Pierre De Fermat 's last Theorem.
The conditions:
x,y,z,n are the integers and >0. n>2.
Proof:
z^n=/x^n+y^n.
We have;
z^3=[z(z+1)/2]^2-[(z-1)z/2]^2
Example;
5^3=[5(5+1)/2]^2-[5(5-1)/2]^2=225-100=125
And
z^3+(z-1)^3=[z(z+1)/2]^2-[(z-2)(z-1)/2]^2
Example;
5^3+4^3=[5(5+1)/2]^2-[(5-2)(5-1)/2]^2=225-36=189
And
z^3+(z-1)^3+(z-2)^3=[z(z+1)/2]^2-[(z-3)(z-2)/2]^2
Example
5^3+4^3+2^3=[5(5+1)/2]^2-[(5-3)(5-2)/2]^2=225-9=216
And
z^3+(z-1)^3+(z-2)^3+(z-3)^3=[z(z+1)/2]^2-[(z-4)(z-3)/2]^2
Example
5^3+4^3+3^3+2^3=[5(5+1)/2]^2-[(5-4)(5-3)/2]^2=225-1=224
General:
z^3+(z-1)^3+....+(z-m)^3=[z(z+1)/2]^2-[(z-m-1)(z-m)/2]^2
We have;
z^3=z^3+(z-m-1)^3 - (z-m-1)^3.
Because:
z^3+(z-m-1)^3=[z^3+(z-1)^3+....+(z-m-1)^3] -
[(z-1)^3+....+(z-m)^3]
So
z^3=[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 -
[z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3.
Similar:
z^3=z^3+(z-m-2)^3 - (z-m-2)^3.
So
z^3=[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 -
[z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3.
....
....
Suppose:
z^n=x^n+y^n
So
z^(n-3)*z^3=x^(n-3)^n*x^3+y^(n-3)*y^3.
So
z^(n-3)*{[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 -
[z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 -
(z-m-1)^3}=x^(n-3)*{[x(x+1)/2]^2-[(x-m-2)(x-m-1)/2]^2 -
[x(x-1)/2]^3+[(x-m-1)(x-m)/2]^2 -
(x-m-1)^3}+y^(n-3)*{[y(y+1)/2]^2-[(y-m-2)(y-m-1)/2]^2 -
[y(y-1)/2]^3+[(y-m-1)(y-m)/2]^2 - (y-m-1)^3}
Similar:
z^(n-3)*{[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 -
[z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 -
(z-m-2)^3=x^(n-3)*{[x(x+1)/2]^2-[(x-m-3)(x-m-2)/2]^2 -
[x(x-1)/2]^3+[(x-m-2)(x-m-1)/2]^2 -
(x-m-2)^3+y^(n-3)*{[y(y+1)/2]^2-[(y-m-3)(y-m-2)/2]^2 -
[y(y-1)/2]^3+[(y-m-2)(y-m-1)/2]^2 - (y-m-2)^3.
....
....
Because it is codified .
So
Impossible all are the integers.
So:
z^n=/x^n+y^n.
ISHTAR.
