Is Fermat's last Theorem a equation metamorphosis?

Answer 1

Attention:

Only (N+).

Fermat's last Theorem z ^ 3 = x ^ 3 + y ^ 3 is capable exists a solution if fully meet the following conditions:

First step:

(1+2+3+4+........+a)^2+(1+2+3+4+........+b)^2=v^2.

In fact, using the computer, this equation has the ability to survive.

Second step:

(1+2+3+4+........+a+1)^2+(1+2+3+4+........+b+1)^2=s^2.

Third step:

v=1+2+3+4+........+c.

In fact, using the computer, third step and first step have the ability to survive in same time.

Fourth step:

s=1+2+3+4+........d.

Fifth step:

d=c+1

If all five steps are satisfied.This equation is capable of existence.

[z(z+1)/2]^2 - [z(z-1)/2]^2=[x(x+1)/2]^2- [x(x-1)/2]^2+[y(y+1)/2]^2 - [y(y-1)/2]^2.

Because:

z^3=[z(z+1)/2]^2 - [z(z-1)/2]^2.

Mean this equation is capable of existence.

z^3=x^3+y^3.

However, too hard to satisfy all five conditions in same time..

And an other solution:

Attention about series of number:

1,3,6,10,15,21,28,36,45........

Recognize:

10 and 15 are two number consecutive which belong this string.

Having:

15^2 - 10^2=5^3.

Or:

z^3=[z(z+1)/2]^2 - [z(z-1)/2]^2.

Impossible in same time exist both:

[z(z+1)/2]^2=[x(x+1)/2]^2+[y(y+1)/2]^2

And

[z(z-1)/2]^2=[x(x-1)/2]^2+[y(y-1)/2]^2

Attention:

All numbers as z(z+1)/2 and x(x+1)/2 and y(y+1)/2 and z(z-1)/2 and x(x-1)/2 and y(y-1)/2 are belong this string and they are Pythagorean

This is main proof:

z^3=[z(z+1)/2]^2 - [z(z-1)/2]^2

Define:

x<x+a<y.

x^3+y^3=[y(y+1)/2]^2 - [x(x-1)/2]^2 - [(x+1)^3+(x+2)^3+........+(x+a-1)^3+(x+a)^3+(x+a+1)^3+........+(y-1)^3]

Suppose:

z^3=x^3+y^3.

Because also:

(x+a)^3= [(x+a)(x+a+1)/2]^2 - [(x+a)(x+a-1)/2]^2.

Therefore a system of equations is generated.

[z(z+1)/2]^2 - [z(z-1)/2]^2=[y(y+1)/2]^2 - [x(x-1)/2]^2 - [(x+a)(x+a+1)/2]^2 + [(x+a)(x+a-1)/2]^2 - [(x+1)^3+(x+2)^3+........+(x+a-1)^3+(x+a+1)^3+........+(y-1)^3]

[z(z+1)/2]^2 - [z(z-1)/2]^2=[y(y+1)/2]^2 - [x(x-1)/2]^2 - [(x+b)(x+b+1)/2]^2 + [(x+b)(x+b-1)/2]^2 - [(x+1)^3+(x+2)^3+........+(x+b-1)^3+(x+b+1)^3+........+(y-1)^3]

[z(z+1)/2]^2 - [z(z-1)/2]^2=[y(y+1)/2]^2 - [x(x-1)/2]^2 - [(x+c)(x+c+1)/2]^2 + [(x+c)(x+c-1)/2]^2 - [(x+1)^3+(x+2)^3+........+(x+c-1)^3+(x+c+1)^3+........+(y-1)^3]

[z(z+1)/2]^2 - [z(z-1)/2]^2=[y(y+1)/2]^2 - [x(x-1)/2]^2 - [(x+d)(x+d+1)/2]^2 + [(x+d)(x+d-1)/2]^2 - [(x+1)^3+(x+2)^3+........+(x+d-1)^3+(x+d+1)^3+........+(y-1)^3].

........

Can not count the number of equations because number (a) can change to infinity.

And finally a great equation metamorphosis:

Using the two formulas by rotation affect each other:

z^3=[z(z+1)/2]^2 - [z(z-1)/2]^2

And

[z(z+1)/2]^2=1^3+2^3+........+z^3.

This method makes the original equation z ^ 3 = x ^ 3 + y ^ 3 is structured like a Robot Bumblebee Transformer.

Similar,this method is used for general case:

z ^ n = x ^ n + y ^ n.

Mean:

z ^ (n-3) ​​* z ^ 3 = x ^ (n-3) ​​* x ^ 3 + y ^ (n-3) ​​* y^ 3. 

You can structure this equation according to your own discretion.

ADIEU.