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# Who is person who solve shortest Fermat?

Updated: 8/20/2019 Wiki User

11y ago

To:

trantancuong21@Yahoo.com

Le dernier théorème de Pierre de Fermat .

(x, y, z, n) l'ensemble de ( N+ )^4.

n> 2.

( a ) l'ensemble de Z

F est la fonction de (a).

F (a) = [a (a +1) / 2] ^ 2

F (0) = 0 et F (-1) = 0.

Considérons deux équations.

F (z) = F (x) + F (y)

F (z-1) = F (x-1) + F (y-1)

Nous avons une inférence chaîne

F (z) = F (x) + F (y) équivalent F (z-1) = F (x-1) + F (y-1)

F (z) = F (x) + F (y) en déduire F (z-1) = F (x-1) + F (y-1)

F (z-x-1) = F (x-x-1) + F (y-x-1) en déduire F (z-x-2) = F (x-x-2) + F (y-x-2)

nous voyons

F (z-x-1) = F (x-x-1) + F (y-x-1)

F (z-x-1) = F (-1) + F (y-x-1)

F (z-x-1) = 0 + F (y-x-1)

donner

z = y

et

F (z-x-2) = F (x-x-2) + F (y-x-2)

F (z-x-2) = F (-2) + F (y-x-2)

F (z-x-2) = 1 + F (y-x-2)

donner z = / = y.

de sorte

F (z-x-1) = F (x-x-1) + F (y-x-1) ne pas en déduire F (z-x-2) = F (x-x-2) + F (y-x-2)

de sorte

F (z) = F (x) + F (y) ne pas en déduire F (z-1) = F (x-1) + F (y-1)

de sorte

F (z) = F (x) + F (y) n'est pas équivalente F (z-1) = F (x-1) + F (y-1)

Donc avoir deux cas.

[F (x) + F (y)] = F (z) et F (x-1) + F (y-1)] = / = F (z-1)

ou vice versa

de sorte

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

Ou

F (x)-F (x-1) + F (y)-F (y-1) = / = F (z)-F (z-1).

nous voyons

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

=(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4).

=x^3.

F(y)-F(y-1) =y^3.

F(z)-F(z-1) =z^3.

de sorte

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

n> 2. . Similaire.

Nous avons une inférence chaîne

G (z) * F (z) = G (x) * F (x) + G (y) * F (y) équivalente G (z) * F (z-1) = G (x) * F (x -1) + G (y) * F (y-1)

G (z) * F (z) = G (x) * F (x) + G (y) * F (y) en déduire G (z) * F (z-1) = G (x) * F (x -1) + G (y) * F (y-1)

G (z) * F (z-x-1) = G (x) * F (x-x-1) + G (y-x-1) * F (y) en déduire G (z) * F (z-x-2) = G ( x) * F (x-x-2) + G (y) * F (y-x-2)

nous voyons

G (z) * F (z-x-1) = G (x) * F (x-x-1) + G (y) * F (y-x-1)

G (z) * F (z-x-1) = G (x) * F (-1) + G (y) * F (y-x-1)

G (z) * F (z-x-1) = 0 + G (y) * F (y-x-1)

donner z = y.

et

G (z) * F (z-x-2) = G (x) * F (x-x-2) + G (y) * F (y-x-2)

G (z) * F (z-x-2) = G (x) * F (-2) + G (y) * F (y-x-2)

G (z) * F (z-x-2) = G (x) + G (y) * F (y-x-2)

x> 0 en déduire G (x)> 0.

donner z = / = y.

de sorte

G (z) * F (zx-1) = G (x) * F (xx-1) + G (yx-1) * F (y) ne pas en déduire G (z) * F (z-x-2) = G (x) * F (x-x-2) + G (y) * F (y-x-2)

de sorte

G (z) * F (z) = G (x) * F (x) + G (y) * F (y) ne pas en déduire G (z) * F (z-1) = G (x) * F (x-1) + G (y) * F (y-1)

de sorte

G (z) * F (z) = G (x) * F (x) + G (y) * F (y) n'est pas équivalente G (z) * F (z-1) = G (x) * F (x-1) + G (y) * F (y-1)

Donc avoir deux cas

[G (x) * F (x) + G (y) * F (y)] = G (z) * F (z) et [G (x) * F (x-1) + G (y) * F (y-1)] = / = G (z-1) * F (z-1)

ou vice versa.

de sorte

[G (x) * F (x) + G (y) * F (y)] - [G (x) * F (x-1) + G (y) * F (y-1)] = / = G (z) * [F (z)-F (z-1)].

Ou

G (x) * [F (x) - F (x-1)] + G (y) * [F (y)-F (y-1)] = / = G (z) * [F (z) -F (z-1).]

nous voyons

x ^ n = G (x) * [F (x)-F (x-1)]

y ^ n = G (y) * [F (y)-F (y-1)]

z ^ n = G (z) * [F (z)-F (z-1)]

de sorte

x ^ n + y ^ n = / = z ^ n

Le bonheur et la paix

Tran Tan Cuong Wiki User

11y ago   Earn +20 pts
Q: Who is person who solve shortest Fermat?
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