The n-th root of a prime number is irrational 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 $n\ge2$
 A: 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 $n$.
A: Wlog, suppose $x=\frac{a}{b}$ such that $a, b\in \mathbb{Z}, b\neq 0$ and $(a,b)=1$. Then $pb^n=a^n$ so that $p|a$. Hence, we can write $a=a_1p$, $a_1\in \mathbb{Z}$. It follows that $pb^n=p^na_1^n$ so that $b^n=p^{n-1}a_1^n$. We see that $p|b$, which contradicts the fact that $(a,b)=1$.
A: By Eisenstein's criterion, the polynomial $X^n-p\in\mathbb{Z}[X]$ is irreducible, so it's irreducible also in $\mathbb{Q}[X]$ by Gauss' lemma; therefore it can't have roots in $\mathbb{Q}$.
A: if possible, Let the n-th root of p prime number be rational. Therefore we can write p1/n as
p1/n = a/b 
p = an/bn
since a and b are coprime, an/bn will be integer if bn = 1
therefore, p = an    .................................equation1
dividing both sides by a, we get:
p/a = an-1
Since a is integer, an-1 is also integer,
therefore, p/a is also integer. p is prime, so p/a is integer if a =1 or a=p
put 1 in equation1, we get a = 1 which is not possible. we then put a=p, and get n=1 but first root means the number itself. since there is no value of a/b for which p1/nn is rational, it can be concluded that p1/n is always irrational.
