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[p] Modified Fermat's theorem by akdevaraj 2:09 am
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.

[p] modidied Fermat's theorem by pahio Jun 21
Nice theorem! How do you prove it?

[p] Modified Fermat's theorem by akdevaraj 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.

[p] conjecture pertaining to Gaussian integers by akdevaraj Apr 27
Happy 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 27
Happy 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 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}

[p] conjecture pertaining to Gaussian integers by akdevaraj 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,

[p] conjecture pertaining to Gaussian integers by akdevaraj 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,

[p] conjecture pertaining to Gaussian integers by akdevaraj 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,

[p] conjecture pertaining to Gaussian integers by akdevaraj 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,

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

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

[p] Fermat theorem works by pahio Apr 20
Hi 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 19
What 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?