There is something about him that really captures me, the gentle smile and eyes, the quiet observation and the manner he explains his ideas. Amazing person, Mr. Wiles is.
@@mariangloser8382 Seems to come from the same artist although I'm not certain if it's the full thing. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-Ite_9m4xf60.html
Can anyone imagine how does it feel like to study for a degree at Oxford, then 40 years later they name a new building of the department after you, and you go to work there everyday.
He will be remembered for as long as humans are still around and do mathematics. When aliens arrive they will compare their proof and Wiles', and find that they are the same.
After 3 centuries a vietnamese old man solve Fermat in only page and any body can understand it solve Fermat in digital age, just one internet site I think I've found the real face of piere de Fermat in a dreaming math Define Sx=1+2^2+3^2+4^2+....+x^2.=x(x+1)(2x+1)/6=(2x^3+3x^2+x)/6 Sy=1+2^2+3^2+4^2+....+y^2=y(y+1)(2y+1)/6=(2y^3+3y^2+y)/6 Sz=1+2^2+3^2+4^2+....+z^2=z(z+1)(2z+1)/6=(2z^3+3z^2+z)/6 So 2x^3=6Sx-3x^2-x 2y^3=6Sy-3y^2-y 2z^3=6Sz-3z^2-z so x^3=3Sx-3/2x^2-x/2 y^3=3Sy-3/2y^2 - y/2 z^3=3Sz -3/2z^2-z/2 supose x^3+y^3=z^3 3Sx-3/2x^2-x/2+3Sy-3/2y^2 - y/2 - (3Sz -3/2z^2-z/2)=0 or 2Sx-x^2-x/3+2Sy-y^2 - y/3 - (2Sz -z^2-z/3)=0 or 2Sx+2Sy-2Sz-(x^2+y^2-z^2) =(x/3+y/3-z/3) because 2Sx+2Sy-2Sz-(x^2+y^2-z^2) is integer so (x/3+y/3-z/3) is also integer or x=3k y=3h and z=3g K,h,g are integers So 27k^3+27h^3=27g^3. or k^3+h^3=g^3 had had condition x ^ 3 + y ^ 3 = z ^ 3 Unable to meet the two conditions in the same time except x=k,y=h and z=g but x=3k and k=x so x=3x this is impossible conclusive x^3+y^3=/z^3 general Z^n=/x^n+y^n using formular 1^a+2^a+3^a+4^a+....+n^a
Thank you very much You had written How can you assert that 2Sx+2Sy-2Sz-(x^2+y^2-z^2) is integer ? :) 2Sx+2Sy-2Sz-(x^2+y^2-z^2) is integer because Sx=1+2^2+3^2+....+x^2 so it is integer. Similar with Sy and Sz. And condition give x,y,z are integers Old writing have mistake. I am sorry. . I write again which have no mistake. Please read it. Define Sx=1+2^2+3^2+4^2+....+x^2.=x(x+1)(2x+1)/6=(2x^3+3x^2+x)/6 Sy=1+2^2+3^2+4^2+....+y^2=y(y+1)(2y+1)/6=(2y^3+3y^2+y)/6 Sz=1+2^2+3^2+4^2+....+z^2=z(z+1)(2z+1)/6=(2z^3+3z^2+z)/6 So 2x^3=6Sx-3x^2-x 2y^3=6Sy-3y^2-y 2z^3=6Sz-3z^2-z so x^3=3Sx-3/2x^2-x/2 y^3=3Sy-3/2y^2 - y/2 z^3=3Sz -3/2z^2-z/2 supose x^3+y^3=z^3 3Sx-3/2x^2-x/2+3Sy-3/2y^2 - y/2 - (3Sz -3/2z^2-z/2)=0 or 2Sx-x^2-x/3+2Sy-y^2 - y/3 - (2Sz -z^2-z/3)=0 or 2Sx+2Sy-2Sz-(x^2+y^2-z^2) =(x/3+y/3-z/3) Sx+S(x-1)+Sy+S(y-1) -Sz -S(z-1)=(x/3+y/3-z/3) Define the function f(x) is Sx+S(x-1) So f(y)=Sy+S(y-1) f(z)=Sz+S(z-1) And g(x )is x/3 so g(y)=y^3 g(z)=z^3 So f(x)+f(y)-f(z)=g(x)+g(y)-g(z) homogeneous them so f(x)=g(x) But this is wrong So x^3+y^3 impossible =z^3
