Discussion:
A Proof of Fermat's Last Theorem (Version 59)
robersi
2017-02-18 16:28:26 UTC
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A Proof of Fermat's Last Theorem (Version 59)

assume there is at least one solution for,

(1.) a^p + b^p + c^p = 0,

where a, b, and c are non-zero pairwise co-prime

integers and p >= 3 is an odd integer.

(2.1.1) (a + b) | (a^p + b^p).

(2.1.2) (a + b) | (-c^p).

(2.2.1) (a + c) | (a^p + c^p).

(2.2.2) (a + c) | (-b^p).

(2.3.1) (b + c) | (b^p + c^p).

(2.3.2) (b + c) | (-a^p).

(2.1.2), (2.2.2), and (2.3.2) yield,

(2.4) (a + b)(a + c)(b + c) | (abc)^p.

definition of Y:

(3.1) Y = (b + a)(b + c)(a + c).

(2.4 and 3.1->3.2) Y | (abc)^p.

< snip: I don't understand why the following section (section 8) is unnecessary in this proof.>

Edit: there may be a certain set of general equations (undefined by me) that fall in a catory that make this method of proof effective.

section 8.

expansion of Y:

(8.1) Y = (b^2 + bc + ab + ac)(a+ c).

(8.2) Y = b^2a + b^2c + abc + bc^2 + ba^2 + abc + a^2c + ac^2.

(8.3) Y = ab(a +b) + ac(c + a) + bc(c + b) + 2abc.

(8.4) Y = ab(a + b + c - c) + ac(a + c + b - b) + bc(b + c + a - a) + 2abc.

(8.5) Y = (a + b + c)(ab + ac +bc) - abc.

(8.6) Y = (ab + ac + bc)(a + b + c) - abc.

end of section 8.

definition of X:

(5.1) X = Y + abc.

(3.2 and 5.1->5.2) (X - abc) | (abc)^p.

let B be such that,

(6.1) B = gcd(X, abc).

let F be such that,

(6.2) BF = X.

(5.3 and 6.1->6.3) (X/B - abc/B) | (abc)^p.

(6.3 and 6.2->6.4) (F - abc/B) | (abc)^p.

(6.1 and 6.2->6.5) 1 =gcd(F, (abc)/B) .

(6.5->6.6) 1 = gcd((F - (abc)/B), (abc)/B).

(6.6->6.7) 1 = gcd((F - (abc)/B), (abc)^p/B).

clearly,

(6.8) (abc)^p/B | (abc)^p/B,

and,

(6.9) (abc)^p/B = (abc)^p/B.

This may seem as if the givens have little (are independent) of the proof. I conjecture instep, that the general class of equations yield similar if not identical proofs in which (or whereby),

(U - V) | W,

GCD(U,V) = 1

except THAT, V | W, U | W (and also UV | W),

where |UV| > W. (oops, contradiction).

There may exist a given set of equations that can be proved false under this general method of proof.

(6.7, 6.8 and 6.4 ->6.10)

[(F - abc/B)(abc)^p/B] | (abc)^p/B.

(6.10 -> 6.11) [(F - abc/B)(abc)^p/B] > (abc)^p/B.

(recognizing Y=/= 0 yields a contradiction)

End of proof.

Simon C. Roberts.
a***@gmail.com
2017-02-18 17:20:32 UTC
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I lost it much earlier; can you give a precis?
Post by robersi
[(F - abc/B)(abc)^p/B] | (abc)^p/B.
(6.10 -> 6.11) [(F - abc/B)(abc)^p/B] > (abc)^p/B.
(recognizing Y=/= 0 yields a contradiction)
End of proof.
Simon C. Roberts.
robersi
2017-02-18 18:49:15 UTC
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A Proof of Fermat's Last Theorem (Version 60)

assume there is at least one solution for,

(1.) a^p + b^p + c^p = 0,

where a, b, and c are non-zero pairwise co-prime

integers and p >= 3 is an odd integer.

(2.1.1) (a + b) | (a^p + b^p).

(2.1.2) (a + b) | (-c^p).

(2.2.1) (a + c) | (a^p + c^p).

(2.2.2) (a + c) | (-b^p).

(2.3.1) (b + c) | (b^p + c^p).

(2.3.2) (b + c) | (-a^p).

(2.1.2), (2.2.2), and (2.3.2) yield,

(2.4) (a + b)(a + c)(b + c) | (abc)^p.

definition of Y:

(3.1) Y = (b + a)(b + c)(a + c).

(2.4 and 3.1->3.2) Y | (abc)^p.

note.

