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

• Math100
In summary: I'm glad I could help you with your proof. In summary, given a prime number p and an integer a, if p divides a to the power of n, then p to the power of n also divides a to the power of n. This can be proven by showing that if p divides a, then p to the power of n also divides a to the power of n.

#### 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 ##.

Last edited by a moderator:
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. ##

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 ##.

Last edited:
Math100 said:
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.
Math100 said:
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.##

Math100 said:
... 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 ##.

fresh_42
Thank you!