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 $p$:
 * $q_p \left({2}\right) = \dfrac {2^{p - 1} - 1} p$

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

Proof
When $p = 3$:
 * $q_3 \left({2}\right) = \dfrac {2^{3 - 1} - 1} 3 = 1$

which is square.

When $p = 7$:
 * $q_7 \left({2}\right) = \dfrac {2^{7 - 1} - 1} 7 = \dfrac {63} 7 = 9$

which is square.