Discussion:
Proof of Fermat's Last Theorem.
(too old to reply)
Simon Roberts
2016-08-23 22:33:06 UTC
Permalink
Version (2016.08.23:18:31)

August, 23, 2016 approx: 6:30 pm (New York)

Fermat's Last Theorem
(proof by contradiction)
--------------------------------------------------
Assume,

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

(1.a) a, b, and c are non-zero, pairwise co-prime.

(1.b) p is an odd prime.

-----------------------------------------

(2.a) (a+b) || (a^p + b^p = -c^p).

(2.b) (a+c) || (a^p + c^p = -b^p).

(2.c) (b+c) || (b^p + c^p = -a^p).

From (2.a), (2.b), and (2.c)

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

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

(5.) Y || (abc)^p

---------------------------------------------------------

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

Y = b^2a + b^2c + abc + bc^2 + ba^2 + abc + a^2c + ac^2

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

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

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

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

(7.) Make X = (ab + ac + bc)(a + b + c) = Y + abc

(8.) (X - abc) || (abc)^p

---------------------------------------------

Assume X || (abc)^p

XR = (abc)^p for some R

[(abc)^p]/R - abc = (abc)^p

[(abc)^p] - (abc)R = [(abc)^p]R

[(abc)^p][1 - R] = (abc)R

[(abc)^p][1 - R]X = (abc)RX

[(abc)^p][1 - R]X = (abc)RX

XR = (abc)^p

[1 - R]X = (abc)

X || (abc)


(7.) X = (ab + ac + bc)(a + b + c) = Y + abc

X = Y + abc

(5.) Y || (abc)^p

Y + abc || (abc)

|Y + abc| =< |abc|

---------------------------

If |Y + abc| = |abc|

then Y + abc = +/- abc

but, Y =/= 0

so, Y = -2abc

----------------------

If |Y + abc| < |abc|

then

-2abc < Y < 0

2abc > |Y| > 0

Recall,

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

|Y| = |(b+a)(b+c)(a+c)| < 2abc

--------------------------------

but,

It is always a case that,

2abc < 0

contradiction.

(10.) X !| (abc)^p

--------------------------------------------

(10.) X !| (abc)^p

(5.) Y || (abc)^p

(7.) X = (ab + ac + bc)(a + b + c) = Y + abc

X = Y + abc

(X - Y = abc) || (abc)^p

(10.) and (5.) => (11.)

(11.) (X - Y) !| (abc)^p

(12.) (X -Y) || (abc)^p

Contradiction.

The assumption(s) of (1.), (1.a) and/or (1.b) are false.

-Simon Roberts
***@gmail.com
quasi
2016-08-23 23:06:04 UTC
Permalink
Post by Simon Roberts
Version (2016.08.23:18:31)
August, 23, 2016 approx: 6:30 pm (New York)
Fermat's Last Theorem
(proof by contradiction)
--------------------------------------------------
Assume,
(1.) a^p + b^p + c^p = 0,
(1.a) a, b, and c are non-zero, pairwise co-prime.
(1.b) p is an odd prime.
-----------------------------------------
(2.a) (a+b) || (a^p + b^p = -c^p).
(2.b) (a+c) || (a^p + c^p = -b^p).
(2.c) (b+c) || (b^p + c^p = -a^p).
From (2.a), (2.b), and (2.c)
(3.) (a+b)(a+c)(b+c) || (abc)^p
Incorrect use of the symbol "||".
Post by Simon Roberts
(4.) Y = (b+a)(b+c)(a+c)
(5.) Y || (abc)^p
Incorrect use of the symbol "||".
Post by Simon Roberts
---------------------------------------------------------
Y = (b^2 + bc + ab + ac)(a+ c)
Y = b^2a + b^2c + abc + bc^2 + ba^2 + abc + a^2c + ac^2
Y = ab(a +b) + ac(c + a) + bc(c + b) + 2abc
Y = ab(a + b + c - c) + ac(a + c + b - b) + bc(b + c + a - a) + 2abc
Y = (a + b + c)(ab + ac +bc) - abc
(6.) Y = (ab + ac + bc)(a + b + c) - abc
(7.) Make X = (ab + ac + bc)(a + b + c) = Y + abc
(8.) (X - abc) || (abc)^p
Incorrect use of the symbol "||".

