Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Hint: Look at prime factors. – Brett Frankel Apr 5 '13 at 0:29
You can start with the fundamental theorem of arithmetic. – 1015 Apr 5 '13 at 0:30
Consider this. Might I call yhis a duplicate? Since they end up in asking the same question. – awllower Apr 5 '13 at 13:32
@awllower: you're right that it's a duplicate, but I think this one has better answers. – Javier Apr 5 '13 at 14:16
@JavierBadia Probably because the previous one prohibited the use of GCD and UFD. – awllower Apr 5 '13 at 15:01
up vote 6 down vote accepted

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 $a \mid b$.

share|cite|improve this answer

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 $\rm\:b/a\in\Bbb Z\:\Rightarrow\:a\mid b.$

share|cite|improve this answer
@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 5 '13 at 0:41
Yes, that was my idea. I was awfully unclear, sorry. – Pedro Tamaroff Apr 5 '13 at 0:44

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 $n$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.