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

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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
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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
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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 $b$.

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