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

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 square $p = 3$ or $p = 7$.

Proof
When $p = 3$:
 * $\map {q_3} 2 = \dfrac {2^{3 - 1} - 1} 3 = 1$

which is square.

When $p = 7$:
 * $\map {q_7} 2 = \dfrac {2^{7 - 1} - 1} 7 = \dfrac {63} 7 = 9$

which is square.

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 square.

Then $2^{p - 1} - 1 = p x^2$ for some integer $x$.

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 squares.

In particular we have:
 * $\exists k \in \Z: 2^n + 1 = k^2$

by 1 plus Power of 2 is not Perfect Power except 9, the only solution to the equation above is:
 * $n = k = 3$

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

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 squares.

In particular we have:
 * $\exists k \in \Z: 2^n - 1 = k^2$

For $n > 1$:
 * $2^n - 1 \equiv 3 \pmod 4$

which by Square Modulo 4 is never a square.

Hence we must have $n = 1$.

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

We have already shown that $p = 3, 7$ gives square Fermat quotients.

Hence the result.