Discussion:
Can these be completed (if correct) ; Fermat's?
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Simon Roberts
2018-01-26 13:31:22 UTC
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Hello and a plea,


Can these be completed (if correct) ; Fermat's?


If so then_please you are welcome to finish and post if willing.



A link to a directory (folder) of two copies of hand written pages related and potentially pertinent to Fermat's Last Theorem:



https://www.dropbox.com/sh/602u4zmdr1b4goz/AACKp6RaA3KD7H6i4Ci15QxPa?dl=0

Thank You,


Simon C. Roberts
Simon Roberts
2018-01-26 15:38:45 UTC
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On Friday, January 26, 2018 at 8:31:29 AM UTC-5, Simon Roberts wrote:


Yes finally: THIRD PAGE INDUCED (PLUS THE ASIDE) (ROUGH PROOF COMPLETE)


https://www.dropbox.com/sh/602u4zmdr1b4goz/AACKp6RaA3KD7H6i4Ci15QxPa?dl=0

Thanks

Simon
Post by Simon Roberts
Hello and a plea,
Can these be completed (if correct) ; Fermat's?
If so then_please you are welcome to finish and post if willing.
https://www.dropbox.com/sh/602u4zmdr1b4goz/AACKp6RaA3KD7H6i4Ci15QxPa?dl=0
Thank You,
Simon C. Roberts
Simon Roberts
2018-01-26 15:43:17 UTC
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Post by Simon Roberts
Yes finally: THIRD PAGE INDUCED (PLUS THE ASIDE) (ROUGH PROOF COMPLETE)
https://www.dropbox.com/sh/602u4zmdr1b4goz/AACKp6RaA3KD7H6i4Ci15QxPa?dl=0
Thanks
Simon
Post by Simon Roberts
Hello and a plea,
Can these be completed (if correct) ; Fermat's?
If so then_please you are welcome to finish and post if willing.
https://www.dropbox.com/sh/602u4zmdr1b4goz/AACKp6RaA3KD7H6i4Ci15QxPa?dl=0
Thank You,
Simon C. Roberts
ASIDE OR APPENDIX

Version 3.1

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

p_j = 1 (mod q) where,

a^q - b^q = (a - b)[Product_j (p_j)],

GCD(a,b) = 1,

q is an odd prime and,

p_j are odd primes.

*****************************************

Proof.

(0.)

(a^q - b^q)/(a - b) is a positive odd integer.

(I leave this conclusion for the reader).

(1.)

(a^q - b^q)/(a - b) =

a^(q-1) + ba^(q-2) + b^2a^(q-3) + ... +

a^2b^(q-3) + ab^(q-2) + b^(q-1).

(2.)

[(a^q - b^q)/(a - b)] - qa^(q-1) =

a^(q-1) - a^(q-1) + (b-a)a^(q-2) + (b^2-a^2)a^(q-3) + ... +

a^2(b^(q-3)-a^(q-3)) + a(b^(q-2)-a^(q-2) + (b^(q-1)-a^(q-1)).

(a-b) divides the RHS of (2.).

Therefore,

(3.)

(a-b) | [(a^q - b^q)/(a - b) - qa^(q-1)].

But,

(4.)

GCD( (a^q - b^q)/(a - b), qa^(q-1) ) = 1.

Therefore,

(5.)

GCD((a^q - b^q)/(a - b), (a-b)) = 1.

(6.)

(a^q - b^q) = (a-b) * PRODUCT_j[p_j] =

PRODUCT_i[r_i] * PRODUCT_j[p_j] where,

a - b = PRODUCT_i[r_i], r_i are odd primes and,

(a^q - b^q)/(a - b) = PRODUCT_j[p_j], p_j are odd primes.

(7.)

For any j and for any i, p_j =/= r_i. This follows from

(5.)