The notation u|v means v = u*w for some integer w.

The notation (u^k) || v means (u^k)|v but u^(k+1) does not
divide v.
Post by Simon Roberts
---------------------------------------------
Assume X || (abc)^p
Once again, your use of the symbol "||" is not correct.
Post by Simon Roberts
XR = (abc)^p for some R
[(abc)^p]/R - abc = (abc)^p
False claim.

You can't claim equality.

What you can claim is:

(((abc)^p)/R - abc) | (abc)^p

quasi
Pfsszxt
2016-08-24 13:34:42 UTC
Permalink
Post by Simon Roberts
Version (2016.08.23:18:31)
August, 23, 2016 approx: 6:30 pm (New York)
Fermat's Last Theorem
(proof by contradiction)
--------------------------------------------------
Assume,
(1.) a^p + b^p + c^p = 0,
(1.a) a, b, and c are non-zero, pairwise co-prime.
(1.b) p is an odd prime.
-----------------------------------------
--------------
Post by Simon Roberts
-Simon Roberts
************************
Now let someone who knows some math show you your error(s)!
Peter Percival
2016-08-24 13:49:41 UTC
Permalink
Post by Simon Roberts
Post by Simon Roberts
Version (2016.08.23:18:31)
August, 23, 2016 approx: 6:30 pm (New York)
Fermat's Last Theorem
(proof by contradiction)
--------------------------------------------------
Assume,
(1.) a^p + b^p + c^p = 0,
(1.a) a, b, and c are non-zero, pairwise co-prime.
(1.b) p is an odd prime.
-----------------------------------------
--------------
Post by Simon Roberts
-Simon Roberts
************************
Now let someone who knows some math show you your error(s)!
Mr quasi--with commendable patience--is doing so.
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
Simon
2016-08-24 16:49:47 UTC
Permalink
Post by Simon Roberts
Version (2016.08.23:18:31)
August, 23, 2016 approx: 6:30 pm (New York)
Fermat's Last Theorem
(proof by contradiction)
--------------------------------------------------
Assume,
(1.) a^p + b^p + c^p = 0,
(1.a) a, b, and c are non-zero, pairwise co-prime.
(1.b) p is an odd prime.
-----------------------------------------
(2.a) (a+b) || (a^p + b^p = -c^p).
(2.b) (a+c) || (a^p + c^p = -b^p).
(2.c) (b+c) || (b^p + c^p = -a^p).
From (2.a), (2.b), and (2.c)
(3.) (a+b)(a+c)(b+c) || (abc)^p
(4.) Y = (b+a)(b+c)(a+c)
(5.) Y || (abc)^p
---------------------------------------------------------
Y = (b^2 + bc + ab + ac)(a+ c)
Y = b^2a + b^2c + abc + bc^2 + ba^2 + abc + a^2c + ac^2
Y = ab(a +b) + ac(c + a) + bc(c + b) + 2abc
Y = ab(a + b + c - c) + ac(a + c + b - b) + bc(b + c + a - a) + 2abc
Y = (a + b + c)(ab + ac +bc) - abc
(6.) Y = (ab + ac + bc)(a + b + c) - abc
(7.) Make X = (ab + ac + bc)(a + b + c) = Y + abc
(8.) (X - abc) || (abc)^p
---------------------------------------------
Assume X || (abc)^p
XR = (abc)^p for some R
[(abc)^p]/R - abc = (abc)^p
[(abc)^p] - (abc)R = [(abc)^p]R
[(abc)^p][1 - R] = (abc)R
[(abc)^p][1 - R]X = (abc)RX
[(abc)^p][1 - R]X = (abc)RX
XR = (abc)^p
[1 - R]X = (abc)
X || (abc)
(7.) X = (ab + ac + bc)(a + b + c) = Y + abc
X = Y + abc
(5.) Y || (abc)^p
