User talk:Enrique Santos L.

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Please remember to sign your messages on talk pages by typing four tildes ( ~ ); this will automatically insert your username and the date. If you need help, check out Questions, ask me on my talk page, or to ask for help on your talk page, and a volunteer should respond shortly. Again, welcome! RJFJR (talk) 03:30, 26 October 2015 (UTC)

Carmichael function
Your last two edits assume that if b p &equiv; 1 (mod q), then p is the Carmichael function of q. This is wrong. p may divide the Carmichael function of q. — Arthur Rubin (talk) 05:31, 27 May 2019 (UTC)


 * Thank you, Arthur. But my edits also includes the condition p is prime, which is the key point to say that p is the Carmichael function of q. Notice the original proof in Mersenne prime, which also does not cite a source, says: "Since p is prime and q is not a factor of 21 − 1, p is also the smallest positive integer x such that q is a factor of 2x − 1", and that is exactly the definition of Carmichael function. I mean, if my proof is wrong, the previous proof is also wrong, as they are both the same proof, but the original one includes indirectly the proof of Carmichael function from the Fermat's little theorem, which is what I simplify. Enrique Santos L. (talk) 12:45, 27 May 2019 (UTC)


 * The Carmichael function, &lambda;(n), is the least positive &lambda; such that if m and n are relatively prime, then m&lambda; &equiv; 1 (mod n). Hence 2$&lambda;(n)$ &equiv; 1 (mod n).  The reverse is not true; it is not necessarily the case that if 2$m$ &equiv; 1 (mod n), then &lambda;(n) divides m.  — Arthur Rubin  (talk) 14:36, 27 May 2019 (UTC)


 * You are rigth!, sorry, it is not &lambda;(n), but a divisor of &lambda;(n). Even so, just modifying this part the proof is correct. Anyway, the other properties on Repunit has no proof, and Wikipedia recomends not writing proofs unless "they expose or illuminate the concept or idea" (MOS:MATH). So it would be better to eliminate the proof, or just refering to the Mersenne prime similar proof (which I tried to simplify). About the proof in Mersenne prime, my motivation was that the original proof is written in a somewhat redundant way, not so clear, but maybe my proof is not clearer for others, because Carmichael Function is not so known as Fermat's Little Theorem. Enrique Santos L. (talk) 18:53, 27 May 2019 (UTC)
 * $$\operatorname{gcd} \left( a^n-1, a^m-1\right) = a^{\operatorname{gcd}(n, m)}-1$$
 * should have a name, and should be somewhere.... — Arthur Rubin (talk) 10:56, 28 May 2019 (UTC)


 * I see what you mean. I watched that property somewhere in Wikipedia, and I remember that the first reference for it, with proof, was this book of Donald Knuth, but I think that he wrote it just for base 2 (look here, Lemma 2).
 * A less general property based on it, which must have the same proof, is the last one in Repunit: "if gcd(m, n) = 1, then gcd(Rm(b), Rn(b)) = R1(b) = 1". The same is applied to base 2 in theorem number 6 of Mersenne_prime.
 * But for a more general name, I searched and found this: Divisibility_sequence, where Mersenne and repunit numbers are examples.