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Topic: FLT proof where n | x
Replies: 7   Last Post: Jan 14, 2014 1:45 PM

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Posts: 54
Registered: 1/8/14
FLT proof where n | x
Posted: Jan 8, 2014 7:08 PM
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Fermat;s Last Theorem.

Proof for the case n | x;

Assume (x,y,z)=1?(1)

Assume n is prime; n? yz?(2)

Assume x^n + y^n = z^n ?(3)

Assume n| x?(4)

then z-y = 0 mod n?.(5)

x^n = z^n - y^n =(z-y)(z^(n-1)+ [z^(n-2)]y + [z^(n-3)]y^2 +?+ z^(n-1))?(6)

per Note 1 and Note 2,z-y= n(r^n)?(7)

Per Note 2,(z^(n-1)+ yz^(n-2)+ (y^2)(z^(n-3))+ ? + z^(n-1))
= [n^(2fn-1)](d^n)?(8)

Per Note 2,(z-y)^(n-1)+nyz(z-y)^(n-2)+?+ n[y^((n-1)/2)][z^((n-1)/2)]
= [n^(2fn-1)]d^n?(9)

Now using (7) and modulo n^2 of (9),we get
n[y^((n-1)/2)][z^((n-1)/2)] = 0 mod n^2?(10)

This implies that n | y or n |z and violates (4)

So our assumption of x^n + y^n = z^n is false if n divides x and n is prime.
End of Proof.

Note 1: if (z-y) and (z^(n-1)+z^(n-2) y+?+y^(n-1) ) have a common factor p,then p=n

assume (z-y) and (z^(n-1)+ z^(n-2)y + ? + y^(n-1) ) have common factor p?.(1.1)

But z= y mod p;nz^(n-1)=0 mod p;this implies p divide n or z; or p=n?.(1.2)

p?n as n is prime;p ?z as then p divides z-y and hence p | x which means p |y? (1.3)

If p=n,then n | (y+z) and n | x? (1.4)

Note 2: On (z-y) and (z^(n-1)+z^(n-2) y+?+y^(n-1)) when n | x?

If n |x,then z-y=0 mod n? (2.1)

(z^(n-1)+z^(n-2) y+?+y^(n-1) )=ny^(n-1) mod n=0 mod n.

But does n^2 divide y+z?

For this let us take a step back;
z^3 - y^3=(z-y)((z-y)^2 + 3yz))? (2.2)

z^5 - y^5 =(z-y)((z-y)^4 + 5yz(z-y)^2 + 5y^2 z^2 )?(2.3)

z^7 - y^7 =(z-y)((z-y)^6 + 7yz(z-y)^4 + 14y^2 z^2(z-y)^2 +7y^3 z^3 )? (2.4)

Then based on pattern seen in (2.2),(2.3) and (2.4),
(z^(n-1)+yz^(n-2) +?+ y^(n-1))=
(z-y)^(n-1)+ nyz[(z-y)^(n-3)]+ ?+ n[y^((n-1)/2)][z^((n-1)/2)] ... (2.5)
(proof left as an exercise for reader)

Let z-y = 0 mod n^2? (2.6)

Each term of (2.5) is divisible by n^2 except last term:

n[y^((n-1)/2)][z^((n-1)/2))]/n^2 =[y^((n-1)/2)][z^((n-1)/2))]/n?(2.7)

This implies that n|y or n|z, which is false per (2)

So this means that (y+z)=0 mod n is ok

but y+z=0 mod n^2,0 mod n^3 etc lead to contradictions.

Now we can write y+z = nr^n

Only one n is y+z; All the other n come from the second term

(z-y)^(n-1)+nyz(z-y)^(n-2)+?.+ ny^((n-1)/2) z^((n-1)/2)=n^(2fn-1) d^n

where f is a positive integer and gcd(r,d)=1

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