How is newest proof Fermat?

Answer 1

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.