$ be prime,  the $\gcd(2, 2k+1) =1$ and $\frac{(2-1)(2k+1 -1)}2 = k$ is prime.
Case 3:  $\frac {q-1}2 = -1$
$q = -1$ and $p= k + 1$ for any prime.  i.e. $\gcd(k+1,-1)=1$ and $\frac{(k+1-1)(-1 -1)}2 = -k$ is prime.
Case 4:  $\frac{q-1}2=1$
$q = 3$ and $p = k +1$ for any prime $k=3$ or $k \equiv 1 \mod 3$
So options are $(2,2k+1)$ for any prime $k$.
$(-1, k+1)$ for any prime.
$(3,4)$ and $(3,k+1)$ for any prime of the form $k \equiv 1 \mod 3$.
