# A simple doubt in number theory problem.

I considered an even number $$n\geq 10$$, where it is divisible by some positive integer $$k$$.

Also, $$k$$ divides $$\frac{n}{2}$$. Then $$n = kq\implies k\cdot\frac{q}{2} = \frac{n}{2}$$.

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

PS.

From the answer given by Bill, it does not satisfy if we take $$n=18,k=3,j=9.$$ Kindly help. Let me rectify where I am going wrong. Please help.

$$\quad\ \ \ \ \ \ \ \ \$$ If $$\,\rm\ J,K\mid N\,\$$ then $$\rm\,\ \color{#c00}K\ {\LARGE \mid} {\Large \frac{N}{\color{#0a0}J}}\! \iff\color{#0a0}J\ {\LARGE \mid} {\Large \frac{N}{\color{#c00}K}}\,$$ [$$\rm\!\!\iff\!\! JK\mid N\,$$]

This part is not satisfied.

• Regarding the PS, the logical if-and-only-if assertion $3\mid2\iff9\mid6\iff27\mid18$ is vacuously true because all three parts of it, $3\mid2$, $9\mid6$, and $27\mid18$, are false. – Barry Cipra Apr 16 at 11:08
• I am still unable to grasp it. I think my mind has stopped working :( – monalisa Apr 16 at 11:44
• It looks like you are under the misimpression that a logical statement of the form $P\iff Q$ is true if and only if $P$ and $Q$ are both true. That's not correct. $P\iff Q$ is also true when $P$ and $Q$ are both false. – Barry Cipra Apr 16 at 12:01

## 4 Answers

It's a special case of the below divisor $$\rm\color{#c00}{recip}\color{#0a0}{rocity}\ % divisor reciprocity$$ (put $$\rm\ J = 2).\,$$ Here $$\rm\, x\mid y\,$$ means $$\,\rm x$$ divides $$\rm y$$.

$$\ \rm Suppose\,\ \ J,K\mid N.\,\ Then \ \rm\ \ \color{#c00}K\ {\LARGE \mid} {\Large \frac{N}{\color{#0a0}J}}\! \!\iff\!\color{#0a0}J\ {\LARGE \mid} {\Large \frac{N}{\color{#c00}K}} \rm\!\!\iff\!\! JK\mid N$$

$${\bf Proof}\ \ \text{Note the fractions}\ \rm\ (N/J)/K =\rm\, (N/K)/J = N/(KJ)\,$$ are equal. Each divisibility above is equivalent to the fraction below it being an integer. $$\$$ QED $$\ \$$ You have $$\,\rm J = 2\,$$ so it speciallizes to

$$\ \rm Suppose\,\ \ 2,K\mid N.\$$ Then $$\,\rm\ \ \color{#c00}K\ {\LARGE \mid} {\Large \frac{N}{\color{#0a0}2}}\! \!\iff\!{\color{#0a0}2\ {\LARGE \mid} \Large \frac{N}{\color{#c00}K}}\!=\!Q$$

\begin{align}\rm\text{Hence, we conclude that }\,\ \ \ &\rm \color{#c00}K\ {\LARGE \mid} { \frac{N}{\color{#0a0}2}}\ \Longrightarrow\ \dfrac{N}K\! =\! Q\,\ is\ even\ \ [this\ question]\\[.8em] &\rm\color{#c00}K\ {\LARGE \nmid} { \frac{N}{\color{#0a0}2}}\ \Longrightarrow\ \dfrac{N}K\! =\! Q\ \,is\ odd\ \ \ [next\ question]\end{align}

• $\quad\ \ \ \ \$ Thus $\large \,\rm\ 2,K\mid N\,\$ and $\large \rm\,\ \color{#c00}K\ {\LARGE \mid} {\Large \frac{N}{\color{#0a0}2}} \Longrightarrow\ \color{#0a0}2\ {\LARGE \mid} {\Large \frac{N}{\color{#c00}K}} = Q$ $\quad$ – Bill Dubuque Apr 12 at 14:33
• Can you please provide more info about divisor reciprocity. Please – monalisa Apr 16 at 9:30
• @monalisa $\rm\,P\!\iff\! Q\,$ is true precisely when $\rm P$ and $\rm Q$ have the same truth value, i.e. when they are both true or both false. You chose values where all the divisibilities are false, i.e. $\,\rm 3\mid 2\iff 9\mid 6\iff 27\mid 18.\$ Generally both divisibilities are equivalent to $\, \rm JK\mid N,\,$ i.e. both are true if $\, \rm JK\mid N,\,$ and both are false if $\, \rm JK\nmid N,\,$ as follows by the chain of $3$ equivalences in my answer. – Bill Dubuque Apr 16 at 14:42
• @monalisa For example consider when $\,P\!\iff\! Q\$ is $\ 2\mid a\!\iff\! 2\mid a\!+\!2n$. When $P,Q$ are both true this says $\,a\,$ is even $\iff a\!+\!2n\,$ is even. When $P,Q$ are both false it says $\,a\,$ is odd $\iff a+2n\,$ is odd. They are negations of each other since $\,P\!\iff\! Q\,$ is equivalent to $\,\lnot P\!\iff\! \lnot Q.\,$ Which is more natural (or convenient) depends on the context (e.g. even vs. odd above, i.e. divisible by $2,\,$ or not divisible by $2)\ \$ – Bill Dubuque Apr 16 at 14:58
• @monalisa In the prior comment, that $P$ and $Q$ have the same truth value means that $\,a\,$ and $\,a+2n\,$ have the same parity, i.e. either both are divisible by $2$ (both even), or both are not divisible by $2$ (both odd). – Bill Dubuque Apr 16 at 15:06

