Discussion:
Proof of Fermat's Last Theorem.
Simon Roberts
2016-08-23 22:33:06 UTC
Raw Message
Version (2016.08.23:18:31)

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

Fermat's Last Theorem
--------------------------------------------------
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

(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

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
Raw Message
Simon Roberts wrote:
>
>Version (2016.08.23:18:31)
>
>August, 23, 2016 approx: 6:30 pm (New York)
>
>Fermat's Last Theorem
>--------------------------------------------------
>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 "||".

>(4.) Y = (b+a)(b+c)(a+c)
>
>(5.) Y || (abc)^p

Incorrect use of the symbol "||".

>---------------------------------------------------------
>
>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.

>---------------------------------------------
>
>Assume X || (abc)^p

Once again, your use of the symbol "||" is not correct.

>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
Raw Message
On 8/23/2016 5:33 PM, Simon Roberts wrote:
> Version (2016.08.23:18:31)
>
> August, 23, 2016 approx: 6:30 pm (New York)
>
> Fermat's Last Theorem
> --------------------------------------------------
> 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.
>
> -----------------------------------------
--------------

>
> -Simon Roberts
> ***@gmail.com
>
************************
Now let someone who knows some math show you your error(s)!
Peter Percival
2016-08-24 13:49:41 UTC
Raw Message
Pfsszxt wrote:
> On 8/23/2016 5:33 PM, Simon Roberts wrote:
>> Version (2016.08.23:18:31)
>>
>> August, 23, 2016 approx: 6:30 pm (New York)
>>
>> Fermat's Last Theorem
>> --------------------------------------------------
>> 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.
>>
>> -----------------------------------------
> --------------
>
>>
>> -Simon Roberts
>> ***@gmail.com
>>
> ************************
> 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
Raw Message
On Tuesday, August 23, 2016 at 6:33:13 PM UTC-4, Simon wrote:
> Version (2016.08.23:18:31)
>
> August, 23, 2016 approx: 6:30 pm (New York)
>
> Fermat's Last Theorem
> --------------------------------------------------
> 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
>
>
> (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
>
>
> The assumption(s) of (1.), (1.a) and/or (1.b) are false.
>
> -Simon Roberts
> ***@gmail.com

This version is garbage.
Simon Roberts
2016-11-24 14:41:59 UTC
Raw Message
On Wednesday, August 24, 2016 at 12:49:55 PM UTC-4, Simon Roberts wrote:
> On Tuesday, August 23, 2016 at 6:33:13 PM UTC-4, Simon wrote:
> > Version (2016.08.23:18:31)
> >
> > August, 23, 2016 approx: 6:30 pm (New York)
> >
> > Fermat's Last Theorem
> > --------------------------------------------------
> > 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
> >
> >
> > (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
> >
> >
> > The assumption(s) of (1.), (1.a) and/or (1.b) are false.
> >
> > -Simon Roberts
> > ***@gmail.com
>
> 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
Raw Message
On Tuesday, August 23, 2016 at 6:33:13 PM UTC-4, Simon Roberts wrote:
> Version (2016.08.23:18:31)
>
> August, 23, 2016 approx: 6:30 pm (New York)
>
> Fermat's Last Theorem
> --------------------------------------------------
> 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
>
>
> (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
>
>
> The assumption(s) of (1.), (1.a) and/or (1.b) are false.
>
> -Simon Roberts
> ***@gmail.com

This one is perfect. Sincere.

Simon
Simon Roberts
2016-11-25 22:56:32 UTC
Raw Message
On Thursday, November 24, 2016 at 9:45:27 AM UTC-5, Simon Roberts wrote:
> On Tuesday, August 23, 2016 at 6:33:13 PM UTC-4, Simon Roberts wrote:
> > Version (2016.08.23:18:31)
> >
> > August, 23, 2016 approx: 6:30 pm (New York)
> >
> > Fermat's Last Theorem
> > --------------------------------------------------
> > 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
> >
> >
> > (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
> >
> >
> > The assumption(s) of (1.), (1.a) and/or (1.b) are false.
> >
> > -Simon Roberts
> > ***@gmail.com
>
> This one is perfect. Sincere.
>
> Simon

wrong, incomplete and incorrect, has potetial
Simon Roberts
2018-02-13 00:50:03 UTC
Raw Message
On Tuesday, August 23, 2016 at 6:33:13 PM UTC-4, Simon Roberts wrote:
> Version (2016.08.23:18:31)
>
> August, 23, 2016 approx: 6:30 pm (New York)
>
> Fermat's Last Theorem
> --------------------------------------------------
> 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>

>
> 2abc < 0
>
>
> (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
>
>
> The assumption(s) of (1.), (1.a) and/or (1.b) are false.
>
> -Simon Roberts
> ***@gmail.com

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
Raw Message
On Monday, February 12, 2018 at 7:50:17 PM UTC-5, Simon Roberts wrote:
> On Tuesday, August 23, 2016 at 6:33:13 PM UTC-4, Simon Roberts wrote:
> > Version (2016.08.23:18:31)
> >
> > August, 23, 2016 approx: 6:30 pm (New York)
> >
> > Fermat's Last Theorem
> > --------------------------------------------------
> > 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>
>
> >
> > 2abc < 0
> >
> >
> > (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
> >
> >
> > The assumption(s) of (1.), (1.a) and/or (1.b) are false.
> >
> > -Simon Roberts
> > ***@gmail.com
>
> 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
Raw Message
On Monday, February 12, 2018 at 8:17:21 PM UTC-5, Simon Roberts wrote:
> On Monday, February 12, 2018 at 7:50:17 PM UTC-5, Simon Roberts wrote:
> > On Tuesday, August 23, 2016 at 6:33:13 PM UTC-4, Simon Roberts wrote:
> > > 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>
> >
> > >
> > > 2abc < 0
> > >
> > >
> > > (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
> > >
> > >
> > > The assumption(s) of (1.), (1.a) and/or (1.b) are false.
> > >
> > > -Simon Roberts
> > > ***@gmail.com
> >
> > 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
Raw Message
On Tuesday, August 23, 2016 at 6:33:13 PM UTC-4, Simon Roberts wrote:
> Version (2016.08.23:18:31)
>
> August, 23, 2016 approx: 6:30 pm (New York)
>
> Fermat's Last Theorem
> --------------------------------------------------
> 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.

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