Use a particular method to prove if $a^n \mid b^n$ then $a\mid b$ A question in my tutorial asks to use the fact (can be used without proof) that 

$(a,b)=(a,c)=1 \Rightarrow (a,bc)=1$ 

to prove: 

if $a^n \mid b^n$ then $a\mid b$.

I did the following, but my tutor marked it as wrong. I have tried to find the error without success. 

May I know where is the error? Sincere thanks.



Assume $a^n \mid b^n$.

$b^n=a^n\cdot k$ , $k\in \mathbb{Z}$

Let $d:=(a,b)$.

$(\frac{a}{d},\frac{b}{d})=1$

By the fact above, $(\frac{a}{d},(\frac{b}{d})^{n-1})=1$.

Note that since $(\frac{b}{d})^n=(\frac{a}{d})^n\cdot k$, so $\frac{a}{d}\mid(\frac{b}{d})(\frac{b}{d})^{n-1}$. This, in addition to $(\frac{a}{d},(\frac{b}{d})^{n-1})=1$, yields $\frac{a}{d}\mid\frac{b}{d}$. (Euclid's Lemma)

Hence $a\mid b$.
Your proof is correct, viz. by cancelling $\rm\:(a,b)\:$ one reduces to the case $\rm\:(a,b) = 1,\:$ which follows immediately by inductive application of Euclid's Lemma.

The proof is a special monic binomial case of the analogous proof of the Rational Root Test (RRT), since  for $\rm\:c = b/a\:$ we have $\rm\:c^n = k\in \Bbb Z,\:$ so $\rm\:c\:$ is a root of $\rm\:x^n - k\:$ so RRT implies $\rm\:c\in \Bbb Z.\:$ 

In fact, more generally, the Freshman's Dream for gcds is true $\rm\:(a^n,b^n) = (a,b)^n$
If $(a,bc)\neq 1$, then there is a prime number $p$ such that $p \mid a$ and $p \mid bc$.
But $p \mid bc \Rightarrow p \mid b$ or $p \mid c$. Then we have $p \mid a$ and $p \mid b, \Leftrightarrow (a,b) \neq 1$ or $p\mid a$ and $p\mid c, \Leftrightarrow (a,c) \neq 1$, contradiction.
Sorry I proved what we don't need.

If $a$ does not divide $b$. then there is $p$ prime such that $p \mid a$ but 
$p$ does not divide $b$. Then $p^n \mid a^n$ but $p^n$ does not divide $a^n$. Then $a^n$ does not divide $b^n$. Contradiction.