TITLE: Proof by induction; $a^n$ divides $b^n$ implies $a$ divides $b$
QUESTION [7 upvotes]: I want to prove by induction that $a^n \mid b^n$ implies that $a \mid b$ holds for all integers $n\geq 1$.
Clearly for $n=1$ this is true, since if $a \mid b$, then $a \mid b$.
Suppose this is true for some $n = k$. Then $a^k \mid b^k$, so $a|b$.
$a^k \mid b^k$ means there exists some integer $m$ such that $b^k = ma^k$, and $a \mid b$ means there exists an integer $r$ such that $b=ra$.
Then we can multiply both sides of the above equality by any constant, namely $b$. 
Then $b \cdot b^k = bma^k$
so $b^{k+1} = (ra)ma^k = rma^{k+1}$
which means that $a^{k+1} \mid b^{k+1}$ completing the induction step, and thus proving that for all integers $n \geq1$ the original statement holds.
Are there any logical holes in the proof?

REPLY [6 votes]: Let $a,b$ be integers and let $P(k)$ be the statement "$a^k \mid b^k$ implies $a \mid b$."  You are trying to prove by induction that $P(k)$ is true for all $k\in \mathbb{N}$.
You start out by proving the base case $P(1)$, which is fine.  Then you say "suppose this is true for some $k$," by which I mean you are supposing that $P(k)$ is true.  But then you say, "then $a^k \mid b^k$."  This does not follow from $P(k)$, since $P(k)$ says nothing about whether $a^k$ divides $b^k$ or not, just about what happens if it does.
What you need to do is show that if $P(k)$ is true, then $P(k+1)$ is also true.  Others have already talked about how to approach this, I just wanted to clarify a bit more what goes wrong with your attempt.