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.

$p = 2$.

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:

which is impossible.

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


 * $r \divides \paren {p - 1}$
 * $r \divides \paren {p - 1}$

By Sum of Geometric Sequence:
 * $A = 1 + p + p^2 + \cdots + p^{q - 1} \equiv q \pmod r$

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.