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This question already has an answer here:

I considered an even number $n\geq 9$, where it is divisible by some positive integer $k$.

Also, $k$ does not divide $\frac{n}{2}\ $ (negation of the hypothesis in my prior question).

Let $n = kq$.

Can we say that $q$ is even in this case? Can anyone help to get it theoretically? Thanks a lot for the help.

My attempt:

Let $n = kq$.

Since $k$ does not divide $\frac{n}{2}$, we have

$n/2 = k.q_1 + r$ for $0<r\leq k-1$.

How to conclude the even or odd property of $k$ and $q$ here?

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marked as duplicate by lulu, Bill Dubuque elementary-number-theory Apr 15 at 13:51

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    $\begingroup$ This is not clear. Say $k=n$. Then of course $k$ divides $n$, $k$ does not divide $\frac n2$ but $q=1$ is odd. Is that a counterexample to your claim? $\endgroup$ – lulu Apr 15 at 11:17
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    $\begingroup$ Try a couple of examples. What if $n=10$? Or $12$? What can $k$ possibly be? What can $q$ possibly be? $\endgroup$ – Arthur Apr 15 at 11:17
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    $\begingroup$ Strongly related: math.stackexchange.com/questions/3184984/… $\endgroup$ – Dirk Apr 15 at 11:19
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    $\begingroup$ @monalisa Can you clarify your question? As stated there is no difficulty coming up with counterexamples. I assume that you intended something else? $\endgroup$ – lulu Apr 15 at 11:21
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    $\begingroup$ Follows immediately from my answer in your prior question, namely by divisor $\rm\color{#c00}{recip}\color{#0a0}{rocity}\ % divisor reciprocity $ $$ {\rm If} \ \,\rm\ \large 2,K\mid N\ \ \ {\rm then} \ \ \rm\,\ \color{#c00}K\ {\LARGE \mid} {\large \frac{N}{\color{#0a0}2}}\! \iff{\large \color{#0a0}2}\ {\LARGE \mid} {\large \frac{N}{\color{#c00}K}}\!=\!Q\,\ [\rm\!\!\iff\!\ 2K\mid N\,]$$ Thus $\,\rm Q\,$ is even $\!\rm \iff\! K\mid N/2,\,$ which by hypothesis is false here, and true in your prior question. This is just the negation of your prior question. $\endgroup$ – Bill Dubuque Apr 15 at 14:07
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As a counter example, take $n = 14$ and $k = 2$. Then $2$ does not divide $14/2 = 7$ and $q = 7$ is not an even number.

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  • $\begingroup$ I tried numerically but theoretically, I am unable to prove it. $\endgroup$ – monalisa Apr 15 at 11:32
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    $\begingroup$ @monalisa: You can't prove it because it's false. You must disprove it, which means giving a (numerical) counterexample. $\endgroup$ – Cameron Buie Apr 15 at 11:48
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    $\begingroup$ @CameronBuie Actually we can do better than a single counterexample, we can prove that it is always false, i.e. $q$ is always odd - see my comment on the question. $\endgroup$ – Bill Dubuque Apr 15 at 14:09
  • $\begingroup$ @Bill: Fair point $\endgroup$ – Cameron Buie Apr 15 at 17:10

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