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 \( a \in \mathbb{ N } _ 1 \)