Is there a way, in general, to tell whether the nth root of a integer is rational? Is there a way, in general, to tell whether the $n^{th}$ root of a integer is rational?
More explicitly, is it possible to elegantly determine whether the result of $k^{1/n}$ is rational for $k,n \in \mathbb{Z}$?
Obviously, one could attempt to factor $k$ into various rationals to the $n^{th}$ power, but it seems there must be a more elegant solution, right? If there isn't, I'd appreciate any explanation why this is impossible.
 A: The $n$-root of an integer $N$ is rational iff $N$ is an $n$-th power.
There is a fast algorithm for testing this that does not rely on factoring:

Daniel J. Bernstein,
Detecting perfect powers in essentially linear time,
Math. Comp. 67 (1998), 1253–1283

Testing whether an integer is a perfect power is an important first step in the AKS primality test.
A: As others have mentioned, the $n^{th}$ root of a integer is either an integer or irrational.
Working with non-negative integers only (as the other case easily follows), we can exploit the fact that $x\mapsto x^n$ is strictly increasing to get the statement: 

If, for integer $a,b$ it holds that $a^n<b<(a+1)^n$, then $\sqrt[n]{b}$ is irrational.

Then, we just apply our favorite numeric algorithm to find bounds for $\sqrt[n]{b}$ by two consecutive integers. We can do it by bisection with $\log(b)$ complexity - more sophisticated root-finding algorithms can likely yield a better bound.
A: Consider a rational $p/q$ in reduced form ($p$ and $q$ coprime and $q$ positive).  If $q\neq 1$ then $p^n/q^n$ is not an integer, which can be seen by using uniqueness of prime factorization.  Thus the $n$-th root of a rational number is either an integer or irrational.  This reduces the problem to factoring $k$ over the integers.  
A: As other answerers said, it's either irrational, or an integer.
That being said, looking at the prime factorization of $k$ gives an answer to the rationality of $k^{\frac{1}{n}}$ without actually computing roots:  
If the amount of each distinct prime factor of $k$ is a multiple of $n$, then the nth root of $k$ is rational.  For example, 49 = 7x7, so the square root of 49 is rational because there are two of each prime factor.  The cube root of 54 is irrational, since 54 = 3x3x3x2, which doesn't have enough 2s.  The cube root of 216 is rational, since 216 = 3x3x3x2x2x2.  etc.
A: Yes, there is an elegant way. It's not hard to prove that $k^{1/n} \in \mathbb{Z}$ if and only if $k$ is a $n$th power itself. If you know the proof that shows $\sqrt2$ is irrational, then the strategy is the same. Essentially, you go by contradiction. Assume $k^{1/n} = \frac{a}{b}$, with coprime $a,b \in \mathbb{Z}$. Then you rewrite the equation as $k \cdot b^n = a^n$. This is a contradiction and there are a few different ways to realize that. One is by considering the uniqueness of the prime factorization of $a^n$. 
A: In general, the nth roots of integers are irrational except when that integer is a perfect power of some other integer to the n. When speaking of say n=3, only  1, 8, 27, 64, ... have rational cube roots - all the other cube roots of integers are irrational. This holds for any n and has been stated well enough in the previous answers. What's lacking and what's hard to find is a proof that doesn't leave things out and is somewhat easy to understand, in other words heuristic. I have one somewhere in my notes, which I will try to find and post.
