Question

If \(p | a^m\), how does \(p|a\)

Answer 1

https://brilliant.org/assessment/techniques-trainer/rational-root-theorem/

Look at \(\textbf{Theorem 1}\)'s proof.

Answer 2

Does your question still stand?

Answer 3

Yes.

Answer 4

It probably has something to do with the fact that p is prime.

Answer 5

Okay... I got it. This is modulo stuff again... :D

Answer 6

We know that
\[\huge a^m \equiv0(mod \ p)\]

Answer 7

We could take the m'th root of both sides, and we get...
\[\huge a \equiv c(mod \ p)\]
We don't know what c is... What we do know is...

Answer 8

\[\huge c^m=0(mod \ p)\]
p is prime, so for the mth power of c to divide p, it must be 0, or some multiple of p.
\[\huge c=pk\] for some integer k.
Then...
\[\huge a\equiv pk(mod \ p)\]Which can only mean
\[\huge a\equiv 0(mod \ p)\]
Which can only mean...
\[\huge a = pr \ \ \ r\in\mathbb{Z}\]

Which finally means
\[\huge p|a\]

Answer 9

you can use this lemma inductively
http://en.wikipedia.org/wiki/Euclid's_lemma