Y + abc || (abc)
|Y + abc| =< |abc|
---------------------------
If |Y + abc| = |abc|
then Y + abc = +/- abc
but, Y =/= 0
so, Y = -2abc
----------------------
If |Y + abc| < |abc|
then
-2abc < Y < 0
2abc > |Y| > 0
Recall,
(4.) Y = (b+a)(b+c)(a+c)
|Y| = |(b+a)(b+c)(a+c)| < 2abc
--------------------------------
but,
It is always a case that,
2abc < 0
contradiction.
(10.) X !| (abc)^p
--------------------------------------------
(10.) X !| (abc)^p
(5.) Y || (abc)^p
(7.) X = (ab + ac + bc)(a + b + c) = Y + abc
X = Y + abc
(X - Y = abc) || (abc)^p
(10.) and (5.) => (11.)
(11.) (X - Y) !| (abc)^p
(12.) (X -Y) || (abc)^p
Contradiction.
The assumption(s) of (1.), (1.a) and/or (1.b) are false.
-Simon Roberts
This version is garbage.
Simon Roberts
2016-11-24 14:41:59 UTC
Permalink
Post by Simon
Post by Simon Roberts
Version (2016.08.23:18:31)
August, 23, 2016 approx: 6:30 pm (New York)
Fermat's Last Theorem
(proof by contradiction)
--------------------------------------------------
Assume,
(1.) a^p + b^p + c^p = 0,
(1.a) a, b, and c are non-zero, pairwise co-prime.
(1.b) p is an odd prime.
-----------------------------------------
(2.a) (a+b) || (a^p + b^p = -c^p).
(2.b) (a+c) || (a^p + c^p = -b^p).
(2.c) (b+c) || (b^p + c^p = -a^p).
From (2.a), (2.b), and (2.c)
(3.) (a+b)(a+c)(b+c) || (abc)^p
(4.) Y = (b+a)(b+c)(a+c)
(5.) Y || (abc)^p
---------------------------------------------------------
Y = (b^2 + bc + ab + ac)(a+ c)
Y = b^2a + b^2c + abc + bc^2 + ba^2 + abc + a^2c + ac^2
Y = ab(a +b) + ac(c + a) + bc(c + b) + 2abc
Y = ab(a + b + c - c) + ac(a + c + b - b) + bc(b + c + a - a) + 2abc
Y = (a + b + c)(ab + ac +bc) - abc
(6.) Y = (ab + ac + bc)(a + b + c) - abc
(7.) Make X = (ab + ac + bc)(a + b + c) = Y + abc
(8.) (X - abc) || (abc)^p
---------------------------------------------
Assume X || (abc)^p
XR = (abc)^p for some R
[(abc)^p]/R - abc = (abc)^p
[(abc)^p] - (abc)R = [(abc)^p]R
[(abc)^p][1 - R] = (abc)R
[(abc)^p][1 - R]X = (abc)RX
[(abc)^p][1 - R]X = (abc)RX
XR = (abc)^p
[1 - R]X = (abc)
X || (abc)
(7.) X = (ab + ac + bc)(a + b + c) = Y + abc
X = Y + abc
(5.) Y || (abc)^p
Y + abc || (abc)
|Y + abc| =< |abc|
---------------------------
If |Y + abc| = |abc|
then Y + abc = +/- abc
but, Y =/= 0
so, Y = -2abc
----------------------
If |Y + abc| < |abc|
then
-2abc < Y < 0
2abc > |Y| > 0
Recall,
(4.) Y = (b+a)(b+c)(a+c)
|Y| = |(b+a)(b+c)(a+c)| < 2abc
--------------------------------
but,
It is always a case that,
2abc < 0
contradiction.
(10.) X !| (abc)^p
--------------------------------------------
(10.) X !| (abc)^p
(5.) Y || (abc)^p
(7.) X = (ab + ac + bc)(a + b + c) = Y + abc
X = Y + abc
(X - Y = abc) || (abc)^p
(10.) and (5.) => (11.)
(11.) (X - Y) !| (abc)^p
(12.) (X -Y) || (abc)^p
Contradiction.
The assumption(s) of (1.), (1.a) and/or (1.b) are false.
-Simon Roberts
This version is garbage.
Happy Thanksgiving. This one takes the cake. It's really good. That's what campbell's soup is, MM MM good. Pi on my face.

