If $\gcd(a,b)=1$ and $p \mid a$ then $p \nmid b$. But how one can show that $\gcd(p^k,b)=1$? And can one show that if $p^k \nmid b$, then $\gcd(p^k, b)=1$? And can one show that if $\gcd(p^k,b)=1$, then $p^k \nmid b$?
Can one show that if $\gcd(p^k,b)=1$, then $p^k \nmid b$?
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elementary-number-theory
1 Answers
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Since $\gcd(p^k,b)=1$ doesn't hold in general, I'll assume that you intended the first question to share the assumptions of the first sentence, $\gcd(a,b)=1$ and $p\mid a$. In that case, since $p\nmid b$ and $p$ is the only prime factor in $p^k$, it follows that $p^k$ and $b$ have no factors in common, and thus $\gcd(p^k,b)=1$.
The answer to the second question is no: $p^2\nmid p$, but $\gcd(p^2,p)\ne1$.
The answer to the third question is yes, for if $p^k\mid b$, then $p^k\mid\gcd(p^k,b)$, so $\gcd(p^k,b)\ne1$.
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0@alvoutila: I showed this in my answer above. Since you seem to be lacking$a$rather basic understanding of factorization, it's very difficult for me to know what part of the explanation requires more detail. Please be as specific as you can in your questions and refer specifically to the part of the answer that you don't understand. Also, I would suggest that you read a basic text that covers things like unique factorization and gcds, as the answers to your questions will probably become apparent to you once you've understood those concepts. – 2011-10-29