User:Palaniswami.k/sandbox

Dr.K.Palaniswami               31/05/2017 To The Editor, Publication Committee, Wikipedia. Sir, Sub: Requesting to review & publish the article-reg I am submitting article given below for review & possible publication. I request you to do the need for review & communicate the comments to my email id: kpswaami@gmail.com and oblige.

Topic: Generalizing Pierre de Fermat's final theorem and de-mystifying Andrew Wiles' puzzle with KPS' algorithm.

Palaniswami.K *

Pierre de Fermat's final theorem can be modified as a generalized theorem (equation) as a*n + b*(n+1) = c*n ...(1), selecting a=b=(2*n)-1 and c= 2a ( call KPS' algorithm),('a'power n is represented as a*n) such that a,b and c all of them are divisible by a common number, 'a', justifies the equation (1) for all values of 'n'.

Proof:

Let a*n + b*(n+y') = c*n ...(2) Substituting KPS' algorithm, we get (2*n -1)*n +(2*n -1)*(n+y')=[2(2*n -1)]*n ...(3) Dividing on both the sides by (2*n -1)*n we get 1+(2*n -1)*y'=2*n ...(4) ie., (2*n -1)*y'=(2*n -1)*1 ..(5) In equation (5) bases are equal and hence powers should be equal. ie., y' = 1 ...(6) Substituting equation(6) in equation(2), we get modified and generalized Pierre de Fermat's theorem rewritten as          a*n + b*(n+1)=c*n  ...(7) is unique,for all values of 'n'.

Thus,Andrew Wiles' million dollar puzzle is unraveled,solved and demystified henceforth. And also, all the three numbers a,b and c are equally divisible by a common number 'a' satisfying the requirement observed by Andrew Wiles. [ Proof: values of a,b and c are 1,1,2 for n=1; 3,3,6 for n=2; 7,7,14 for n=3,all are divisible by a common number and so on ].

Corollary 1 : In KPS'algorithm, 'a' and 'b' are odd numbers and 'c' is an even number.

Corollary 2 : In equation (7), when bases are equal, powers are not equal and while powers are equal,bases are not equal.

Addendum:

In KPS' algorithm, it had been suggested to assume a=b and    c= 2a or (a + b ). Re-arranging the values as a=b and making c=a+b( becomes 2a since a= b)   by simple algebra, it is a well known fact that a*2 + b*2 ≠ (a + b)*2 ... (8)         a*3 + b*3 ≠ (a + b)*3 ... (9) extending, we get a*n +b*n ≠ (a +b)*n  ... (10) Equation (10) may be rewritten as         a*n + b*n ≠ c*n ... (11) It substantiates and justifies the proof for modified de Fermat's final Theorem without any ambiguity. _______________________________________________________________ Ex.Professor in ECE,PSG Tech(1996-99),Coimbatore, Ex.Professor,HOD and Director in Karunya University( founded by Dr.D.G.S.Dhinakaran,Jesus Calls,India(1999-2010),Coimbatore.
 * Dr.Palaniswami.K, D.Sc(corllins).,Ex.Faculty,IIT,Madras(1978-87),