Simon C. Roberts
***@gmail.com
Simon Roberts
2016-11-24 14:45:15 UTC
Permalink
Post by Simon Roberts
Version (2016.08.23:18:31)
August, 23, 2016 approx: 6:30 pm (New York)
Fermat's Last Theorem
(proof by contradiction)
--------------------------------------------------
Assume,
(1.) a^p + b^p + c^p = 0,
(1.a) a, b, and c are non-zero, pairwise co-prime.
(1.b) p is an odd prime.
-----------------------------------------
(2.a) (a+b) || (a^p + b^p = -c^p).
(2.b) (a+c) || (a^p + c^p = -b^p).
(2.c) (b+c) || (b^p + c^p = -a^p).
From (2.a), (2.b), and (2.c)
(3.) (a+b)(a+c)(b+c) || (abc)^p
(4.) Y = (b+a)(b+c)(a+c)
(5.) Y || (abc)^p
---------------------------------------------------------
Y = (b^2 + bc + ab + ac)(a+ c)
Y = b^2a + b^2c + abc + bc^2 + ba^2 + abc + a^2c + ac^2
Y = ab(a +b) + ac(c + a) + bc(c + b) + 2abc
Y = ab(a + b + c - c) + ac(a + c + b - b) + bc(b + c + a - a) + 2abc
Y = (a + b + c)(ab + ac +bc) - abc
(6.) Y = (ab + ac + bc)(a + b + c) - abc
(7.) Make X = (ab + ac + bc)(a + b + c) = Y + abc
(8.) (X - abc) || (abc)^p
---------------------------------------------
Assume X || (abc)^p
XR = (abc)^p for some R
[(abc)^p]/R - abc = (abc)^p
[(abc)^p] - (abc)R = [(abc)^p]R
[(abc)^p][1 - R] = (abc)R
[(abc)^p][1 - R]X = (abc)RX
[(abc)^p][1 - R]X = (abc)RX
XR = (abc)^p
[1 - R]X = (abc)
X || (abc)
(7.) X = (ab + ac + bc)(a + b + c) = Y + abc
X = Y + abc
(5.) Y || (abc)^p
Y + abc || (abc)
|Y + abc| =< |abc|
---------------------------
If |Y + abc| = |abc|
then Y + abc = +/- abc
but, Y =/= 0
so, Y = -2abc
----------------------
If |Y + abc| < |abc|
then
-2abc < Y < 0
2abc > |Y| > 0
Recall,
(4.) Y = (b+a)(b+c)(a+c)
|Y| = |(b+a)(b+c)(a+c)| < 2abc
--------------------------------
but,
It is always a case that,
2abc < 0
contradiction.
(10.) X !| (abc)^p
--------------------------------------------
(10.) X !| (abc)^p
(5.) Y || (abc)^p
(7.) X = (ab + ac + bc)(a + b + c) = Y + abc
X = Y + abc
(X - Y = abc) || (abc)^p
(10.) and (5.) => (11.)
(11.) (X - Y) !| (abc)^p
(12.) (X -Y) || (abc)^p
Contradiction.
The assumption(s) of (1.), (1.a) and/or (1.b) are false.
-Simon Roberts
This one is perfect. Sincere.

