User talk:Becker-Sievert

Recent edit to 2014 West Africa Ebola virus outbreak[edit]

Hello, and welcome to Wikipedia. I wanted to let you know that I removed one or more external links you added to the 2014 West Africa Ebola virus outbreak article, because they seemed to be inappropriate for an encyclopedia. You may find our linking guidelines helpful in this regard. If you think I made a mistake, or if you have any questions, you can leave me a message on my talk page. Thank you! Kev ^(talk) 10:21, 26 August 2014 (UTC)[reply]

Becker-Sievert formula for Pythagorean triplets[edit]

Would you please stop adding your method to actual articles, until it is described in an article published by someone other than yourself. Even then, it would be better if you requested addition, rather than adding it yourself, per WP:COI, specifically WP:SELFCITE. To begin with there is no potential reason to include both the $a > b$ and $b > a$ formulations. And, as I noted in Talk:Pythagorean triplets, it's a variant of Dickson's method, so you would need to justify its importance. — Arthur Rubin (talk) 14:58, 22 November 2014 (UTC)[reply]

Proof of Fermat's Theorem on sum of two squares[edit]

Please do not add original research to articles. Especially not when it is at best unclear and more likely nonsense. Your "Identity proof" does not show in any way that a prime that is congruent to one modulo 4 can be written as a sum of two squares. Magidin (talk) 14:57, 8 April 2016 (UTC)[reply]

For sure[edit]

Every Prime ℙ written as $4n+1=x^{2}+y^{2}$ ∨ $x^{2}-y^{2}$ with x>y, >0 in ℕ

Proof $(x^{4}-y^{4})=(x^{2}-y^{2})(x^{2}+y^{2})$

Talk: Fermat's last Theorem[edit]

For the third time: stop adding original research (even if trivial) to articles, and most especially to talk pages. Talk pages are for discussion of improvements of the article, not for you to post equations, your research, or mathematical trivialities in isolation. If you have specific improvements to propose to the articles, then propose them: make a case for them, and specify explicitly what you are proposing and why. Right now, all you are doing is vandalising the pages. Magidin (talk) 17:23, 27 April 2017 (UTC)[reply]

Please stop your disruptive editing. If you continue to violate Wikipedia's no original research policy by adding your personal analysis or synthesis into articles, as you did at Talk:Fermat's Last Theorem, you may be blocked from editing. —David Eppstein (talk) 17:39, 27 April 2017 (UTC)[reply]

You may be blocked from editing without further warning the next time you violate Wikipedia's no original research policy by inserting unpublished information or your personal analysis into an article, as you did at Talk:Fermat's Last Theorem. Note that you also are close to a violation of Wikipedia's rules against edit-warring, another blockable offense. —David Eppstein (talk) 17:47, 27 April 2017 (UTC)[reply]

You have been blocked temporarily from editing for abuse of editing privileges. Once the block has expired, you are welcome to make useful contributions. If you think there are good reasons why you should be unblocked, you may request an unblock by first reading the guide to appealing blocks, then adding the following text to the bottom of your talk page: {{unblock|reason=Your reason here ~~~~}}.

—David Eppstein (talk) 17:51, 27 April 2017 (UTC)[reply]

This user's unblock request has been reviewed by an administrator, who declined the request. Other administrators may also review this block, but should not override the decision without good reason (see the blocking policy).

Becker-Sievert (block log • active blocks • global blocks • contribs • deleted contribs • filter log • creation log • change block settings • unblock • checkuser (log))

Request reason:

your reason here Becker-Sievert (talk) 10:36, 28 April 2017 (UTC)[reply]

Decline reason:

No reason given to consider an unblock. Yamla (talk) 10:52, 28 April 2017 (UTC)[reply]

If you want to make any further unblock requests, please read the guide to appealing blocks first, then use the {{unblock}} template again. If you make too many unconvincing or disruptive unblock requests, you may be prevented from editing this page until your block has expired. Do not remove this unblock review while you are blocked.

This user's unblock request has been reviewed by an administrator, who declined the request. Other administrators may also review this block, but should not override the decision without good reason (see the blocking policy).

