For How Large an 'n' can the Carmichael Function of n be 2?

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The discussion focuses on determining the largest integer n for which the Carmichael function λ(n) equals 2. It concludes that n must be of the form 6a, as all primes p not equal to 2 or 3 are congruent to 1 or 5 modulo 6. Additionally, it asserts that all primes less than or equal to the square root of n must be included in n, leading to the conclusion that no primes greater than 5 can satisfy the condition. Ultimately, the largest n for which λ(n) equals 2 is identified as 24. The conversation invites further questions on the topic of number theory.
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What is the largest n such that λ(n)=2? Is there such a bound? This isn't a homework question. I'm just interested.
 
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If ##p## is not in ##n## than ##(p,n)=1 ##.
Therefore for ##p^2\equiv 1mod(n)## to hold ##n=6a## since all ##p## other than 2 and 3 are of the form ##6b\pm 1##
Also all ##p\leq \sqrt{n}## must be in ##n## since otherwise ##p^2<n##.
so we need a ##p\sharp## such that ##p_{c+1}>\sqrt{p\sharp}##.
It's easy to see that this does not hold for ##p>5##.
Therefore the answer is 24.
[EDIT:-##p\sharp=p_1\times p_2 \times p_3\times...\times p_c##.]
 
Did I scare you off ? I assumed you were familiar with some number theory terminology. If you have any questions, please ask.
 
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