Simon
Simon Roberts
2016-11-25 22:56:32 UTC
Permalink
Post by Simon Roberts
Post by Simon Roberts
Version (2016.08.23:18:31)
August, 23, 2016 approx: 6:30 pm (New York)
Fermat's Last Theorem
(proof by contradiction)
--------------------------------------------------
Assume,
(1.) a^p + b^p + c^p = 0,
(1.a) a, b, and c are non-zero, pairwise co-prime.
(1.b) p is an odd prime.
-----------------------------------------
(2.a) (a+b) || (a^p + b^p = -c^p).
(2.b) (a+c) || (a^p + c^p = -b^p).
(2.c) (b+c) || (b^p + c^p = -a^p).
From (2.a), (2.b), and (2.c)
(3.) (a+b)(a+c)(b+c) || (abc)^p
(4.) Y = (b+a)(b+c)(a+c)
(5.) Y || (abc)^p
---------------------------------------------------------
Y = (b^2 + bc + ab + ac)(a+ c)
Y = b^2a + b^2c + abc + bc^2 + ba^2 + abc + a^2c + ac^2
Y = ab(a +b) + ac(c + a) + bc(c + b) + 2abc
Y = ab(a + b + c - c) + ac(a + c + b - b) + bc(b + c + a - a) + 2abc
Y = (a + b + c)(ab + ac +bc) - abc
(6.) Y = (ab + ac + bc)(a + b + c) - abc
(7.) Make X = (ab + ac + bc)(a + b + c) = Y + abc
(8.) (X - abc) || (abc)^p
---------------------------------------------
Assume X || (abc)^p
XR = (abc)^p for some R
[(abc)^p]/R - abc = (abc)^p
[(abc)^p] - (abc)R = [(abc)^p]R
[(abc)^p][1 - R] = (abc)R
[(abc)^p][1 - R]X = (abc)RX
[(abc)^p][1 - R]X = (abc)RX
XR = (abc)^p
[1 - R]X = (abc)
X || (abc)
(7.) X = (ab + ac + bc)(a + b + c) = Y + abc
X = Y + abc
(5.) Y || (abc)^p
Y + abc || (abc)
|Y + abc| =< |abc|
---------------------------
If |Y + abc| = |abc|
then Y + abc = +/- abc
but, Y =/= 0
so, Y = -2abc
----------------------
If |Y + abc| < |abc|
then
-2abc < Y < 0
2abc > |Y| > 0
Recall,
(4.) Y = (b+a)(b+c)(a+c)
|Y| = |(b+a)(b+c)(a+c)| < 2abc
--------------------------------
but,
It is always a case that,
2abc < 0
contradiction.
(10.) X !| (abc)^p
--------------------------------------------
(10.) X !| (abc)^p
(5.) Y || (abc)^p
(7.) X = (ab + ac + bc)(a + b + c) = Y + abc
X = Y + abc
(X - Y = abc) || (abc)^p
(10.) and (5.) => (11.)
(11.) (X - Y) !| (abc)^p
(12.) (X -Y) || (abc)^p
Contradiction.
The assumption(s) of (1.), (1.a) and/or (1.b) are false.
-Simon Roberts
This one is perfect. Sincere.
Simon
wrong, incomplete and incorrect, has potetial
Simon Roberts
