Fermat Quotient of 2 wrt p is Square iff p is 3 or 7/Generalization

Theorem

Let $p$ be a prime number.

The Fermat quotient of $2$ with respect to $p$:

$\map {q_p} 2 = \dfrac {2^{p - 1} - 1} p$

is a perfect power if and only if $p = 3$ or $p = 7$.

Proof

To show that these are the only ones, we observe that since $p$ is an odd prime, write:

$p = 2 n + 1$ for $n \ge 1$.

Let $\map {q_p} 2$ be a perfect power.

Then $2^{p - 1} - 1 = p x^y$ for some integers $x, y$.

Note that:

$2^{p - 1} - 1 = 2^{2 n} - 1 = \paren {2^n - 1} \paren {2^n + 1}$

and we have:

$\gcd \set {2^n - 1, 2^n + 1} = \gcd \set {2^n - 1, 2} = 1$

so $2^n - 1$ and $2^n + 1$ are coprime.

Hence there are $2$ cases:

Case $1$: $p \divides 2^n - 1$

By Divisor of One of Coprime Numbers is Coprime to Other:

$\gcd \set {\dfrac {2^n - 1} p, 2^n + 1} = 1$

Hence both the numbers are perfect $y$th powers.

In particular we have:

$\exists k \in \Z: 2^n + 1 = k^y$

by 1 plus Power of 2 is not Perfect Power except 9, the only solution to the equation above is:

$n = k = 3, y = 2$

This gives $p = 2 n + 1 = 7$.

$\Box$

Case $2$: $p \divides 2^n + 1$

By Divisor of One of Coprime Numbers is Coprime to Other:

$\gcd \set {\dfrac {2^n + 1} p, 2^n - 1} = 1$

Hence both the numbers are perfect $y$th powers.

In particular we have:

$\exists k \in \Z: 2^n - 1 = k^y$

By 1 plus Perfect Power is not Prime Power except for 9, this equation has no solution for $n, k, y > 1$.

Hence we must have $n = 1$.

This gives $p = 2 n + 1 = 3$.

$\Box$

In the main theorem we have already shown that $p = 3, 7$ gives perfect power Fermat quotients.

Hence the result.

$\blacksquare$