Released by: Giegling Germany Release/catalogue number: FORUM 01 Release date: Sep 16, 2013 All rights reserved for the producers of these tracks.
Пікірлер: 28
@beincheekym88 жыл бұрын
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
@0cer04 ай бұрын
Here, ladies and gentlemen, we are leaving the domain of entertainment music and moving deep into the realm of art. Deep.
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 .