# 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 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:

$2^q - 1 \equiv q + 1 \equiv 0 \pmod r$

$q \divides r - 1$

and

$r \divides q + 1$

which is impossible.

Hence both $p$ and $q$ are odd primes.

$r \divides \paren {p - 1}$
$A = 1 + p + p^2 + \cdots + p^{q - 1} \equiv q \pmod r$

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 N.M. Stephens in $1971$, by means of a computer search.

Note that David Wells, in his Curious and Interesting Numbers, 2nd ed. of $1997$, misreports this result, stating it about the common divisor of $p^p - 1$ and $q^q - 1$.

This 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 $2$.