User talk:Becker-Sievert

Recent edit to 2014 West Africa Ebola virus outbreak
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)

Becker-Sievert formula for Pythagorean triplets
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)

Proof of Fermat's Theorem on sum of two squares
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)

For sure
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
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)

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)

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)

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:. —David Eppstein (talk) 17:51, 27 April 2017 (UTC)


 * Unblock reason:

Factorisation of $$x^p-y^p$$
−		 −		 −		 −	x,y,p in N and p is prime number p − −	⇒ Factorisation of $$(x^p-y^p)$$ −		 −		 −	$$(x^2-y^2)=(x-y)(x^1y^0+x^0y^1)$$ −		 −	$$(x^3-y^3)=(x-y)(x^2y^0+x^1y^1+x^0y^2)$$ −		 −	$$(x^5-y^5)=(x-y)(x^4y^0+x^3y^1+x^2y^2+x^1y^3+x^0y^4)$$ −		 −	$$(x^p-y^p)=(x-y)(x^(p-1)y^0+.......+x^0y^(p-1))$$ −		 −		 −		 −		 −		 −	⇒ −	FLT $$(x^p-y^p)$$≠ $$(x-y)^p$$ −		 −	--Becker-Sievert (talk) 09:07, 27 April 2017 (UTC) −		 −

Factorisation of $$x^p+y^p$$
−		 −		 −		 −		 −		 −	x,y,p in N and p is prime number p − −	⇒ Factorisation of $$(x^p+y^p)$$ −		 −		 −	$$(x^2+y^2)=(x^2+y^2)$$ −		 −	$$(x^3+y^3)=(x+y)(x^2y^0-x^1y^1+x^0y^2)$$ −		 −	$$(x^5+y^5)=(x+y)(x^4y^0-x^3y^1+x^2y^2-x^1y^3+x^0y^4)$$ −		 −	$$(x^p+y^p)=(x+y)(x^(p-1)y^0-.......+x^0y^(p-1))$$ −		 −		 −		 −		 −		 −	⇒ −	FLT $$(x^p+y^p)$$≠ $$(x+y)^p$$ −		 −	(Becker-Sievert (talk) 12:24, 27 April 2017 (UTC)) Becker-Sievert (talk) 10:59, 28 April 2017 (UTC).

The Truth is still the Truth
⇒ 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
x,y,p in N and p is prime number p ⇒ Factorisation of  $$(x^p-y^p)$$ $$(x^2-y^2)=(x-y)(x^1y^0+x^0y^1)$$ $$(x^3-y^3)=(x-y)(x^2y^0+x^1y^1+x^0y^2)$$ $$(x^5-y^5)=(x-y)(x^4y^0+x^3y^1+x^2y^2+x^1y^3+x^0y^4)$$ $$(x^p-y^p)=(x-y)(x^(p-1)y^0+.......+x^0y^(p-1))$$ ⇒ FLT $$(x^p-y^p)$$≠ $$(x-y)^p$$ --Becker-Sievert (talk) 09:07, 27 April 2017 (UTC)

Factorisation of $$x^p+y^p$$
x,y,p in N and p is prime number p ⇒ Factorisation of  $$(x^p+y^p)$$ $$(x^2+y^2)=(x^2+y^2)$$ $$(x^3+y^3)=(x+y)(x^2y^0-x^1y^1+x^0y^2)$$ $$(x^5+y^5)=(x+y)(x^4y^0-x^3y^1+x^2y^2-x^1y^3+x^0y^4)$$ $$(x^p+y^p)=(x+y)(x^(p-1)y^0-.......+x^0y^(p-1))$$ ⇒ −	FLT $$(x^p+y^p)$$≠ $$(x+y)^p$$

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


 * 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)

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))
 * 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)

@ 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
Identity $$x^(2m)+y^(2m)$$

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

⇒

$$x^(2m)+y^(2m)$$≠ $$(x^2+y^2)^(2m)$$

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

Factorisation $$a^n$$ ∧ $$b^n$$
$$n>2$$

⇒

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

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

⇒

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

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

Stop adding original research to talk pages
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)

– 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
−	$$p>2$$ −		 −	$$a^p= (a-1)(a^(p-1)+a^(p-2)...+a^1)+a $$ ≠ $$x^p+y^p$$ −		 −		 −	(Becker-Sievert (Diskussion) 09:25, 5. Mai 2017 (CEST)). --Becker-Sievert (talk) 11:57, 5 May 2017 (UTC)

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

Talk page access revoked
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)
 * 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)