< snip: I don't understand why the following section (section 8) is unnecessary in this proof.>

end note.

section 8.

expansion of Y:

(8.1) Y = (b^2 + bc + ab + ac)(a+ c).

(8.2) Y = b^2a + b^2c + abc + bc^2 + ba^2 + abc + a^2c + ac^2.

(8.3) Y = ab(a +b) + ac(c + a) + bc(c + b) + 2abc.

(8.4) Y = ab(a + b + c - c) + ac(a + c + b - b) + bc(b + c + a - a) + 2abc.

(8.5) Y = (a + b + c)(ab + ac +bc) - abc.

(8.6) Y = (ab + ac + bc)(a + b + c) - abc.

end of section 8.

definition of X:

(5.1) X = Y + abc.

(3.2 and 5.1->5.2) (X - abc) | (abc)^p.

let B be such that,

(6.1) B = gcd(X, abc).

let F be such that,

(6.2) BF = X.

(5.3 and 6.1->6.3) (X/B - abc/B) | (abc)^p.

(6.3 and 6.2->6.4) (F - abc/B) | (abc)^p.

(6.1 and 6.2->6.5) 1 =gcd(F, (abc)/B) .

(6.5->6.6) 1 = gcd((F - (abc)/B), (abc)/B).

(6.6->6.7) 1 = gcd((F - (abc)/B), (abc)^p/B).

clearly,

(6.8) (abc)^p/B | (abc)^p/B,

and,

(6.9) (abc)^p/B = (abc)^p/B.

note.

This may seem as if the givens have little (are independent) of the proof. I conjecture therefore, that the general class of equations yield simalar if not identical proofs in which (or whereby),

(U - V) | V^m

GCD(U,V) = 1,

GCD(U,V^m) = 1

except THAT,

V^m | V^m and U | V^m => UV^m | V^m,

=> |UV^m| =< |V^m| except,

This outline of a method can be genralised much further.

There may exist a given set of equations that can be proved false under this general method or more general method of proof.

end note.

(6.7, 6.8 and 6.4 ->6.10) [(F - abc/B)(abc)^p/B] | (abc)^p/B.

(6.10 -> 6.11) |(F - abc/B)(abc)^p/B| > |(abc)^p/B|.

recognizing Y/B = (F - abc/B) =/= 0 yields the contradiction,

(7.1) |(F - abc/B)(abc)^p/B| < |(abc)^p/B|.

End of proof.

Simon C. Roberts
***@gmail.com
robersi
2017-02-18 19:55:15 UTC
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some extras:

