# Solution

Show that if $a$ and $b$ are integers such that $a \mid b$, then $a^n \mid b^n$ for every positive integer $n$

If $a$ and $b$ are integers such that $a \mid b$, then $b = ak_1$ for some integer $k_1$. Now, raising both sides to the $n^\textrm{th}$ power, we have \begin{align}b^n &= (ak_1)^n\\ &= a^n {k_1}^n\end{align} However, ${k_1}^n$ must be an integer due to the closure of the integers under multiplication. So, defining $k_2 = {k_1}^n$, we see that $b^n = a^n k_2$ for some integer $k_2$. Hence, $a^n \mid b^n$.

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