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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?

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I've deleted my comment, which was incorrect and apologize to Portland for my flippancy. – Harry Gindi Mar 26 '10 at 2:29
@QY, turn the comment into an answer and let Portland accept it. – Mariano Suárez-Alvarez Mar 26 '10 at 3:24
up vote 17 down vote accepted

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.

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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 '10 at 3:53
I've just check all pairs $p,q\leq 1000$th prime out of curiosity... – Mariano Suárez-Alvarez Mar 26 '10 at 4:00
Very useful, thank you Qiaochu. – Portland Mar 26 '10 at 4:06

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