http://math.stackexchange.com/users/286478/simon-roberts

http://mathforum.org/kb/forumcategory.jspa?categoryID=16

https://www.dropbox.com/sh/c6feaogrtmiggd8/AACvVc1WCz5h_x1YnOJciHK6a?dl=0

robersi
2017-02-18 21:31:27 UTC
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All proofs in this thread, at least, are not complete, and therefore not correct.
Simon Roberts
2017-02-22 10:06:16 UTC
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Post by robersi
A Proof of Fermat's Last Theorem (Version 59)
WRONG.
Post by robersi
assume there is at least one solution for,
(1.) a^p + b^p + c^p = 0,
where a, b, and c are non-zero pairwise co-prime
integers and p >= 3 is an odd integer.
(2.1.1) (a + b) | (a^p + b^p).
(2.1.2) (a + b) | (-c^p).
(2.2.1) (a + c) | (a^p + c^p).
(2.2.2) (a + c) | (-b^p).
(2.3.1) (b + c) | (b^p + c^p).
(2.3.2) (b + c) | (-a^p).
(2.1.2), (2.2.2), and (2.3.2) yield,
(2.4) (a + b)(a + c)(b + c) | (abc)^p.
(3.1) Y = (b + a)(b + c)(a + c).
(2.4 and 3.1->3.2) Y | (abc)^p.
< snip: I don't understand why the following section (section 8) is unnecessary in this proof.>
Edit: there may be a certain set of general equations (undefined by me) that fall in a catory that make this method of proof effective.
section 8.
(8.1) Y = (b^2 + bc + ab + ac)(a+ c).
(8.2) Y = b^2a + b^2c + abc + bc^2 + ba^2 + abc + a^2c + ac^2.
(8.3) Y = ab(a +b) + ac(c + a) + bc(c + b) + 2abc.
(8.4) Y = ab(a + b + c - c) + ac(a + c + b - b) + bc(b + c + a - a) + 2abc.
(8.5) Y = (a + b + c)(ab + ac +bc) - abc.
(8.6) Y = (ab + ac + bc)(a + b + c) - abc.
end of section 8.
(5.1) X = Y + abc.
(3.2 and 5.1->5.2) (X - abc) | (abc)^p.
let B be such that,
(6.1) B = gcd(X, abc).
let F be such that,
(6.2) BF = X.
(5.3 and 6.1->6.3) (X/B - abc/B) | (abc)^p.
(6.3 and 6.2->6.4) (F - abc/B) | (abc)^p.
(6.1 and 6.2->6.5) 1 =gcd(F, (abc)/B) .
(6.5->6.6) 1 = gcd((F - (abc)/B), (abc)/B).
(6.6->6.7) 1 = gcd((F - (abc)/B), (abc)^p/B).
clearly,
(6.8) (abc)^p/B | (abc)^p/B,
and,
(6.9) (abc)^p/B = (abc)^p/B.
This may seem as if the givens have little (are independent) of the proof. I conjecture instep, that the general class of equations yield similar if not identical proofs in which (or whereby),
(U - V) | W,
GCD(U,V) = 1
except THAT, V | W, U | W (and also UV | W),
where |UV| > W. (oops, contradiction).
There may exist a given set of equations that can be proved false under this general method of proof.
(6.7, 6.8 and 6.4 ->6.10)
[(F - abc/B)(abc)^p/B] | (abc)^p/B.
(6.10 -> 6.11) [(F - abc/B)(abc)^p/B] > (abc)^p/B.
(recognizing Y=/= 0 yields a contradiction)
End of proof.
Simon C. Roberts.
Simon Roberts
2018-02-13 05:09:07 UTC
Raw Message
Post by Simon Roberts
Post by robersi
A Proof of Fermat's Last Theorem (Version 59)
WRONG.
Post by robersi
assume there is at least one solution for,
(1.) a^p + b^p + c^p = 0,
where a, b, and c are non-zero pairwise co-prime
integers and p >= 3 is an odd integer.
(2.1.1) (a + b) | (a^p + b^p).
(2.1.2) (a + b) | (-c^p).
(2.2.1) (a + c) | (a^p + c^p).
(2.2.2) (a + c) | (-b^p).
(2.3.1) (b + c) | (b^p + c^p).
(2.3.2) (b + c) | (-a^p).
(2.1.2), (2.2.2), and (2.3.2) yield,
(2.4) (a + b)(a + c)(b + c) | (abc)^p.
(3.1) Y = (b + a)(b + c)(a + c).
(2.4 and 3.1->3.2) Y | (abc)^p.
< snip: I don't understand why the following section (section 8) is unnecessary in this proof.>
Edit: there may be a certain set of general equations (undefined by me) that fall in a catory that make this method of proof effective.
section 8.
(8.1) Y = (b^2 + bc + ab + ac)(a+ c).
(8.2) Y = b^2a + b^2c + abc + bc^2 + ba^2 + abc + a^2c + ac^2.
(8.3) Y = ab(a +b) + ac(c + a) + bc(c + b) + 2abc.
(8.4) Y = ab(a + b + c - c) + ac(a + c + b - b) + bc(b + c + a - a) + 2abc.
(8.5) Y = (a + b + c)(ab + ac +bc) - abc.
(8.6) Y = (ab + ac + bc)(a + b + c) - abc.
end of section 8.
(5.1) X = Y + abc.
(3.2 and 5.1->5.2) (X - abc) | (abc)^p.
let B be such that,
(6.1) B = gcd(X, abc).
let F be such that,
(6.2) BF = X.
(5.3 and 6.1->6.3) (X/B - abc/B) | (abc)^p.
(6.3 and 6.2->6.4) (F - abc/B) | (abc)^p.
(6.1 and 6.2->6.5) 1 =gcd(F, (abc)/B) .
(6.5->6.6) 1 = gcd((F - (abc)/B), (abc)/B).
(6.6->6.7) 1 = gcd((F - (abc)/B), (abc)^p/B).
clearly,
(6.8) (abc)^p/B | (abc)^p/B,
and,
(6.9) (abc)^p/B = (abc)^p/B.
This may seem as if the givens have little (are independent) of the proof. I conjecture instep, that the general class of equations yield similar if not identical proofs in which (or whereby),
(U - V) | W,
GCD(U,V) = 1
except THAT, V | W, U | W (and also UV | W),
where |UV| > W. (oops, contradiction).
There may exist a given set of equations that can be proved false under this general method of proof.
(6.7, 6.8 and 6.4 ->6.10)
[(F - abc/B)(abc)^p/B] | (abc)^p/B.
(6.10 -> 6.11) [(F - abc/B)(abc)^p/B] > (abc)^p/B.
(recognizing Y=/= 0 yields a contradiction)
End of proof.
Simon C. Roberts.
the error is at (6.7) can't do that.