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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.

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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.

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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 $n-1<n$ 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: $b\in\{-1, 0, 1\}$.

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