How to prove that $p^k|ab$ if and only if $p^k|a$ and $p^k|b$? [closed]

Let p be prime, $k \in$ N and let $a,b \in$ Z such that gcd(a,b)=1. How to prove that $p^k|ab$ if and only if $p^k|a$ and $p^k|b$? Trying: (<=) $p^k |a$ and $p^k|b$. Then $a=p^kq$ and $b=p^kq'$ => $ab=p^kqp^kq'=p^kq''$ => $p^k|ab$

-

closed as unclear what you're asking by Najib Idrissi, Olivier Bégassat, Mathmo123, amWhy, user1729Jul 31 at 11:50

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

It seems you're simply ignoring comments people have given to your questions, e.g. here and here‌​. If you disagree with their comments, please reply so the issues can be discussed; simply insisting on your posting style without engaging with criticism is not a solution. Also, the English abbreviation is "gcd"; please post in English as far as possible so as many people as possible can understand you. –  joriki Sep 12 '11 at 7:34
Surely it should be $p^k | a$ or $p^k | b$. –  JSchlather Sep 12 '11 at 7:41
alvoutila, what you have tried in your edit is fine, but it is the "only if" part that does not work. See my answer and Jacob's comment. –  Dan Brumleve Sep 12 '11 at 8:08

The statement is false. For example, let $p=2$, $k=1$, $a=2$, and $b=1$. $p^k$ divides $a*b$ but $p^k$ does not divide $b$.