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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11$\begingroup$ See en.wikipedia.org/wiki/Feit-Thompson_conjecture . $\endgroup$– Qiaochu YuanCommented Mar 26, 2010 at 2:12
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2$\begingroup$ I've deleted my comment, which was incorrect and apologize to Portland for my flippancy. $\endgroup$– Harry GindiCommented Mar 26, 2010 at 2:29
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1 Answer
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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
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4$\begingroup$ 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. $\endgroup$ Commented Mar 26, 2010 at 3:53
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$\begingroup$ I've just check all pairs $p,q\leq 1000$th prime out of curiosity... $\endgroup$ Commented Mar 26, 2010 at 4:00