1
$\begingroup$

Not sure if my thinking is correct. For the problem "$a$ divides $b$ if and only if $a$ divides $b^2$." So far my proof goes: since $a$ divides $b$ there exists an integer $n$ such that $b=an$. Then $b^2=a^2*n^2=a(an^2)$. Hence $a$ divides $b^2$. I am having a problem proving the converse.

$\endgroup$
1
  • 1
    $\begingroup$ The statement is true if $a$ is squarefree (or prime), but false without this assumption. $\endgroup$
    – vadim123
    Commented Sep 10, 2015 at 15:31

2 Answers 2

3
$\begingroup$

It is incorrect. $16$ divides $16=4^2$, but $16$ doesn't divide $4$.

$\endgroup$
0
$\begingroup$

The converse isn't true. let's say $$an=b^2$$ for some $n$. Now let's say $a>b$. Then $n<b$ and thus: $$a>b$$. which means $a$ doesn't divide $b$, but $a$ does divide $b^2$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .