Number divides Number iff Square divides Square

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

Then:
 * $a^2 \mathrel \backslash b^2 \iff a \mathrel \backslash b$

where $\backslash$ denotes integer divisibility.

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

is a geometric progression.

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

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

Thus:

Let $a \mathop \backslash b$.

Then: