I just wanted to make sure about an issue.
Let $m, p= \text{prime}∈ ℕ-\{0\}$
If $$m≤p^k,$$
does the following exist? $$\gcd(m,p)≠1⟺ \gcd(m,p^k)≠1$$ and does the restriction $$m≤p^k$$ matter to the existence?
Thanks in advance!!!
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CodeLearner
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EDITED THE MISTAKE – CodeLearner Oct 05 '20 at 09:45
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If $m$ and $p$ are distinct primes then $gcd(m,p)=1$, or if $p=m$ we have $gcd(p,m)=m$ and $gcd(m,p^{k})=m$ for $k\geq 1$. – Alessio K Oct 05 '20 at 09:47
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@Äres can you see the edit pls? thanks! – CodeLearner Oct 05 '20 at 09:53
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Sorry I can't see what you've edited. – Alessio K Oct 05 '20 at 10:01
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@Äres m is not necessarily a prime, only p – CodeLearner Oct 05 '20 at 10:10
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By the dupe: $,a_1,a_2,\ldots a_k,$ are all coprime to $m\iff $ their product is coprime to $m$. Yours is special case all $,a_i = p\ \ $ – Bill Dubuque Oct 05 '20 at 10:42
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@BillDubuque TY! – CodeLearner Oct 05 '20 at 17:03
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$m \le p^k$ is not required. If gcd$(m,p) \neq1$ then $m$ and $p$ must share a common factor, because the only factor of $p$ is $p$ (and one, but we will not need to worry about that) then at least $p$ must be a factor of $m$. If we now look at gcd$(m,p^k)$ $p^k$ will have a factor of $p$ and we already know that $m$ has a factor of $p$ and therefore gcd$(m,p^k) \neq1$ as it must at least be equal to $p$. At no point have we required $m \le p^k$.
DBruwel
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