$a^n\mid b^n$ if and only if $a\mid b$.

Question

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.

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

GDumphart · Answer 1 · 2014-10-14T21:34:21.487

2

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

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.

edited Oct 14 '14 at 21:34

answered Oct 14 '14 at 21:26

GDumphart

2,300

You need to be careful: $4 \mid -16$ and $-16 = -4 \cdot 4$. But taking the square root of $-4$ does not work here. – Mathmo123 Oct 14 '14 at 21:29
furthermore, the first root of $\frac{1}{2}$ is $\frac{1}{2}$ which is neither an integer nor irrational, so your statement is invalid. – JMoravitz Oct 14 '14 at 21:30
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@JMoravitz I think it's clear that he meant non-trivial roots. – Mathmo123 Oct 14 '14 at 21:31
@Mathmo123 Question starts with "Suppose a, b, n are positive" :) – GDumphart Oct 14 '14 at 21:32
@JMoravitz Of course $n>1$ is implicitly assumed because the claim for $n=1$ is trivial – Hagen von Eitzen Oct 14 '14 at 21:32
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@GDumphart ah so it does. Perhaps you should replace $k \in \mathbb Z$ with $k \in \mathbb N$ in your answer. – Mathmo123 Oct 14 '14 at 21:32
Haha leave me alone with the $n = 1$ case. – GDumphart Oct 14 '14 at 21:33
@Mathmo123 True, I'll do that. Thank you. – GDumphart Oct 14 '14 at 21:33

score 0 · Answer 2 · answered May 31 '24 at 12:38

Let's prove the easy direction $a \mid b \Rightarrow a^n \mid b^n$ first. By assumption there is a $k$ such that $a k = b$. We then have $$ a^n (k^n) = (a k)^n = b^n $$ and therefore $a^n \mid b^n$.

For the other direction, let $g = \mathsf{gcd}(a, b)$ be the greatest common divisor of $a$ and $b$. We then know that $a_\ast := a/g$ and $b_\ast := b/g$ are coprime integers. We are assuming that $a^n \mid b^n$ so there is a $k$ with $a^n k = b^n$. Note now that we have $$ a^n k = b^n \implies (g a_\ast)^n k = (g b_\ast)^n \implies a_\ast^n k = b_\ast^n $$ Assuming $n > 0$ this actually shows $a_\ast \mid b_\ast^{n-1} b_\ast$. Given that $a_\ast$ and $b_\ast$ are coprime, Euclid's Lemma allows us to conclude that $a_\ast \mid b_\ast^{n-1}$. By iterating this, we arrive at $a_\ast \mid b_\ast$, from which we can now easily conclude $a \mid b$.

$a^n\mid b^n$ if and only if $a\mid b$.

2 Answers2