GCD((a^q - b^q)/(a - b), (a-b)) = 1.

`````````````````````````````````````````````````````````````````````
Inductive Proof.

(8.0)

a =/= b (mod p_i) for all p_i.

(8.1)

a^q = b^q (mod p_i) for any p_i.

(8.2.1) from (8.0 and 8.1)

a^(q+1) =/= b^(q+1) (mod p_i) for all p_i.

(8.2.2) from (8.0 and 8.1)

a^(q-1) =/= b^(q-1) (mod p_i) for all p_i.

(8.3)

a^2 =/= b^2 (mod p_i) for all p_i.

This follows from (8.2.1) and/or (8.2.2).

That is, 2 divides either or both (q-1) or (q+1).

(8.3.1)

a^(q-2) =/= b^(q-2) (mod p_i) for all p_i

(this follows from (8.3) and (8.1)).

(8.3.2)

a^(q+2) =/= b^(q+2) (mod p_i) for all p_i

(this also follows from (8.3) and (8.1)).

Now,

3 divides [(q-2) or (q-1) or (q+1) or (q+2)].

(3 does not divide q unless q =3. Then proof would be near complete).

Therefore,

(8.3.3)

a^3 =/= b^3 (mod p_i) for all p_i.

................................................

Induction.

Assume,

(8.m)

a^m =/= b^m (mod p_i) for all p_i

and m < q.

Also it is also assumed, at this point,

a^k =/= b^k (mod p_i) for k: 1 =< k =< m < (q-1).

(8.m.1)

a^(q-k) =/= b^(q-k) (mod p_i) for all p_i

(this follows from (8.m) and (8.1)).

(8.m.2)

a^(q+k) =/= b^(q+k) (mod p_i) for all p_i

(this follows from (8.m) and (8.1)).

(8.m.3)

[(m+1)|(q+k)] OR [(m+1)|(q-k)] for all k: 1 =< k =< m < (q-1);

(q-k) to (q+k) has 2m + 1 values in succession from (q-m) to (q+m),

including q.

(m+1) must divide one of the values in this range.

Therefore,

(8.[m+1])

a^(m+1) =/= b(m+1) (mod p_i) for all p_i.

End of induction.

........................................
``````````````````````````````````````````

(9.0)

a^n =/= b^n (mod p_i) for all p_i

n: 1 =< n < q,

Also,

(10.0)

a^n =/= b^n (mod p_i) for all p_i

n: n =/= wq, w is any positive integer.

Recall,

(6.)

(a^q - b^q) = PRODUCT_i[r_i] * PRODUCT_j[p_j].

a - b = PRODUCT_i[r_i].

Now,

(11.)

a^(p_j-1) = b^(p_j-1) (mod p_j) for all p_j,

(this follows from Fermat's Little Theorem).

(12.)

p_j - 1 =/= n;

if p_j - 1 = n, then

a^(n) = b^(n) (mod p_j) for all p_j. Contradiction.

n is equal to any positive integer other than wq.

w is any positive integer.

This implies,

p_j - 1 = wq

and therefore,

(13.)

p_j = 1 (mod q) for all p_j.

(6.1)

(a^q - b^q)/(a-b) = PRODUCT_j[p_j]

End Proof.

*******************************************
--------------------------------------------
-Simon Roberts
***@gmail.com
Simon Roberts
2018-01-30 13:51:19 UTC
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Post by Simon Roberts
Yes finally: THIRD PAGE INDUCED
update if you have been following the "proof" ( I may have near 60 failed proofs up to this point "crying wolf" in reverse )


typo meant "INCLUDED and roughly done"

(PLUS THE ASIDE or if you like "Appendix" as text)

now a fourth page (.pdf) that shows p^2 | (a + b + c).

I intend to eventually and properly type all of this [four items or files pieced together in one .pdf ) up and also include how
x + y + z can be made to = [equal] 0 "exactly".

(awaiting my eye glasses, my myopia is brutally bad)
Post by Simon Roberts
Post by Simon Roberts
Simon C. Roberts
Simon Roberts
2018-01-31 05:52:07 UTC
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Here it is:
pieced together
almost proper
error fixed
typos gone but some more aadded
.pdf format.
"look ma' no eyes"

https://www.dropbox.com/s/g3aifbanng6h44j/A%20Proof%20of%20Fermat%27s%20Last%20Theorem%20%28Y.1%29.pdf?dl=0
bassam king karzeddin
2018-01-31 09:52:06 UTC
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Post by Simon Roberts
pieced together
almost proper
error fixed
typos gone but some more aadded
.pdf format.
"look ma' no eyes"
https://www.dropbox.com/s/g3aifbanng6h44j/A%20Proof%20of%20Fermat%27s%20Last%20Theorem%20%28Y.1%29.pdf?dl=0
If no responses from any professional mathematicians was posted yet, then implies immediately that is worth to investigate it, for sure

BKK
Simon Roberts
2018-01-31 14:34:05 UTC
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Fermat's Last Theorem.


