Given that p is a prime? (Review/verify this proof)?

[Math100]

Homework Statement

Given that $p$ is a prime and $p\mid a^n$, prove that $p^n \mid a^n$.

Relevant Equations

None.

Proof:

Suppose that p is a prime and $p \mid a^n$.
Note that a prime number is a number that has only two factors,
1 and the number itself.
Then we have (p*1)$\mid$a*$a^{(n-1)}$.
Thus p$\mid$a, which implies that pk=a for some k$\in\mathbb{Z}$.
Now we have $a^n$=$(pk)^n$
=$p^n k^n$.
This means $p^n \mid a^n$.
Therefore, given that p is a prime and $p \mid a^n$,
we have proven that $p^n \mid a^n$.

 

[fresh_42]

You don't have to encapsulate every single letter or symbol. ## stands for "begin math mode" and "end math mode", so it is ##a^n = (pk)^n ## to get $a^n=(pk)^n$ or ##p \mid a^n ## to get $p \mid a^n.$

Here is your corrected code:

Proof:        Suppose that ##p## is a prime and ##p\mid a^n ##.
                 Note that a prime number is a number that has only two factors,
                 ##1## and the number itself.
                 Then we have ##(p*1)\mid a* a^{n-1} ##.
                 Thus ##p\mid a##, which implies that ## pk=a## for some ##k \in\mathbb{Z}##.
                 Now we have ## a^n = (pk)^n = p^nk^n ##.
                 This means ## p^n \mid a^n ##.
                 Therefore, given that p is a prime and ##p\mid a^n ##,
                 we have proved that ## p^n \mid  a^n ##.

 

[fresh_42]

Homework Statement:: Given that $p$ is a prime and $p\mid a^n$, prove that $p^n \mid a^n$.
Relevant Equations:: None.

Proof: Suppose that $p$ is a prime and $p\mid a^n$.
Note that a prime number is a number that has only two factors,
$1$ and the number itself.
Then we have $(p*1)\mid a* a^{n-1}$

We have $p\,|\,a^n.$

The easiest way to continue is to use the definition of a prime: If it divides a product, then it already divides a factor. This gives us directly $p\,|\,a.$ Etc.

Another way is to write $a=p_1\cdot \ldots \cdot p_m$ as a product of primes. Then $p\,|\,p_1^n\cdot\ldots\cdot p_m^n$ and $p=p_1$ without loss of generality. Etc.

Thus $p\mid a$,...

Why? It is true, but why? You cannot conclude $p\,|\,a$ from $p\cdot 1= a\cdot a^{n-1}$ without explanation. E.g. $4\cdot 1 \,|\, 6\cdot 6^2$ but $4\nmid 6.$

... which implies that $pk=a$ for some $k \in\mathbb{Z}$.
Now we have $a^n = (pk)^n = p^nk^n$.
This means $p^n \mid a^n$.
Therefore, given that p is a prime and $p\mid a^n$,
we have proved that $p^n \mid a^n$.

 

[Math100]

I'll use/apply the first way. Here's my revised proof:

Suppose that $p$ is a prime and $p\mid a^n$.
Note that if a prime number divides a product of integers,
then it must divide one of the factors from a product of integers.
This gives us $p\mid a$, which implies that $pk=a$ for some $k \in\mathbb{Z}$.
Then we have $a^n = (pk)^n= p^nk^n$.
Thus $p^n \mid a^n$.
Therefore, given that p is a prime and $p\mid a^n$,
we have prove that $p^n\mid a^n$.

 

[Math100]

Thank you!