Let $a, b, n$ be elements of $\mathbb N $ such that $ a^n\mid b^n $. Show that $a\mid b$. [closed]
Hint $\rm\ \ (b/a)^n = k \in \Bbb Z\ \Rightarrow\ b/a\in \Bbb Z\ $ by the Rational Root Test, i.e. if $\,x\,$ is a rational root of the polynomial $\rm \,\color{#c00}1\cdot x^n-k\,$ then its least-terms denominator divides $\color{#c00}1,\,$ so $\,x\,$ is an integer.
Hint: let $p_1^{\alpha_1}\cdot…\cdot p_k^{\alpha_k}=a$, $q_1^{\beta_1}\cdot…\cdot q_m^{\beta_m}=b$ be prime decompositions. Then $a^n=p_1^{n\alpha_1}\cdot…\cdot p_k^{n\alpha_k}$, $b^n=q_1^{n\beta_1}\cdot…\cdot q_m^{n\beta_m}$. Now write up what $a^n|b^n$ means in term of prime powers.