Divisors of One

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

That is:
 * $a \divides 1 \iff a = \pm 1$

Necessary Condition
Let $a = \pm 1$.

From Integer Divides Itself‎ we have that $1 \divides 1$.

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

Sufficient Condition
Let $\exists a \in \Z: a \divides 1$.

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

From Absolute Value Function is Completely Multiplicative we have that:
 * $\size a \cdot \size c = \size 1$

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

So $\size a \ge 1$ and $\size c \ge 1$.

But if $\size a > 1$ then:
 * $\size a \cdot \size c > \size c$

and so:
 * $\size a \cdot \size c > 1$

So:
 * $\size a = 1$

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