Is there a method for determining whether a number has an n'th integer root? Does 249344 have such a root? What is it? I've tried a few possible roots. But now my head hurts. Please help.  
 A: Start by finding the prime factorization. Here, it is $249344= 2^9\cdot 487^1$.
If $249344$ were an $n$th power, all exponents would be multiples of $n$. But $\gcd(9,1)=1$, so no luck with this number.
Of course, a very partial factorization suffices or at leat allows you to reduce the trial search: If you are not sure if $487$ is prime or not, even knowing that $2^9||249344$ suffices that at most $\sqrt[3]\cdot$ or $\sqrt[9]\cdot$ can be integer (and since $3|9$, the latter will automatically fail if the former fails).
So, factoring out all $2,3,5$ will typically boil the possible roots down a lot.
But what if no small prime factors can be found? Say you have a ten digit number and it is not divisible by any prime $\in\{2,3,5,7,11,13,17,19\}$? Then at leat you can stop trying roots as soon as the (noninteger) result gets smaller than the next prine, $23$, which happens after $\sqrt[8]\cdot$ at this size, i.e. here you need only check $\sqrt\cdot,\sqrt[3]\cdot,\sqrt[5]\cdot,\sqrt[7]\cdot$.
