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Message: <blockquote data-quote="luxmen" data-source="post: 21172487" data-attributes="member: 409399"><p>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.</p></blockquote><p></p>

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