Show that none of the elements in $\{2, 3, \ldots , p− 2\}$ modulo $p$ are self-invertible.
Self-invertible means $a^{−1} ≡ a \pmod n$.
Let's assume, for the purpose of contradiction, that $a$ is self-invertible, where $a \in \{2, 3, \ldots , p− 2\}$.
That is,
$$ a^{-1} \equiv a \pmod p$$
By definition of a multiplicative inverse,
$$ a \times a^{-1} \equiv 1 \pmod p $$
Since $a^{-1} \equiv a \pmod p$
$$ \begin{align} a \times a & \equiv 1 \pmod p \\ \\ a^2 -1 & \equiv 0 \pmod p \\ \\ (a+1)(a-1) & \equiv 0 \pmod p \end{align}$$
By Proposition (3.14)(a) we have $(a +1) \equiv 0 \pmod p$ or $(a -1) \equiv 0 \pmod p$.
That is, at least one of the following is true
- $a \equiv 1 \pmod p$
- $a \equiv -1 \equiv p-1 \pmod p$
Neither of these are in $\{2, 3, \ldots , p− 2\}$ modulo $p$, which contradicts the assumption that elements in $\{2, 3, \ldots , p− 2\}$ are self-invertible.
So we conclude that none of the elements in $\{2, 3, \ldots , p− 2\}$ are self-invertible.
Alternatively, from Exercise (3.3).11, $a$ is only self-invertible in modulo $p$ if $a \equiv \pm 1 \pmod p$. That is $a \equiv 1 \pmod p$ or $a \equiv -1 \equiv p-1 \pmod p$. But these $a$ are not in $\{2, 3, \ldots , p− 2\}$, and so elements of this set are not self-invertible.