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Jul 27, 2016 at 2:04 comment added Ted Mao Agree. A quote from a facebook friend of mine: "(to say that it is too short to publish in a paper) because its proof is homework-level. The result is that if $6k+1, 18k+1,54k^2+12k+1$ is prime, then their product $n$ is a Carmichael number. But one can prove this directly using Chinese remainder and little Fermat. Just need to show that $(n-1)$ is divisible by $6k, 18k, 54k^2+12k$." One can generate many similar criteria with a program.
Jul 27, 2016 at 1:45 history edited Maksym Voznyy CC BY-SA 3.0
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Jul 26, 2016 at 15:41 review Low quality posts
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Jul 26, 2016 at 15:51
Jul 26, 2016 at 15:15 history answered Maksym Voznyy CC BY-SA 3.0