ALL EDITED WITH POTENTIAL PLAGIARISM AND TYPOS

https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0

The name Birkhoff and the phrase "primative roots" rings a bell.
Simon Roberts
2018-01-31 14:36:24 UTC
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Post by Simon Roberts
Fermat's Last Theorem.
ALL EDITED WITH POTENTIAL PLAGIARISM AND TYPOS
https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0
The name Birkhoff and the phrase "primative roots" rings a bell.
typo: "primitive roots"

URGENT!

Simon
Simon Roberts
2018-02-02 16:08:14 UTC
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Here it is.


A (Dropbox) link to one (decent easy to read) .pdf file.

A Proof of Fermat's Last Theorem (version Y [.45]):


https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0

I think_this_one it's correct and without typos.

Simon Christopher Roberts
of Watertown, New York.

seriously final (or extremely close)
Simon Roberts
2018-02-13 00:57:10 UTC
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Post by Simon Roberts
Here it is.
A (Dropbox) link to one (decent easy to read) .pdf file.
https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0
I think_this_one it's correct and without typos.
Simon Christopher Roberts
of Watertown, New York.
seriously final (or extremely close)
Complete for now as version 55.1
Simon Roberts
2018-02-13 12:47:41 UTC
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(best version so far {2018 02 13 7:45am NEW YORK})

A Dropbox link to the folder of one easy to read 7 page .pdf file of

'A Proof of Fermat's Last Theorem ([ver.] YXZ.70)':

https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0

Simon Christopher Roberts
of Watertown, New York.
Simon Roberts
2018-02-13 14:25:01 UTC
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On Tuesday, February 13, 2018 at 7:47:49 AM UTC-5, Simon Roberts wrote:
(best version so far {2018 02 13 9:22am NEW YORK})
Post by Simon Roberts
A Dropbox link to the folder of one easy to read 7 page .pdf file of
https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0
Simon Christopher Roberts
of Watertown, New York.
Simon Roberts
2018-02-13 21:21:39 UTC
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(best version so far {2018 02 13 4:21pm NEW YORK})
Post by Simon Roberts
A Dropbox link to the folder of one easy to read 7 page .pdf file of
'A Proof of Fermat's Last Theorem ([ver.]YXZ.79)':
https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0
Post by Simon Roberts
Simon Christopher Roberts
of Watertown, New York.
Simon Roberts
2018-02-13 21:25:13 UTC
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Post by Simon Roberts
(best version so far {2018 02 13 4:21pm NEW YORK})
A Dropbox link to the folder of one easy to read 6 page .pdf file of
Post by Simon Roberts
https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0
Post by Simon Roberts
Simon Christopher Roberts
of Watertown, New York.
Simon Roberts
2018-02-13 23:04:17 UTC
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Post by Simon Roberts
Post by Simon Roberts
(best version so far {2018 02 13 4:21pm NEW YORK})
A Dropbox link to the folder of one easy to read 6 page .pdf file of
Post by Simon Roberts
https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0
Post by Simon Roberts
Simon Christopher Roberts
of Watertown, New York.
I was just wondering about step
(10.) d^(p_j -1) - e^(p_j - 1) = (d - e)\prod{p_j}. it's fine. neither p_j divides d nor p_j divides e. Cannot.
for all j. phew, that was close.
Simon Roberts
2018-02-13 23:10:10 UTC
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panic.


I was just wondering about step
(10.) d^(p_j -1) - e^(p_j - 1) (mod q_j). it's fine. p_j neither divides d nor does p_j divide e. Cannot.

