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Suppose $p$ is a prime and $a$ and $k$ are positive integers. Prove that, if $p \mid a^k$, then $p^k\mid a^k$ also.

I have already proven that if $a,b,n\in\mathbb{N}$ and if $a^n\mid b^n$, then $a\mid b$. I tried using induction:

$k=1 \Rightarrow p\mid a^1 \Rightarrow p^1\mid a^1$. Assume the statement holds for some $k$. Then if $p\mid a^{k+1}\Rightarrow p\mid a^ka$. Since $p$ is prime, $p\mid a$ or $p\mid a^k$. If $p\mid a$, we are done. If $p\mid a^k$ then by assumption, $p^k\mid a^k$ and so by the lemma, $p\mid a$ and so $p^{k+1}\mid a^{k+1}$.

Does this work?

jitender
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user132095
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4 Answers4

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By the Euclid's lemma: if $p|ab$ then $p|a$ or $p|b$. Using this lemma we have $$p|a\cdot a^{k-1}\implies (p|a)\lor (p|a^{k-1})$$ and if $p|a$ then we are done and if $p|a^{k-1}$ then by descending induction we prove that $p|a$. Finally $p|a\implies p^k|a^k$.

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Hint: Try to prove that: $p\:|\:a^k$ implies $p\:|\:a$.

user35603
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As $ a^{k} = a\cdot ....\cdot a $ and $ a = p_{1}^{e_{1}}\cdot p_{2}^{e_{2}}\cdot .... \cdot p_{n}^{e_{n}} $ for some $ p_{1},....,p_{n} $ all prime and $ e_{1},....,e_{n} \in \mathbb{N} $, it follows that $ a^{k} = p_{1}^{ke_{1}}\cdot .... \cdot p_{n}^{ke_{n}} $. Thus if $ p|a^{k} $ for $ p $ a prime then $ p \in \{p_{1},....,p_{n}\} $. So suppose $ p = p_{1} $ (otherwise relabel), then $ \frac{a^{k}}{p^{k}} = p_{1}^{ke_{1}-k} \cdot .... \cdot p_{n}^{ke_{n}} $ and so $ p^{k} | a^{k} $ as required.

cfairwea
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I would use induction and the property for prime elements to show $p \mid a$, so we get $p^{k}\mid a^{k}$.

Celsius
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  • I'm confused... did I not do that in the inductive step? If $p|a^k$, then $p|a^{k-1}a$ so $p|a^{k-1}$ or $p|a$ and continue in this fashion? – user132095 Jul 29 '14 at 17:59
  • No, you supposed $p^{k} \mid a^{k}$ as the induction requirement. – Celsius Jul 29 '14 at 18:04