Carmichael numbers that remain "absolute pseudoprimes" in any given number field are known to exist. See

[url=http://www.pseudoprime.com/pseudo3.pdf]There are Infinitely Many Perrin Pseudoprimes[/url]

Also, Carmichael numbers which remain "absolute pseudoprimes" in [i]all[/i] number fields up to a given degree are thought to exist, based on heuristics. However, this is only known to be true for degree 2. For examples in this case, see

[url=http://www.ams.org/journals/mcom/2000-69-232/S0025-5718-00-01225-4/S0025-5718-00-01225-4.pdf]HIGHER-ORDER CARMICHAEL NUMBERS[/url]

Carmichael numbers
 

[QUOTE=Nick;442830]And in the ring of Eisenstein integers?[/QUOTE]

Thanks to a tip on use of pari given by Charles Greathouse IV I can show that 561, although a Carmichael number, is only a pseudoprime in the ring of Eisenstein integers. Unfortunately the computations cannot be copied and pasted on Mersenneforum.org .

Details: Norm of (2 - row) is 7. Hence 561 is coprime with 7, Therefore((2-row)^560 -1) should yield an integer in this ring which the operation does not.

[QUOTE=devarajkandadai;471847]Thanks to a tip on use of pari given by Charles Greathouse IV I can show that 561, although a Carmichael number, is only a pseudoprime in the ring of Eisenstein integers. Unfortunately the computations cannot be copied and pasted on Mersenneforum.org .

Details: Norm of (2 - row) is 7. Hence 561 is coprime with 7, Therefore((2-row)^560 -1) should yield an integer in this ring which the operation does not.[/QUOTE]

sure they can, use the mark command or click the right way to highlight then hit enter. then ctrl+v ( or other ways to paste) when you focus on the forums reply box.

[QUOTE=devarajkandadai;471847]Thanks to a tip on use of pari given by Charles Greathouse IV I can show that 561, although a Carmichael number, is only a pseudoprime in the ring of Eisenstein integers. Unfortunately the computations cannot be copied and pasted on Mersenneforum.org .[/quote]That's funny, I do it all the time.

[quote]Details: Norm of (2 - row) is 7. Hence 561 is coprime with 7, Therefore((2-row)^560 -1) should yield an integer in this ring which the operation does not.[/QUOTE]
Your "hence" is a non-sequitur. I think you mean, "561 is coprime with [tex]2-\rho[/tex]." And of course [tex](2-\rho)^{560}[/tex] is an integer in [tex]\mathbb{Z}[\rho][/tex], it just isn't congruent to 1 (mod 561). Even if you take the exponent to be 561^2 - 1 (561^2 being the norm of 561), you [i]still[/i] don't get an integer congruent to 1 (mod 561). The problem is the prime factor 17 of 561.