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Re: Andrew Beal owes Archimedes Plutonium $1 million in prize money for proving Generalized Fermat's Last Theorem

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  Re: Andrew Beal owes Archimedes Plutonium $1 million in prize money for proving Generalized Fermat's Last Theorem bursejan@gmail.com - 2018-09-10 13:14 -0700
    Re: Archimedes "Total Failure" Plutonium flunked the Math Test of a lifetime-generation test moroney@world.std.spaamtrap.com (Michael Moroney) - 2018-09-11 03:09 +0000

#423322 — Re: Andrew Beal owes Archimedes Plutonium $1 million in prize money for proving Generalized Fermat's Last Theorem

Whats better countable infinite many wheel chairs:

    |_    |_    |_
    |0|   |1|   |2|   ....
    * *   * *   * * 

Or a single wheel chair:

    |_
    |_|
    * *

Am Samstag, 28. Juli 2018 15:18:32 UTC+2 schrieb Archimedes Plutonium:
> Unless of course, well, if Andrew Beal with proxy American Mathematical Society of Charles Fefferman, Ron Graham, Daniel Mauldin, have stipulated that the winner of Beal's donation has to be Jewish. I did not read all the terms of the prize payout. And there is no way anyone in the world can steal AP's proof, for you have to mention at least once-- the word "rectangle". Because a proof of Generalized FLT is all about a rectangle replacing a number.
> 
> Besides, AP is throwing in another proof for free-- That the Ellipse is never a conic section, always a cylinder section. And it gives the AMS really -- some math to do at their place, other than what they do at this moment in time-- nothing mathematical, but only corruption of mathematics.
> 
> The Generalized FLT was proven by AP as long ago as 2014
> 
> AP proves Generalized FLT in 2014, and today is 28JUL2018, over 4 years for anyone at AMS to digest this proof.
> 
> 
> Newsgroups: sci.math
> Date: Wed, 8 Jan 2014 02:32:14 -0800 (PST)
> Complaints-To: groups...@google.com
> Injection-Info: glegroupsg2000goo.googlegroups.com; posting-host=166.137.110.8;
>  posting-account=fsC03QkAAAAwkSNcSEKmlcR-W_HNitEd
> NNTP-Posting-Host: 166.137.110.8
> User-Agent: G2/1.0
> MIME-Version: 1.0
> Message-ID: <0930d831-785f-4440-bf56-132cc17c5707@googlegroups.com>
> Subject: taking on both FLT and Beal in one proof; using Condensed Rectangles
>  #1388 Correcting Math
> From: Archimedes Plutonium <plutonium....@gmail.com>
> Injection-Date: Wed, 08 Jan 2014 10:32:14 +0000
> 
> 
> taking on both FLT and Beal in one proof; using Condensed Rectangles #1388 Correcting Math
> 
> Alright, I was curious about something which lead to the Beal conjecture which lead to the idea that if I can prove FLT, the same method should prove Beal, sounds reasonable to a logical person.
> 
> So I looked up the smallest triple that solves Beal and it is 2^3 + 2^3 = 2^4 and the next smallest is 3^3 + 6^3 = 3^5. So, unlike FLT where I am looking for 1+2B=C in conjunction with A+B=C, in Beal we are looking for 2B=C in conjunction with A+B=C.
> 
> So, I thought and thought about it, and realized I was looking at the wrong geometric figure to base the FLT proof upon. I cannot do the Beal on angles and triangles. What I can do the Beal on is what I call, _condensed rectangles_.
> 
> A condensed rectangle is that of given a number, how to compact the number as unit squares into the most smallest sides of a rectangle. For instance, given 27,216,243 the most condensed rectangle for each of those three numbers is 3x9, then 9x24, then 9 by 27. In other words, given a number you take the unit squares of that number and try to make those unit squares into a square as much as possible even though most times it will be a rectangle. So given the number 36, we could make a 2 by 18 rectangle or a 3 by 12 rectangle or a 4 by 9 rectangle or a 6 by 6 rectangle and the most condense of those is a 6x6 or a square. Now most numbers of their condensed rectangle will not be a square. Many numbers will be just a 1 by something rectangle such as 9 is 1x9.
> 
> So, what I am going to do is drop the angle and triangle and apply _condensed rectangle_ to both FLT and Beal.
> 
> Now I will use these tables for FLT:
> 
> exp3 {1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, . .} 
> 
> exp4 {1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000, . .} 
> 
