so hypnotic, perfect vocal, simple efficient, you picture a crowd standing bumping around, not talking, everybody in a mind trip, with this voice telling you to concentrate on... your body.... your body....... your body
In a dream in a best midnight , i see ghost Fermat say that:" i have a magnificent spaceship in galaxy outside universe, it contain the solution about root equation Fermat. He say to me :Let's do it immediately ! Original equation of genius Pierre De Fermat is really just the trouble !. z^n=x^n+y^n. Relationships about exponents between three and two and vice versa . First formula. z^3=[z(z+1)/2]^2 - [z(z-1)/2]^2 . Example: 3^3=6^2-3^2. Second formula. [z(z+1)/2]^2=1^3+2^3+........+z^3 Example: 6^2=1^3+2^3+3^3. Third formula. x^3+y^3=[y(y+1)/2]^2 - [x(x-1)/2]^2 - [(x+1)^3+(x+2)^3+........+(y-1)^3] Example: 2^3+4^3=10^2-1^2-3^3. Add the variable (a) arbitrary : x^3+y^3=[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] Fourth formula. [b(b+1)/2]^2 - [a(a-1)/2]^2=a^3+(a+1)^3+........+(b-1)^3+b^3. Example: 10^2-1^2=2^3+3^3+4^3. Root Fermat:: z^3=x^3+y^3. The second generation of the Fermat equation generated by the following two formulas First z^3=[z(z+1)/2]^2 - [z(z-1)/2]^2 Second x^3+y^3=[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] After root Fermat was altered ,it born a systems composed of many small equations : [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]. ........ The require about absolute integer is impossible ! Ishtar .
so hypnotic, perfect vocal, simple efficient, you picture a crowd standing bumping around, not talking, everybody in a mind trip, with this voice telling you to concentrate on... your body.... your body....... your body
Here, ladies and gentlemen, we are leaving the domain of entertainment music and moving deep into the realm of art. Deep.
on the ground. chewing terrain. dope
Danke Dich Dettmann
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Everytime listening to this 🥵
Sevgilerimle. Danimarka śûper 🩷💙💐💙🙏🙏💙💙🩷🩷💐💙🩷🩷🩷🩷🙏🙏🙏⭐⭐💚💚💚💚💚⭐⭐💚💚🩵💚⭐⭐🩵🩵🩵
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amazing
3:16 nice part
In a dream in a best midnight , i see ghost Fermat say that:" i have a magnificent spaceship in galaxy outside universe, it contain the solution about root equation Fermat. He say to me :Let's do it immediately !
Original equation of genius Pierre De Fermat is really just the trouble !.
z^n=x^n+y^n.
Relationships about exponents between three and two and vice versa .
First formula.
z^3=[z(z+1)/2]^2 - [z(z-1)/2]^2 .
Example:
3^3=6^2-3^2.
Second formula.
[z(z+1)/2]^2=1^3+2^3+........+z^3
Example:
6^2=1^3+2^3+3^3.
Third formula.
x^3+y^3=[y(y+1)/2]^2 - [x(x-1)/2]^2 - [(x+1)^3+(x+2)^3+........+(y-1)^3]
Example:
2^3+4^3=10^2-1^2-3^3.
Add the variable (a) arbitrary :
x^3+y^3=[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]
Fourth formula.
[b(b+1)/2]^2 - [a(a-1)/2]^2=a^3+(a+1)^3+........+(b-1)^3+b^3.
Example:
10^2-1^2=2^3+3^3+4^3.
Root Fermat::
z^3=x^3+y^3.
The second generation of the Fermat equation generated by the following two formulas
First
z^3=[z(z+1)/2]^2 - [z(z-1)/2]^2
Second
x^3+y^3=[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]
After root Fermat was altered ,it born a systems composed of many small equations :
[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].
........
The require about absolute integer is impossible !
Ishtar .
good job
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nemoj me JEBAT da je ovo tako dobro
goodbye... gooodbye... goodbye
Drama Queen Magnet
DUBAI :D
they want efx...no?
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