## A number theory question related to the Feit-Thompson theorem

In the book by Gelbaum and Olmsted on counterexamples in mathematics, they quote the following as being an open question as of 1990.

The problem comes from a long series of papers by Feit and Thompson on the solvability of groups of odd order (I couldn’t find where exactly, but for example their lemma 34.2 of chapter 5 certainly looks related). Not that I know anything about it all of course. The question as stated by Gelbaum and Olmsted is:

given two different primes $p$ and $q$, are the integers $N_p(q):=\sum_{k=0}^{q-1}p^k=\frac{p^q-1}{p-1}$ and $N_q(p):=\sum_{i=0}^{p-1}q^i$ relatively prime ?

and they comment that solving it in the affirmative would make the Feit-Thompson proof “considerably shorter”.

This looks a very natural question to answer even without the group theory background. Since nothing so obvious shows up here I tried to do some reverse engineering, namely to find if the integers $u,v\in\mathbb{Z}$ satisfying $uN_p(q)+vN_q(p)=1$ have some recognizable pattern (and without loss of generality assumed p smaller than q).

For $p=2$ we obtain for $q=3,5,7,11,13$ respectively the couples $(u,v)=(-1,2)$, $(1,-5)$, $(-1,16)$, $(-5,853)$, $(1,-585)$,

The numbers $N_2(q)$ form the sequence 3, 7, 31, 127, 2047, 8191,… of Mersenne numbers, but it’s just a name, and the OEIS unfortunately didn’t help saying anything about the sequence 2, -5, 16, 853, -585…

So, no progress at all on this Feit-Thompson question.

Update
: Apparently $N_p(q)$ is called a quantum integer (or q-bracket) and pops up in many areas, see this paper by Nathanson and this book by Kac and Cheung.