> exp5 {1, 32, 243, 1024, 3125, 7776, 16807, 32,768, 59,049, 100,000, 161,051, 248,832, 371,293, . .} 
> 
> And for Beal, I will use a conglomerated table composed of all those exponents into one set starting with exp3
> 
> {1, 8, 16, 27, 32, 64, 81, 125, 243, 256, . .} 
> Notice in Beal we immediately get the primitive triple of 2B=C, whereas in FLT the primitive triple of 1+2B=C occurs in exponent2 and nowhere else. In both Beal and FLT we want to know of A+B=C. 
> 
> In Beal we have solutions of A+B=C but want to know if any solutions without a prime factor for A,B,C. It is this requirement that made me realize I was chasing after the wrong geometry figure of a angle and triangle and must chase after just _condensed rectangles_.
> 
> In FLT, no A,B,C exists for A+B=C because no 1+2B=C exists in exp3 or higher. In Beal, 2B=C exists and then A+B=C exists but with that requirement of a prime factor.
> --       
> 
> Recently I re-opened the old newsgroup of 1990s and there one can read my recent posts without the hassle of mockers and hatemongers.        
> 
> https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe        
> 
> Archimedes Plutonium
> 
> Here is the latest update on Generalized FLT, for I sort of held back for some years now, because it troubled me to go from proof of Generalized to proof of FLT itself, but now, I accept those final few lines of proof-- see Karl Runge website.
> 
> 
> Newsgroups: sci.math 
> Date: Tue, 24 Jul 2018 21:47:12 -0700 (PDT) 
> 
> Subject: Review of my Generalized FLT proof and its corollary FLT 
> From: Archimedes Plutonium <plutonium....@gmail.com> 
> Injection-Date: Wed, 25 Jul 2018 04:47:13 +0000 
> 
> Review of my Generalized FLT proof and its corollary FLT 
> I need a complete review of this proof-- especially the part where it is said-- once you have generalized FLT, you easily have FLT. I do not see the fallout as some on the Internet claim. 
> 
> 
> Well the sci.math has a complete history of my adventures with Fermat's Last Theorem starting with August 1993, and developing many twists and turns. Several times I had to throw out my prior attempts. 
> 
> What emerged was decades later, a final true proof of FLT. It has one huge starting surprise to it. Hugely Startling Surprise. The surprise is that a class of math proofs cannot be proven until a "more general theorem that overlays the "lesser theorem tucked away inside" the more general theorem is proven FIRST" 
> 
> So that-- there never exists a proof of FLT until you first prove true the Generalized FLT 
> 
> What is Generalized FLT? Glad you asked, for it is: 
> 
> A^x + B^y = C^z where A,B,C,x,y,z are positive integers with x,y,z greater than 2 
> 
> FLT is of course 
> 
> A^n + B^n = C^n where A,B,C, n are positive integers and n is greater than 2 
> 
> Newsgroups: sci.math 
> Date: Thu, 9 Jun 2016 13:38:48 -0700 (PDT) 
> 
> Subject: Proof of Generalized FLT & Fermat's Last Theorem #178 Correcting Math 4th ed 
> From: Archimedes Plutonium <plutonium....@gmail.com> 
> Injection-Date: Thu, 09 Jun 2016 20:38:49 +0000 
> 
> 
> Proof of Generalized FLT & Fermat's Last Theorem #178 Correcting Math 4th ed 
> 
> Proof of Generalized FLT & Fermat's Last Theorem #178 Correcting Math 4th ed 
> 
> PROOFS of Generalized FLT & FERMAT's LAST THEOREM 
>   
> 
> PROOFS of Generalized FLT & FERMAT's LAST THEOREM as corollary 
> 
> Editing and commentary corner: Once Generalized FLT and FLT are turned into a geometrical perspective-- condensed rectangles, then the proof is as easy as making homemade cherry pie. And the importance of proving Generalized FLT first is that FLT never has a proof until Generalized FLT is proven first. 
> 
> Summary of text: Both proofs of Generalized FLT and FLT are based on a fact of geometry, that you can represent a number with its cofactors as the sides of a rectangle. And to prove either Generalized FLT or FLT is a simple matter of stacking two rectangles that have equal sides, A and B to produce a third rectangle C which has a side equalling the _shared side_ of A and B. 
> 
> TEXT: 
> 
> I decided I needed to stiffen up the proof of Generalized FLT below. To a logical person, they often streamline their proofs and do not bother with adding some information which people not used to logic have a difficult time in seeing. For me, it is redundant to include the distributive law into the proof of Generalized FLT, but to those who do not do math but only on occasion, they would not see the logic unless I explain those very details. So below is a more stiffened up proof of Generalized FLT. 
