Under what conditions does an integer dividing the square of n imply that the integer must divide n?

Question

Under what conditions does an integer dividing the square of n imply that the integer must divide n?

I believe it's that the integer must be prime, but maybe not...I can see counterexamples for certain composites, like 12 divides the square of 6, but 12 does not divide 6, but I it seems like if 6 (also a composite) divides the square of a number, it will necessarily divide the number. What's different about 6 and 12? Is it that 6's prime factorization has an only odd powers?

I ask because of a response under a question about proving that if n squared is even, then n is even. One proof says essentially, "given that 2 divides square of n, since 2 is prime, 2 must divide n," but I'm not seeing so much the "since 2 is prime" part...

Answer 1

the simplest condition is that the number that does the dividing be squarefree.

Answer 2

The inverse is easier to understand: under what conditions does $m$ divide $n^2$ but fail to divide $n$? In other words, when does the prime factorization of $m$ fail to show up in $n$ when it shows up in $n^2$? For this to happen $m$ must be composite (because if $n$ and $n^2$ have the same unique prime factors-- if $m$ equals one of them, it'll show up in both). Hopefully this helps to explain the bit about "because 2 is prime. . .".

Here's the generalization of the first answer: if there are "more" of a particular factor in the denominator than in the numerator (for a factor that shows up in both), you might be able to divide the square of the numerator by the denominator, but not the numerator itself.

Basically, you want to match prime factors in the denominator to prime factors in the numerator. If there are more prime factors in the numerator than in the denominator, it won't divide (this is where your 6/12 problem fails).