# Dividing a Power of a Prime

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.

• Do you know the Fundamental Theorem of Arithmetic and are you allowed to use it? – David K Nov 1 '16 at 18:26
• @DavidK Sure. I am allowed to use that. – user193319 Nov 1 '16 at 18:30
• Some of the answers use the theorem. – David K Nov 1 '16 at 19:39

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 $m \le n$.
• 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 $m \le n$. – user193319 Nov 1 '16 at 18:59
• 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$, which is a contradiction...Does this seem right? – user193319 Nov 1 '16 at 19:02 • @user193319, I had in mind this property: If$q$is a prime and$q$divides$ab$, then$q$divides$a$or$b$. – lhf Nov 1 '16 at 19:17 • Oh, yes. I see. Does my proof work, though? – user193319 Nov 3 '16 at 0:13 Hint: Factor$a$into distinct primes. Which of those primes can divide$p^n$? How many distinct primes are there in the factorization of$a$, really? Hint:$p^n$is the unique prime factorization of the number$k=p^n$. What form must the divisors of$k$have then? If$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....?