Prove that if $a^n\mid b^n$ then $a\mid b$

Solution 1:

Suppose that $x=\frac pq$ where $p$ and $q$ are relatively prime; that is, there are $a$ and $b$ so that $$ ap+bq=1\tag{1} $$ Suppose that there are integers $c_k$ so that $$ x^n+\sum_{k=0}^{n-1}c_kx^k=0\tag{2} $$ Equation $(1)$ says $x=\frac{1-bq}{aq}$, and substituting into $(2)$ yields $$ \begin{align} 0 &=\left(\frac{1-bq}{aq}\right)^n+\sum_{k=0}^{n-1}c_k\left(\frac{1-bq}{aq}\right)^k\\ &=\frac1{(aq)^n} +\sum_{k=0}^{n-1}\binom{n}{k}\frac1{(aq)^k}\left(-\frac ba\right)^{n-k} +\sum_{k=0}^{n-1}c_k\left(\frac{1-bq}{aq}\right)^k\tag{3} \end{align} $$ Multiply $(3)$ by $a^nq^{n-1}$ and move $\frac1q$ to the left side $$ \frac1q =b\sum_{k=0}^{n-1}\binom{n}{k}(-bq)^{n-k-1} -a\sum_{k=0}^{n-1}c_k(1-bq)^k(aq)^{n-k-1}\tag{4} $$ Everything on the right side of $(4)$ is an integer, so $\frac1q$ must also be an integer. Thus, $x=\frac pq$ is also an integer.

Therefore, any rational number $x$ which satisfies $(2)$ with integer $c_k$ must be an integer.

Finish off as awllower does. Let $x=\frac ba$. We are given that $x^n-k=0$ for some integer $k$. The argument above says that $x=\frac ba$ is an integer.

Solution 2:

If $a^n|b^n$, then $b^n/a^n=(\frac{b}{a})^n$ is an integer. Since $\frac{b}{a}$ is rational, this implies that $\frac{b}{a}$ is an ordinary integer, by this theorem.
Q.E.D.
notice that the use of the theorem is suggested by @CalvinLin .