November 23rd, 2016, 09:41 PM  #11  
Senior Member Joined: Dec 2012 Posts: 925 Thanks: 23  Quote:
If you think you know ALL, pls, stay far from here in you hottest quiet Hilton's room. I've to relize that it's more easy to explain how I solve both this problems to a young student, that to, unfortunately, many PHD won't to abandon their certainties. Pls study how works my two hands clock, than you will probably come back with some excuses. But I repeat, I lost my hope with you. Ciao Stefano  
November 23rd, 2016, 10:44 PM  #12 
Senior Member Joined: Dec 2012 Posts: 925 Thanks: 23 
I'm able now to prove that another answer to Fermat's the last can be also: It has no solution since $y'=nx^{(n1)}$ is a continuous derivate that can be squared in the Integers and in the Rationals, due to what is known as the "telescopic sum property", I rediscover and write as: $\displaystyle A^n = \sum_{1}^{A} (X^n(X1)^n)$ To arrive to Fermat infinite descent (that is not exactly the Newton's one, but the one known as Riemann integral method) I have to understand how to join the integers Sum from 1 to A, to the integral from 0 to A. And I made that discovering the trick (works just for A Integers): Calling the Complicate Modulus in the integers: $M_n=(X^n(X1)^n)$ the Complicate Modulus in the Rationals. $\displaystyle M_{n,K}= {n \choose 1}x^{n1}/K + {n \choose 2}x^{n2}/K^2 + {n \choose 3}x^{n3}/K^3 +... +/ \frac{1}{K^n} $ Than the squaring of the area below the derivate (In the integer / rationals / infimus) lead to the same result so to the identity: $\displaystyle A^n = \sum_{1}^{A} M_n = \sum_{1/K}^{A} M_{n,K} = \lim_{K\to\infty} \sum_{x=\frac{1}{K}}^{A} M_{n,K} = \int_{0}^{A} n x^{(n1)} dx $ While we have to cut one or more parts if we have $A\in Q$, where the integer sum doesn't work, or $A\in (RQ)$ where just the integral works. To be all the clear just take reference to the simple n=2 case: Here for n=3 (sorry old pictures with some error in the labels): Make a reason of it in your brain, I know it's like a garlic for someone, and a frog or a hat for others... but you've to digest it: your math make it like that and I just discover it. (please no others with pedigree...)  Fermat's equation has no integer solutions from n=3 since it fix an Irrational upper limit cannot be rised with an Integer Derivate (and I'm able to add the Rational also), that vice versa from the continuous one, IS NOT INVERTIBLE, so there are no ways to take out from the Integer / Rational Sum the irrational $1/2^{1/n}$ part. As shown with my method, this cause of course no problem when we return to the infimus and to the integral where the derivate return to be a continuous monotone invertible function... I turn around here for 8 years before blind the loop, please respect if don't understand it immediately...  This kind of solution show that squaring derivates area with Gnomons (rectangular columns) lead to have a finite area made by a known height function depending on a known fixed base. Regarding Beal equation I can prove is a minor consequence, once my method was understood:  Since any power of integer / rational can be build or dismounted with such Gnomons (like bricks) and since each brick has a common base we decide (1 or 1/K),  Since I show and prove any Powers of Integer / rational can be represented by a linear Sum of (2X1) gnomons that can be represented on a Cartesian plane as a sum of columns with all the top roof that fits (in media) on a trapezoid  Since we have just 3 trapezoidal area to be equal 2 vs 1  The biggest area $C^z$, to be equal to the sum of the other 2 ($A^x+B^y$), must have a common Base width given exactly by Base of A, plus Base of B This is the first of the two conditions to let the Beal equation works given by the fact we are using the same "Power measuring system" that differs from 1,2,3,4... N just in the form, but I prove is an it's bijection... The second is that the 2 linear distribution of the brick's height must follow the Linear rule Y= K(2X1) Here the fact that each power can be transformed in a Sum of linear roof columns: Any Nth power of integers (from n>=3) is equal to a Sum of Linear Terms: If n is ODD : $\displaystyle A^{n} = \sum_{X=1}^{A^{(n1)/2}}{(2XAA)} $ If n is EVEN: $\displaystyle A^{n} = \sum_{X=1}^{A^{n/2}}{(2X1)} $ The tangent of the linear roof can differs since we accept different power (x,y,z), and this allow to rise the solution if and only if the missed/ exceeding area of all the gnomons of A, or B will be equal to the missing one. Here an example: This is what I call the "butterfly solution", But there are just other 2 possible combination of the trapezoids:  when $A^n$ perfectly cut of part of $C^n$  when $A^x,B^y,C^z$, once linearized, have a the same Reduced Common base: No other combinations are possible since we decide all 3 must be Powers of integers, so must be measured with the same instrument I call "Two Hands Clock"... Yes, I know, you wanna/accept just ...rigor... mortis... Thanks Ciao Stefano Last edited by skipjack; November 24th, 2016 at 12:35 AM. 
