Prove that if $n|(mn+k)$ then $n|k$

Question

My attempt at a proof:

Let $x=mn+k$, as $n|x$,by the Distributive Property and the fact that division is the inverse of multiplication $x=n(m+k/n)$ . It follows from the definition of divisibility that $(m+k/n)$ is an integer. As the integers are closed under addition, therefore $k/n$ must be an integer and thus $n|k$.

This last sentence seems as though I'm missing something important, or that I am making a logical error. Any assistance in correcting my proof is greatly appreciated.

Basically yes, but I personally don't think you've written it in a way that is easy to read. — Adam Rubinson, Feb 15 '22 at 22:42
I'm not sure what logical error you think you are making... care to clarify? — Adam Rubinson, Feb 15 '22 at 22:48
I was worried about assuming part of what I am trying to prove. Namely that if m+n is an integer, and m is an integer, then a must be an integer. — user437485, Feb 15 '22 at 23:00
In your comment, I assume you meant to say something like, "if $m+z$ is an integer, and $m$ is an integer, then $z$ must be an integer." And I don't think this is what you were trying to prove though, so I don't see the problem. — Adam Rubinson, Feb 15 '22 at 23:04
What if $n=0$? The statement is still true if $n=0$, but the argument you gave doesn't work in this case. — Kyle Miller, Feb 15 '22 at 23:07
Does this answer your question? How to formally prove: if $d\mid da+b$, then $d\mid b$?. Note this was found using Approach0. — John Omielan, Feb 16 '22 at 02:35

Adam Rubinson · Answer 1 · 2022-02-15T23:29:55.273

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Yes, but it's harder to parse than necessary. You don't need to state every axiom you are using at every step (if you did this, you would struggle to get far in maths because you would fail to see the wood from the trees). This is how I would have written it:

If $n=0$ then $k=0$ and the result is immediate. So suppose $n\neq 0 $ and let $x=mn+k=n(m+k/n).$ Since $n|x,$ it follows that $(m+k/n)$ is an integer and since $m$ is an integer, $k/n$ must be an integer also. Therefore, $n|k$.

edited Feb 15 '22 at 23:29

answered Feb 15 '22 at 22:46

Adam Rubinson

24,300

Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Feb 16 '22 at 08:28

score 1 · Answer 2 · answered Feb 15 '22 at 22:50

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there is an integer $t$ with $$ nt = mn + k$$ $$ nt - nm = k $$ $$n(t-m) = k $$ As $t-m$ is an integer, this means...

answered Feb 15 '22 at 22:50

Will Jagy

146,052

Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Feb 16 '22 at 08:28

score 0 · Accepted Answer · edited Feb 15 '22 at 23:03

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$n \mid (mn + k) \Longrightarrow \exists t, \; tn = mn + k \Longrightarrow tn - mn = k \Longrightarrow n(t - m) = k \Longrightarrow n \mid k. \tag{1}$

edited Feb 15 '22 at 23:03

John Omielan

52,653

answered Feb 15 '22 at 22:49

Robert Lewis

72,871

Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Feb 16 '22 at 08:28

score -1 · Answer 4 · answered Feb 15 '22 at 23:19

It's nice using rationals to prove things about integers, but in this case there's an awkwardness because $n$ might be zero, and you can't divide by that.

Here's a proof:

Proposition. Suppose $m$, $n$, and $k$ are integers. If $n\mid (mn+k)$ then $n\mid k$.

Proof. Suppose $n\mid (mn+k)$. Let $h\in\mathbb{Z}$ be such that $hn=mn+k$, using the definition of divisibility. Then $(h-m)n=k$, so $n\mid k$ by the definition of divisibility. QED

Another proof makes use of two nice lemmas that you should make sure to know: (1) if $a,b,c$ are integers such that $a\mid b$ then $a\mid bc$, and (2) if $a,b,c$ are integers such that $a\mid b$ and $a\mid c$, then $a\mid (b+c)$. Both are proved in a similar way to the above proposition.

Using these, here's the proof:

Proof. Suppose $n\mid (mn+k)$. Since $n\mid n$, then $n\mid (-mn)$. Then using the assumption, $n\mid ((mn+k)+(-mn))$, so $n\mid k$. QED

Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. — Bill Dubuque, Feb 16 '22 at 08:28
That's a very nice approach using the lemmas you sited. I'm doing a review of number theory and I appreciate your insights. — user437485, Feb 17 '22 at 00:55

Prove that if $n|(mn+k)$ then $n|k$

4 Answers4