TITLE: The n-th root of a prime number is irrational
QUESTION [7 upvotes]: If $p$ is a prime number, how can I prove by contradiction that this equation  $x^{n}=p$ doesn't admit solutions in $\mathbb {Q}$ where $n\ge2$

REPLY [7 votes]: Let $p=a^n/b^n$ as above, so $pb^n=a^n$.  The number of the prime factors of $a^n$ is a multiple of $n$, whereas the number of prime factors of $pb^n$ isn't a multiple of $n$.