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I need help with the following:

Show that:

If $p$ is prime such that $p$ divides $a^n$

Then $p^n$ divides $a^n$

I know that if $p$ is a prime and divides a square number $a$ then $p$ also divides $a$ but I'm not sure how to apply this to the given problem.

hejseb
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Benji_Bombadill
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  • Have you considered induction? – abiessu Oct 09 '13 at 15:48
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    Try to show that $p$ divides $a$, and then conclude that $p^n$ divides $a^n$. – Arthur Oct 09 '13 at 15:50
  • @abiessu I don't think induction is relevant here; one has to use properties of primes. For example, $4$ divides $2^2$, but $4^2$ doesn't divide $2^2$. So there must be some number-theoretic property being used that doesn't hold for $4$. – user43208 Oct 09 '13 at 15:53
  • @user43208: my thought was something along the lines of "since $p$ is prime and $p|a^{n+1}$, then $p|a$ or $p|a^n$" as an inductive step. But you're right, induction would be the wrong direction in this case. – abiessu Oct 09 '13 at 18:39
  • So with this help, I know that if p divides a^n then a's prime factorization contains some p. therefore p divides a and p^n divides a^n. – Benji_Bombadill Oct 09 '13 at 19:54

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Notice that if $p$ is a prime and $p$ divides $a^n$, then $p$ divides $a$. Why this is follows from the fact that if $p$ is a prime and does not divide $a$ (but divides $a^n$), then $p$ cannot be a prime. So it's a contradiction, and $p$ must divide $a$. Let $q$ denote the product of all other primes that divide $a$. Then we can write $$ a=pq $$ and hence $$ a^n=(pq)^n=p^nq^n. $$ So $p^n$ divides $a^n$.

hejseb
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Prove by induction that if $p\mid a^n$, then $p\mid a$. It's then obvious that, if $x\mid y$, then $x^m\mid y^m$.

egreg
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Write $a$ as a product of prime numbers

Then $a^n$ is a product of $n$th-powers of prime numbers

One of these is the $n$th-power of $p$, since $p$ divides $n$ (and the fundamental theorem of arithmetic)

So $p^n$ divides $a^n$

Henry
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  • The proof of the uniqueness part of FTA that I know uses the fact that if $p$ is a prime and $p|ab\implies p|a\mbox{ or } p|b$. So using FTA may not be circular but you are going around the block to get to the corner. – Baby Dragon Oct 09 '13 at 16:35