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Who do understand Fermat?

Updated: 8/20/2019
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12y ago

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ultimo teorema di Pierre De Fermat

(x,y,z, n) elemento della (N +)^ 4

n> 2

(a) elemento della Z

F è la funzione (a).

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

F (0) = 0 e F (-1) = 0

Si considerino due equazioni.

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

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

Abbiamo una catena di inferenza

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

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

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

vediamo

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)

conclusione

z = y

e

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)

conclusione

z = / = y.

conclusione

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

conclusione

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

conclusione

F (z) = F (x) + F (y) non sono equivalenti di F (z-1) = F (x-1) + F (y-1)

Pertanto, i due casi.

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

o viceversa

conclusione

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

Or.

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

vediamo

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

= (X ^ 4 +2 x ^ 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

conclusione

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

n> 2. risolvere simili

Abbiamo una catena di inferenza

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

vediamo

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)

conclusione

z = y.

e

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 (x-y-2)

x> 0 conclusioni G (x)> 0

conclusione

z = / = y.

conclusione

G (z) * F (z-x-1) = G (x) * F (x-x-1) + G (y-x-1) * F (y) alcuna conclusione G (z) * F (z-x-2) = G (x) * F (x-x-2) + G (y) * F (y-x-2)

conclusione

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

conclusione

G (z) * F (z) = G (x) * F (x) + G (y) * F (y) non sono equivalenti di G (z) * F (z-1) = G (x) * F ( x-1) + G (y) * F (y-1)

Pertanto, i due casi.

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

o viceversa

conclusione

[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)].

o

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

vediamo

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)]

conclusione

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

Felicità e la pace

Cuong Tran

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