# Integer to Power of p-1 over 2 Modulo p

## Theorem

Let $a \in \Z$.

Let $p$ be an odd prime.

Let $b = a^{\frac {\paren {p - 1} } 2}$.

Then one of the following cases holds:

$b \bmod p = 0$

which happens exactly when $a \equiv 0 \pmod p$, or:

$b \bmod p = 1$

or:

$b \bmod p = p - 1$

where:

$b \bmod p$ denotes the modulo operation
$x \equiv y \pmod p$ denotes that $x$ is congruent modulo $p$ to $y$.

## Proof

By definition of congruence modulo $p$:

$\forall x, y \in \R: x \equiv y \pmod p \iff x \bmod p = y \bmod p$

We have that:

$b = a^{\frac{\paren {p - 1} } 2}$

and so:

$b^2 = a^{p - 1}$

Let $a \equiv 0 \pmod p$.

Then by definition of congruence modulo $p$:

$p \divides a$

and so:

$p \divides a^{\frac{\paren {p - 1} } 2}$

where $\divides$ denotes divisibility.

Thus by definition of congruence modulo $p$:

$b \equiv 0 \pmod p$

and so:

$b \bmod p = 0$

$\Box$

Otherwise, from Fermat's Little Theorem:

$b^2 \equiv 1 \pmod p$

That is:

$b^2 - 1 \equiv 0 \pmod p$
$b^2 - 1 = \paren {b + 1} \paren {b - 1}$

So either:

$p \divides b + 1$

or:

$p \divides b - 1$

Aiming for a contradiction, suppose both $p \divides b + 1$ and $p \divides b - 1$.

$p \divides \paren {b + 1} - \paren {b - 1} = 2$

But $p$ is an odd prime.

So it cannot be the case that $p \divides 2$.

From this contradiction it follows that

Note that $p$ cannot divide both $b + 1$ and $b - 1$.

So either:

 $\ds \paren {b - 1}$ $\equiv$ $\ds 0$ $\ds \pmod p$ $\ds \leadsto \ \$ $\ds b$ $\equiv$ $\ds 1$ $\ds \pmod p$ Modulo Addition is Well-Defined

or:

 $\ds \paren {b + 1}$ $\equiv$ $\ds 0$ $\ds \pmod p$ $\ds \leadsto \ \$ $\ds b$ $\equiv$ $\ds -1$ $\ds \pmod p$ Modulo Subtraction is Well-Defined $\ds$ $\equiv$ $\ds p - 1$ $\ds \pmod p$ Negative Number is Congruent to Modulus minus Number

Hence the result.

$\blacksquare$