Number divides Number iff Square divides Square

Theorem
Let $a, b \in \Z$.

Then:
 * $a^2 \divides b^2 \iff a \divides b$

where $\divides$ denotes integer divisibility.

Proof
From Between two Squares exists one Mean Proportional:
 * $\tuple {a^2, ab, b^2}$

is a geometric progression.

Let $a, b \in \Z$ such that $a^2 \divides b^2$.

Then from First Element of Geometric Progression that divides Last also divides Second:
 * $a^2 \divides a b$

Thus:

Let $a \divides b$.

Then: