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## Latest Messages

Jun 23
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,