> 
> Proof of the Generalized FLT conjecture that proves Fermat's Last Theorem 
> 
> What is Generalized FLT? Glad you asked, for it is: 
> 
> A^x + B^y = C^z where A,B,C,x,y,z are positive integers with x,y,z greater than 2 
> 
> FLT is of course 
> 
> A^n + B^n = C^n where A,B,C, n are positive integers and n is greater than 2 
> 
> I proved Generalized FLT and FLT (Fermat's Last Theorem) before I discovered that Logical Material Implication Table of T,F,T,T is incorrect and should be that of T,F,undefined, undefined. What this does is remove reductio ad absurdum as a mathematics proof technique as only a probability technique and no longer a deduction. It means the only valid proof method for mathematics, is construction proofs. My Generalized FLT and FLT are construction proofs. Another feature of the true Logic Implication operator, is that corollaries of mathematics cannot be proven as "stand alone conjectures". And that a corollary requires the over-arching- theorem attending the corollary be proven beforehand. This implies that Wiles's FLT is not true for in order to truly prove FLT, a proof of Generalized FLT had to come first. 
> 
> We see the relationship between a Theorem and its Corollary very easily from Generalized FLT and FLT, in that we have all these cases to worry about. The history of FLT was a case study, prove it in the case of exponent this and that, but never a proof of all of FLT, and that is because corollaries are never proven as stand alone theorems and must have their theorem (Generalized FLT) proven first. This is because Logic Implication is not T,F,T,T where false proofs hide behind those last two T, T when they should be undefined, undefined. The reason Implication must be T,F,undefined, undefined is because mathematics has 2/0 and 0/0 where division by 0 is undefined, and that Old Logic with its T,F,T,T does not allow for division by zero. 
> 
> Detailed Proofs Generalized FLT conjecture with its FLT corollary 
> 
> 
> Both proofs of Generalized FLT and FLT are based on a fact of geometry, that you can represent a number with its cofactors as the sides of a rectangle. And to prove either Generalized FLT or FLT is a simple matter of stacking two rectangles that have equal sides, A and B to produce a third rectangle C which has a side equalling the _shared side_ of A and B. 
> ________________________ 
> DETAILED PROOF OF Generalized FLT 
> ________________________ 
> 
> It is a constructive proof. 
> 
> We make the table of all the numbers possible in the Generalized FLT Conjecture as the conglomeration of exponents of 3 or larger as this set: 
> 
> {1, 8, 16, 27, 32, 64, 81, 125, 128, 216, 243, 256, . .} 
> 
> Here we have conglomerated exp3 and exp4 and exp5 etc etc into one set. 
> 
> We know Generalized FLT has solutions of A+B=C in that set for here are three examples: 
> 
> 2^3 + 2^3 = 2^4 with prime divisor 2 
> 3^3 + 6^3 = 3^5 with prime divisor 3 
> 7^3 + 7^4 = 14^3 with prime divisor 7 
> 
> What we need to prove is that all solutions have a prime divisor in common, ie all three rectangles of A, B, C so that A+B=C, have one shared side equal to one another. 
> 
> Definition of Condensed-Rectangle: given any number in the set of conglomerated exponents, we construct rectangles of that number from its unit squares whose sides are cofactors of the number. For instance, rectangle of 216 units as either 12x18 units, or 9x24 units, or 6x36 units or 3x72 units, or 2x108, but never a 1x216 units. We exclude 1 times the number as a condensed rectangle. So a condensed-rectangle is one in which it is composed of cofactors of the number in question, except for 1, and the number itself for 1x216 units is not a condensed-rectangle. 
> 
> Now for the constructive proof that Generalized FLT solutions must have a common prime divisor. 
> 
> We stack Condensed-Rectangles of the number-space that Generalized FLT conjecture applies: 
> 
> Solution Number Space for Generalized FLT now becomes these condensed-rectangles: 
> 
> { 2*4, 2*8, 4*4, 3*9, 2*16, 4*8, 2*32, 4*16, 8*8,. . . }   
> 