2018-02-13 00:50:03 UTC
Permalink
Post by Simon Roberts
Version (2016.08.23:18:31)
August, 23, 2016 approx: 6:30 pm (New York)
Fermat's Last Theorem
(proof by contradiction)
--------------------------------------------------
Assume,
(1.) a^p + b^p + c^p = 0,
(1.a) a, b, and c are non-zero, pairwise co-prime.
(1.b) p is an odd prime.
-----------------------------------------
(2.a) (a+b) || (a^p + b^p = -c^p).
(2.b) (a+c) || (a^p + c^p = -b^p).
(2.c) (b+c) || (b^p + c^p = -a^p).
From (2.a), (2.b), and (2.c)
(3.) (a+b)(a+c)(b+c) || (abc)^p
(4.) Y = (b+a)(b+c)(a+c)
(5.) Y || (abc)^p
---------------------------------------------------------
Y = (b^2 + bc + ab + ac)(a+ c)
Y = b^2a + b^2c + abc + bc^2 + ba^2 + abc + a^2c + ac^2
Y = ab(a +b) + ac(c + a) + bc(c + b) + 2abc
Y = ab(a + b + c - c) + ac(a + c + b - b) + bc(b + c + a - a) + 2abc
Y = (a + b + c)(ab + ac +bc) - abc
(6.) Y = (ab + ac + bc)(a + b + c) - abc
(7.) Make X = (ab + ac + bc)(a + b + c) = Y + abc
(8.) (X - abc) || (abc)^p
---------------------------------------------
Assume X || (abc)^p
XR = (abc)^p for some R
[(abc)^p]/R - abc = (abc)^p
[(abc)^p] - (abc)R = [(abc)^p]R
[(abc)^p][1 - R] = (abc)R
[(abc)^p][1 - R]X = (abc)RX
[(abc)^p][1 - R]X = (abc)RX
XR = (abc)^p
[1 - R]X = (abc)
X || (abc)
(7.) X = (ab + ac + bc)(a + b + c) = Y + abc
X = Y + abc
(5.) Y || (abc)^p
Y + abc || (abc)
|Y + abc| =< |abc|
---------------------------
If |Y + abc| = |abc|
then Y + abc = +/- abc
but, Y =/= 0
so, Y = -2abc
----------------------
If |Y + abc| < |abc|
then
-2abc < Y < 0
2abc > |Y| > 0
Recall,
(4.) Y = (b+a)(b+c)(a+c)
|Y| = |(b+a)(b+c)(a+c)| < 2abc
--------------------------------
but,
It is always a case that ,
without loss of generality a, b and c can always be such that <2018 FEB 12 edit>
Post by Simon Roberts
2abc < 0
contradiction.
(10.) X !| (abc)^p
--------------------------------------------
(10.) X !| (abc)^p
(5.) Y || (abc)^p
(7.) X = (ab + ac + bc)(a + b + c) = Y + abc
X = Y + abc
(X - Y = abc) || (abc)^p
(10.) and (5.) => (11.)
(11.) (X - Y) !| (abc)^p
(12.) (X -Y) || (abc)^p
Contradiction.
The assumption(s) of (1.), (1.a) and/or (1.b) are false.
-Simon Roberts
All good. Could have probably stopped at the statement

