Divisors of One

Theorem
The only divisors of $1$ are $1$ and $-1$.

Proof
From Integer Divides Itself‎ we have that $1 \mathop \backslash 1$.

From Integer Divides its Negative we have that $-1 \mathop \backslash 1$.

Now suppose $\exists a \in \Z: a \mathop \backslash 1$.

Then $\exists c \in \Z: a c = 1$.

From Absolute Value Function is Completely Multiplicative we have that:
 * $\left \vert {a}\right \vert \cdot \left \vert {c}\right \vert = \left \vert {1}\right \vert$

Neither $a$ nor $c$ can be zero, from Integers form Integral Domain.

So $\left \vert {a}\right \vert \ge 1$ and $\left \vert {c}\right \vert \ge 1$.

But if $\left \vert {a}\right \vert > 1$ then $\left \vert {a}\right \vert \cdot \left \vert {c}\right \vert > \left \vert {c}\right \vert$ and so $\left \vert {a}\right \vert \cdot \left \vert {c}\right \vert > 1$.

So:
 * $\left \vert {a}\right \vert = 1$

that is:
 * $a = 1$ or $a = -1$