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Suppose $a$, $b$, $n$ are positive. Prove that $a^n\mid b^n$ if and only if $a\mid b$.

I know that this can be proved through prime factorization, but I want to prove it using other methods. I understand that the if and only if statement requires you to prove the two cases:

1. If $a^n\mid b^n$ then $a\mid b$.
2. If $a\mid b$ then $a^n\mid b^n$.

but I have no idea on how to approach it.

Michael Hardy
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    Can you at least do one of the directions? – hmakholm left over Monica Oct 14 '14 at 21:21
  • In other words: (2) follows by a direct application of the definitions. – PhoemueX Oct 14 '14 at 21:23
  • +1 with Henning. I'll wait before giving out the answer, but will remind you the definitions of divisibility which will hopefully give you ideas on how to set it up. Remember that $a|b$ means that $b = ak$ for some $k\in \mathbb{Z}$. As for proving the first direction, try proof by contrapositive. – JMoravitz Oct 14 '14 at 21:24
  • You might find this intresting http://math.stackexchange.com/questions/4467/sqrt-a-is-either-an-integer-or-an-irrational-number – Harto Saarinen Oct 14 '14 at 21:25

1 Answers1

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$a|b \ :\Leftrightarrow \exists k\in\mathbb{N} : k \cdot a = b$

1.) $$a^n|b^n \ \ \ \Rightarrow \ \ \ b^n = \hat{k} \cdot a^n $$

$$\sqrt[n]{\hat{k}} = \frac{b}{a}$$

Since $a,b$ are positive integers, $\frac{b}{a} \in \mathbb{Q}$. For $n>1$, the $n$-th root of a positive non-zero rational number is either an integer or irrational, but in this case it cannot be irrational because the RHS is rational. Therefore, $k' := \sqrt[n]{\hat{k}} \in \mathbb{N}$ and $b = k' \cdot a$, thus $a|b$.

2.) $$a|b \ \ \ \Rightarrow \ \ \ b = k \cdot a \ \ \ \Rightarrow \ \ \ b^n = k^n \cdot a^n \ \ \ \Rightarrow a^n | b^n$$ because $k^n \in \mathbb{N}$.

P.S.: The approach for 1.) is beating around the bush, you should really use the prime factorization as the main argument.

GDumphart
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