|Y| = |(b+a)(b+c)(a+c)| < 2abc

as a contradiction . ! ?
Simon Roberts
2018-02-13 01:17:13 UTC
Permalink
Post by Simon Roberts
Post by Simon Roberts
Version (2016.08.23:18:31)
August, 23, 2016 approx: 6:30 pm (New York)
Fermat's Last Theorem
(proof by contradiction)
--------------------------------------------------
Assume,
(1.) a^p + b^p + c^p = 0,
(1.a) a, b, and c are non-zero, pairwise co-prime.
(1.b) p is an odd prime.
-----------------------------------------
(2.a) (a+b) || (a^p + b^p = -c^p).
(2.b) (a+c) || (a^p + c^p = -b^p).
(2.c) (b+c) || (b^p + c^p = -a^p).
From (2.a), (2.b), and (2.c)
(3.) (a+b)(a+c)(b+c) || (abc)^p
(4.) Y = (b+a)(b+c)(a+c)
(5.) Y || (abc)^p
---------------------------------------------------------
Y = (b^2 + bc + ab + ac)(a+ c)
Y = b^2a + b^2c + abc + bc^2 + ba^2 + abc + a^2c + ac^2
Y = ab(a +b) + ac(c + a) + bc(c + b) + 2abc
Y = ab(a + b + c - c) + ac(a + c + b - b) + bc(b + c + a - a) + 2abc
Y = (a + b + c)(ab + ac +bc) - abc
(6.) Y = (ab + ac + bc)(a + b + c) - abc
(7.) Make X = (ab + ac + bc)(a + b + c) = Y + abc
(8.) (X - abc) || (abc)^p
---------------------------------------------
Assume X || (abc)^p
XR = (abc)^p for some R
[(abc)^p]/R - abc = (abc)^p
[(abc)^p] - (abc)R = [(abc)^p]R
[(abc)^p][1 - R] = (abc)R
[(abc)^p][1 - R]X = (abc)RX
[(abc)^p][1 - R]X = (abc)RX
XR = (abc)^p
[1 - R]X = (abc)
X || (abc)
(7.) X = (ab + ac + bc)(a + b + c) = Y + abc
X = Y + abc
(5.) Y || (abc)^p
Y + abc || (abc)
|Y + abc| =< |abc|
---------------------------
If |Y + abc| = |abc|
then Y + abc = +/- abc
but, Y =/= 0
so, Y = -2abc
----------------------
If |Y + abc| < |abc|
then
-2abc < Y < 0
2abc > |Y| > 0
Recall,
(4.) Y = (b+a)(b+c)(a+c)
|Y| = |(b+a)(b+c)(a+c)| < 2abc
--------------------------------
but,
It is always a case that ,
without loss of generality a, b and c can always be such that <2018 FEB 12 edit>
Post by Simon Roberts
2abc < 0
contradiction.
(10.) X !| (abc)^p
--------------------------------------------
(10.) X !| (abc)^p
(5.) Y || (abc)^p
(7.) X = (ab + ac + bc)(a + b + c) = Y + abc
X = Y + abc
(X - Y = abc) || (abc)^p
(10.) and (5.) => (11.)
(11.) (X - Y) !| (abc)^p
(12.) (X -Y) || (abc)^p
Contradiction.
The assumption(s) of (1.), (1.a) and/or (1.b) are false.
-Simon Roberts
All good. <snip>
I don't even know the fundamentals?

