Suppose
Z^3=X^3+Y^3.
=>
Z=(X^3+Y^3)^1/3.
Define F(X,Y).
F(X,Y)= (X^3+Y^3)^1/3 – [(X-X-1)^3+(Y-X-1)^3]^1/3.
=>
F(X,Y)=Z- [(Y-X-1)^3-1]^1/3.
Because.
(Y-X-1) is an integer so
[(Y-X-1)^3-1]^1/3. and [(Y-X-1)^3-1]^2/3. are two irrational numbers.
=>
{ [(Y-X-1)^3-1]^1/3 – [ Z - F(X,Y)] }^3=0.
=>
[(Y-X-1)^3-1] +3[(Y-X-1)^3-1]^2/3*Z +V=0.
Because
3[(Y-X-1)^3-1]^2/3*Z is an irrational number So
V is an irrational number.
Because
3Z[(Y-X-1)^3-1]^2/3 = – {[(Y-X-1)^3-1] +.V}
So
an integer * an irrational number=an integer+an irrational number.
Unreasonable.
So
Z is an irrational number.
So
Z^3=/X^3+Y^3.
Similar.
Z^N=/X^N+Y^N.
ISHTAR.

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

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