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Indirect proof of FLT ???
Posted:
Oct 15, 2013 6:34 PM


Dear Professionals and Amateurs ! I would like to present in brief my new thread of FLT proof and just complete it for n=3. ( Could it be already known ? ) Lets complete certain ideal set of natural numbers prototypes of X;Y;Z supposed to fulfill equality: X^n +Y^n = Z^n for n>=3 prime numbers. Now we should notice as P = X^n +Y^n +Z^n is divided by sum Q of these ideal numbers Q = X+Y+Z . How it is going for regular natural numbers ? Lets check relations of P = A^n +B^n +C^n divided by Q = A+B+C : Here for n=3 let A=x; B=x+a; C=x+b where x is some changing value and a;b natural value of parameters ; then once A<B<C so a<b . A^3 +B^3 +C^3 will appear as polynomial P = 3x^2 + 3(a+b)x^2 +3(a^2 +b^2)x +a^3 +b^3 once Q = 3x+a+b so divisibility of P by Q will demand in the first step a+b=3c ;then once a<b we could suppose: a=k ; b=2k ; and for such input : P = 3x^3 +9kx^2 +15k^2 x +9k^3 ; Q = 3x +3k ; and then P is divided in whole once: P = (x^2 +2kx +3k^2)(3x +3k) There are infinite numbers of such solutions P(3)/Q with following alterations: 1o) x=1 or every bigger natural number m ; x=m ; 20) a=1 or every bigger natural number k ; a=k ; but then b=2k Now see certain ideal natural numbers prototypes from FLT for n=3 from so called by me TAB developments : I) for Y/3 : A=3^(3u1) a^3 ; B=b^3 ;X+Y=t^3 once X^3 +Y^3 =Z^3 so X^3 +Y^3 +Z^3 = 2Z^3 where Z=t*z =X+YT =t^3 3^u abt =t(t^2 3^u ab) P=2Z^3 =2t^3 *z^3 ; Q =X+Y+Z =t^3 +tz = t(t^2 +z) here P and Q has common term t and we extract it: P'=2t^2 z^3 ; Q' =t^2 +z then once t is prime to z so also t prime to t^2 +z and division should be find in: P''=2z^3 and also once z=t^2 3^u ab : P''=2t^6 2*3^(u+1)abt^4 +2*3^(2u+1)a^2b^2t^2 2*3^3ua^3b^3 and Q'=2t^2 3^uab and really: P''=(t^4 2*3^uabt^2 +2*3^2ua^2b^2)(2t^2 3^uab) II) for Z/3 : A=a^3; B=b^3; X+Y=3^(3u1) t^3; Z=3^u tz; P=2Z^3 =2*3^3u t^3z^3 ; Q=X+Y+Z =3^(3u1) t^3 +tz ; Q=3^u t[3^(2u1) t^2 +z] then for: Z=X+YT=3^(3u1)t^3 3^u abt =3^u t[3^(2u1)t^2 ab] z=3^(2u1) t^2 ab and so on quite similar we'll achieve whole division of P by Q Once divisibility P/Q occurs in whole for ideal numbers so their values should be chosen from our infinite sets of regular A;B;C numbers. But here b=2k means let ZX=A and then YX = AB = a=k ; from these A=2k and AB=k we'll come to B=k ; now once A should be prime to B so for k=1 and B=1 reminds A=2 what is inconsistent to prototype of A: A =a^3 or A =3^(3u1)a^3 Q.E.D for n=3
Thank You very much for Your attention and sorry for big amounts of very brief explainations of my developments... If true so will be worth of any necessary corrections and addictions With Best Regards RoBin



