[p] **conjecture pertaining to Gaussian integers** by akdevaraj Apr 27Happy to report that "Nick", on mersenneforum.org, has stated
that my conjecture can be taken as proved.

[p] **conjecture pertaining to Gaussian integers** by akdevaraj Apr 27Happy to report that "Nick", on mersenneforum.org, has stated
that my conjecture can be taken as proved.

[p] **conjecture pertaining to Gaussian integers** by akdevaraj Apr 26A 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}

[p] **conjecture pertaining to Gaussian integers** by akdevaraj Apr 24Let 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,

[p] **conjecture pertaining to Gaussian integers** by akdevaraj Apr 24Let 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,

[p] **conjecture pertaining to Gaussian integers** by akdevaraj Apr 24Let 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,

[p] **conjecture pertaining to Gaussian integers** by akdevaraj Apr 24Let 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,

[p] **Fermat theorem works** by akdevaraj Apr 24Hi Jussi. Thanks will try.

[p] **Fermat theorem works** by akdevaraj Apr 24Hi Jussi. Thanks will try.

[p] **Fermat theorem works** by pahio Apr 20Hi Deva, perhaps the entry
'theorem on sums of two squares by Fermat'
may explain it or help this problem,
Jussi

[p] **Fermat's theorem works when the base is a Gaussian integer** by akdevaraj Apr 19What puzzles me is that the theorem works when the base is a prime in the ring
of Gaussian integers and the exponent is a prime of shape 4m + 1
but does not work when the exponent is a prime of shape 4m+3.Can
any one throw some light on this?

[p] **On Measurement Assessment and Division Matrices** by ProfHasan Apr 9http://jsaer.com/download/vol-3-iss-6-2016/JSAER2016-03-06-233-237.pdf

[p] **Division of Matrices** by ProfHasan Apr 9http://jsaer.com/download/vol-3-iss-5-2016/JSAER2016-03-05-101-104.pdf

[p] **Division of Matrices** by ProfHasan Apr 9http://jsaer.com/download/vol-3-iss-5-2016/JSAER2016-03-05-101-104.pdf