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 2017-11-08, 03:45 #1 devarajkandadai     May 2004 22×79 Posts Modified Fermat's theorem Works when the base is a Gausian integer as well as Z + Z*I*sqroot(7). Members may recall that Modified Fermat's theorem as a^(p^2-1) = = 1 (mod p) where p is prime of shape 3m + 1 or 4m+3.
2017-11-10, 03:58   #2

May 2004

1001111002 Posts

Quote:
 Originally Posted by devarajkandadai Works when the base is a Gausian integer as well as Z + Z*I*sqroot(7). Members may recall that Modified Fermat's theorem as a^(p^2-1) = = 1 (mod p) where p is prime of shape 3m + 1 or 4m+3.
Of course a and p have to be coprime.

 2017-11-10, 11:33 #3 devarajkandadai     May 2004 1001111002 Posts Modified Fermat's theorem This has been practically proved for Gaussian integer and bases a + b*sqrt(5). see (Hardy's intro to number theory and Pollard's intro to algebraic number theory.For the rest of quadratic algebraic integers I do not know about proofs.However I can, with the help of pari, say what it works for. In my next post will give a few for which this conjecture seems to be valid.
2017-11-12, 03:47   #4

May 2004

13C16 Posts

Quote:
 Originally Posted by devarajkandadai This has been practically proved for Gaussian integer and bases a + b*sqrt(5). see (Hardy's intro to number theory and Pollard's intro to algebraic number theory.For the rest of quadratic algebraic integers I do not know about proofs.However I can, with the help of pari, say what it works for. In my next post will give a few for which this conjecture seems to be valid.
Mft works up to p = 89.Above this beyond pari(my version's capability ).

2017-11-12, 04:26   #5
CRGreathouse

Aug 2006

135448 Posts

Quote:
 Originally Posted by devarajkandadai Mft works up to p = 89.Above this beyond pari(my version's capability ).
Really? Could you give an example of a calculation just beyond its ability?

 2017-11-12, 10:07 #6 bhelmes     Mar 2016 5·89 Posts Perhaps this might be helpful for you: http://devalco.de/#104 Greetings from the primes Bernhard
2017-11-12, 11:23   #7

May 2004

4748 Posts

Quote:
 Originally Posted by CRGreathouse Really? Could you give an example of a calculation just beyond its ability?
Sure: consider Mod(x,x^2+97). Let this be %1. You have to calculate ((2+%1)^(101^2-1)/101.
Thank you.

Last fiddled with by devarajkandadai on 2017-11-12 at 11:30 Reason: Typo

2017-11-12, 11:34   #8
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

23·3·5·71 Posts

Quote:
 Originally Posted by devarajkandadai Sure: consider Mod(x,x^2+97). Let this be %1. You have to calculate ((2+%1)^(101^2-1)/101. Thank you.
no second ending parentheses on the other side of things ??

 2017-11-12, 15:39 #9 Dr Sardonicus     Feb 2017 Nowhere 3×7×311 Posts The obvious "generalization of Fermat's theorem" is the generalization of Euler's theorem to number fields. This, in turn, is a special case of the result that, if G is a finite group, g is an element of G, and |G| the number of elements in G, then g|G| = 1, the identity of G. This is a consequence of Lagrange's theorem, applied to the cyclic group generated by g. The application to number fields is, R is the ring of algebraic integers of a number field K, I is a non-zero ideal of R, and G = (R/I)x the multiplicative group of invertible elements mod I (which is finite).
2017-11-12, 16:50   #10
Batalov

"Serge"
Mar 2008
San Diego, Calif.

5·2,081 Posts

Quote:
 Originally Posted by bhelmes Perhaps this might be helpful for you: http://devalco.de/#104 Greetings from the primes Bernhard
Mod warning: Another irrelevant plug for your webpages and you will receive a ban.

What the heck does it even have to do with the topic of this thread, huh?!

Greetings from the composites!
Have a nice day!

2017-11-12, 17:09   #11
bhelmes

Mar 2016

1101111012 Posts

Quote:
 Originally Posted by Batalov Mod warning: Another irrelevant plug for your webpages and you will receive a ban. What the heck does it even have to do with the topic of this thread, huh?! You just go from thread to thread and spam with "your website".
As far as i understood the OP has tried to deal with a+bI ( a complex number) and a+b*sqrt (A) and the cycle construction concerning the primes

In the given link you find a detailled version to the different cycle construction.
This was a gentle and completely correct mathematic link.
Besides you will not find this detailled information some where else.

The link i have given is a part of nice mathematic and programmed skill.

It is not nice to shoot with big guns, without any reason.

By the way, i have dealt since some times with primes,
and i have spent a lot of work to give a clear information about some prime topics on my website.

You do not seem to appriciate my own work.

Primes are very beautiful flowers
Greetings from the primes
Bernhard

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