# Variant of Stephens result $\gcd(\Phi_p(q),\Phi_q(p))=2pq+1$ for $p=17$, $q=3313$

Let $\Phi_n(t)$ be the $n$-th cyclotomic polynomial over the rationals. Stephens proved in 1971 that $\Phi_p(q)$ and $\Phi_q(p)$ are not always coprime for primes $p,q$ since one has

$$gcd(\Phi_p(q),\Phi_q(p)) = 2pq+1$$

for

$$p=17, q=3313$$

See the wiki page about the Feit-Thompson Conjecture.

Question: Are there more solutions to

$$\gcd(m,n)=1$$

and

$$\gcd(\Phi_m(n),\Phi_n(m)) = 2mn+1\ ?$$

None found for small $m,n.$

UPDATE: I have a new solution: $m=464$, $n=21$ for which I do not know why $r=2mn+1$ is also prime !

After several weeks of computations there are no more new solutions with $$m \leq 13400.$$

You may want to look at Karl Dilcher and Joshua Knauer, On a conjecture of Feit and Thompson, in the book, High Primes and Misdemeanours: Lectures in Honour of the 60th Birthday of Hugh Cowie Williams, 169–178, Fields Inst. Commun., 41, Amer. Math. Soc., Providence, RI, 2004, MR2076245 (2005c:11003). In addition to a thorough discussion of the state of the art as regards the Feit-Thompson conjecture, they discuss the case you appear to be interested in, where the variables are not necessarily prime. They find, for example, that if $p$ is an odd prime, then under various conditions on $k$ we get $\gcd(\Phi_2(kp),\Phi_{kp}(2))=kp+1$.
• I just checked the Mathscinet and Zentralblatt reviews. No $m,n$ showed as new solution of the equation. I need probably to check the actual paper. Thanks for tip. – Luis H Gallardo Apr 28 '11 at 0:42
Probably none others with $\min(p,q)<10000$ and $\max(p,q)<20000$. For $p,q \lt 4000$ there are $59$ cases of $\Phi_p(q) \mod 2pq+1=0.$ In all those cases $2pq+1$ is prime. Taking that as an additional requirement, there are $201$ cases with the bounds as above but no cases in which both congruences hold. Note that $\Phi_q(p)=\frac{p^q-1}{p-1}$ so to check if this is a multiple of $m=2pq+1$ it is enough to check if $p^q \mod m=1$ and that is a quick calculation.
• In your last two lines: You need of course $q$ prime. – Luis H Gallardo Apr 28 '11 at 11:12
• New solution: $m=464,$ $n=21$, $r=2mn+1=19489$. Here also $r$ is prime. – Luis H Gallardo Apr 28 '11 at 23:11