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I know already that:

If $m = p \cdot q \cdot r$ with $p \neq q \neq r$ prime and $p-1, q-1, r-1 \mid m-1$, then m is a Carmichael number.

And Carmichael numbers are the product of at least 3 different odd prime numbers.

How can I easily find the smallest Carmichael number with this information?

I know that this question is already answered before, but I think there has to be an easier way then it was done there. Why is $ 561 = 3*11*17 $ the smallest Carmichael number?

Laura van Leuven
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  • I mean, a raw search takes very little time. – lulu May 19 '21 at 14:26
  • I did a research. But I wasn't be able to find an easy way to find the smallest Carmichael number. Maybe I am wrong, and there just isn't an easier way. – Laura van Leuven May 19 '21 at 14:56
  • Again, a raw search is extremely easy. Pencil and paper easy. I don't know what more you could want...it's certainly a plausible guess that $3$ divides the smallest one, which happens to be true, but maybe that is pure chance. After all the next $6$ are prime to $3$. – lulu May 19 '21 at 15:05
  • Unless I blundered, the next one divisible by $3$ is the $14^{th}$, $62745$. That surprises me a bit. After all, you get the condition $2,|,(n-1)$ for free. But I guess that's just how it plays out. – lulu May 19 '21 at 15:10
  • @lulu Extremely easy is exaggerated, but to find the smallest is in fact not too diffficult. – Peter May 19 '21 at 16:07
  • @LauravanLeuven Why do you think that there is an easier way ? You already know that we must have only odd prime factors, at least $3$ and a Carmichael number is squarefree. This leaves quite few possibilities not exceeding $561$ which have to be tested. – Peter May 19 '21 at 16:09
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    @Peter True. Well, my guess that it was divisible by $3$ sped things up a lot. But that really seems to have been sheer luck. – lulu May 19 '21 at 16:16
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    What helps is that there cannot be prime factors $p,q$ with $p\mid q-1$. This is an important property of Carmichael numbers ! – Peter May 19 '21 at 16:17
  • Well in the link "Why is 561=3∗11∗17 the smallest Carmichael number?", the person who answered the question did it very extensive. First he assumed that one prime factor is 3, then 5, etc. But I think that the only option left is then indeed to do it by brute force, or not? It's not a pretty way, but it is at least better then the way it was done in that extensive answer. – Laura van Leuven May 19 '21 at 18:22
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    A little brute force is inevitable. Once $561$ is found, we can quickly finished since already $7\cdot 11\cdot 13$ is larger. If the smallest prime factor is $5$, the cofactor must be less than $113$ and $11$ cannot be a prime factor. This only leaves the possibility $5\cdot 7\cdot 13$. A little more work is necessary for the case that the smallest prime factor is $3$. – Peter May 20 '21 at 07:42

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