If

A || C

and

B \| C

does THIS imply

(A = B) \ | C ?
Simon Roberts
2018-02-13 01:29:18 UTC
Permalink
Post by Simon Roberts
Post by Simon Roberts
Post by Simon Roberts
Version (2016.08.23:18:31)
August, 23, 2016 approx: 6:30 pm (New York)
Fermat's Last Theorem
(proof by contradiction)
--------------------------------------------------
Assume,
(1.) a^p + b^p + c^p = 0,
(1.a) a, b, and c are non-zero, pairwise co-prime.
(1.b) p is an odd prime.
-----------------------------------------
(2.a) (a+b) || (a^p + b^p = -c^p).
(2.b) (a+c) || (a^p + c^p = -b^p).
(2.c) (b+c) || (b^p + c^p = -a^p).
From (2.a), (2.b), and (2.c)
(3.) (a+b)(a+c)(b+c) || (abc)^p
(4.) Y = (b+a)(b+c)(a+c)
(5.) Y || (abc)^p
---------------------------------------------------------
Y = (b^2 + bc + ab + ac)(a+ c)
Y = b^2a + b^2c + abc + bc^2 + ba^2 + abc + a^2c + ac^2
Y = ab(a +b) + ac(c + a) + bc(c + b) + 2abc
Y = ab(a + b + c - c) + ac(a + c + b - b) + bc(b + c + a - a) + 2abc
Y = (a + b + c)(ab + ac +bc) - abc
(6.) Y = (ab + ac + bc)(a + b + c) - abc
(7.) Make X = (ab + ac + bc)(a + b + c) = Y + abc
(8.) (X - abc) || (abc)^p
---------------------------------------------
Assume X || (abc)^p
XR = (abc)^p for some R
[(abc)^p]/R - abc = (abc)^p
[(abc)^p] - (abc)R = [(abc)^p]R
[(abc)^p][1 - R] = (abc)R
[(abc)^p][1 - R]X = (abc)RX
[(abc)^p][1 - R]X = (abc)RX
XR = (abc)^p
[1 - R]X = (abc)
X || (abc)
(7.) X = (ab + ac + bc)(a + b + c) = Y + abc
X = Y + abc
(5.) Y || (abc)^p
Y + abc || (abc)
|Y + abc| =< |abc|
---------------------------
If |Y + abc| = |abc|
then Y + abc = +/- abc
but, Y =/= 0
so, Y = -2abc
----------------------
If |Y + abc| < |abc|
then
-2abc < Y < 0
2abc > |Y| > 0
Recall,
(4.) Y = (b+a)(b+c)(a+c)
|Y| = |(b+a)(b+c)(a+c)| < 2abc
--------------------------------
but,
It is always a case that ,
without loss of generality a, b and c can always be such that <2018 FEB 12 edit>
Post by Simon Roberts
2abc < 0
contradiction.
(10.) X !| (abc)^p
--------------------------------------------
(10.) X !| (abc)^p
(5.) Y || (abc)^p
(7.) X = (ab + ac + bc)(a + b + c) = Y + abc
X = Y + abc
(X - Y = abc) || (abc)^p
(10.) and (5.) => (11.)
(11.) (X - Y) !| (abc)^p
(12.) (X -Y) || (abc)^p
Contradiction.
The assumption(s) of (1.), (1.a) and/or (1.b) are false.
-Simon Roberts
All good. <snip>
I don't even know the fundamentals?
If
A || C
and
B \| C
does THIS imply
(A = B) \ | C ?
no counter example

10 || 20

10 - 6 || 20

6 \| 20.

I have only one potential proof left.
Simon Roberts
2018-03-10 07:41:29 UTC
Permalink
Post by Simon Roberts
Version (2016.08.23:18:31)
August, 23, 2016 approx: 6:30 pm (New York)
Fermat's Last Theorem
(proof by contradiction)
--------------------------------------------------
Assume,
(1.) a^p + b^p + c^p = 0,

(1.a) a, b, and c are non-zero, pairwise co-prime.

(1.b) p is an odd prime.

(2.a) (a+b) || (a^p + b^p = -c^p).

(2.b) (a+c) || (a^p + c^p = -b^p).

(2.c) (b+c) || (b^p + c^p = -a^p).

From (2.a), (2.b), and (2.c)

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

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

(5.) Y || (abc)^p

---------------------------------------------------------

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

Y = b^2a + b^2c + abc + bc^2 + ba^2 + abc + a^2c + ac^2

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

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

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

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

(7.) let X = (ab + ac + bc)(a + b + c) = Y + abc

(8.) (X - abc) || (abc)^p

---------------------------------------------

Assume X || (abc)^p

XR = (abc)^p for some R

[(abc)^p]/R - abc = (abc)^p

[(abc)^p] - (abc)R = [(abc)^p]R

[(abc)^p][1 - R] = (abc)R

[(abc)^p][1 - R]X = (abc)RX

[(abc)^p][1 - R]X = (abc)RX

XR = (abc)^p

[1 - R]X = (abc)





(9.) X || (abc)

(assuming X || (abc)^p)




(10. from 7. and 9.) [X = (ab + ac + bc)(a + b + c)] || abc

(11. from 10.) (a + b + c) || abc

GCD((a + b + c), abc) = 1.

contradiction

therefore

(12.) X !| (abc)^p


(7.) let X = (ab + ac + bc)(a + b + c) = Y + abc

(13.) (Y + abc) !| (abc)^p

(5.) Y || (abc)^p

that's it. unfinished and i'm guessing cannot be finished.


<snip>


-Simon Roberts
***@gmail.com

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