Number does not divide Number iff Square does not divide Square

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

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

where $a \nmid b$ denotes that $a$ is not a divisor of $b$.

Proof
Let $a \nmid b$.


 * $a^2 \mathrel \backslash b^2$
 * $a^2 \mathrel \backslash b^2$

where $\backslash$ denotes divisibility.

Then by Number divides Number iff Square divides Square:
 * $a \mathrel \backslash b$

From Proof by Contradiction it follows that $a^2 \mathrel \backslash b^2$ is false.

Thus $a^2 \nmid b^2$.

Let $a^2 \nmid b^2$.


 * $a \mathrel \backslash b$
 * $a \mathrel \backslash b$

Then by Number divides Number iff Square divides Square:
 * $a^2 \mathrel \backslash b^2$

From Proof by Contradiction it follows that $a \mathrel \backslash b$ is false.

Thus $a \nmid b$.