November 24th, 2016, 04:40 AM  #13  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,544 Thanks: 2148 Math Focus: Mainly analysis and algebra  Quote:
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You'd have to ask them. But nonstandard notation will not help.  
November 24th, 2016, 05:56 AM  #14  
Senior Member Joined: Dec 2012 Posts: 925 Thanks: 23  Quote:
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Pushing the index in Rationals opens the hell... Saying I've a proof stop immediately the reading, while the understanding was never starded here... Sorry in the past I was clearly often in wrong, but now the circle is closed and blinded. Pls be specific on a "problem" you see in the proof track I already shown above, and I'll answer you. Thanks Ciao Stefano  
November 24th, 2016, 09:20 AM  #15  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,544 Thanks: 2148 Math Focus: Mainly analysis and algebra  Quote:
That's the point, you have an approximation. But that's not really very important in the context of this discussion. Quote:
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Graphs and Excel tables are not proofs, you need an firm analytical connection to the standard body of mathematics for your work to go anywhere. Quote:
Yes. The fact that you don't understand why this should be is another sign that you don't understand what you are doing. The index effectively counts the terms in the summation. For this you must have it running over a countable domain. Going to irrational indexes would, on the face of it, make the number of terms in the summation uncountable. And addition is a finite operation. We get to countably infinite cases by using the theory of limits. If you want to go uncountable, you need a new theory. As far as I know, nobody knows how to add uncountably many terms. You haven't provided any basis for this part of your work.  
November 24th, 2016, 10:42 AM  #16 
Math Team Joined: Jul 2011 From: Texas Posts: 2,429 Thanks: 1196  
November 24th, 2016, 10:46 AM  #17 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,544 Thanks: 2148 Math Focus: Mainly analysis and algebra 
That's how I feel about popcorn too.

