Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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'$ => $ab=p^kqp^kq'=p^kq''$ => $p^k|ab$

share|improve this 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
add comment

1 Answer 1

up vote 2 down vote accepted

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 $b$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.