Square Free
An integer $$f \in \mathbb{ Z } ^ { \neq 0 }$$ is said to be square free if there is no $$k \in \mathbb{ N } _ 1$$ such that $k ^ 2 \mid f$
Square Free Division Characterization
$$f \in \mathbb{ Z } ^ { \neq 0 }$$ is square free iff for any $$k \in \mathbb{ Z }$$ $k ^ 2 \mid f \implies k \in \left\{ 1, -1 \right\}$
Divisor of Square Free Product is Divisor of Other
Suppose that $$f \in \mathbb{ Z } ^ { \neq 0 }$$ if $$f$$ is square free then $a ^ 2 \mid f b ^ 2 \implies a \mid b$ for any $$a, b \in \mathbb{ Z }$$
If a Square Free Number Divides a Square then It divides the Square Root
Suppose $$f \in \mathbb{ Z } ^ { \neq 0 }$$, if $$f$$ is square free then $f \mid a ^ 2 \implies f \mid a$
Suppose that $$f \mid a ^ 2$$, then we have $$f ^ 2 \mid f a ^ 2$$ therefore we have that $$f \mid a$$
If a Square Free Number Divides the Nth Power then it Divides the Nth Square Root
Suppose that $$f \in \mathbb{ Z } ^ { \neq 0 }$$ is square free then $f \mid a ^ n \implies f \mid a$ for any $$n \in \mathbb{ N } _ 1$$
We assume that $$f \mid a ^ n$$, now if $$n = 1$$ we are done, otherwise $$n \ge 2$$ and therefore we know that $$2n \ge 2 + n$$ so that $$2n - 2 \ge n \ge 2$$, thus we have that $$f \mid a ^ { 2 \left( n - 1 \right) }$$ therefore by assumption we know that $$f \mid a ^ { n - 1 }$$, if $$n - 1 = 1$$ we are done, otherwise $$n - 1 \ge 2$$ and we repeat this process finitely many times until $$n - j = 1$$ at which point we've deduced that $$f \mid n$$ as needed.
Power of a Square free Number Divides the Power of a Number Implies the Square Free Number Divides the Other
Suppose $$f \in \mathbb{ Z } ^ { \neq 0 }$$ is square free, then $f ^ f \mid a ^ a \implies f \mid a$ for any $$a \in \mathbb{ N } _ 1$$