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

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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 if and only if $p = 3$ or $p = 7$.


Generalization

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

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 $p = 2 n + 1 = 7$.

$\Box$


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 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 $p = 2 n + 1 = 3$.

$\Box$


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

Hence the result.

$\blacksquare$


Sources

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