# What other prime numbers have been ruled out as counterexamples to the Feit-Thompson conjecture?

Given distinct primes $p$ and $q$, $$\frac{p^{\,q} - 1}{p - 1}$$ is never a divisor of $$\frac{q^{\,p} - 1}{q - 1}.$$ Or so we believe. If $p = 2$, then it's clear that no odd prime $q$ can make a counterexample. If $p = 3$, then we have $$\frac{3^q - 1}{2}$$ and $$\frac{q^3 - 1}{q - 1}.$$ I am confident that with time I can prove there's no counterexample here either, but it's of course quite laborious to go through the primes one by one. Has anyone been able to rule out a bunch of primes at once?

• It appears that for odd $a$ we have $\dfrac{(6x+1)^a-1}{6x}\equiv a\bmod6$ and $\dfrac{(6x-1)^a-1}{6x-2}\equiv 1\bmod6$, which would imply for instance that $p\equiv q\bmod6$ for $p,q>3$. – Jaycob Coleman Feb 16 '15 at 20:56
• By elementary means if $p=3<q$ is a counterexample then $r=q^2+q+1$ is also prime and $3^q-1$ is divisible by $r$. – DanielWainfleet Sep 8 '15 at 2:25
• a) Hmm, in wikipedia they focus divisibility of one cyclotomic expression by the other one (and no counterexample is known) and in Weissstein's mathworld they focus on common factors (gcd()) and document one known case as counterexample. Which is the original Feit-Thompson-conjecture? b) How should ever $(q^3-1)/(q-1)$ be divisible by $(3^q-1)/(3-1)$ when $q$ is larger than $3$? Didn't you confuse the direction of divisibility here? – Gottfried Helms Mar 5 '17 at 19:35
• Ah, I found in the wp-linked article Feit, Walter, and John G. Thompson. “A Solvability Criterion for Finite Groups and Some Consequences.” that $p \gt q$ was assumed and divisibility was focused. – Gottfried Helms Mar 5 '17 at 19:46

First, note that $(3^q-1)(q-1)>2(q^3-1)$ for every $q>3$. This is because $3$ is the point where the two lines meet, and $3^q$ grows faster than $q^3$.
Now, we simply let $q$ and $n$ be prime and natural (respectively) such that $n(\frac{3^q-1}{2})=\frac{q^3-1}{q-1}$.
Then, $n(3^q-1)=2(\frac{q^3-1}{q-1})$, and thus $n(3^q-1)(q-1)=2(q^3-1)$. This means that $q\leq3$. Of course, because the final equation also holds for both $2$ and $3$, this means that there are no numbers greater than 3 where this is true. So, there are no counterexamples.