prime number divided by another prime number Let $p, q\in\mathbb{N}$ prime numbers, such that there exists a $a\in\mathbb{Z}$ with $p^n=aq$ for some$n\in\mathbb{N}$. Does there exist $\hat{a}\in\mathbb{Z}$ with $p=\hat{a}q$? Sorry for the stupid question, I'm confused right now..
 A: Yes, it does. Try to show that if $p^n = aq$ for an integer $a$, then we have $p=q$. From this, you will of course get the existence of $\hat{a}=1$.
A: The prime decomposition of $aq$ is $p^n$. $q$ is prime so we also know that $q$ is part of the prime decomposition of $aq$. Given that the prime decomposition of $aq$ only has one prime in it ($p$), it follows that $p=q$ (and $a=p^{n-1}$).
So $\hat{a}$ exists, and $\hat{a}=1$.
A: The Fundamental Theorem of Arithmetic (FTA): Let $a,b,c$ be integers with $c\ne 0.$ If $c$ is a divisor of $ab$ and if $c$ is co-prime to $a$ (i.e. $\gcd (c,a)=1$) then $c$ is a divisor of $b.$
Corollary. If $p,q$ are prime and $n\in \mathbb N$ and $q$ is a divisor of $p^n$ then $q=p.$
Proof: Let $p,q$ be unequal primes. We show that $q$ does not divide  $p^n$ for any $n\in \mathbb N.$
(i).  $q,p$ are co-prime. 
(ii). If $n_0$ is the $least$ $n\in \mathbb N$ such that $q$ divides $p^n$ then $n_0>1$  (because $q$ does not divide $p$),so $n_0-1\in \mathbb N$ and $q$ $does$ $not$ divide $p^{n_0-1}.$ 
(iii). By FTA with $a=p, b=p^{n_0-1},c=q,$ we find by (i) and (ii) that $q$ $does$ divide $p^{n_0-1}.$ Therefore the assumption that $n_0$ exists leads to a paradox.
