# $\langle ab + 1 : a,b \text{ prime}\rangle$ is not a finitely generated subsemigroup of $\Bbb{Z}^{\times}$.

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?

• What's the source of this problem, please? Nov 24 '15 at 6:59
• @GerryMyerson prime numbers and all the unsolved problems therein. Nov 24 '15 at 7:02
• $ab+1$ is even and hence composite, unless $a$ or $b$ is $2$. In the latter case, you have Sophie Germain primes. It is conjectured that there are infinitely many of those. Nov 24 '15 at 7:06
• In other words, I'm out of luck... :'| Nov 24 '15 at 7:10

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 $S$.
• What's $\pi_2(x)$? Nov 24 '15 at 7:11
• The number of semiprimes less than $x$. See math.stackexchange.com/a/841159/1284 Nov 24 '15 at 7:17
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 ...
• I like this one. So you're saying pick any prime $p$ and some multiple $m$ such that $pm - 1 = ab$. Not sure how to do that though. Nov 24 '15 at 7:19
• Pick a prime $p$ and consider all solutions to $ab = -1 \pmod p$? Nov 24 '15 at 7:24