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[Ref] isolated square free numberDefinitio by peruchoFeb 19
[Ref] multiplication and division of fractions by pahioJan 7
[Ref] field arising from special relativity by pahio15-10-31
[Ref] Carmichael numbers - 561 (contd) by akdevaraj15-07-13
[Ref] Carmichal number 561 by akdevaraj15-07-12
[Ref] Carmichal number 561 by akdevaraj15-07-12
[Ref] Carmichal number 561 by akdevaraj15-07-12
[Ref] Carmichal number 561 by akdevaraj15-07-12
[Ref] Failure functions by akdevaraj15-05-27
[Ref] induction proof of fundamental theorem of arithmet... by pahio15-03-29
[Ref] Birkhoff Recurrence Theorem by Filipe15-03-20
[Ref] Multiple Recurrence Theorem by Filipe15-03-20
[Ref] convergence in the mean by pahio15-01-30
[Ref] definition of vector space needs no commutativity by pahio15-01-20
[Ref] multiplication and division of fractions by pahioJan 7
[Ref] field arising from special relativity by pahio15-10-31
[Ref] Carmichael numbers - 561 (contd) by akdevaraj15-07-13
[Ref] Carmichal number 561 by akdevaraj15-07-12
[Ref] Carmichal number 561 by akdevaraj15-07-12
[Ref] Carmichal number 561 by akdevaraj15-07-12
[Ref] Carmichal number 561 by akdevaraj15-07-12
[Ref] Failure functions by akdevaraj15-05-27
[Ref] induction proof of fundamental theorem of arithmet... by pahio15-03-29
[Ref] Birkhoff Recurrence Theorem by Filipe15-03-20
[Ref] Multiple Recurrence Theorem by Filipe15-03-20
[Ref] convergence in the mean by pahio15-01-30
[Ref] definition of vector space needs no commutativity by pahio15-01-20
Latest Messages
Jun 23
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Apr 24
I don't see such a proof. Can you please write it in PlanetMath?
Jun 23
Pahio, happy to say "Nick" of Mersenneforum.org has given a simple
proof.
Jun 23
Pahio, happy to say "Nick" of Mersenneforum.org has given a simple
proof.
Jun 23
Pahio, happy to say "Nick" of Mersenneforum.org has given a simple
proof.
Jun 23
Pahio, happy to say "Nick" of Mersenneforum.org has given a simple
proof.
Jun 23
Pahio, happy to say "Nick" of Mersenneforum.org has given a simple
proof.
Jun 23
Pahio, happy to say "Nick" of Mersenneforum.org has given a simple
proof.
Jun 22
Before replying to Pahio's call for proof would like to add that
I forgot to add the condition: a and p should be co-prime.
Jun 21
Nice theorem! How do you prove it?
Jun 21
Modified Fermat's theorem: Let a belong to the ring of Gaussian integers
Then a^(p^2-1)= = 1 (mod p). Here p is a prime number with shape 4m+1 or 4m+3.
Apr 27
Happy to report that "Nick", on mersenneforum.org, has stated
that my conjecture can be taken as proved.
Apr 27
Happy to report that "Nick", on mersenneforum.org, has stated
that my conjecture can be taken as proved.
Apr 26
A couple of examples given below:Reading GPRC: gprc.txt ...Done.
GP/PARI CALCULATOR Version 2.6.1 (alpha)
i686 running mingw (ix86/GMP-5.0.1 kernel) 32-bit version
compiled: Sep 20 2013, gcc version 4.6.3 (GCC)
(readline v6.2 enabled, extended help enabled)
Copyright (C) 2000-2013 The PARI Group
PARI/GP is free software, covered by the GNU General Public License, and comes WITHOUT ANY WARRANTY WHATSOEVER.
Type ? for help, \q to quit.
Type ?12 for how to get moral (and possibly technical) support.
parisize = 4000000, primelimit = 500000
(10:53) gp > ((2+I)^8-1)/3
%1 = -176 - 112*I
(10:54) gp > ((2+I)^48-1)/7
%2 = -8220080432083104 - 2221404619138848*I
(10:55) gp > ((2+I)^120-1)/11
%3 = 48335053046044394818188476307133621695792 - 62299385456398106436997673432684416797456*I
(10:55) gp >\begin{flushright}
\end{flushright}
Apr 24
Let the base be a Gaussian integer = a + ib. Let p be a prime
number of shape 4m + 3. Then ((a + ib)^(p^2-1) - 1) == 0 (mod p).
This is subject to the base, a + ib and p being co-prime,
Latest Messages
Jun 23
Jun 23
Jun 23
Jun 23
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Apr 27
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Apr 24
I don't see such a proof. Can you please write it in PlanetMath?
Jun 23
Pahio, happy to say "Nick" of Mersenneforum.org has given a simple
proof.
Jun 23
Pahio, happy to say "Nick" of Mersenneforum.org has given a simple
proof.
Jun 23
Pahio, happy to say "Nick" of Mersenneforum.org has given a simple
proof.
Jun 23
Pahio, happy to say "Nick" of Mersenneforum.org has given a simple
proof.
Jun 23
Pahio, happy to say "Nick" of Mersenneforum.org has given a simple
proof.
Jun 23
Pahio, happy to say "Nick" of Mersenneforum.org has given a simple
proof.
Jun 22
Before replying to Pahio's call for proof would like to add that
I forgot to add the condition: a and p should be co-prime.
Jun 21
Nice theorem! How do you prove it?
Jun 21
Modified Fermat's theorem: Let a belong to the ring of Gaussian integers
Then a^(p^2-1)= = 1 (mod p). Here p is a prime number with shape 4m+1 or 4m+3.
Apr 27
Happy to report that "Nick", on mersenneforum.org, has stated
that my conjecture can be taken as proved.
Apr 27
Happy to report that "Nick", on mersenneforum.org, has stated
that my conjecture can be taken as proved.
Apr 26
A couple of examples given below:Reading GPRC: gprc.txt ...Done.
GP/PARI CALCULATOR Version 2.6.1 (alpha)
i686 running mingw (ix86/GMP-5.0.1 kernel) 32-bit version
compiled: Sep 20 2013, gcc version 4.6.3 (GCC)
(readline v6.2 enabled, extended help enabled)
Copyright (C) 2000-2013 The PARI Group
PARI/GP is free software, covered by the GNU General Public License, and comes WITHOUT ANY WARRANTY WHATSOEVER.
Type ? for help, \q to quit.
Type ?12 for how to get moral (and possibly technical) support.
parisize = 4000000, primelimit = 500000
(10:53) gp > ((2+I)^8-1)/3
%1 = -176 - 112*I
(10:54) gp > ((2+I)^48-1)/7
%2 = -8220080432083104 - 2221404619138848*I
(10:55) gp > ((2+I)^120-1)/11
%3 = 48335053046044394818188476307133621695792 - 62299385456398106436997673432684416797456*I
(10:55) gp >\begin{flushright}
\end{flushright}
Apr 24
Let the base be a Gaussian integer = a + ib. Let p be a prime
number of shape 4m + 3. Then ((a + ib)^(p^2-1) - 1) == 0 (mod p).
This is subject to the base, a + ib and p being co-prime,
