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CTK Exchange
neat_maths
Member since Aug-22-03
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Oct-12-09, 10:00 AM (EST) |
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"Sophie Germain and Fermat's Last Theorem"
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i) I can prove that all the prime factors of (x^n + y^n) / (x+y), where x, y, (x+y) and n are mutually prime and n is a prime >2
are of the form p = 1+ 2*n*k where k is some positive integer
ii) I can prove that (x+y) is mutually prime to (x^n + y^n) / (x+y) with x, y and n as defined above
which means that (x+y) is mutually prime to all p and (-x/y) mod(p) cannot equal 1 mod(p)
iii) In the equation x^n + y^n + z^n = 0, where x, y, z and n are mutually prime and n is any prime > 2,
I can prove that (-x/y)^n = 1 mod(p) where p is any (or every) prime factor of z not contained in (x+y)
This means that (-x/y) is the nth root of 1 in mod(p) or (-x/y)^n mod(p) = (a^2k)^n = 1 for all integers a from 1 to (p-1)
iv) as z = 0 mod(p), I can prove that (-x/y) must be an nth power in mod(p) because x and y must be nth powers in mod(p)
Therefore we have 3 statements
1) (-x/y) cannot be 1 in mod(p)
2) (-x/y) = a^(2k) in mod(p) where a is some integer from 1 to (p-1) and (-x/y) is an nth root of 1 mod(p)
3) (-x/y) = b^n in mod(p) where b is some integer from 1 to (p-1) and (-x/y) is an nth power in mod(p)
It is clear that the only integer that can be an nth power mod(p) and an nth root of 1 mod(p) at the same time is 1 mod(p)
Statements 1, 2 and 3 cannot be true at the same time.
This is due to the very nature of a prime number that a^(p-1) = 1 = a^(2nk) in mod(p) where 2nk = (p-1)
This appears to prove Case I of Fermat's last theorem that x^n + y^n + z^n = 0 is impossible, where x, y, z and n are mutually prime, n prime>2
It is along the ideas of Sophie Germain with her definition of an "auxiliary" prime modified to include all the prime factors of x, y and z not included in (x+y), (y+z) or (x+z)
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neat_maths
Member since Aug-22-03
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Oct-15-09, 06:39 AM (EST) |
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4. "Sophie Germain and Fermat's Last Theorem"
In response to message #3
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If n is any prime > 2 and p = 1 + 2*n*k where k is every integer from 1 to infinity which makes p prime, It is clear that the only integer that can be an nth power mod(p) and an nth root of 1 in mod(p) at the same time is 1 mod(p) And 1 mod(p) is precisely the only excluded value. This is so neat I am certain that it is worthy of
"cuius rei demonstrationem mirabilem sane detexi" please consider |
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