I don't think you can assert that f(x)+f(y)-f(z)=g(x)+g(y)-g(z) implies that f(x) = g(x) and f(y) = g(y) and f(z) = g(z) :( Yes you right. I am sorry I think and write again completely. Define Sx=1+2^2+3^2+4^2+....+x^2.=x(x+1)(2x+1)/6=(2x^3+3x^2+x)/6 Sy=1+2^2+3^2+4^2+....+y^2=y(y+1)(2y+1)/6=(2y^3+3y^2+y)/6 Sz=1+2^2+3^2+4^2+....+z^2=z(z+1)(2z+1)/6=(2z^3+3z^2+z)/6 So 2x^3=6Sx-3x^2-x 2y^3=6Sy-3y^2-y 2z^3=6Sz-3z^2-z so x^3=3Sx-3/2x^2-x/2 y^3=3Sy-3/2y^2 - y/2 z^3=3Sz -3/2z^2-z/2 supose x^3+y^3=z^3 3Sx-3/2x^2-x/2+3Sy-3/2y^2 - y/2 - (3Sz -3/2z^2-z/2)=0 or 2Sx-x^2-x/3+2Sy-y^2 - y/3 - (2Sz -z^2-z/3)=0 or 2Sx+2Sy-2Sz-(x^2+y^2-z^2) =(x/3+y/3-z/3) So (2Sx+2Sy-2Sz) - (x/3+y/3-z/3)=(x^2+y^2-z^2) this tell that (x^2+y^2-z^2) is a funtion of [ (x/3+y/3-z/3 and (2Sx+2Sy-2Sz)] And a other way to know x^2+y^2-z^2 follow x+y- z because (x+y)^2=x^2+y^2+2xy (x+y-z)^2=(x+y)^2+z^2-2z(x+y)=x^2+y^2+2xy+z^2-2z(x+y=x^2+y^2-z^2+2xy-2z(x+y +3z^2 So (x^2+y^2 - z^2)=(x+y-z)^2+2xy-2z(x+y +3z^2 This tell that (x^2+y^2 - z^2) is a function of [ (x+y-z)^2 and 2xy-2z(x+y +3z^2] (x^2+y^2-z^2) is a funtion of [ (x/3+y/3-z/3 and (2Sx+2Sy-2Sz)] And (x^2+y^2 - z^2) of [ (x+y-z)^2 and 2xy-2z(x+y +3z^2] Can not satify two conditions in a same time. note first function have no xy,zx and zy Second funtion have
Well, I think you're not very sure of your results, trying this and that approach. If you think you are really in the verge of a major result (which would be astonishing, I should say) I advise you to work on that proof carefully and not just throw attempts. ;) This is not a try/error endeavor, or at least not in a youtube comments box, were typically people doesn't have so much free time to work on partial and not so thought results. But hey; go for it.
2:57 man, for a blink moment I thought it was him, thinking like "how could one man possess so much awesomeness"... guess he's awesome enough already. :D
Andrew wiles real genius But i just prove fo happy in math Please read it, that is a short message about Flt Define Sx=1+2^2+3^2+4^2+....+x^2.=x(x+1)(2x+1)/6=(2x^3+3x^2+x)/6 Sy=1+2^2+3^2+4^2+....+y^2=y(y+1)(2y+1)/6=(2y^3+3y^2+y)/6 Sz=1+2^2+3^2+4^2+....+z^2=z(z+1)(2z+1)/6=(2z^3+3z^2+z)/6 So 2x^3=6Sx-3x^2-x 2y^3=6Sy-3y^2-y 2z^3=6Sz-3z^2-z so x^3=3Sx-3/2x^2-x/2 y^3=3Sy-3/2y^2 - y/2 z^3=3Sz -3/2z^2-z/2 suppose x^3+y^3=z^3 3Sx-3/2x^2-x/2+3Sy-3/2y^2 - y/2 - (3Sz -3/2z^2-z/2)=0 or 2Sx-x^2-x/3+2Sy-y^2 - y/3 - (2Sz -z^2-z/3)=0 or 2Sx+2Sy-2Sz-(x^2+y^2-z^2) =(x/3+y/3-z/3) So (2Sx+2Sy-2Sz) - (x/3+y/3-z/3)=(x^2+y^2-z^2) this tell that (x^2+y^2-z^2) is a function of [ (x/3+y/3-z/3 and (2Sx+2Sy-2Sz)] because (x+y)^2=x^2+y^2+2xy (x+y-z)^2=(x+y)^2+z^2-2z(x+y)=x^2+y^2+2xy+z^2-2z(x+y=x^2+y^2-z^2+2xy-2z(x+y +3z^2 So (x^2+y^2 - z^2)=(x+y-z)^2+2xy-2z(x+y +3z^2 This tell that (x^2+y^2 - z^2) is a function of [ (x+y-z)^2 and 2xy-2z(x+y +3z^2] Had had first (x^2+y^2-z^2) is a funtion of [ (x+y-z)/3 and (2Sx+2Sy-2Sz)] and more have another function (x^2+y^2 - z^2) is a funtion of [ (x+y-z)^2 and 2xy-2z(x+y )+3z^2] note first function have no xy,zx and zy Second funtion have. It point out that (x ^ 2 + y ^ 2 - z ^ 2) is not only followas the (x, y and z) but also (x ^ 2 + y ^ 2 - z ^ 2) acording individual x. Cannot satisfy two functions different in a same time on this case.
The Universe has a prime-cofactor quantization superposition cause-effect format, with/to which we respond by/to recognise mathematical concepts, altogether.
