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I have attempted to use the division theorem, but I am unsure exactly where to begin. Any hints/direction would be appreciated.

Gertrude Porter
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  • Proved as part of the proofs in the linked dupe. Said symmetrically $\bmod n!:\ $ if $,k\equiv m,$ then $,k\equiv 0\iff m\equiv 0,$ follows immediately because cogruence is an equivalence relation. – Bill Dubuque Feb 18 '21 at 03:25
  • It´s pretty easy, only you need use the definition.

    Assume that $k=nl+m$ and that $n|k$, then by definition $k=nq$ where $q\in \mathbb{Z}$ then $nq=nl+m \Rightarrow nq-nl=m \Rightarrow n(q-l)=m$ and then $n|m$

    Can you take frome here

     –  Feb 18 '21 at 03:29
  • $k = n(l + \frac mn)$. If $\frac mn$ is an integer then $n|k$. And $\frac kn = l+\frac mn$. If $\frac mn$ is not an integer then neither is $\frac kn$> – fleablood Feb 18 '21 at 09:20

2 Answers2

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As a hint, consider why

  • if $n$ divides $m$ then $n$ divides $nl+m$
  • if $n$ divides $k$ then $n$ divides $k-nl$

To make things even easier in the first case, if $n$ divides $m$ then $a=\frac mn$ is an integer and $m=na$, so the question becomes: why if $n$ divides $m$ then $n$ divides $nl+na$? Similarly with the second case.

Henry
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Use the definitions directly: $$ n \mid m \quad\Longrightarrow\quad m = nr \text{ for some } r \in \mathbb{Z} \quad\Longrightarrow\quad k = nl + nr = n(l + r) \quad\Longrightarrow\quad n \mid k $$ since $l + r \in \mathbb{Z}$ if $l, r \in \mathbb{Z}$.

To prove the other direction, rewrite $$ m = k - nl, $$ and argue in an analogous manner.

Sammy Black
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