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Carmichael numbers

  1. Mar 5, 2013 #1
    1. The problem statement, all variables and given/known data

    A Carmichael number is a composite integer n greater than or equal to 2 such that [itex] b^{n-1} \equiv 1 [/itex] (mod n) for all integers b that re relatively prime to n.

    Let n be a Product of at least 3 distinct odd primes. Prove that if [itex](p-1)\mid (n-1)[/itex] for every prime divisor p of n, then n is a Carmichael number.

    2. Relevant equations



    3. The attempt at a solution

    My first question actually comes up on the proof itself, I have that by fermats little theorem [itex] b^{p_i-1}\equiv 1 (\textrm{mod} p_i)[/itex] for all the [itex]p_i[/itex] that divide n. but then I have that this means that [itex] b^{n-1}\equiv 1 (\textrm{mod} p_i) [/itex] and I don't know why that's true.

    My second question is for ILS if he sees this post, can I apply Eulers theorem to do this proof instead?
     
  2. jcsd
  3. Mar 5, 2013 #2

    I like Serena

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    Hey Arcana! o:)


    I don't see how Euler's theorem can be used, but I certainly see how the Chinese Remainder Theorem can be used. :cool:

    Suppose ##\displaystyle n=\prod_{i=1}^m p_i## with ##p_i## distinct primes.
    Then according to CRT, we have that ##\psi## given by:
    $$\psi: (\mathbb Z/n \mathbb Z)^* \to (\mathbb Z/p_1 \mathbb Z)^* \times (\mathbb Z/p_2 \mathbb Z)^* \times ... \times (\mathbb Z/p_m \mathbb Z)^*$$
    $$\psi: (x \text{ mod } n) \mapsto (x \text{ mod } p_1, ~x \text{ mod } p_2, ~ ... ,~x \text{ mod } p_m)$$
    is an isomorphism.

    Now what if we fill in ##x=b^{n-1}##?



    (In other words: ##\text{Fermat} \prec \text{Euler} \prec \text{CRT}##. :biggrin:)
     
    Last edited: Mar 5, 2013
  4. Mar 6, 2013 #3
    How do you know everything ILS? I have not yet asked a question which you did not seem to be an expert on. Surely even if you did learn everything the knowledge must leak out of your brain after a while; you can't possible always remember everything you've learned... How do you do it?
     
  5. Mar 6, 2013 #4

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    I keep thinking what an amazing coincidence it is that you seem to be interested in exactly the topics that I know a little about. :wink:
     
  6. Mar 6, 2013 #5

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    Since ##(p_i - 1)|(n-1)## it follows that there is some integer k such that ##k(p_i - 1)=(n-1)##.
    So
    $$b^{n-1} \equiv b^{k(p_i - 1)}\equiv (b^{p_i - 1})^k \equiv 1^k \equiv 1 \pmod{p_i}$$

    The next step would be that:
    $$b^{n-1} = 1 + k_1 p_1$$
    $$b^{n-1} = 1 + k_2 p_2$$
    $$...$$
    $$b^{n-1} = 1 + k_m p_m$$

    Now what can you conclude from that?
     
  7. Mar 6, 2013 #6
    Thats okay, I know how to end the proof. It was that one step I didn't follow. Thanks for spelling it out :)
     
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