We are given that

• $$\sf{n=qk}$$ as $$\sf{k\mid n}$$, and

• $$\sf{\frac n2=rk}\implies n=2rk$$ as $$\sf{k\mid\frac n2}$$ for some integers $$\sf{q,r}$$.

Equating gives $$\sf{qk=2rk}$$ and thus $$\sf{q=2r}$$ since $$\sf{k\ne0}$$. As $$\sf{r}$$ is an integer, $$\sf{q}$$ must be divisible by $$\sf{2}$$ and is thus even.

It seems that you have misunderstood the claim in Bill Dubuque's answer. You are correct that the different (fragment) claim $$\sf{j,k\mid n}\implies k\mid\frac nj$$ is not necessarily true by itself. Let

• $$\sf n=\prod\limits_{i=1}^n p_i^{a_i}$$ where $$\sf p_i$$ are distinct prime numbers and $$\sf a_i$$ are non-negative integers,

• $$\sf j=\prod\limits_{i=1}^n p_i^{b_i}$$ where $$\sf b_i$$ are non-negative integers such that for each $$\sf i$$, $$\sf b_i\le a_i$$, and

• $$\sf k=\prod\limits_{i=1}^n p_i^{c_i}$$ where $$\sf c_i$$ are non-negative integers such that for each $$\sf i$$, $$\sf c_i\le a_i$$.

Then clearly the condition that $$\sf j,k\mid n$$ is satisfied, since $$\sf{\prod\limits_{i=1}^n p_i^{a_i-b_i},\prod\limits_{i=1}^n p_i^{a_i-c_i}\in\Bbb Z^+}$$. The statement that $$\sf k\mid\frac nj$$ means that for some integer $$\sf{d}$$, $$\sf\prod\limits_{i=1}^n p_i^{a_i-b_i}=d\prod\limits_{i=1}^n p_i^{c_i}\implies d=\prod\limits_{i=1}^n p_i^{a_i-b_i-c_i}$$ but the inequality that $$\sf a_i-b_i-c_i\ge 0\implies b_i+c_i\le a_i$$ may not always hold.

However, Bill's complete claim is that if $$\sf{j,k\mid n}$$ then $$\sf{ k\mid\frac nj\!\iff\! j\mid\frac nk}$$. This means that

• if $$\sf j,k\mid n$$ then $$\sf k\mid\frac nj$$ if $$\sf j\mid\frac nk$$ (equivalent to: if $$\sf j,k\mid n$$ and $$\sf j\mid\frac nk$$ then $$\sf k\mid\frac nj$$), and

• if $$\sf j,k\mid n$$ then $$\sf j\mid\frac nk$$ if $$\sf j\mid\frac nk$$ (equivalent to: if $$\sf j,k\mid n$$ and $$\sf k\mid\frac nj$$ then $$\sf j\mid\frac nk$$).

It can be easily proven that these two statements are true, and $$\sf jk\mid n$$ follows. In general the statement $$\sf x\implies y\iff z$$ means that $$\sf x$$ implies $$\sf y$$ is true if $$\sf z$$ is true, and that $$\sf x$$ implies $$\sf z$$ is true if $$\sf y$$ is true.

Now you give the example $$\sf n=18, k=3, j=9$$. This is not a valid counterexample as it in fact satisfies neither $$\sf j\mid \frac nk$$ (needed for the first bullet point) nor $$\sf k\mid \frac nj$$ (needed for the second bullet point).

• Why the downvote? – TheSimpliFire Apr 23 at 16:44

We can. Probably better to start with the second part, though: $$\frac{n}{2} = k p$$ where $$p$$ is integer. Therefore $$kq = n = k\cdot (2p)$$, so $$q = 2p$$ and therefore $$q$$ is even.

Yes. Since $$k \mid \frac{n}{2}$$ then $$\frac{n/2}{k} = \frac{q}{2}$$ is an integer, so $$q$$ must have a factor of 2, and by definition is even.