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Let $a, b, n$ be elements of $\mathbb N$ such that $ a^n\mid b^n $. Show that $a\mid b$.

[P.S. Use the axioms of natural numbers.]

Are we using the properties of divisibility and afterwards induction? I kind of have an idea but I am not quite sure.

Shaun
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a1bcdef
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    If you use the unique factorisation theorem, it follows pretty smoothly. – user88595 Dec 15 '13 at 16:09
  • i am not allowed to use the unique factorization theorem,because i also thought of using that @user88595 – a1bcdef Dec 15 '13 at 16:14
  • You have zero accept answers so far. Please read about accepting answers here and here. – Git Gud Dec 15 '13 at 16:31
  • To prove this normally requires using something that is essentially equivalent to uniqueness of prime factorizations (e.g. see the many equivalents listed here). Do you have any of these available? If not, you will need to prove one of them (or, essentially, inline their (inductive) proofs in your proof). – Bill Dubuque Dec 15 '13 at 18:37
  • If you kind of have an idea, you should include the details, so that we know how to effectively help you. – Greg Martin Dec 15 '13 at 19:31
  • All of the divisibility theorems + Bezu's theorem are allowed to be used?How can i connect them with the question more concretely?@BillDubuque – a1bcdef Dec 18 '13 at 00:52
  • @d.i... Bezout's theorem immediately yields Euclid's Lemma, which immediately yields the Rational Root Test. It also immediately yields the Prime Divisor Property $, p\mid ab\Rightarrow p\mid a,$ or $,p\mid b,,$ which yields uniqueness of prime factorizations. Thus, from Bezout, you can quickly deduce the results needed in either my answer or Michael's. – Bill Dubuque Dec 26 '13 at 15:26

2 Answers2

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Hint $\rm\ \ (b/a)^n = k \in \Bbb Z\ \Rightarrow\ b/a\in \Bbb Z\ $ by the Rational Root Test, i.e. if $\,x\,$ is a rational root of the polynomial $\rm \,\color{#c00}1\cdot x^n-k\,$ then its least-terms denominator divides $\color{#c00}1,\,$ so $\,x\,$ is an integer.

Community
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Bill Dubuque
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  • From a foundational perspective (and possibly relevant to the OP's class), it's interesting to consider how the rational root test compares to the uniqueness of prime factorizations. The proof of the rational root test uses two number theory facts: (1) if $a$ and $b$ are coprime, then so are $a$ and $b^n$; (2) if $a$ and $b$ are coprime and $a$ divides $bc$, then $a$ divides $c$. Both of these can be proved from Bézout's identity alone, without using prime factorizations. – Greg Martin Dec 15 '13 at 19:30
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Hint: let $p_1^{\alpha_1}\cdot…\cdot p_k^{\alpha_k}=a$, $q_1^{\beta_1}\cdot…\cdot q_m^{\beta_m}=b$ be prime decompositions. Then $a^n=p_1^{n\alpha_1}\cdot…\cdot p_k^{n\alpha_k}$, $b^n=q_1^{n\beta_1}\cdot…\cdot q_m^{n\beta_m}$. Now write up what $a^n|b^n$ means in term of prime powers.

Michael
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