GingerAle

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Nombre de usuarioGingerAle
Puntuación2511
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Respuestas 740

 #16
avatar+2511 
+4

WTF! This looks like a bloody inquisition. 

 

You plop in here as an imperious, fucking know-it-all –and you are anonymous because ...well the AssHoleMan doesn’t want to present any kind of identity. But of course, we know who you are by your erudite syntax and presentation of pure condescending BULLSHIT!

 

(Ahem) What do you think about the answerers who don't bother to check their answers or read the question correctly? Don't you think that it can be irresponsible towards the users that come here to learn?

This line is particularly condescending:

Especially when this site hosts many knowledgeable mathematicians and has earned a grear reputation?

You can take your patronizing superiority and shove it up your ass!

 

If you were truly knowledgeable about this site, you’d know there are very few knowledgeable mathematicians on this forum.  To you, anyone who knows the multiplication tables is a knowledgeable mathematician, so it’s just a matter of perspective. In any case, you are not qualified to make the statement. 

 

I’ve seen your work on here, and it’s clear you are a fucking idiot. You think of yourself as a knowledgeable mathematician and educator because you’ve provided solutions to a few junior high school math questions. I’m sure you’ll start having wet dreams when you remember how to solve quadric equations. I am personally overwhelmed by your majestic presence. You belong in the forums “BB” class. BB stands for Blarney Bag and it includes Bullshit Bugs and Blarney Bankers.  This is the league of Extraordinary Batshit Stupid Gentlemen: those who’ve retired with a pension, hardened arteries, atrophied brains, and an extreme desire prove their virile intellect via mathematics.

 

I’m sure you belong in this class. Let me guess, you are a recently retired what ... a salesman ...insurance maybe.  Whatever... You’ve not practiced any math for forty years or more, and now that you have time, and are bored-out-of-your-gourd, you’ve decided you want to help the children of the current generation.

 

How fucking noble! We are soooo honored to have you. To prove it, we will shit marble every time you post.

 

 

GA

6 nov 2019
 #11
avatar+2511 
+4

Solution: 

This method uses Euler's totient and the modulo inverse functions.

 

\(\begin{array}{rcll} n &\equiv& {\color{red}1} \pmod {{\color{green}4}} \\ n &\equiv& {\color{red}1} \pmod {{\color{green}3}} \\ n &\equiv& {\color{red}2} \pmod {{\color{green}5}} \\ \text{Set } m &=& 4\cdot 3\cdot 5 = 60 \\ \end{array} \)

 

\(\small{ \begin{array}{l} n = {\color{red}1} \cdot {\color{green}3\cdot 5} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}3 \cdot 5) }^{\varphi({\color{green}4}) -1 } \pmod {{\color{green}4}} ] }_{=\text{modulo inverse }(3\cdot 5) \mod 4 } }_{=(3\cdot 5)^{2-1} \mod {4}} }_{=(3\cdot 5)^{1} \mod {4}} }_{=(15\pmod{4})^{1} \mod {4}} }_{=(15)^{1} \mod {4}} }_{= 3} + {\color{red}1} \cdot {\color{green}4\cdot 5} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}4\cdot 5) }^{\varphi({\color{green}3}) -1} \pmod {{\color{green}3}} ] }_{=\text{modulo inverse } (4\cdot 5) \mod {3}} }_{=(4\cdot 5)^{2-1} \mod {3}} }_{=(4\cdot 5)^{1} \mod {3}} }_{=(20\pmod{3})^{1} \mod {3}} }_{=(2)^{1} \mod {3}} }_{=2} + {\color{red}2} \cdot {\color{green}4\cdot 3} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}4\cdot 3) }^{\varphi({\color{green} 5}) -1 } \pmod {{\color{green}5}} ] }_{=\text{modulo inverse } (4\cdot 3) \mod 5 } }_{=(4\cdot 3)^{4-1} \mod { 5}} }_{=(4\cdot 3)^{3} \mod {5}} }_{=(12\pmod{5})^{3} \mod {5}} }_{=(2)^{3} \mod {5}} }_{=3}\\ \\ n = {\color{red}1} \cdot {\color{green}3\cdot 5} \cdot [3] + {\color{red}1} \cdot {\color{green}4\cdot 5} \cdot [2] + {\color{red}2} \cdot {\color{green}4\cdot 3} \cdot [3] \\ n = 45+ 40 + 72 \\ n = 157 \\\\ n \pmod {m} = 157 \pmod {60} \\ n = 37+ k\cdot 60 \; \; k \in Z\\ \mathbf{n_{min}} \; \mathbf{=} \; \mathbf{37} \end{array} }\)

 

 

 

 

GA

.
5 nov 2019
 #8
avatar+2511 
+3

Well, you are not alone in your absence of understanding.

 

The Andrew Wiles and Wiles-Taylor proof of Fermat’s Last Theorem is not conveniently accessible to casual post-graduate Ph.D. students.  To understand and hold on to Wiles’ proof of FLT, requires months of study for the elite, number theory, Ph.D. level mathematicians. This assumes these mathematicians have a thorough prerequisite understanding of the epsilon conjecture and Ribet's proof, defining the relationships of three-dimensional surfaces to that of two-dimensional elliptical curves, along with the properties of modular forms associated with Galois representations of vector spaces over a field and free modules over rings of prime numbers.

 

There were still major bridges to build to connect these and other proofs to FLT. One notable bridge is in Ribet’s proof of the Epsilon conjecture. It’s known that elliptic curves with given conductors of (gN) produce stable prime elliptic curves in higher dimensions with rational Fourier coefficients. However, some of the curves are unstable –producing catastrophe curves that do not have rational Fourier coefficients. These curves are not elliptic curves; they belong to the higher-dimensional Abelian selection. 

 

Because these curves are not elliptic curves, Andrew Wiles realized they weren’t needed for the FLT proof. Using Kolyvagin–Flach approach to adapt the Iwasawa theory, he circumvented the higher-dimensional Abelian selections, while maintaining the integrity of the proof of the Epsilon conjecture.

 

None of the mathematics for these theories is conveniently accessible to lower-level life forms, but I still think it is very cool!.Yep! Very Cool!

 

GA

28 oct 2019
 #4
avatar+2511 
-1
25 oct 2019