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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
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