1. trantancuong says

April 4, 2012

I have the solve for FLT in special case exponent n=3.
We can write:
A^3=A*(A+1)^2 / 4 – (A-1)*A^2 / 4.
B^3=B*(B+1)^2 / 4 – (B-1)*B^2 / 4.

A^3+B^3 = A(A+1)^2/4+B (B+1)^/4 – {(A-1)A^2/4+(B-1)B^2/4}}.

X^3=X(X+1)^2/4-(X-1)X^2/4

I find X.
If:
X(X+1)^2/4-(X-1)X^2/4 = A(A+1)^2/4+B (B+1)^/4 – {(A-1)A^2/4+(B-1)B^2/4}}.
Infer
A^3+B^3=X^3..
the function F(X)={X(X+1)}^2
G(X)={(X-1)*X}^2.
If F(X) =F(A)+F(B) and G(X)=G(A)+G(B) infer X^3=A^3+B^3.
F(X) not =F(A)+F(B) or G(X) not = G(A)+G(B) infer X^3 not=A^3+B^3
Here( P and Q )infer R .
So not (P and Q) infer Not R.
I use theorem De Morgan:
Not(P and Q)=(Not P) or (not Q).
Can saying:
(not P) or( not Q) infer not R.
Attention : can not saying:
R infer ( P and Q..)
Now i must proving:
Two consecutive integers squared don’t equal two consecutive integers squared plus two consecutive integers squared..
For three consecutive integers.
1,2,3,
I have.
{1 (1 +1)} ^ 2 = 4
{2 (2 +1)} ^ 2 = 36
{3 (3 +1)} ^ 2 = 144.
We see;
(4 +36) <144. and 144-40 = 104
Or another.
2,3,4
{2 (2 +1} ^ 2 = 36
{3 (3 +1)} ^ 2 = 144
{4 (4 +1)} ^ 2 = 400
(36 +144) <400 and (400-180) = 120.
We see 104 <120.
At X (X +1) ^ 2 – {A (A +1) ^ 2 + B (B +1) ^ 2} .Thís difference Increased from 104 to 120.
X-2, X-1, X are three consecutive integers
If the (X-1) is a fixed .
( X-2) increasingly smaller to (X-3), (X-4)… and X grows to (X+1), (X+2)…..
We see at X (X +1) ^ 2 – { (X-2)*(X-1) ^ 2 +( X-1)*(X) ^ 2 }. This difference increased .
Inferred;
With three any numbers (A,B,X) belong N^3..
The equation {A(A+1)^2+{B(B+1)}2 not={X(X+)}^2
Or :X^3 not =A^3+B^3.
So : x^n+y^n not=z^n.
Happy&Peace.
Cường

2. trantancuong says

December 25, 2011

I dream Sir.Pierre De fermat. He said to me a formula. That is 3 cube + 4 cube + 5 cube= 6 cube. He said more “I have discovered a truly marvelous proof of this theorem, which this doodle is too small to contain.”. If x is not equal 3 . This formula: B exponent x+C exponent x+D exponent x = E exponent x.. With BCDE is the numbers consecutive or any number. The equation have no answers.

3. TRANTANCUONG says

November 15, 2011

Problem one.

