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