# What are the conditions for a and b if given a | b^2, it follows that a | b? [duplicate]

Given a | b, it must be true that a | b^2. However, I was wondering about the conditions necessary for a and b such that given a | b^2, it must be true that a | b.

• – Gone Mar 9 at 22:20

The condition is that $$a$$ be squarefree, that is, that there be no prime $$p$$ such that $$p^2$$ divides $$a$$.

If $$a$$ divides $$b^2$$, and $$a$$ is squarefree, then $$a$$ divides $$b$$. If, for all $$b$$ such that $$a$$ divides $$b^2$$, $$a$$ divides $$b$$, then $$a$$ is squarefree.

Proofs: Let $$a$$ divide $$b^2$$, and let $$a$$ be squarefree. Then any prime $$p$$ dividing $$a$$ must divide $$b^2$$, whence it must divide $$b$$, whence $$a$$ must divide $$b$$.

Now suppose that for all $$b$$ such that $$a$$ divides $$b^2$$, $$a$$ divides $$b$$, and further suppose there is a prime $$p$$ such that $$p^2$$ divides $$a$$. Then $$a$$ divides $$(a/p)^2$$, but $$a$$ doesn't divide $$a/p$$, contradiction. Hence, $$a$$ is squarefree.

Let me change your question a little bit. You may want to ask:

1. What are the conditions on $$a$$ so that $$a\mid b^2\implies a\mid b$$ for some $$b$$?
2. What are the conditions on $$a$$ so that $$a\mid b^2\implies a\mid b$$ for every $$b$$?
3. What are the conditions on $$b$$ so that $$a\mid b^2\implies a\mid b$$ for some $$a$$?
4. What are the conditions on $$b$$ so that $$a\mid b^2\implies a\mid b$$ for every $$a$$?

This is really four different questions. To me, only the question 2 yields somewhat interesting answer, the rest of the questions all yield trivial answers of some sort.

Question 1: Any $$a$$ has this property. (Pick $$b=a$$.)

Question 2: $$a=0, 1, -1$$ satisfy this in a trivial way. Apart from that, any number $$a$$ that is a product of different primes ("square-free" integer: https://en.wikipedia.org/wiki/Square-free_integer) would do. This is because, if $$a$$ is square-free, then every prime factor of $$a$$ would divide $$b^2$$, therefore it would divide $$b$$, and then $$a\mid b$$. If $$a$$ is not square-free, then it's got one prime factor $$p$$ with exponent $$n\ge 2$$, and take $$b=a/p$$: the exponent of $$p$$ in $$b$$ is $$n-1 but is $$2(n-1)\ge n$$ in $$b^2$$, so $$a\mid b^2$$ but $$a\not\mid b$$.

Question 3: Any $$b$$ would do: take $$a=b$$

Question 4: As this must be satisified by $$a=b^2$$ too, we would have $$b^2\mid b$$, i.e. $$kb^2=b$$ for some $$k$$. Thus, either $$b=0$$ or $$kb=1$$, in which case $$b=\pm 1$$. Thus, the condition on $$b$$ is: $$b\in\{-1, 0, 1\}$$.