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Does $a^n \mid b^n$ imply $a\mid b$? I think it does but haven't been able to prove it. I don't know much number theory so an elementary answer would be great.

TMM
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Javier
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    Hint: Look at prime factors. – Brett Frankel Apr 05 '13 at 00:29
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    You can start with the fundamental theorem of arithmetic. – Julien Apr 05 '13 at 00:30
  • Consider this. Might I call yhis a duplicate? Since they end up in asking the same question. – awllower Apr 05 '13 at 13:32
  • @awllower: you're right that it's a duplicate, but I think this one has better answers. – Javier Apr 05 '13 at 14:16
  • @JavierBadia Probably because the previous one prohibited the use of GCD and UFD. – awllower Apr 05 '13 at 15:01
  • Here is a simple proof. Find $ r,s \in \mathbb{N} $ that satisfy $ \gcd(r,s) = 1 $ and $ \dfrac{b}{a} = \dfrac{s}{r} $. If $ a^{n} | b^{n} $ so that $ b^{n} = k a^{n} $ for some $ k \in \mathbb{N} $, then $ s^{n} = k r^{n} $ also. Now, we must have $ r = 1 $, otherwise $ p | r $ for some prime number $ p $, which would imply that $ p | s^{n} $ and consequently $ p | s $; it would then follow that $ \gcd(r,s) \geq p $, which contradicts our earlier assumption that $ \gcd(r,s) = 1 $. Hence, $ k = s^{n} $, which yields $ b^{n} = s^{n} a^{n} $ and consequently $ b = s a $. Therefore, $ a | b $. – Transcendental Jan 24 '17 at 04:30

3 Answers3

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Hint $\ $ Either examine exponents in unique prime factorizations, or, by the Rational Root Test, the reduced rational root $\rm\:x = b/a\:$ of $\rm\:x^n = c\in\Bbb Z\:$ must be integral, so $\rm\:b/a\in\Bbb Z\:\Rightarrow\:a\mid b.$

Math Gems
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  • @Peter For that, after cancelling any common factor, one needs only $\rm:(a,b)=1:\Rightarrow:(a,b^n)=1,:$ true by iterating Euclid's Lemma. Thus $\rm:1 < a\nmid b^n,:$ so $\rm:a^n\nmid b^n.\ \ $ – Math Gems Apr 05 '13 at 00:41
  • Yes, that was my idea. I was awfully unclear, sorry. – Pedro Apr 05 '13 at 00:44
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If you can assume the fundamental theorem of arithmetic (that each integer has a unique factorization in prime numbers), you can write: $$ \begin{align*} a &= p_1^{e_1} p_2^{e_2} \ldots p_r^{e_r} \\ b &= q_1^{d_1} q_2^{d_2} \ldots q_s^{d_s} \end{align*} $$ Here the $p_i$, $q_i$ are primes, and $e_i$ and $d_i$ are all greater than 0. If $a^n \mid b^n$, then $p_i^{n e_i}$ must have a counterpart in a $q_j^{n d_j}$, in that $p_i = q_j$ and $n e_i \le n d_j$, so it must then also be that $e_i \le d_j$; and this means $a \mid b$.

vonbrand
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Hint: $p$ is a prime factor of $k$ if and only if $p^n$ is a factor of $k^n$. This holds for any prime $p$, integer $k$, and positive integer $n$.

Cameron Buie
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