Becker-Sievert (block log • active blocks • global blocks • contribs • deleted contribs • filter log • creation log • change block settings • unblock • checkuser (log))

Request reason:

Reason moved outside unblock request for technical reasons. Optimist on the run (talk) 11:31, 28 April 2017 (UTC)[reply]

Decline reason:

I am declining your unblock request because it does not address the reason for your block, or because it is inadequate for other reasons. To be unblocked, you must convince the reviewing administrator(s) that

the block is not necessary to prevent damage or disruption to Wikipedia, or
the block is no longer necessary because you
1. understand what you have been blocked for,
2. will not continue to cause damage or disruption, and
3. will make useful contributions instead.

Please read the guide to appealing blocks for more information. Optimist on the run (talk) 11:31, 28 April 2017 (UTC)[reply]

If you want to make any further unblock requests, please read the guide to appealing blocks first, then use the {{unblock}} template again. If you make too many unconvincing or disruptive unblock requests, you may be prevented from editing this page until your block has expired. Do not remove this unblock review while you are blocked.

The following discussion is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

Unblock reason

Factorisation of $x^{p}-y^{p}$ [edit]

− − − − x,y,p in N and p is prime number p

− − ⇒ Factorisation of $(x^{p}-y^{p})$

− − − $(x^{2}-y^{2})=(x-y)(x^{1}y^{0}+x^{0}y^{1})$

− − $(x^{3}-y^{3})=(x-y)(x^{2}y^{0}+x^{1}y^{1}+x^{0}y^{2})$

− − $(x^{5}-y^{5})=(x-y)(x^{4}y^{0}+x^{3}y^{1}+x^{2}y^{2}+x^{1}y^{3}+x^{0}y^{4})$

− − $(x^{p}-y^{p})=(x-y)(x^{(}p-1)y^{0}+.......+x^{0}y^{(}p-1))$

− − − − − − ⇒

− FLT $(x^{p}-y^{p})$ ≠ $(x-y)^{p}$

− − --Becker-Sievert (talk) 09:07, 27 April 2017 (UTC)[reply]

− −

Factorisation of $x^{p}+y^{p}$ [edit]

− − − − − − x,y,p in N and p is prime number p

− − ⇒ Factorisation of $(x^{p}+y^{p})$

− − − $(x^{2}+y^{2})=(x^{2}+y^{2})$

− − $(x^{3}+y^{3})=(x+y)(x^{2}y^{0}-x^{1}y^{1}+x^{0}y^{2})$

− − $(x^{5}+y^{5})=(x+y)(x^{4}y^{0}-x^{3}y^{1}+x^{2}y^{2}-x^{1}y^{3}+x^{0}y^{4})$

− − $(x^{p}+y^{p})=(x+y)(x^{(}p-1)y^{0}-.......+x^{0}y^{(}p-1))$

− − − − − − ⇒

− FLT $(x^{p}+y^{p})$ ≠ $(x+y)^{p}$

− − (Becker-Sievert (talk) 12:24, 27 April 2017 (UTC))[reply]

Becker-Sievert (talk) 10:59, 28 April 2017 (UTC).[reply]

The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

The Truth is still the Truth[edit]

⇒ Please contact Mr. Arthur Rubin Erdös No.1 for proofing my FLT Identities

This Identities are true so FLT is for sure proofed with $(x+y)$ ∨,∧ $(x-y)$ be a odd prime number[edit]

x,y,p in N and p is prime number p

⇒ Factorisation of $(x^{p}-y^{p})$

$(x^{2}-y^{2})=(x-y)(x^{1}y^{0}+x^{0}y^{1})$

$(x^{3}-y^{3})=(x-y)(x^{2}y^{0}+x^{1}y^{1}+x^{0}y^{2})$

$(x^{5}-y^{5})=(x-y)(x^{4}y^{0}+x^{3}y^{1}+x^{2}y^{2}+x^{1}y^{3}+x^{0}y^{4})$

$(x^{p}-y^{p})=(x-y)(x^{(}p-1)y^{0}+.......+x^{0}y^{(}p-1))$

⇒

FLT $(x^{p}-y^{p})$ ≠ $(x-y)^{p}$

--Becker-Sievert (talk) 09:07, 27 April 2017 (UTC)[reply]

