Let $T \equiv PP + 1 \equiv \{ ab + 1 : a,b \text{ are prime }\} \subset \Bbb{Z}^{\times}$. Consider the subsemigroup generated by $T$. How can I show that it is not finitely generated, by that I mean there doesn't exist a finite set of integers $\{z_i\}$ such that each element of $\langle T \rangle$ can be written $t = z_1^{e_1}\cdots z_n^{e_n}$ for some $e_j \geq 0$?
I suppose I could do it by showing that there are infinitely many primes $p = 1 + ab$, but how do I do that?
Hint: you only need to show that there are infinitely many primes dividing numbers in $T$. To start, let $S$ be a finite set of primes and compare $\pi_2(x)$ with how many numbers less than $x$ are products of primes in $S$.
One can show that for each prime $p$ there is a number $ab+1\in T$ with $p\mid ab+1$. As this implies that a generating set must contain a multiple of $p$, $T$ cannot be finitely generated. So, given $p$ pick aprime $b\ne p$ and see what simple condition for $a$ you get ...
