Necessity of coprimeness in Carmichael numbers

Asked May 02 '20 at 01:17

Active May 02 '20 at 01:17

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From http://oeis.org/wiki/Carmichael_numbers, Carmichael numbers are (1) "composite numbers $n$ which divide $b^n − b$ for every integer $b$ . OR equivalently (2) Carmichael numbers have the property that $b^{n − 1} ≡ 1 \pmod n$ for all bases $b$ coprime to $n$."

So essentially my question is: why does the second definition requires $b$ to be coprime to $n$ and the first doesn't?

I tried to go from $b^{n-1} \equiv 1 \pmod n$ (with $b$ coprime) to $b^n \equiv b \pmod n$ for ANY $b$ but I hit a lot of dead ends.

asked May 02 '20 at 01:17

Matías Santurio

It's simple, really. Given $b^n\equiv b$ you get $b^{n-1}\equiv 1$ through dividing by $b$. But, that is allowed only if $b$ has a multiplicative inverse $\bmod n$ so the division becomes a multiplication, and this requires $b, n$ to be relatively prime. – Oscar Lanzi May 02 '20 at 01:29
Compare to Fermat's little theorem, where the same distinction arises. – Greg Martin May 02 '20 at 01:31
2

@Oscar Of course it's simple if you choose the trivial direction of the equivalence. The other direction is not so simple, and is what the OP actually asks about. – Bill Dubuque May 02 '20 at 01:45
@Matías Santurio: See corollary $4$ in this article:$;$https://kconrad.math.uconn.edu/blurbs/ugradnumthy/carmichaelkorselt.pdf – quasi May 02 '20 at 02:10
That's it thank you very much @quasi – Matías Santurio May 02 '20 at 09:24

Necessity of coprimeness in Carmichael numbers

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