> We convert each of those numbers into Condensed-Rectangle, except 1 of course, and where many numbers have several condensed rectangles so the Solution Space of Numbers increases by a large amount. If an A and B as condensed-rectangles have the same side such as 3x9 units and 9x24units 
> wherein you stack them on their shared side of 9 and which matches another number of its condensed-rectangle such as 9x27 units, then you have a Generalized FLT solution of A+B=C. For if we were to take the 9 by 27 condensed rectangle it decomposes into 3x9 and 9x24. 
> 
> Now, the question is, are all A+B equal to a C, form stacked condensed rectangles that share a common side? 
> 
> All stackable condensed-rectangles must have one side the same for the two rectangles to stack, in the case above it is the side 9 with its common divisor of the prime 3. 
> 
> If any other solution to Generalized FLT had A stacked upon B without a common side between them, then the figure formed cannot be a overall new rectangle but something that looks like this: 
> 
> HHHHHH 
> HHHHHH 
> HHHHHHHHHH 
> HHHHHHHHHH 
> 
> That is not a condensed rectangle and all the numbers of the Solution Space, except the number 1, are condensed rectangles. That above figure is 6-sided figure. That is a 6-sided figure, yet a rectangle is only a 4 sided figure. That is not a Condensed Rectangle, represented by its cofactors. So that equality can not be achieved by any other stacking than condensed rectangles equalling condensed rectangles. 
> 
> Then the question is, can we have an A and B with a shared common side equal to a C that is a condensed rectangle with no common side shared with A and B? 
> 
> The question is, how can I be sure that all the A, B, C such that A + B = C have a common prime divisor? 
> 
> As if the question is asking whether the Condensed Rectangles covered the question by forbidding any equality unless there is a common shared side in A in B and in C. 
> 
> Well, the answer is easily enough covered for the Condensed Rectangles eliminates any possibility of A,B,C where A+B = C and not have a shared side by all three of A, B, C. It does this by the Distributive law of integers. 
> 
> All the numbers in the solution space of Generalized FLT have condensed rectangles, except the number 1. 
> 
> All the A, B, C are written as condensed rectangles of (s*t) + (r*p) = (u*v). 
> 
> For example (9*3) + (9*24) = (9*27) which is 3^3 + 6^3 = 3^5  which is 27 + 216 = 243. 
> 
> In order for (s*t) + (r*p) = (u*v) then the s, r, u have to be equal sides. Have to be (9*t) + (9*p) = (9*v) in the example of 27 + 216 = 243. 
> 
> Distributive Law (9*3)/9 + (9*24)/9 = (9*27)/9 where 3 + 24 = 27 
> 
> So, give me three numbers A, B, C, chosen arbitrarily from the Solution Space of Generalized FLT {1, 8, 16, 27, 32, 64, 81, 125, 128, 216,  243, 256, . .} except of course 1, and can those three chosen arbitrarily A, B, C obey A + B = C? Only if they obey the Distributive Law can the A + B equal to the C when we write the A, B, C as condensed rectangles. 
> 
> QED 
> 
> 
> _________________________________________ 
> Detailed Proof of FLT using condensed-rectangles 
> _________________________________________ 
> 
> It is a construction proof method for we show that it is impossible to construct A+B = C inside of a specific exponent. 
> 
> Fermat's Last Theorem FLT conjecture says there are no solutions to the equation a^y + b^y = c^y where a,b,c,y are positive integers and y is greater than 2. 
> 
> The number Space that governs FLT is this: 
> 
> exp3 {1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, . .} 
> 
> exp4 {1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000, . .} 
> 
> exp5 {1, 32, 243, 1024, 3125, 7776, 16807, 32,768, 59,049, 100,000, 161,051, 248,832, 371,293, . .} 
> 
> exp6 ..... 
> . 
> . 
> . 
> . 
> 
> So in FLT we ask whether there are any triples, A,B,C in any one of those _specific exponents_ such that A+B=C. In FLT, our solution space is only one particular exponent such as 3 or 4, or 5 to hunt down and find a A,B,C to satisfy A+B=C. 
> 
> In the proof we use Condensed-Rectangles which is defined as a rectangle composed of unit squares of the cofactors of a number, except for 1 x number itself. So the number 27 in exp3 has Condensed Rectangles of 3x9 only. The number 125 in exp3 has condensed rectangles of 5x25 only, and the number 81 in exp4 has condensed rectangles of 3x27 and 9x9. 
