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5

3

It might be well-known (and sorry if it is), but a quick search did not return the answer.

Consider prime numbers $p \neq q$.

Are $\displaystyle \frac{p^q-1}{p-1}$ and $\displaystyle \frac{q^p-1}{q-1}$ relatively prime?

edited Mar 26 2010 at 2:29

Harry Gindi
8,868●2●33●96

asked Mar 26 2010 at 1:30

Portland
1,904●3●16

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See en.wikipedia.org/wiki/Feit-Thompson_conjecture . – Qiaochu Yuan Mar 26 2010 at 2:12

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I've deleted my comment, which was incorrect and apologize to Portland for my flippancy. – Harry Gindi Mar 26 2010 at 2:29

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@QY, turn the comment into an answer and let Portland accept it. – Mariano Suárez-Alvarez Mar 26 2010 at 3:24

Qiaochu Yuan · Accepted Answer · 2010-03-26 03:33:15Z UTC

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The answer is no. As the Wikipedia article in my comment states, the counterexample $p = 17, q = 3313$ was found by Stephens in 1971, but the stronger question of whether one can ever divide the other is a famous open problem because its solution would greatly simplify a step in the proof of the Feit-Thompson theorem.

answered Mar 26 2010 at 3:33

Qiaochu Yuan
38.4k●5●116●301

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See also Karl Dilcher and Joshua Knauer, On a conjecture of Feit and Thompson, in High primes and misdemeanors, Fields Inst Commun 41 (2004) 169-178, MR 2005c:11003, where it is confirmed that the Stephens example is still the only one known (as of the date of the publication, of course), and where details are provided about the scope of the search for more examples. – Gerry Myerson Mar 26 2010 at 3:53

I've just check all pairs $p,q\leq 1000$th prime out of curiosity... – Mariano Suárez-Alvarez Mar 26 2010 at 4:00

Very useful, thank you Qiaochu. – Portland Mar 26 2010 at 4:06

Prime numbers that lead to relatively prime

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