Since d^p - e^p = (d - e)\prod{q_j} and d and e are coprime. say q_j | d then q_j must divide e. contradiction.
Post by Simon Roberts
for all j. phew, that was close.
Simon Roberts
2018-02-13 23:36:40 UTC
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Post by Simon Roberts
panic.
I was just wondering about step
(10.) d^(p_j -1) - e^(p_j - 1) (mod q_j). it's fine. p_j neither divides d nor does p_j divide e. Cannot.
Since d^p - e^p = (d - e)\prod{q_j} and d and e are coprime. say q_j | d then q_j must divide e. contradiction.
Post by Simon Roberts
for all j. phew, that was close.
rejected in mere 7 and 3/4 hour from can. math. bulletin. never heard back from u of illinois and project iclick, and run around with ams, they're interested in my biography she wanted me to type it (all proper like) up for review (OMG), i'm not even gonna try ArXiv as they have little shits that are extremely rude in order to deal with the likes of me. ANY suggestions wise guy? must I BEHAVE? A math prof. dissed me at "a second kick at the can". i was fawning and being very respectful previous to this response where he also basically and close to bluntly called me a "crank" and said if I was "willing to pay him to find my errors" [sarc.] he would. An attempted and correct FLT is "wrong" unless you are a nice guy with a good PEER affiliation and can deal with the infinite and complex numbers [sarc.]. His proof seems not all that difficult to follow after watching a general outline on you tube (well done). Kinda neat actually. This situation was expected (i have a similar track record in ALMOST all aspects of my life; things for getting much better personally and outside) and is very easy for me to cope with but i am angered but not surprised {now, I gotta figa the best way to wash my dishes and handle food in orda to have a good hot dinner [supper], get off my boney ass, and get to it. good clean sleep and all. i have built my own prison (a good one) with my pet pest}

good day, i'm wiped, as we used to say on LI in my formidable years.

Simon. "who loves ya' baby?" - kojak
Simon Roberts
2018-02-14 10:47:07 UTC
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Post by Simon Roberts
panic.
I was just wondering about step
(10.) d^(p_j -1) - e^(p_j - 1) (mod q_j). it's fine. p_j neither divides d nor does p_j divide e. Cannot.
Since d^p - e^p = (d - e)\prod{q_j} and d and e are coprime. say q_j | d then q_j must divide e. contradiction.
Post by Simon Roberts
for all j. phew, that was close.
better.
Simon Roberts
2018-02-14 10:46:33 UTC
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Post by Simon Roberts
Post by Simon Roberts
Post by Simon Roberts
(best version so far {2018 02 13 4:21pm NEW YORK})
A Dropbox link to the folder of one easy to read 6 page .pdf file of
Post by Simon Roberts
https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0
Post by Simon Roberts
Simon Christopher Roberts
of Watertown, New York.
I was just wondering about step
(10.) d^(p_j -1) - e^(p_j - 1) = (d - e)\prod{p_j}. it's fine. neither p_j divides d nor p_j divides e. Cannot.
for all j. phew, that was close.
poorly written.
Simon Roberts
2018-02-21 01:08:09 UTC
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[A Proof] of Fermat's Last Theorem.


'Fermat_sLastTheoremYXZ80.pdf'

'Lemma_A.pdf'

at

https://www.dropbox.com/sh/iu33ve2mfli3b9p/AABXNfj1AxTp7FNJJYmJqRAWa?dl=0

{latest as of 2 - 20 - 2018}


Simon C. Roberts
Simon Roberts
2018-02-21 05:15:30 UTC
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Post by Simon Roberts
[A Proof] of Fermat's Last Theorem.
'Fermat_sLastTheoremYXZ80.pdf'
'Lemma_A.pdf'
at
https://www.dropbox.com/sh/iu33ve2mfli3b9p/AABXNfj1AxTp7FNJJYmJqRAWa?dl=0
{latest as of 2 - 20 - 2018}
Simon C. Roberts
mistake: lead you on a wild goose chase.

