I was playing around and thought of the following question:
If given $a \not=1 \in \mathbb{R}^{+}$, prove that there are infinitely many integers $n$, such that $\sqrt[n]a$ is irrational.
I have proven a very simple case:
If $1 < m \in\mathbb{N}$, then there are infinitely many positive integer $n$ such that $\sqrt[n]m$ is irrational. To show this, I argue by noting the polynomial, $$ p(x) = x^n -m$$ is irreducible in $\mathbb{Z}$[x] for $n > m$, hence it is irreducible in $\mathbb{Q}[x]$ as it is also primitive. So there are no rational solutions to $x^n = m$. And $\sqrt[n]m$ is irrational.
How do I prove the general case? As it does seem quite intuitive.
The “simple” case is all you need.
Suppose $a>0$ is irrational; then $\sqrt[n]{a}$ is irrational as well, otherwise $a=\bigl(\sqrt[n]{a}\bigr)^n$ would be rational.
For a general rational $a=p/q$, consider that $$ \sqrt[n]{a}=\frac{\sqrt[n]{pq^{n-1}}}{q} $$ Therefore $\sqrt[n]{a}$ is rational if and only if $\sqrt[n]{pq^{n-1}}$ is rational.