Factorisation of $x^{p}+y^{p}$ [edit]

x,y,p in N and p is prime number p

⇒ Factorisation of $(x^{p}+y^{p})$

$(x^{2}+y^{2})=(x^{2}+y^{2})$

$(x^{3}+y^{3})=(x+y)(x^{2}y^{0}-x^{1}y^{1}+x^{0}y^{2})$

$(x^{5}+y^{5})=(x+y)(x^{4}y^{0}-x^{3}y^{1}+x^{2}y^{2}-x^{1}y^{3}+x^{0}y^{4})$

$(x^{p}+y^{p})=(x+y)(x^{(}p-1)y^{0}-.......+x^{0}y^{(}p-1))$

⇒

− FLT $(x^{p}+y^{p})$ ≠ $(x+y)^{p}$

(Becker-Sievert (talk) 07:08, 28 April 2017 (UTC))[reply]

That's not what Fermat's Last Theorem states. That's like saying that the Pythagorean identity asserts that

(x+y)^{2}=x^{2}+y^{2}

, which it most certainly does not. Fermat's Last Theorem does not merely say that

x^{p}+y^{p}

is not equal to

(x+y)^{p}

nor to

(x-y)^{p}

, it asserts that it is not equal to the

p

th power of any integer. You would need to show that there does not exist an integer

k

such that

x^{p}+y^{p}=(x+y+k)^{p}

, and you have most certainly not done so. Regardless of the (complete lack of) correctness of your statements, they don't belong on a Wikipedia page, or a Talk page for an article. So stop adding them. Magidin (talk) 16:38, 28 April 2017 (UTC)[reply]

For sure is here the special case proofed $(x-y)$ ∨ $(x+y)$ is a odd prime $q$ ⇒ $(x^{p}$ ± $y^{p})$ ≠ $(q)^{p}$ (Becker-Sievert (talk) 18:03, 28 April 2017 (UTC))[reply]

First: the correct spelling is proved, not "proofed". Second, no, you aren't proving a damn thing except that you have no clue. Magidin (talk) 19:19, 28 April 2017 (UTC)[reply]

@ Magidin proved sorry. But irrelevant for this FLT proof.

Relevant is the missing Identity :

Factorisation of FLT $x^{(}2m)+y^{(}2m)$ , m be odd number in ℕ[edit]

Identity $x^{(}2m)+y^{(}2m)$

$x^{(}2m)+y^{(}2m)=(x^{2}+y^{2})(x^{(}2m)-x^{(}2m-2)y^{2}+....x^{m}y^{m}-x^{(}2)y^{(}2m-2)+y^{(}2m))$

⇒

$x^{(}2m)+y^{(}2m)$ ≠ $(x^{2}+y^{2})^{(}2m)$

(Becker-Sievert (talk) 06:04, 29 April 2017 (UTC))[reply]

Factorisation $a^{n}$ ∧ $b^{n}$ [edit]

$n>2$

⇒

$a^{n}=(a-1)(a^{(}n-1)+a^{(}n-2)+....a^{1})+a$

$b^{n}=(b-1)(b^{(}n-1)+b^{(}n-2)+....b^{1})+b$

⇒

$a^{n}+b^{n}=[(a-1)(a^{(}n-1)+a^{(}n-2)+....a^{1})+a]+[(b-1)(b^{(}n-2)+b^{(}n-1)+....b^{1})+b]$

(Becker-Sievert (talk) 07:11, 2 May 2017 (UTC))[reply]

Stop adding original research to talk pages[edit]

Please do not add original research or novel syntheses of published material to articles as you apparently did to Talk:Fermat's Last Theorem. Please cite a reliable source for all of your contributions. Thank you. Magidin (talk) 16:55, 2 May 2017 (UTC)[reply]

This user's unblock request has been reviewed by an administrator, who declined the request. Other administrators may also review this block, but should not override the decision without good reason (see the blocking policy).

