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01-14-2017, 04:12 AM

http://vixra.org/pdf/1601.0053v1.pdf
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01-14-2017, 04:16 AM

Quote:
Originally Posted by mag123 View Post
those who write in this elakiri forum alone says your proof is flowed, now that you published this more than a year ago, but any body ever said this is correct ? and most people in elakiri says your proof has flows. If some one is silent after you said some thing that alone means he does not want to argue with you because you are a kind of person who do not see the truth, but say the same thing over and over again.
Please give priority to Maths. Not to your thoughts or my thoughts or elakiri thoughts. WHAT EVER, IF YOU FORWARD , ACCORDING TO MATHS, THEN I ACCEPT, what I see here, when I gave a answer mathematically, they have to accept, and silent.
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01-14-2017, 04:17 AM

Just go to University of Colombo Mathamatics Department and meet Dr Dr. T. U. Hewage , Head of the Department or any other professor . He /they will tell you your proof is correct or not.
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01-14-2017, 04:21 AM

Quote:
Originally Posted by mag123 View Post
Just go to University of Colombo Mathamatics Department and meet Dr Dr. T. U. Hewage , Head of the Department or any other professor . He /they will tell you your proof is correct or not.
Thanks very much, I accept genuine answers like this answer.
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01-24-2017, 07:24 PM

Lankave siyaluma university vala inna Maths professors lata meya peneva.
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01-24-2017, 07:24 PM

VERY IMPORTANT--------MOST IMPORTANT PART FOR THIS MY PROOF NOW, I EXPOSE TO THE WORLD NOW . YOU CAN SEE FROM INEQUALITY [1] ----c^n/2 <a^n/2 + b^n/2--then from that inequality , I took when n=2, c<a+b---[2]----SO MY PROOF IS FOR WHEN c^2=a^2+b^2 then there can not be a,b,c whole numbers for n>2 for c^n=a^n+b^n [a,b,c common for any n there] . NOW LETS CONSIDER WHEN FOR ANY n when n is not equal to 2----c^n =a^n +b^n----[A]. You can take it as [c^n/2]^2=[a^n/2]^2 + [b^n/2]^2------[A] .SO there should be a^n/2 and b ^n/2 should be fractions[ALREADY PROVED].So a ^n/2 and b^n/2 should be fractions. FOR THAT a and b should be fractions. SO I PROVED FERMAT LAST THEOREM VERY CLEARLY.

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01-24-2017, 07:29 PM

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