If $a$, $b$, and $n$ are positive integers, prove that $a^n|b^n$ if and only if $a|b$.
So far I've done one way of the proof...
I've proved that if $a|b$, then $b=ak$, $k$ is an integer, then raising each to the power of $n$, I get $b^n=k^na^n$, and since $k^n$ is an integer, then $a^n|b^n$. Im wondering two things. Firstly, is this a correct proof. Secondly, how do I prove the iff statement, the other way?
Thanks
Your "if" proof looks correct. What about "only if", here are my suggestions:
$a^n|b^n \Rightarrow b^n = ka^n, k \in \mathbb N \Rightarrow k = (\frac ba)^n$. If $\frac ba$ is not an integer, then $(\frac ba)^n, n \in \mathbb N$ can't be an integer as well*, but $k \in \mathbb N$, so $\frac ba \in \mathbb N \Rightarrow a|b$.
*I decided to add a proof of statement marked above.
So we need to prove that $(\frac ba)^n \in \mathbb N \Rightarrow \frac ba \in \mathbb N, n \in \mathbb N$.
Let $\frac ba \not\in \mathbb N$ and $a^* = \frac{a}{\gcd(a,b)}$, $b^* = \frac{b}{\gcd(a,b)}$. Obviously, $\frac{b^*}{a^*} = \frac ba$ and $a^*$ is relatively prime with $b^*$. We supposed that $a \nmid b \Rightarrow a^* > 1$. Let $$a^* = a_1^{\alpha_1} * a_2^{\alpha_2} * ... * a_N^{\alpha_N}$$ $$b^* = b_1^{\beta_1} * b_2^{\beta_2} * ... * b_M^{\beta_M}$$
Then $$(\frac {b^*}{a^*})^n = \frac{b_1^{\beta_1n} * b_2^{\beta_2n} * ... * b_M^{\beta_Mn}}{a_1^{\alpha_1n} * a_2^{\alpha_2n} * ... * a_N^{\alpha_Nn}}$$ Every $a_i$ and $b_i$ are prime, so $a_i^{\alpha_in}$ and $b_i^{\beta_in}$ are prime as well. Thus, this fraction can't be simplified. As $a > 1$, this number is not an integer. However, $(\frac ba)^n = (\frac {b^*}{a^*})^n$ and $(\frac ba)^n$ is an integer. We came to a contradiction, so our suggestion about $a \nmid b$ was not true $\Rightarrow \frac ba \in \mathbb N \Rightarrow a|b$.