Proof: if $n=pq$ then $p-1\mid q-1$ and $q-1\mid p-1$

Now I'm asking my first question to understand a specific proof:

Let $n=pq$ and $q,p \in \mathbb{P}$. Then we get $p-1\mid n-1$ and $q-1\mid n-1$, because there are prime integers mod $p$ and mod $q$. Further we get $n-1=pq-1=p(q-1)+p-1$. To this step everything is clear. Now the author says: from $p(q-1)+p-1$ it follows $q-1\mid p-1$ and $p-1\mid q-1$. I don't have a clue how he gets there. Any help is appreciated. Thanks :-)

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You get the right spacing if you use \mid instead of |. Also "mod" is usually not italicized. –  joriki Oct 25 '11 at 12:31
Thanks for \mid –  ulead86 Oct 25 '11 at 12:46
I also would like to know who the author is. Carmichael numbers are cool. I would also suggest making the title of the problem more tidy so it fits on the main page. –  JavaMan Oct 25 '11 at 12:48

3 Answers

This is to show that if $n=pq$ and if $p-1$ divides $n-1$, then $p-1$ divides $q-1$.

To wit, $p-1$ divides $n-1$ hence $p-1$ divides $(n-1)-(p-1)=p(q-1)$ as well. But $p-1$ and $p$ are relatively prime hence $p-1$ divides $q-1$.

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You have a mistake somewhere. Take $p=5,q=3$ then $n=pq=15$ but $n-1=14$ is not divisible by $p-1=4$.

Who is the author?

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It's not the complete proof, it's one step in the proof of the Korselt-Theorem including Carmichael-Numbers. If n is a CMN <=> for all $p \in \mathbb{P}$ with $p \mid n$ we get $(p-1) \mid (n-1)$ –  ulead86 Oct 25 '11 at 12:44
Note that stating this in your question itself will help you get helpful answers. –  Gadi A Oct 25 '11 at 13:01
I know, sorry for the inconvenience, I'll be more specific nexttime, I promise –  ulead86 Oct 25 '11 at 13:54
Daniel, you can go back and edit your question to include relevant information –  The Chaz 2.0 Oct 25 '11 at 18:43

$p\!-\!1\mid\! \overbrace{pq}^n\!\!-\!1\iff p\!-\!1\mid q\!-\!1,\$ by  mod $\ p\!-1\!:\ \ \color{#c00}{p\equiv 1}\,\Rightarrow\, \color{#c00}pq\!-\!1 \equiv q\!-\!1\ \$ QED

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