> 
> Now in the proof of Generalized FLT solutions of A + B = C require a common divisor. 
> 
> 2^3 + 2^3 = 2^4 with prime divisor 2 
> 3^3 + 6^3 = 3^5 with prime divisor 3 
> 7^3 + 7^4 = 14^3 with prime divisor 7 
> 
> Here in FLT, we need not even worry about exponents because all solutions to Generalized FLT encompass exponents so that if these were FLT solutions: 
> 
> 2^? + 2^? = 2^? with prime divisor 2 
> 3^? + 6^? = 3^? with prime divisor 3 
> 7^? + 7^? = 14^? with prime divisor 7 
> 
> So in that construction of a solution to FLT there is a common divisor and so now we divide the equation by the common divisor, and we get this: 
> 
> 1^y + 1^y = 1^y 
> 1^z + 2^z = 1^z 
> 1^w + 1^w = 2^w 
> 
> And those are impossible constructions. So the proof of FLT requires Generalized FLT be proven first in order to display that a Generalized FLT proof cannot co-exist unless FLT has no solutions. 
> 
> QED 
> 
> AP 
> 
> 
> Newsgroups: sci.math 
> Date: Tue, 24 Jul 2018 22:48:48 -0700 (PDT) 
> 
> Subject: Re: Review of my Generalized FLT proof and its corollary FLT 
> From: Archimedes Plutonium <plutonium....@gmail.com> 
> Injection-Date: Wed, 25 Jul 2018 05:48:49 +0000 
> 
> 
> The condensed rectangles is good, for it allows us to recognize 2^3 as 8, in 8+8 = 16, where 16 is 2^4 
> 
> So you have 
> ___________ 
> |                 |2 
> |__________| 
>          4 
> ___________ 
> |                 | 
> |__________| 
> 
> Numbers as Rectangles 
> 
> and thus 8 rectangle stacked on 8 rectangle 
> 
> is a larger rectangle of 4 by 4 
> 
> So numbers as rectangles is totally feasible 
> 
> And makes the chore of finding triples as solutions in the generalized set 
> 
> TABLE of numbers in Generalized FLT 
> the table of all the numbers possible in the Generalized FLT Conjecture as the conglomeration of exponents of 3 or larger as this set: 
> {1, 8, 16, 27, 32, 64, 81, 125, 128, 216, 243, 256, . .} 
> 
> So, we browse through that set looking for Triples A,B, C that is A+B= C, we did one above of 8+8= 16 
> 
> Now we ask ourselves, when will we find solutions and do all solutions have a common feature, a common characteristic? And the answer is that in order to be equal, for A+B to equal C, we have to have a common shared side to stack. In the case above, the 4 is common side to each rectangle. 
> 
> In essence, that is the proof, and nothing more needs to be said about Generalized FLT. If two rectangles do not share a common side (a common factor) then they add up to not being a rectangle of 4 sides but a 
> figure with 6 sides. 
> 
> 
> HHHHHH 
> HHHHHH 
> HHHHHHHHHH 
> HHHHHHHHHH 
> 
> When you turn all the numbers in the Generalized FLT space into rectangles, only those two rectangles that add up to a new larger rectangle are solutions to a triple. 
> 
> So Generalized FLT was simple and easy. 
> 
> Now going from Generalized to the Corollary (in prior posts I called it a lemma by mistake, it is a corollary), of FLT, and here is where I have some difficulty. 
> 
> According to a number of people on Internet, once you proven Generalized FLT, they claim you automatically without fuss have a proof of FLT itself. 
> 
> I find that hard to swallow. 
> 
> Because, say that for exp3, that D,E,F were solutions, and those on Internet would say that D,E,F carry a common factor, due to Generalized FLT. But, I would retort-- so what, from the distributive law, if DEF was a solution, it has an infinite number of solutions due to distributive law where kD, kE, kF are infinitely many solutions as k can be any constant. So why fuss and bother about one factor solution from Generalized FLT. 
> 
> So, either the Internet folks have some missing logic, or, AP is dense about this movement from Generalized FLT to FLT itself. And I rather think the former. 
> 
> AP 
> 
> Newsgroups: sci.math 
> Date: Wed, 25 Jul 2018 00:17:48 -0700 (PDT) 
> 
> Subject: Re: Review of my Generalized FLT proof and its corollary FLT 
> From: Archimedes Plutonium <plutonium....@gmail.com> 
> Injection-Date: Wed, 25 Jul 2018 07:17:48 +0000 
> 
> On Wednesday, July 25, 2018 at 2:07:29 AM UTC-5, Archimedes Plutonium wrote: 
> > Now here is one of several websites discussing how a proof of Generalized FLT would immediately prove FLT 
> > 
> > —- quoting KarlRunge.com 
> > 
