If 'a' is divisible by 'b^2' then 'a' is divisible by 'b'.

Question

It's something that I've never really thought about before but it makes sense nonetheless. Bearing in mind that 'a' and 'b' are both positive integers, what would be the best way to go about proving this statement? Which method of proof , for example, would be the best way to solve this problem? I'm just looking for a starting block to set me going :)

Answer 1

If $a$ is divisible by $b^{2}$, then the factors of $b^{2}$ are factors of $a$. But $b$ is a factor of $b^{2}$, so $b$ is a factor of $a$.

What I mean is this: If $a$ is divisible by $b^{2}$, then $a = b^{2} k$ for some positive integer $k$ (if both $a$ and $b$ are positive integers). This is exactly the definition of divisibility.

That implies $a = b(bk)$, so $a = bj$ for some positive integer $j$ (namely, for the integer $j = bk$), so $a$ is divisible by $b$ by the very definition of divisibility.

Answer 2

By the definition of divisibility you have $$b^2|a\Rightarrow a=mb^2=(mb)b=nb$$ for an $m \in \mathbb{Z}$ which means that $$b|a$$