# Can one show that if $\gcd(p^k,b)=1$, then $p^k \nmid b$?

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 $p^k \nmid b$?

-

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 $\gcd(p^k,b)\ne1$.
How you show that if $gcd(a,b)=1$ and $p \mid a$, then $p \nmid b$? And can you show intermediate stages from $gcd(a,b)=1$ and $p \mid a$ to $gcd(p^k, b)=1$? To me it doesn't look as trivial but requires explanation in language of math if possible. – alvoutila Oct 29 '11 at 14:23
@alvoutila: I get the impression (from this comment but also from the question as a whole) that you should review your understanding of the greatest common divisor. If $p\mid a$ and $p\mid b$, then $p$ is a common factor of $a$ and $b$, so their greatest common divisor is at least $p$. – joriki Oct 29 '11 at 14:25
But how would you show that then $gcd(p^k,b)=1$ using previous assumptions and definition(s) of gcd? – alvoutila Oct 29 '11 at 14:45