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tech / sci.math / Re: Bassam Karzeddin hidden Answer published on Qoura regarding the greatest fuss written about (Galois, Ruffuni & Able) NO theorems

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Re: Bassam Karzeddin hidden Answer published on Qoura regarding the greatest fuss written about (Galois, Ruffuni & Able) NO theorems

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Subject: Re: Bassam Karzeddin hidden Answer published on Qoura regarding the
greatest fuss written about (Galois, Ruffuni & Able) NO theorems
From: b.karzed...@yahoo.com (bassam karzeddin)
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 by: bassam karzeddin - Wed, 30 Aug 2023 13:19 UTC

On Thursday, January 23, 2020 at 10:49:57 AM UTC+2, bassam karzeddin wrote:
> Learn freely the hidden trick (in a few minutes) of the insolvability of Quantic equation and higher by radicals
>
> Q: Why do mathematicians illegally make real root solution for this non-solvable Diophantine equation (n5−m5=nm4) , where (n, m) are positive coprime integers?
>
> full hidden topic (with unseen image of a one line formula) on this link below:
>
> https://www.quora.com/Why-do-mathematicians-illegally-make-real-root-solution-for-this-non-solvable-Diophantine-equation-n-5-m-5-nm-4-where-n-m-are-positive-coprime-integers/answer/Bassam-Karzeddin-1
>
> Is it possible up to this date that my question isn't well-received by public academic mainstream mathematicians despite teaching repeatedly the way to the answer? Wonder!, let us rewrite this insolvable Diophantine Equation here again
> $(n^5 - m^5 = nm^4)$ Eqn. 1
>
> Never mind, we know in advance that there can't be any existing integers $n, m$ that can ever satisfy this Diophantine Equation, (Right?)
> Do you know why? Wonder!
>
> The reason is too simple especially for the beginners in number theory for sure
> Just because $n$ or $m$ divides exactly the RHS of Eqn. 1, But so, unfortunately, doesn't divide exactly the LHS of the same Eqn. 1, hence contradiction, and no solution ever exists and no matter however large your $(n, m)$ are chosen, where also such ratio ($x = \frac{n}{m}$) doesn't exist, for sure
>
> **Note that the word “divides exactly” means the quotient or the result is a full integer, whereas the term “doesn’t divide exactly” means the quotient or the result isn’t a full integer but a fraction or rational number that isn’t a full integer**
>
> But, let me play this magic trick and keep the so innocent people wondering about solutions for the rest of their entire lives and many centuries to come until someone would eventually come and rescue them by revealing the magician’s true secret to keeping the innocent people away from his magic deeds and intentions against humanity as a whole, where then (she/he) would enjoy the outcome alone without being noticed
>
> And what was that magician's trick? Wonder!
> It is unbelievably too simple, just divide the whole Eqn.1 by the term $m^5$, and set( $x = \frac{n}{m})$, then rearrange the terms where one wonderfully gets this incredible and irreducible famous polynomial of odd fifth degrees
>
> $(x^5 - x - 1 = 0)$
>
> that must have at least one real algebraic root (not necessarily by radicals) and 4 other dependent complex roots (derived strictly from the same real root) in accordance with the well-designed fundamental theorem of algebra a few centuries back
>
> Where the innocent people would keep wondering aimlessly why such quintic polynomials and higher are impossible to solve by real radicals, and they would keep trying to understand this mind puzzle in order to solve it by real radicals at least for one real root, (where the untold, hidden fact that there isn’t any real root to be represented by anything or by radicals or be simply classified as real algebraic number solution)
> But we secretly know (together) "now" that no such real root ever exists as per our above simple proof, adding the absolute fact of the impossibility of geometric exact construction of such an imaginable real root (in human minds only)
>
> And please don't complicate the issue anymore by asking like smart questions about the other 4 roots that are strictly associated and derived from that non-existing real number root
>
> Or asking when drawing on the graph where it must cross the (X-Axis), since we did explain them repeatedly and very well earlier in many relevant answers if you do remember for sure
>
> please look carefully into my collapsed answers also in this very important issue to the whole mathematical sciences, since I didn’t delete them where no valid reasons were legally stated behind their suppressing, but because they were simply contradicting the common global educational mathematics among the mainstream academics generally
>
> Where continuity of real numbers was simply proven false beside those imaginary numbers (in other published answers) and in half-page for continuum hypothesis and one page for imaginary numbers
>
> But if we truly insist to have something like an approximated real root for any other non-mathematical purpose, then Wolfram Alpha is good enough, or many other well-known approximation methods like (Cauchy sequences, Dedekind cuts, Newton's approximations, Limits, Convergence, intermediate theorem, …., etc.)
> But in any case, I found this formula as the best applicable for such approximations, with a little manipulation by making $x = \frac{1}{y}$, where the applied polynomial by this old formula would be directly for $(y^5 + y^4 = 1)$, then get your needed approximations (but never as you like but as you are capable to make it up to some integer index $N$ in accordance with this following self-proved formula:
>
> Noting importantly that none of the above-mentioned methods of approximations would be considered valid in defining even a single true real irrational (constructible) number like $\sqrt{2}$ for example, simply because they are incapable to get many uncountable constructible numbers that are strictly less than $\sqrt{2}$ where such situations are independent of any imaginable technology of computations and absolutely perpetual, just because you can’t simply equate exactly any decimal rational number (with its endless field) with truly existing (hence constructible) irrational number like $\sqrt{2}$, where it is existence is independent on rationales as a diagonal distance of a square with unity side, that may also exist in many uncountable triangles (with exactly known sides)
>
> The whole theme of this repeated with many similar answers I did publish for free is to well-understand the impossibilities of solving polynomials by radicals in a few minutes and not to waste your precious life aimlessly into many other much longer reasons that are in total doubt even among the expert mathematicians until this date
>
> I truly want to help others in simple natural forms and without throwing them with made-up riddles by human magicians and with no need at all to understand (Galois, Ruffini, and Able) the so complicated theorems, and you may like to save it, since many others may not like it where they would secretly report it to waste all my efforts to help others again and again for no spoken or any valid reason so far,
>
> And other related published answers of mine are also helpful to get this simple trick completely behind insolvability of polynomials by radicals
> I hope this is helpful
>
> Regards
>
> Copyrights (c), 2020
> Bassam Karzeddin
>
> **Noting that this topic had been well-explained with few elementary proofs based on INTEGERS analysis (that never lies like humans usually do)and well-documented to the vast majorities of the mainstream academic idiots here on sci. math and much earlier related topics (on my hidden and STOLEN profile from the DOOMED math forum),
> where I republished them on moron moderated sites like SE or Qoura, since one knows exactly their unlimited hate towards every superior thinking
>
>
>
> Reference: The Formula -Solution of equations by power series (1995), (earlier posts)
> Public published answers (Bassam Karzeddin)
> Wikipedia (General)
>
> This is purely for future natural historical purposes

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