Feit-Thompson Conjecture/Stronger
Conjecture
There exist no distinct prime numbers $p$ and $q$ such that:
- $\dfrac {p^q - 1} {p - 1}$ and $\dfrac {q^p - 1} {q - 1}$ are not coprime.
Refutation
Let:
- $A = \dfrac {p^q - 1} {p - 1}$
- $B = \dfrac {q^p - 1} {q - 1}$
Suppose there exists a prime number $r$ such that:
- $r \divides A$ and $r \divides B$
where $\divides$ denotes divisibility.
Aiming for a contradiction, suppose $p = 2$.
Then:
- $r \divides 2^q - 1$
and:
- $r \divides q + 1$
From Factor of Mersenne Number $M_p$ is of form $2 k p + 1$:
- $2 q \divides r - 1$
From Absolute Value of Integer is not less than Divisors:
\(\ds r\) | \(\le\) | \(\ds q + 1\) | ||||||||||||
\(\ds 2 q\) | \(\le\) | \(\ds r - 1\) |
which is impossible.
Hence both $p$ and $q$ are odd primes.
Aiming for a contradiction, suppose:
- $r \divides \paren {p - 1}$
- $A = 1 + p + p^2 + \cdots + p^{q - 1} \equiv q \pmod r$
and so $r = q$.
But $q \nmid B$.
Hence:
- $r \nmid \paren {p - 1}$
Similarly:
- $r \nmid \paren {q - 1}$
It follows that:
- $p^q \equiv 1 \pmod r$
from which:
- $q \divides \paren {r - 1}$
It follows then that:
- $(1): \quad r = 2 \lambda p q + 1$
for some $\lambda \in \Z_{>0}$.
It then remains to test the residues of $p^q$ and $q^p$ modulo $r$ for all odd primes $p$ and $q$ for all $r$ of the form given in $1$ in a suitable range.
The counterexample $p = 17$, $q = 3313$ was discovered using the above technique in $1971$, using the ranges $p < 443$, $pq < 200 \, 000$ and $r < 400 \, 000$.
The common divisor was found to be $2 p q + 1 = 112 \, 643$.
It is believed that the exercise has not been reported on since, and that no other counterexamples to the conjecture are known.
Source of Name
This entry was named for Walter Feit and John Griggs Thompson.
Historical Note
The stronger form of the Feit-Thompson Conjecture was refuted by Nelson Malcolm Stephens in $1971$, by means of a computer search.
Note that David Wells, in his Curious and Interesting Numbers of $1986$, misreports this result, stating it about the common divisor of $p^q - 1$ and $q^p - 1$.
Then in his Curious and Interesting Numbers, 2nd ed. of $1997$, he misreports this result again, in a different way, stating it about the common divisor of $p^p - 1$ and $q^q - 1$.
The latter is obviously spurious, as, given odd primes $p$ and $q$, $p^p - 1$ and $q^q - 1$ are both even, and hence have a common divisor of $2$.
Sources
- July 1971: N.M. Stephens: On the Feit-Thompson Conjecture (Math. Comp. Vol. 25, no. 115: p. 625) www.jstor.org/stable/2005226