If the radicand of a square root is a non-square (making the root an irrational), and if the non-square is either a prime number, or a composite number that does not have a square divisor (other than 1), does this mean that the square root is not divisible by an integral divisor (that it does not have an integral factor)?
For example, $\sqrt{200} = \sqrt{100\times2} = \sqrt{100}\times\sqrt{2}=10\sqrt{2}$, so $\sqrt{200}$ has an integral factor of 10 (is divisible by 10). However, $\sqrt{6} = \sqrt{3\times2} = \sqrt{3}\times\sqrt{2}$
I'm mainly wondering for the purpose of reducing fractions that contain radicals and integers.
The phrase "does not have an integral factor" is a little imprecise, since anything has an integral factor: we have $x=17\left(\frac{x}{17}\right)$.
So we rephrase the question. Suppose that $n$ is a square-free integer. Do there exist integers $d$ and $k$, with $k\gt 1$, such that $$\sqrt{n}=k\sqrt{d}?\tag{1}$$ Once we express the question like that, the answer is quick. Suppose to the contrary that there are integers $k,d$, with $k\gt 1$, such that (1) holds. Squaring both sides, we get $n=k^2d$, so $n$ is divisible by a square greater than $1$.