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I am trying to prove that all Carmichael numbers are odd by using contradiction. Here’s what I have but I’m not sure if it’s good enough or if I need to add: Assume there is an even Carmichael number 2n. Then $2n=2*p_{1}*p_{2}…p_{k}$ such that $p_{i}\neq p_{j}$ then using Korselt’s theorem $2-1 \mid 2n-1$ is true because $1\mid 2n-1$ but $p_{1}-1 \nmid 2n-1$ because $p_{1}-1$ will be even, and $2n-1$ is odd. This is a contradiction, therefore all Carmichael numbers must be odd.

Bill Dubuque
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Allie
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    I don't see any problems, but in my opinion it's an overkill to use Korselt's theorem. See here: https://math.stackexchange.com/questions/708644/show-that-every-carmichael-number-is-odd – Tri Nov 25 '24 at 01:37
  • Hint: that contradiction follows from any element $,a,$ of even order (we don't need a generator (primitive elt) of $,\Bbb Z_p^*,$ as in Korselt's proof), since then by the Order Theorem $,a^{2n-1} \equiv 1,\Rightarrow, {\rm ord}(a)=2k\mid 2n-1,,$ contradiction; e.g. we can choose $,a\equiv -1.\ \ $ – Bill Dubuque Nov 25 '24 at 03:23
  • So it is a special case of: $ $ if a group has odd order then so does every element, so $,a^2=1\Rightarrow a=1.\ \ $ – Bill Dubuque Nov 25 '24 at 03:32
  • For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Nov 25 '24 at 03:34
  • @Tri While the Question you linked to asks about the same problem, it barely supports your opinion that Korselt's theorem is an overkill. The first Answer there tersely critiques the "solution verification" posed by the Question, and does hint at a correct proof. The second Answer also critiques the OP's solution, and then ends with "use the Korselt's criterion." – hardmath Nov 25 '24 at 17:33
  • @hardmath Yes, I understand it's just my own thought. That's why I confirmed I saw no problem in OP's solution in my comment before giving a link hinting to another approach, which I personally think is easier. I didn't mean to be impolite, but I am sorry if linking that answer makes you feel uncomfortable. – Tri Nov 25 '24 at 17:51
  • @Tri ... and for my part I'm sorry my remark gave an impression of impoliteness! Your conclusion is good, and I'd invite you to post what may seem to you quite an obvious Answer on the other Question. – hardmath Nov 25 '24 at 22:04
  • @hard "barely supports"??? The linked post strongly supports the claim that using Korselt's theorem here is overkill since it contains an answer (Andre's) with a much simpler proof (accessible at a much more elementary level than Korselt). As I mentioned in the above comments, deriving a contradiction from any even order element (e.g. Andre's $-1)$ can be viewed as a simplification of the idea in Korselt's proof, but this simple (basic) idea can be discovered without any knowledge of Korselt's proof. $\ \ $ – Bill Dubuque Nov 25 '24 at 22:31

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