> 
> Now I am going to replace a name with the words "Generalized FLT" from this quote simply because I cringe at the idea of money hucksterism advertisements in mathematics 
> 
> > It is easy to show that if Generalized FLT(sic) is true then it implies Fermat's Last Theorem: since for x = y = z > 2 , assume a solution A, B, C and then divide the equation the common factor= [gcd(A,B,C)]x, and you get a new solution with gcd(A',B',C') = 1, in contradiction of Generalized FLT(sic). Thus there can be no solutions for x = y = z > 2. This was conjectured by Fermat in the 1600's.... 
> 
> 
> Alright, I am beginning to see what "other mathematicians" feel is a proof of FLT once Generalized is begot. They think that a solution of FLT contradicts Generalized FLT. 
> 
> I think what they think, that reduction of the solution, by factoring out, leaves the new triple without any factors and hence violates the Generalized. 
> 
> So, I accept their assessment. That Generalized is really fully in charge and can easily conquer FLT. 
> 
> AP 
> 
> Newsgroups: sci.math 
> Date: Wed, 25 Jul 2018 09:54:09 -0700 (PDT) 
> 
> Subject: why it bothered me for years Re: Review of my Generalized FLT proof 
>  and its corollary FLT 
> From: Archimedes Plutonium <plutonium....@gmail.com> 
> Injection-Date: Wed, 25 Jul 2018 16:54:09 +0000 
> 
> why it bothered me for years Re: Review of my Generalized FLT proof and its corollary FLT 
> 
> - show quoted text - 
> I am going to "give in" and accept the above as the final proof of FLT as a corollary of Generalized FLT, although I feel a bit what we used to call "hornswaggled". I get a sort of feeling of the proof is "rigged", for it does not seem to flow free and clear. 
> 
> Suppose A,B,C has a A^3 + B^3 = C^3, then because of Generalized factor out its common factor gives D^3 + E^3 = F^3, now, because it has no common factor it violates Generalized FLT. 
> 
> You see, that bothers me-- you use Generalized one moment, next moment you use it again to say it is violated. 
> 
> Troubling, it bothers me, and the reason I have hesitated for years in accepting it. 
> 
> AP 
> 
> On Wednesday, July 25, 2018 at 4:22:24 PM UTC-5, Archimedes Plutonium wrote: 
> FLT needed a superstructure to solve it Re: Review of my Generalized FLT proof and its corollary FLT 
> 
> - show quoted text - 
> Bother me as it did, I accept it now fully, especially the GCD greatest common divisor part. 
> 
> And the importance of FLT, is that it appears to be the first theorem in all of mathematics that needs a overarching theorem to prove it as a corollary. In other words, some proofs are "tiny proofs" that need a larger structure to get hands around. I am not aware of any other mathematical proof that needed a superstructure to penetrate it. 
> 
> AP 
> 
> ==========
> 
> AP is aware of the Mickey Mouse game the AMS is playing with this prize-- with their obnoxious demands of having one of their own idiots publish a proof, then have themselves reveiw the published proof-- all self serving corruption. For what AP says-- those fools cannot even realize the Ellipse was never a conic section which AP proved by 2016 and yet, here it is 2018 and the American Mathematical Society under leadership of Charles Fefferman, Ron Graham, Daniel Mauldin, have not opened that discussion to the world, but rather, has ignored it, even though sci.math is full of the discussion-- ellipse is not a conic-- and that due to corruption Charles Fefferman, Ron Graham, Daniel Mauldin, are wanting to teach to everyone in schools in books, the texts, that the ellipse is a conic-- When, when even a bright High School Student can experiment with a cylinder can and a circle lid, that the lid will be an oval inside the cone and be a ellipse inside the cylinder. One can even imagine the experiment in the "mind eye" and know it is true.
> 
> So, Andrew Beal, and AMS with Charles Fefferman, Ron Graham, Daniel Mauldin, kindly pay Archimedes Plutonium what he is rightly due.
> 
> AP

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#423339 — Re: Archimedes "Total Failure" Plutonium flunked the Math Test of a lifetime-generation test

Hey Jan, do you need a better wheelchair for Archie?

Since he's really a mental cripple, let's see if this will work for him.

🧠
♿

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