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PlanetMath is a virtual community which aims to help make mathematical knowledge more accessible. PlanetMath's content is created collaboratively: the main feature is the mathematics encyclopedia with entries written and reviewed by members. The entries are contributed under the terms of the Creative Commons By/Share-Alike License in order to preserve the rights of authors, readers and other content creators in a sensible way. We use LaTeX, the lingua franca of the worldwide mathematical community. On February 13th 2013, was updated to use the new software system Planetary. Some release notes are here. Please report bugs in the Planetary Bugs Forum or on Github.

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[p] Fermat's theorem in k(i) (c0ntd) by akdevaraj Jul 21
This base, however , does not work in the case of primes having shape 4m+3. A base that works is 1 + i. Example ((1+i)^102 + I)/103 = -21862134113449i.

[p] Fermat's theorem in k(i) (c0ntd) by akdevaraj Jul 20
What is the nature of a, the base? When p has the shape 4m+1 a has the shape of a prime factor of a number having the same shape. Example: Let p = 61. Then ((4 + i)^60 - 1 )/61 = -71525089284120116591639000327021600 + 11369162311133702688684197835211600i

[p] Fermat's theorem in k(i) (c0ntd) by akdevaraj Jul 19
Before giving some further generalisations let me give some examples: case a) ((1+I)^30 + I)/31 = -1057i. ((1 + i )^102 + i)/103 = -21862134113449i Case b) ((1 + i)^12 - 1)/13 = -5. (( 1 + i ) ^100 - 1)/101 = -11147523830125

[p] Fermat's theorem in k(i) (c0ntd) by akdevaraj Jul 17
Although Hardy and Wright have formulated the above theorem in their book ("An introduction to he theory of numbers " we can see how it works with the aid of software like pari. The four examples illustrate this. Now for a few genralisations: a) If p is a prime of form 4m+3, then ((a^(p-1)+ I)/p is congruent to 0 (mod(p)). b) If p is a prime of form 4m+1, then ((a^(p-1) - 1)/p is congruent to 0 (mod(p)).

[p] Fermat's theorem in k(i) by akdevaraj Jul 14
There are four unities in k(i) viz 1, -1, i and -i. Four examples are given here to illustrate Fermat's theorem in k(i). a)((2+3i)^2-1)/3 = -2 +4i b) ((3+2i)^2 + 1)/3 = 2 + 4i c) ((10 + i)^2 + i)/3 = 33 + 7i and d) ((14 +i)^2 - i)/3 = 65 + 9i.

how to determinate the unknowns of this simplex tableau by anouarattn Jul 11
Considering 2 simplex tableau encountered when solving a linear program determine the value of each of the following items(unknowns) : p q r that appear in tables And please if someone has another example like this one please give it to me>

[P] I just read your article and by Happy19th Jun 30
I just read your article and it was totally great, it contains a lot of useful ideas, it is also written in organize manner,thanks for sharing this kind of article. <a href="">

[p] A good question by pahio Jun 25
Please see "sophomore's dream" in Wikipedia.

[p] easy exercice by Ron Castillo Jun 23
1.- Is False 2.- Is True 1.- Is False 1.- Is true, iff b is rational. Regards, Ronald.

[P] failure functions - another exampleL by akdevaraj Jun 23
Let our definition of a failure be a composite number which is also a multiple of 11. Let the parent function be 2^n + 7 (n belongs to N ). Then n = 2 + Eulerphi(11) is a failure function. Also n = 2^(1 + Eulerphi(Eulerphi(11)) is also a failure function.

[P] failure functions - another example by akdevaraj Jun 20
Let our definition of a failure be a non-primitive polynomial in x (x belongs to Z ). Let the parent function be the primitive polynomial x^2 + x + 1. Then x generated by any of the failure functions 1 + 3k, 2 + 7k etc when substituted in the parent function yield failures i.e. non primitive polynomials.

[P] A correction by akdevaraj Jun 16
This refers to " Non-linear failure functions and Automorphism. The second-last line should read: When the relevant quotient is divided by 17 we get a remainder = 5, a member of Z_17.

[P] Non-linear failure functions and automorphism by akdevaraj Jun 16
Let our definition of a failure be a composite number. Let the mother function be the quadratic x^2 + 1 ( x belongs to Z ). When x =4, f(x) =17. x = 4 + 17*k is a failure function. This is linear. The non -linear failure function x = 38 + 17^(k+2) generates values of x, which when substituted in f(x) we get multiples of 289. The relevant quotients when divided by 17 yield the remainder 3, a member of Z_17. Here k belongs to W.

[P] failure functions - applications by akdevaraj Jun 15
Failure functions can be applied in the following areas: a) Solving Diaphontine equations ( for copy of paper send request to b) Indirect primality testing c) In proving conjectures (see sketch proof )