November 24th, 2016, 10:59 PM  #18  
Senior Member Joined: Dec 2012 Posts: 925 Thanks: 23  Quote:
Point 1) for the 1200th times... here... FORGOT WE ARE IN NUMBER THEORY, WE ARE TALKING OF THINK ON A CARTESIAN PLANE.  For some unclear reason when I prove you it's necessary that we use Index "X" ... you start to have problems.... It's clear that: $\displaystyle A^2= \sum_{X=1}^{A}(2X1)$ It's A WAY TO SQUARE THE DERIVATE $y'=2X$, till A ? With rectangular integer columns: $M_{2,X} = 1 * (2X1)$ I stick here a picture, forgot it's Flt n=2, just look at the columns and the derivate: Now it's clear that the upper integer limit A, it's just a quesion of the SCALE WE CHOOSE ? In case we multiply our $A$ integer by $10^m$, we just know that our result for $A^2$, so the area we are mesuring, will be $10^{2m}$ times bigger. So in case we have to work with $A=2,1$ we can multiply it by $10^1$ and make our Integer Sum till upper limit 21, than divide the area by $10^2$ to have the result of our Rational Square... IT'S THERE ANY PROBLEM TILL HERE ??? I hope no.... Than we have just to see that when we multiply THE NUMBER OF INDEX by $10^m$, we obtain as result an area $10^{2m}$ bigger (in case of a square), in general $10^{n*m}$ bigger. So in case we would like to let the Sum Hold the Same Result ($A^2$) we must divide it by $10^{n*m}$ That it's equal to:  multiply the upper limit of the Sum by $K=10^m$ and  Divide Each term of the Sum by $K^n= 10^{n*m}$. But now I found that is possible and will be more usefull to make the math trick in this way: Staring from: $\displaystyle A^n = \sum_{X=1}^{A}{[ X^n(X1)^n]} $ We keep as example the case n=3, the term $M_n= {[ X^n(X1)^n]}$ become: $\displaystyle M_3= {( 3X^23X+1 )} $ As easy reminder keep Tartaglia's Terms for $(X1)^n$, remove the first term and change the sign of the other: $\displaystyle A^3 =\sum_{X=1}^{A}{( 3X^23X+1)} $ Now we can divide all the terms by $K^3$ , remembering we have to multiply the Upper limit by $K$ so we have: $\displaystyle A^3 = \frac {K^3*A^3} {K^3} = \sum_{X=1}^{A*K}{( 3X^2/{K^3}3X/{K^3}+1/{K^3})} $ Now we can call: $ x= X/K $ , so changing $X$ with $X= x*K$, if we respect the following conditions: a) if and only if $K$ is a Factor of $A$, or perfectly divide $A$  the Upper limit becomes: $ K*A/K = A$ (with $K,A\in N^*$ )  the Lower limit becomes: $ x=X/K$ so $X= 1$ becomes $x=1/K$ so: $\displaystyle A^3 = \sum_{x=1/K}^{A}{( 3(x*K)^2/{K^3}3(x*K)/{K^3}+1/{K^3}) } $ Now we can simplify to have our new Step Sum, that moves Step $1/K$ form $1/K$ to $A$, so the Index $x$ will be $x =1/K, 2/K,3/K.... A$: $\displaystyle A^3 = \sum_{x=1/K}^{A}{\left( \frac {3x^2}{K} \frac {3x}{K^2}+\frac{1}{K^3}\right)} $ And than, we can make our slices thinner, and thinner... If all works till here, in the next step I will make the limit for $K\to\infty$ and I'll prove it's right... Nobody pay m for that, I'm subtracting time to my work and to my family... pls who think to send to ArXiv (or elsewhere) my work be aware I'll go by legal way... Nice, but not stupid... history teach. Abstractist will talk of a set of index I=(1/K, 2/K...A) and will make all just more confused, harsh, and more difficoult to be sticked on the cartesian plane. Sorry I will repeat this again and again, when it's too much, it's too much... we have to follow the better easy short way, once possible and proven true, of course... Ciao Stefano Last edited by complicatemodulus; November 24th, 2016 at 11:06 PM.  
November 25th, 2016, 01:28 PM  #19 
Senior Member Joined: Dec 2012 Posts: 925 Thanks: 23 
Lost in the weekend, or I've a chance to have an answer ? Thanks Ciao Stefano 
November 26th, 2016, 05:36 PM  #20 
Senior Member Joined: May 2016 From: USA Posts: 577 Thanks: 248 
Yes I agree that $\displaystyle a^2 = \sum_{x=1}^a(2x  1).$ What is y? It is helpful to have some definitions. It seems that $y = x^2 \implies y' = 2x.$ The square of the derivative then would be $(2x)^2 = 4x^2.$ I am already lost. Has a become x? The definition of M is completely obscure, at least if the green in your first graph is supposed to show it. $Mnx = x^n  (x  1)^n \implies M2,1 = 1^2  0^2 = 1 \ne 4.$ 

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