For a Carmichael number $n$, and $p$ a prime that divides $n$, show that $p-1$ divides $\dfrac{n}{p}-1$. I'm not sure on where to start for this proof, so any help would be appreciated!
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Do you mean if $p$ is prime and $p|n$? For instance, since $67*133 = 8911$, you have also have $66|132$ – Umberto P. Mar 15 '15 at 17:50
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Yes, sorry let me edit my OP – user879559 Mar 15 '15 at 17:52
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If $\ n = pk\ $ then $\ n\!-\!1 = (p\!-\!1)k + k\!-\!1\ $ so $\,p\!-\!1\mid n\!-\!1 \iff p\!-\!1\mid k\!-\!1 = \frac{n}p\!-\!1$
Remark $\ $ A complete proof of Korselt's Criterion is in this answer.
Bill Dubuque
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Sorry I don't follow how you went from $n - 1 = (p - 1)k + k - 1$ to $p - 1 \mid n - 1$ could you explain that? – cuppajoeman Jun 19 '24 at 00:44
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1It's this equivalence: $\ \color{#c00}{p!-!1}\mid\overbrace{(\color{#c00}{p!-!1})k+k!-!1}^{\large n-1}!\iff p!-!1\mid k!-!1,,$ or, said equivalently, $!\bmod, p!-!1!:\ n!-!1\equiv k!-!1,\ $ so $\ p!-!1\mid n!-!1\iff p!-!1\mid k!-!1,,$ cf. divisibility mod reduction. @cuppajoeman $\ \ $ – Bill Dubuque Jun 19 '24 at 00:59