1

I'm trying to figure out if the statement is true or not and I need to prove it if so.

Let $p$ be a prime and $a$ be an integer. If $p\mid a^n$ , is it true that $p^n\mid a^n$ ?

I'm not sure how i would approach this problem. I started as $a^n = p \cdot b$ for some $b \in \mathbb{Z}$.

Cookie
  • 13,532
user207546
  • 13
  • 1
    In general either $p$ divides $a$ or $p$ does not divide $a$. Consider the consequences of each when $p$ is prime. – Henry Jan 13 '15 at 07:36
  • 1
    It literally took me 10 seconds to find the duplicate, congratulations to all the people who answered. – Najib Idrissi Jan 13 '15 at 08:51
  • Hint $\ $ If a prime divides a product then it divides some factor (by the Fundamental Theorem of Arithmetic, i.e. existence and uniqueness of prime factorizations). – Bill Dubuque Jan 13 '15 at 16:14

3 Answers3

2

This can be done using a basic property.

Note that if $p|xy$ then $p|x$ or $p|y$.

So $p|a\cdot a^{n-1}$ means $p|a$ (when we are pretty much done) or $p|a^{n-1}$.

So there is a simple induction to show $p|a$ and we are done as before

Mark Bennet
  • 100,194
1

Since $a$ is an integer we may write the canonical prime factorization of $a$ as $a=p_1^{k_1}\cdots p_s^{k_s}$. Then $a^n=(p_1^{k_1}\cdots p_s^{k_s})^n=p_1^{nk_1}\cdots p_s^{nk_s}$. So if $p\in \{p_1,...,p_s\}$, in other words $p|a$, then $p^n|a^n$.

Another proof: Suppose $p|a^n$. Then $p|a(a^{n-1})$. So $p|a$ or $p|a^{n-1}$. Continue by induction to find $p|a$. Thus $a=pb$ for some integer $b$, and $a^n=p^nb^n$. Therefore, $p^n|a^n$.

Eoin
  • 5,809
  • Are all non-prime numbers divisible by prime numbers? – user207546 Jan 13 '15 at 07:39
  • @user207546 I added another proof that might be more applicable. – Eoin Jan 13 '15 at 07:41
  • @user207546 Every number can be written as a finite expansion of prime numbers :) It is beautiful – Eoin Jan 13 '15 at 07:41
  • for your first proof, i see that if p|a then p^n|a^n but is this also true for if p|a^n then p^n|a^n ? sorry this is my first time learning these things – user207546 Jan 13 '15 at 07:43
  • @user207546 Yes. Suppose you can write $a$ by multiplying a few primes together. Then $a\times a$ are those primes times those primes. But we haven't introduced any new primes. We also haven't taken away any. So we can keep going in this sense and eventually get to the $n$th iteration. I think the second proof will be more applicable though. I like it more to be honest with myself. – Eoin Jan 13 '15 at 07:46
  • oh i see what you mean now. thank you so much for your help! – user207546 Jan 13 '15 at 07:47
  • @user207546 No problem! I'm glad I could convince you. – Eoin Jan 13 '15 at 07:47
  • @user207546 It is easy to see that any number is divisible by a prime. Definition of prime natural number is that it has exactly 2 divisors, 1 and itself. Now, take any number. If it is prime, it is obviously divisible by a prime. If it is not, take any divisor that is not 1 or itself. Repeat until you get prime :) – Ennar Jan 13 '15 at 07:53
  • @Ennar: the number $1$ is not prime, but not divisible by any prime number either. With this exception every positive integer is divisible by a prime number (possibly itself). – Marc van Leeuwen Jan 13 '15 at 08:26
  • Well, yes, you are right of course, the algorithm I provide can't work for 1, as you said. – Ennar Jan 13 '15 at 12:43
1

This is easier if you put in an intermediate statement: $p\mid a^n\implies p\mid a\implies p^n\mid a^n$ (this is used in the other answers as well, I just wanted to highlight this a bit more). The first implication follows by induction from Euclid's lemma, the second from the general fact that $(a\mid k)\land (b\mid l)\implies ab\mid kl$.

Marc van Leeuwen
  • 115,048