Matematik

Sophie Germains bevisforsøg for Fermats sidste sætning
Jeg skal ikke forklare nogle lemma eller hjælpesætninger, dem skal jeg bare konstatere, men jeg skal kunne forklare hvad der sker fra skridt til skridt, altså f.eks. fra punkt 4 til 5, hvordan udtrykke kommer fra det ene til det andet. Matematik dele skal ca. fylde 10 sider, og har ikke rigtigt nogen fornemmelse af, hvor meget dette nedenfor vil fylde når jeg skriver det ind, men som udgangspunkt vil jeg tro det er nok med de første 10 punkter (måske er det alt for meget?). Nedenfor er der det link, hvor jeg har fundet beviset. 
http://fermatslasttheorem.blogspot.com/2005/08/sophies-proof.html
meningen var, at man skulle kunne sige: "man kommer fra ___ til ___ fordi det er denne potensregel som siger at... og ved at omskrive bliver det til... og så kan man forkorte således at...." og på den måde skulle jeg kunne forklare hvad der sker fra skridt til skridt.


Theorem: If xn + yn = zn and n is a prime ≥ 3 and 2n+1 is a prime, then n must divide xyz.
1.    We can assume that x,y,z are relatively prime to each other. [See here]

2.    So, let's assume that n doesn't divide xyz.

3.    Since n is odd, we can set z' to -z and get:
xn + yn + (z')n = 0.

4.    We can rewrite this to be [see here for details]
-xn = (y+z)(y(n-1) - y(n-2)z + ... - yz(n-2) + z(n-1))

5.    From this, we can show that:
y+z and y(n-1) - y(n-2)z + ... - yz(n-2) + z(n-1) are relatively prime since:

a.    Assume that they are not relatively prime.

b.    Then, there exists a prime p such that p divides y+z and p divides y(n-1) - y(n-2)z + ... - yz(n-2) + z(n-1).

c.    from this, we know that z ≡ -y (mod p).

d.    Using the fact that they are congruent (see here), we get:
y(n-1) - y(n-2)z + ... - yz(n-2) + z(n-1) ≡
y(n-1) - y(n-2)(-y) + ... -y(-y)(n-2) + (-y)(n-1) ≡
y(n-1) + y(n-1) + ... + y(n-1) + y(n-1) ≡ (n)y(n-1)
e.    So from Euclid's Lemma, p either divides n or it divides y(n-1).

f.    But p doesn't divide n since:

(i)    n is prime so p would equal n (a prime is only divisible by itself or 1)
(ii) And then n would divide (-x)n
(iii) Which means that it would divide x (from Euclid's Generalized Lemma)
(iv) But this impossible by step #2.
g.    And p doesn't divide y(n-1) because then p would divide y (Euclid's Generalized Lemma) and p would also divide z (since it divides y and it divides y+z), but this is impossible by step #1.
(h) So we have a contradiction and we reject our assumption.

6.    From Step #5 (see here), we can conclude that there exists a such that:
y + z = an

7.    But we can make this same argument for z+x and x + y where:
-yn = (x + z)(x(n-1) - x(n-2)z + ... -xz(n-2) + z(n-1))
-zn = (x + y)(x(n-1) - x(n-2)y + ... -xy(n-2) + y(n-1))

8.    So we know that there exists b,c where:
z + x = bn
x + y = cn

9.    We also know that any value un ≡ ±1 or 0 (mod 2n + 1) since:
a.    Assume that 2n+1 doesn't divide un [ we only need to consider the case where it is not congruent to 0]

b.    Since 2n+1 is a prime, we can apply Fermat's Little Theorem to get:
u2n ≡ 1 (mod 2n+1)

c.    And this means that:
(un)2 ≡ 1 (mod 2n + 1)

d.    So that un ≡ ± 1 (mod 2n+1) [See here for review of modular arithmetic]