'Fermat_sLastTheoremYXZ90.pdf'

'Lemma_A.pdf' used in proof of Fermat's Conjecture (Fermat's Last Theorem).

included in the folder at this link (Dropbox)

(warning: you may spontaneously combust when you click on this link)

https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0

{updated yesterday afternoon: February 20, 2108 Eastern Time; mainly for myself to know}

Simon Roberts
retenshun(at)gmail.commerce
Simon Roberts
2018-02-21 11:19:29 UTC
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'Fermat_sLastTheoremYXZ95.pdf' (important edits that I had overlooked)

'Lemma_A.pdf' used in proof of Fermat's Conjecture (Fermat's Last Theorem).

that are included in the folder at this link (Dropbox)

(warning: you may spontaneously combust when you click on this link)

https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0

{updated : February 21, 2108 ~6 am Eastern Time; mainly for myself to know}

Simon Roberts
retenshun(at)gmail.commerce
Simon Roberts
2018-02-21 13:43:32 UTC
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'Fermat_sLastTheoremYXZ100.pdf' (important edits that I had overlooked)

'Lemma_A2.pdf'

that are included in the folder at this link (Dropbox)

https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0

{updated : February 21, 2108 ~8:40 am Eastern Time; mainly for myself to know}

Simon Roberts
retenshun(at)gmail.commerce
Simon Roberts
2018-02-22 00:56:47 UTC
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'Fermat_sLastTheoremYXZ120.pdf' (more important edits that I had overlooked)

is included in the folder at this link (Dropbox)

https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0

{updated : February 21, 2108 ~8:00 pm Eastern Time; mainly noted for myself}

Simon Roberts
retenshun(at)gmail.commerce

thanks.
Simon Roberts
2018-02-22 19:06:13 UTC
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'Fermat_sLastTheoremYXZ130.pdf' (neat important edits that I had overlooked)

and

'Lemma_A5.pdf' (stand alone. if novel would this be a lemma or theorem?)


are included in the folder at this link (Dropbox)

https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0

also Dropbox folder:

https://www.dropbox.com/sh/iu33ve2mfli3b9p/AABXNfj1AxTp7FNJJYmJqRAWa?dl=0

[all sorts of things (items); including link to facebook page(s)

https://www.facebook.com/retenshun (active)

https://www.facebook.com/profile.php?id=100010537686907&ref=br_rs (closed)

{updated : February 22, 2018 ~2:00 pm Eastern Time; mainly noted for myself}

thanks. thank you. good day.

Simon Roberts
'retenshun'(at)gmail.com.merce
Simon Roberts
2018-03-10 16:38:23 UTC
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'Fermat_sLastTheoremYXZ136.pdf'


and


'Lemma_A 11.pdf' (stand alone although included in 'Fermat's, )


are included in the folder at this link (Dropbox)


https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0







also root Dropbox folder at:

https://www.dropbox.com/sh/iu33ve2mfli3b9p/AABXNfj1AxTp7FNJJYmJqRAWa?dl=0







{updated : March 10th, 2018 ~11:30 AM Eastern Time; mainly noted for myself}



thanks. thank you. merci. good day,

Simon Roberts

retenshun(at)gmail.commerce
Simon Roberts
2018-06-20 03:55:22 UTC
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The final write-up is almost finished but I am posting a major(ly) edited version. Some more edits are still required only to make the proof easier to read.

I only just started and completed most of the major edit, repair and fix today.

I am very happy with the proof in this form although I feel like a fool.
(Are there any fool's I can feel?)

Here is the link to the folder where this version of a proof of Fermat's last theorem is located:

https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0


https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0



The pdf file is named '2_Fermat_sLastTheoremYXZ.pdf'

Today is June 19, 2018 and is very close midnight (U.S. Eastern Time Zone).

Thanks (oh boy, oh Man, oh goodness, I mean it),

Simon Christopher Roberts of NY State.
Simon Roberts
2018-07-03 04:16:08 UTC
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Post by Simon Roberts
The final write-up is almost finished but I am posting a major(ly) edited version. Some more edits are still required only to make the proof easier to read.
I only just started and completed most of the major edit, repair and fix today.
I am very happy with the proof in this form although I feel like a fool.
(Are there any fool's I can feel?)
https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0
https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0
The pdf file is named '2_Fermat_sLastTheoremYXZ.pdf'
Latest version named '7_Fermat_sLastTheoremYXZ.pdf

