Modified Fermat's theorem

devarajkandadai · 2017-11-08, 03:45

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.

devarajkandadai · 2017-11-10, 03:58

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.

devarajkandadai · 2017-11-10, 11:33

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.

devarajkandadai · 2017-11-12, 03:47

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 ).

CRGreathouse · 2017-11-12, 04:26

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?

bhelmes · 2017-11-12, 10:07

Perhaps this might be helpful for you:
http://devalco.de/#104

Greetings from the primes
Bernhard

devarajkandadai · 2017-11-12, 11:23

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.

science_man_88 · 2017-11-12, 11:34

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 ??

Dr Sardonicus · 2017-11-12, 15:39

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).

Batalov · 2017-11-12, 16:50

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?!
You just go from thread to thread and spam with "your website".

Greetings from the composites!
Have a nice day!

bhelmes · 2017-11-12, 17:09

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

2017-11-10, 11:33	#3
devarajkandadai May 2004 316₁₀ 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.

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2017-11-08, 03:45	#1
devarajkandadai May 2004 474₈ Posts	Modified Fermat's theorem Works when the base is a Gausian integer as well as Z + ZIsqroot(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.

2017-11-12, 10:07	#6
bhelmes Mar 2016 444₁₀ Posts	Perhaps this might be helpful for you: http://devalco.de/#104 Greetings from the primes Bernhard

2017-11-12, 15:39	#9
Dr Sardonicus Feb 2017 Nowhere 61·107 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).