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

  1. $p<q$

  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.

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  • $\begingroup$ 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 $\endgroup$ – Hagen von Eitzen Oct 1 '18 at 13:31
  • $\begingroup$ Do you have any suggestions on the quickest way to do this using elementary number theory? @HagenvonEitzen $\endgroup$ – JB071098 Oct 1 '18 at 13:32
  • $\begingroup$ 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 $13$. $\endgroup$ – Derek Holt Oct 1 '18 at 13:49
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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.

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