# Sylow $p$-Subgroups and Counting Prime Numbers

Use the Sylow Theorems to find the number of prime numbers $$p$$ and $$q$$ which satisfy the following conditions:

1. $$p

2. $$q\not\equiv1($$mod $$p)$$,

3. $$pq<100$$

How would the Sylow Theorems apply to this problem? I'm really confused on how to use them in this setting. This problem just seems like a number theory problem that could be solved using Fermat's Theorem. I genuinely don't understand how to use Sylow $$p$$-subgroups to solve this problem. Any hints, or explanations would be extremely helpful since I find the Sylow Theorems hard to understand in general.

• I agree. By the condition $pq<100$ (in particular, $p\in\{3,5,7\}$ and $q\le 29$), this is a very finite problem, whereas any attempt based on classification of groups of order $<100$ seems to be utterly over-complicated – Hagen von Eitzen Oct 1 '18 at 13:31
• Do you have any suggestions on the quickest way to do this using elementary number theory? @HagenvonEitzen – JB071098 Oct 1 '18 at 13:32
• I think you should stop worrying about how to do it and just do it. It's really not very hard. For example, if $p=7$, then $q=11$ or $13$. – Derek Holt Oct 1 '18 at 13:49

By Sylow's theorems, this is all such pairs such that all groups of order $$pq$$ are cyclic. I suppose you could then look at a list of all groups of each order to see which of these had only one. I agree it's a perversely stated problem.