(as of July 2nd, 2018)
Post by Simon Roberts
Today is June 19, 2018 and is very close midnight (U.S. Eastern Time Zone).
Thanks (oh boy, oh Man, oh goodness, I mean it),
Simon Christopher Roberts of NY State.
Simon Roberts
2018-07-11 12:22:09 UTC
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Post by Simon Roberts
Post by Simon Roberts
The final write-up is almost finished but I am posting a major(ly) edited version. Some more edits are still required only to make the proof easier to read.
I only just started and completed most of the major edit, repair and fix today.
I am very happy with the proof in this form although I feel like a fool.
(Are there any fool's I can feel?)
https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0
https://www.dropbox.com/sh/cx4r5el0m85xg7u/AABv2uKtcNuEruHqT7uIKWlka?dl=0
The pdf file is named '2_Fermat_sLastTheoremYXZ.pdf'
Edited version 10_Fermat's_last_TheoremYXZ.pdf as of the morning ny time on 7-11th-2018.



Also entire folder ['Hack Man ...'] at

https://www.dropbox.com/sh/gabop77dlx550wg/AADerU5lqy9yiFKkO0JOExzRa?dl=0

I may edit many files within and these edits (not yet done) may eventually be considered required for better comprehension although not necessarily acceptance or agreement of the opinions and statements within by any reader. IOW most content currently likely a difficult read with my BS internet saying I will make all much easier and less difficult for viewers and readers.

Good Day (sooo far)

Simon Roberts I, Esquire.
Post by Simon Roberts
Latest version named '7_Fermat_sLastTheoremYXZ.pdf
(as of July 2nd, 2018)
Post by Simon Roberts
Today is June 19, 2018 and is very close midnight (U.S. Eastern Time Zone).
Thanks (oh boy, oh Man, oh goodness, I mean it),
Simon Christopher Roberts of NY State.
Simon Roberts
2018-08-18 21:20:24 UTC
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Thanks for 'any' interest and

VIEWER DISCRETION IS ADVISED.

-----
Below are listed three current links [as of 08 / 18 / 2018 (New York time)].
( If you are confused by past posted misguided and broken links it may be sufficient to say to you, 'I am also' ).
-----


1. My entire 'public'

Dropbox folder

['Hack Man (Simon) [201...]' ] at

https://www.dropbox.com/sh/gabop77dlx550wg/AADerU5lqy9yiFKkO0JOExzRa?dl=0



2. and within a sub-folder you can follow this link directly to the

current version of 'A Proof of Fermat's Last Theorem'

[FLT_A06.pdf] at

https://www.dropbox.com/s/wc2d1h8wmdub4r9/FLT_A06.pdfhttps://www.facebook.com/retenshun?dl=0


3. Also, my current Facebook address is

<https://www.facebook.com/retenshun>

['Simon Christopher Roberts']





Again: VIEWER DISCRETION IS [well?] ADVISED.


Best Regards (sometimes),
Simon Roberts - Of America ('hee hee').
[ I don't [quite] [really] understand [or get] this one [joke] either [myself]; [ya' know], what is so [freakin'] funny? ]
Simon Roberts
2018-10-06 19:37:00 UTC
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https://www.dropbox.com/s/x4e9uwflagaaegc/FLT_A07.pdf?dl=0

latest October 6^th 2019.

-Si
Simon Roberts
2018-10-07 10:42:13 UTC
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https://www.dropbox.com/s/2u24uycytsbjo09/version%20F002%20FERMAT.pdf?dl=0

latest version:

version F002 FERMAT.pdf

Dropbox

October 07^(th) 2018.

Simon R.
Simon Roberts
2018-10-15 02:55:50 UTC
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many broken links

https://www.dropbox.com/sh/8wcos6yflq8pcux/AACEZdoT7xosEC_Mhwd26Chta?dl=0

-Si.
Simon Roberts
2018-10-17 22:55:08 UTC
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Permalink
https://www.dropbox.com/sh/8wcos6yflq8pcux/AACEZdoT7xosEC_Mhwd26Chta?dl=0

{ new version October 17Th 2018: V13FERM.pdf here: }


-Si.
Simon Roberts
2018-10-18 15:12:57 UTC
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links are busted.

-Si.

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