Does every odd number coprime to its Euler totient divide some Carmichael number?

Importance: Medium ✭✭
Author(s): Michon, Gérard P.
Recomm. for undergrads: yes
Posted by: maxal
on: August 5th, 2007
Conjecture   Every odd number coprime to its Euler totient divides some Carmichael Number.

In December 2007 Joe Crump discovered the following Carmichael multiples (of the numbers with previously unknown Carmichael multiples):

    \item $ 24354644805191195265 $ is a Carmichael multiple of $ 885 $; \item $ 174470770903594881 $ is a Carmichael multiple of $ 2391 $; \item $ 12832546007164521 $ is a Carmichael multiple of $ 2517 $; \item $ 435262925087145321 $ is a Carmichael multiple of $ 2571 $; \item $ 291226428348047343201 $ is a Carmichael multiple of $ 2589 $; \item $ 947087538769733505 $ is a Carmichael multiple of $ 2595 $; \item $ 114593508055911048606465 $ is a Carmichael multiple of $ 2685 $; \item $ 13053581557039793157 $ is a Carmichael multiple of $ 2949 $.


Gérard P. Michon. "Least Carmichael Numbers with Given Divisors (up to 2999)."

Gérard P. Michon and Joseph K. Crump. "Carmichael Multiples of Odd Cyclic Numbers."

W. R. Alford, A. Granville, and C. Pomerance. "There are Infinitely Many Carmichael Numbers." Ann. Math. 139, 703-722, 1994.

Günter Löh and Wolfgang Niebuhr. "A new algorithm for constructing large Carmichael numbers." Math. Comp. 65 (1996), 823-836.

* indicates original appearance(s) of problem.

2^n == 3 (mod n),

Найдено eще 5 чисел ,удовлятворящию условию 2^n == 3 (mod n), 1. n =2^4700063497-3 2. n =2^3468371109448915-3 3. n =2^8365386194032363-3 4. n =2^10991007971508067-3 5. n =2^63130707451134435989380140059866138830623361447484274774099906755 -3 причем n нечётное, не простое и не делится на 3. Что n нечетное и не делится на 3 выходит из 2^k-3. Что n непростое выходит из 2^k == 3 (mod k), т.е. 2^k -3 = 0 (mod k), т.е. делится на k для выще перечисленных значений. Далее находим следующие 5 чисел где n = 2^(2^k-3)-3 и так далее и так далее до бесконесности. Осюда вытекает следующая Теорема: Чисел удовлятворящию условию 2^n == 3 (mod n) бесконечно много. (Neymet MS)


Утверждение неверно. См.

From the conjecture's author...

Thanks to Maxal for posting this old conjecture of mine (c. 1980) with a link to the page of computational results which I put online a few years ago. I found out about Maxal's post when someone triggered that link and landed on my own "Numericana" site...

To the best of my knowledge, nobody else has yet worked on this conjecture, so the field is wide open. If you find anything (theoretical or computational) please let me know and I'll post your results, with due credit, at the very location where the statement of the conjecture first appeared, namely at Thanks.

Gerard P. Michon, Ph.D.

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