Becker-Sievert (block log • active blocks • global blocks • contribs • deleted contribs • filter log • creation log • change block settings • unblock • checkuser (log))

Request reason:

true FLT Proof https://www.researchgate.net/publication/315842038_A_MARVELOUS_SIMPLE_PROOF_OF_FERMAT%27S_LAST_THEOREM_DISCOVERED A MARVELOUS SIMPLE PROOF OF FERMAT'S LAST THEOREM DISCOVERED Article (PDF Available) · April 2017 with 38 Reads (Becker-Sievert (talk) 07:24, 5 May 2017 (UTC))[reply]

Decline reason:

As you have been told a number of times, Wikipedia is not a forum for publishing your own research, discoveries or ideas. Boing! said Zebedee (talk) 10:21, 5 May 2017 (UTC)[reply]

Let me add that it doesn't even matter if you're right at this stage. Your new work would not be included in Wikipedia until after it was reviewed and became accepted by mainstream academia. If you don't show some sign of understanding this soon, your block is likely to be extended. Boing! said Zebedee (talk) 10:26, 5 May 2017 (UTC)[reply]

If you want to make any further unblock requests, please read the guide to appealing blocks first, then use the {{unblock}} template again. If you make too many unconvincing or disruptive unblock requests, you may be prevented from editing this page until your block has expired. Do not remove this unblock review while you are blocked.

@Werner von Basil: – Sperrprüfung gewünscht Wahrheit bleibt Wahrheit truth is still truth == Identitäts Beweis ==

− $p>2$

− − $a^{p}-a=(a-1)(a^{(}p-1)+a^{(}p-2)...+a^{1})$ $=0$ modulo $(a-1)$ und $p$

− − Bsp.:

− − $4^{3}-4=(4-1)(4^{2}+4^{1})$ mit $p=3,a=4$

− −

Identitäts Beweis Fermats Großer Satz[edit]

− $p>2$

− − $a^{p}=(a-1)(a^{(}p-1)+a^{(}p-2)...+a^{1})+a$ ≠ $x^{p}+y^{p}$

− − − (Becker-Sievert (Diskussion) 09:25, 5. Mai 2017 (CEST)). --Becker-Sievert (talk) 11:57, 5 May 2017 (UTC)[reply]

(Becker-Sievert (talk) 11:57, 5 May 2017 (UTC))[reply]

Talk page access revoked[edit]

As you are continuing posting your equations here without listening to anything you have been told, I have revoked your ability to edit this talk page for the duration of the block. If this behaviour continues when the block expires, your next block will probably be indefinite. Boing! said Zebedee (talk) 12:04, 5 May 2017 (UTC)[reply]

For the record, the above is clearly copied from the German Wikipedia, and at de:Benutzer:Becker-Sievert we find the user is indef blocked with talk page and email access revoked, for the reason "Repeated and aimless setting of discussion contributions in the form of formulas" (Google translate). Boing! said Zebedee (talk) 08:17, 6 May 2017 (UTC)[reply]

Recent edit to 2014 West Africa Ebola virus outbreak[edit]

Becker-Sievert formula for Pythagorean triplets[edit]

Proof of Fermat's Theorem on sum of two squares[edit]

For sure[edit]

Talk: Fermat's last Theorem[edit]

Factorisation of x p − y p {\displaystyle x^{p}-y^{p}} [edit]

Factorisation of x p + y p {\displaystyle x^{p}+y^{p}} [edit]

The Truth is still the Truth[edit]

This Identities are true so FLT is for sure proofed with ( x + y ) {\displaystyle (x+y)} ∨,∧ ( x − y ) {\displaystyle (x-y)} be a odd prime number[edit]

Factorisation of x p + y p {\displaystyle x^{p}+y^{p}} [edit]

Factorisation of FLT x ( 2 m ) + y ( 2 m ) {\displaystyle x^{(}2m)+y^{(}2m)} , m be odd number in ℕ[edit]

Factorisation a n {\displaystyle a^{n}} ∧ b n {\displaystyle b^{n}} [edit]

Stop adding original research to talk pages[edit]

Identitäts Beweis Fermats Großer Satz[edit]

Talk page access revoked[edit]

Factorisation of $x^{p}-y^{p}$ [edit]

Factorisation of $x^{p}+y^{p}$ [edit]

This Identities are true so FLT is for sure proofed with $(x+y)$ ∨,∧ $(x-y)$ be a odd prime number[edit]

Factorisation of $x^{p}+y^{p}$ [edit]

Factorisation of FLT $x^{(}2m)+y^{(}2m)$ , m be odd number in ℕ[edit]

Factorisation $a^{n}$ ∧ $b^{n}$ [edit]