Suppose $a, b$ and $n$ are positive integers. Prove that $a^n$ divides $b^n$ if and only if $a$ divides $b$. I think prime factorization is needed for this question: 
Suppose $a$, $b$ and $n$ are positive integers. Prove that $a^n$ divides $b^n$ if and only if a divides $b$.
 A: One side is trivial. Then suppose $b = p_{1}^{m_{1}} \cdot \ldots \cdot p_{k}^{m_{k}}$ and $a = q_{1}^{l_{1}} \cdot \ldots \cdot q_{h}^{l_{h}}$, where $p_{i}$ and $q_{j}$ are primes. Then $b^{n} = p_{1}^{nm_{1}} \cdot \ldots \cdot p_{k}^{nm_{k}}$ and $a^{n} = q_{1}^{nl_{1}} \cdot \ldots \cdot q_{h}^{nh_{l}}$. Since $a^{n}$ divides $b^{n}$ all the primes appearing in the factorization of $a^{n}$ appear in the one of $b^{n}$ and they have a lower exponent. We have $h \leq k$ and we can reorder in a way that $q_{i}=p_{i}$ for $i \leq h$. Then we have $b^{n}=p_{1}^{n\left(l_{1} + \left( m_{1} - l_{1}\right)\right)} \cdot \ldots \cdot p_{h}^{n\left(l_{h} + \left( m_{h} - l_{h}\right)\right)} \cdot p_{h+1}^{nm_{h+1}} \cdot \ldots \cdot p_{k}^{nm_{k}}$. So the qoutient is $p_{1}^{n\left( m_{1} - l_{1}\right)} \cdot \ldots \cdot p_{h}^{n \left( m_{h} - l_{h}\right)} \cdot p_{h+1}^{nm_{h+1}} \cdot \ldots \cdot p_{k}^{nm_{k}}$; notice that every exponent is divisible by $n$, so we can extract the $n^{th}$ root and we're done.
A: Hint $\,\  \dfrac{b^n}{a^n} = k\in \Bbb Z$ $\ \Rightarrow\ $ $x = \dfrac{b}a\ $ root of $\ x^n\!-k$ $\!\!\!\!\underbrace{\Rightarrow\,x\in\Bbb Z}_{\text{ Rational Root Test}}\!\!\!\!\!$ $\,\Rightarrow\, a\mid b $
A: Prime factorization is not needed: we only need the fact that every integer $\ne\pm1$ has a prime divisor.
Define $r=a/b\in\mathbb Q.$ As $b^n\mid a^n,$ we know $r^n\in\mathbb Z.$
Write $r=a'/b'$ with $\gcd(a',b')=1.$ Let $s=r^n.$ Then $r^n=s$ implies that $(a')^n=(b')^n\cdot s.$ This shows that every prime divisor of $b'$ divides $(a')^n;$ by the definition of a prime, this means that every prime divisor of $b'$ divides $a'.$ This contradicts $\gcd(a',b')=1.$ Therefore $b'$ has no prime divisor, and is equal to $\pm1.$ Thus $r=a/b=a'/b'=\pm a'\in\mathbb Z.$ So $b$ divides $a.$  
Hope this helps.  
P.S.
The converse is trivial: if $b$ divides $a,$ then $a/b$ is an integer, so $a^n/b^n=(a/b)^n$ is also an integer and $b^n$ divides $a^n.$
