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Suppose that the positive integer $a$ divides $p^n$, where $n$ is a positive integer and $p$ is a prime. I want to conclude that $a = p^m$ for some $m \le n$, but I am having trouble. I would appreciate some hints.

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  • $\begingroup$ Do you know the Fundamental Theorem of Arithmetic and are you allowed to use it? $\endgroup$ – David K Nov 1 '16 at 18:26
  • $\begingroup$ @DavidK Sure. I am allowed to use that. $\endgroup$ – user193319 Nov 1 '16 at 18:30
  • $\begingroup$ Some of the answers use the theorem. $\endgroup$ – David K Nov 1 '16 at 19:39
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Hint: If a prime $q$ divides $a$, then $q$ divides $p^n$ ando so $q$ divides $p$.

Therefore, $a$ is a power of $p$ and so $a=p^m$ with $m \le n$.

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  • $\begingroup$ Okay, let me see if I follow. Let $ p_1^{\alpha_1} \dots p_k^{\alpha_k} = a$ be the prime factorization of $a$. Since $p_i | a$ for all $i$, then $p_i |p^n$ and therefore $p_i|p$ which implies $p_i = p$. Hence, $a = p^{\alpha_1} \dots p^{\alpha_k} = p^{\alpha_1 + \dots + \alpha_k}$. Define $m := \alpha_1 + \dots + \alpha_k$. If $m > n$, then $a$ couldn't divide $p^n$, so $m \le n$. $\endgroup$ – user193319 Nov 1 '16 at 18:59
  • $\begingroup$ However, I am having a little trouble with your claim, which if I am not mistaken is equivalent to: if a prime $q$ divides $p^n$ and, then $q=p$. Here is my attempt at proof: Suppose that $q|p^n$ yet $q \neq p$. Then $q = kp+r$, where $r \in [0,p)$. Then $q|p^n$ says $p^n = \ell q$ or $p^n = \ell(kp + r)$ or p(p^{n-1}-\ell k) = r$, which says that $p$ divides $r$, which is a contradiction...Does this seem right? $\endgroup$ – user193319 Nov 1 '16 at 19:02
  • $\begingroup$ @user193319, I had in mind this property: If $q$ is a prime and $q$ divides $ab$, then $q$ divides $a$ or $b$. $\endgroup$ – lhf Nov 1 '16 at 19:17
  • $\begingroup$ Oh, yes. I see. Does my proof work, though? $\endgroup$ – user193319 Nov 3 '16 at 0:13
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Hint: Factor $a$ into distinct primes. Which of those primes can divide $p^n$? How many distinct primes are there in the factorization of $a$, really?

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Hint: $p^n$ is the unique prime factorization of the number $k=p^n$. What form must the divisors of $k$ have then?

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If $p,q$ are distinct primes and $n$ is a non-negative integer then $q\not |\; p^n.$

Proof: Obvious for $n=0.$ If false in general, let $n_0$ be the least $n$ such that $q\;|\;p^n.$ Then $q\;|\;(p)(p^{n_0-1})$ with $n_0\geq 1$ (so $p^{n_0-1}$ is an integer) and $\gcd (p,q)=1.$ So by the Fundamental Theorem of Arithmetic we have $q\;|\;p^{n_0-1},$ contradicting the minimality of $n_0.$

Now if $a\;|\;p^n$ and $q$ is any prime divisor of $a,$ then....?

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