3 cube+4 cube+5 cube=6 cube.
problem two.
1 cube….2 cube….3 cube….4 cube….5 cube….6cube..
1square….2 square….3 square….4 square….5 square….6 square….
We have new series with:1 cube-1 square….2 cube-2 square….3 cube-3 square….4 cube -4 square….
0….4….18….48….100….180….
We have new series with :4-0….18-4….48-18….100-48…..
4….14….30….52….
We have new series with 14-4….30-14….
10….16….22….
We have new series with:16-10….22-16….
6….6….
6=1 multiply 2 multiply 3=3!
6-6=0
problem three.
1 cube….2 cube….3 cube….4 cube….5 cube….6 cube….
We have new series with:2 cube-1 cube….3 cube-2 cube….4 cube-3 cube….5 cube-4 cube….
1….7….19….37….61….91….
We have new series with ;7-1….19-7….37-19….61-37….91-61….
6….12….18….24….30….
We have new series with :
12-6….18-12….24-18….30-24….
6….6….6….6
6= 1 multiply 2 multiply 3=3!
6-6=0.
Why both method have consequences same?
problem four.
1 exponent n+2 exponent n+3 exponent n+4 exponent n+……..+a exponent n+…….
I proved this formula with a and n are any number of sets of natural number.
problem five.
the series of natural number:123456789….
Make new series of number with:1+2….1+2+3….1+2+3+4….1+2+3+4+5….
1….3….6….10….15….
Make new series of number with:1+3….1+3+6….1+3+6+10….1+3+6+10+15….
1….4….10….20….35….Name this series is A.
for series of number1 square….2 square….3 square….4 square….
1….4…..9…..16…..25….
Make new series of number with:1-1….4-4….10-9….20-16….35-25….
0….0….1….4….10
Why it again series A?
problem six.
1 exponent n…. 2 exponent n….3 wxponent n….4 exponent n….
Make new series of number with:2 exponent n-1 exponent n….3 exponent n-2 exponent n….4 exponent n-3 exponent n….5 wexponent n-4 exponent n….
From this new series of number with method subtract same same .We have other series of number.
If continue to subtract the end is zero.Why?
prolem seven.
1+2+3+4+5+6+7.+……+n=S1
S2=1 square+2 square+3 square+4 square+….+nsquare….
S3=1 cube+2 cube+3 cube+4 cube+n cube….+
S4-1 exponent 4+2 exponent 4+….+n exponent 4+….
I find S4 followS3 and Sa+1 follow Sa.
I proved Sa=1exponent a+2 exponent a+3 exponent a+4 exponent a+….+n exponent a….
With a and n are two any number belong the sets of number.
problem eight.
for series of number with exponent n
1exponent n….2 exponent n….3 exponent n….4 exponent n….
second seriesof number with :2 exponent n-1 exponent n….3 exponent n-2 exponent n….4 exponent n-3 exponent n….5 exponent n-4exponent n….
From second series of number get big number subtract small number.We have third series of number.
continue and continue the end is zero.
at third series of number four number is function of 3,4 and n.
at x series of number, y number is the function of x,y and n.
example:
1^4…2^4….3^4….4^4….5^4….6^4….7^4….
second series of number is:
15….65….175….
third series of number is:
50….110….
second number off third series of number is 110.110 is the function of 2,3 and 4.2 is second number.3 is third series of number.4 is exponent number.
The and is zero and before zero is the number equal 4!=1 multiply 2 multiply 3 multiply 4.4 is the exponent number.
I understand and conclusions:
Sn=1^n+2^n+3^n+4^n+5^n+….+a^n+….
Sn+1=1^n+1+2^n+1+3^n+1+4^n+1+….+a^n+1+….
ATTENTION:2^n+1 is 2 exponent n+!.
I find Sn+! is the function of Sn or IF WE KNOW Sn WE KNOW Sn+!.
BEGIN with S1= 1+2+3+4+5+6+7+…..+a..
we know S1 inferred..we know S2 from S1 and we know S3 from S2 continue weknow Sx from Sx_-1
inferred WE have the formula about the total of series of number exponent z.z is number belong sets of number
I find this when I try prove fermat formula I think have CONTACT HERE..
PROVE FERMAT AT EXPONENT 3 AND CONTINUE TO 4 AND CONTINUE TO 5 AND TO n .n IS ANY NUMBER WHICH BELONG SETS OF NUMBER
To prove if formula fermat is right with exponent n inferred formula fermat is right with exponent n+1.
.THE CONTACT IS HERE. YOU inference from my problem to fermat formula.
THANK YOU SO MUCH.
FAREWELL FOREVER MY FANTASY MATHMATICS.
AU REVOIR POUR TOUJOURS MON FANTASMES MATHÉMATIQUES.
THE MAGICAL MATHEMATICIAN.TRANTANCUONG

4. vanloc says

August 27, 2011

3 exponent 3+4 exponent 3+5 exponent 3=6 exponent 3
Nam Mo Nhu Lai Mat Nhan Tu Chung Lieu Nghia Chu Bo Tat Van Hanh Thu Lang Nghiem Kinh
Nam Mo Nhu Lai Mat Nhan Tu Chung Lieu Nghia Chu Bo Tat Van Hanh Thu Lang Nghiem Kinh
Tôi nghe như vầy: Lúc bấy giờ tại tịnh xá Kỳ Hoàn thành Thất La Phiệt, Đức Phật và chúng Đại Tỳ Kheo một ngàn hai trăm năm mươi vị
Tôi nghe như vầy: Lúc bấy giờ tại tịnh xá Kỳ Hoàn thành Thất La Phiệt, Đức Phật và chúng Đại Tỳ Kheo một ngàn hai trăm năm mươi vị
. Na ra cẩn trì. Địa rị sắc ni na, bà dạ ma na, ta bà ha. Tất đà dạ, ta bà ha. Ma ha tất đà dạ ta bà ha. Tất đà dũ nghệ, thất bàn ra dạ ta bà ha. Na ra cẩn trì ta bà ha. Ma ra na ra ta bà ha. Tất ra tăng a mục khư da ta bà ha
. Na ra cẩn trì. Địa rị sắc ni na, bà dạ ma na, ta bà ha. Tất đà dạ, ta bà ha. Ma ha tất đà dạ ta bà ha. Tất đà dũ nghệ, thất bàn ra dạ ta bà ha. Na ra cẩn trì ta bà ha. Ma ra na ra ta bà ha. Tất ra tăng a mục khư da ta bà ha

5. vanloc says

August 27, 2011

3 exponent 3+ 4 exponent 3+ 5 exponent 3= 6 expopnent 3
This is my fun fantasy mathematic.
Maybe it have mathematical symbol same in multidimensional mathematic
Same theorem pythagore in triangle pythagore with formula
3 square + 4 square=5 square
farewell my fantasy mathematic forever
au revoir pour toujour mon fantasmes mathemathiques
Nam Mo Nhu Lai Mat Nhan Tu Chung Lieu Nghia Chu Bo Tat Van Hanh Thu Lang Nghiem Kinh
Nam Mo Nhu Lai Mat Nhan Tu Chung Lieu Nghia Chu Bo Tat Van Hanh Thu Lang Nghiem Kinh
Địa rị sắc ni na, bà dạ ma na, ta bà ha.
Địa rị sắc ni na, bà dạ ma na, ta bà ha.