My Math Forum FLT (hope the end of the story...)

 Number Theory Number Theory Math Forum

January 20th, 2016, 09:32 PM   #11
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Quote:
 Originally Posted by Azzajazz I'm sorry, but you're contradiciting yourself. You start off saying $A = aK$ and then your entire proof hinges on taking the limit as $K\to\infty$. But then in your last post you say that $A$ can't approach $\infty$ as $K\to\infty$ because $A$ has to be finite. But it obviously does. As for your argument against my second point, there's an even bigger problem now. Let's fix some $A$ (which is what you seem to be doing) and then choose some $K$ such that $A = aK$, that is, $K|A$. Now, this implies that $K \leq A$. Now, if we take the limit as $K\to\infty$, we will eventually reach a point where $K>A$ since $A$ is fixed. Now it is impossible that $K|A$. This is the last post I'll make on this subject, since it seems hopelessly full of contradictions and errors.

Sorry, too messy.

- First forgot any concerning on A and B.

As I told and you see nothing change if A is an integer in Sum/StepSum/Integral:

$\displaystyle A^3 = \sum_{x= 1}^{A} (3X^2/K^3-3X/K^3+1/K^3) = \lim_{K\to\infty} \sum_{x= 1/K}^{A} (3x^2/K^3-3x/K^3+1/K^3) = \int_{0}{A} 3x^2$

If $\displaystyle A\in Q$ nothing change again (you can't use the standard Sum, so just use the Step Sum):

$\displaystyle A^3 = \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \lim_{K\to\infty} \sum_{x= 1/K}^{A} (3x^2/K^3-3x/K^3+1/K^3) = \int_{0}{A} 3x^2$

We are now talking of another area bellow the first derivate y'=3x^2 that has to satisfy the FLT conditions.

- First since we unerstand that we can square the derivate with colums, WITH AS A CONDITION THAT THE COLUMS MUST CONTINUOSLY TESSELLATE THE AREA, we are forced to say: ok, we accept the condition that there can be a common divisor between A,B (and sorry I'm not talking about a factor, as probably I left somewhere, but of a divisor that can be bigger as you want, just have to divide perfectly A & B, so cannot be for example $\displaystyle \pi$...).

Another story for C since we just suppose it can be divided by the same K, but as we see after, this condition is necessary to say: ok we start dividing the new area without making "holes" between B and B+1/K, so I said nothing about the fact that there is a difference from the first column of width 1/K of the Step Sum :

$\displaystyle \sum_{x= B+1/K}^{C} (3x^2/K-3x/K^2+1/K^3)$

and the integration methos where we start from B directly:

$\displaystyle \lim_{K\to\infty} \sum_{x= B+1/K}^{C} (3x^2/K-3x/K^2+1/K^3) = \int_{B}^{C} 3x^2 dx = C^3 -B^3$

So still if we can "square" the first part of our "new" area $\displaystyle C^3 -B^3$ with the Step Sum or the integral, we FOR SURE, can't reach the upper limit C, if C is an irrational using a Step Sum where the Step ia a rationa because we are always unable to cover the last part of the area itll x=C

Probably will be easy to see the point rewriting the sum shifted back to the same lower limit 1 (than in 1/K than to the limit)

$\displaystyle \sum_{X= B+1}^{C} (3X^2/K^3-3X/K^3+1/K^3) = \sum_{X= 1}^{C-B} (3(X+B)^2/K^3-3(X+B)/K^3+1/K^3)$

Than re-transform it in a Step Sum with x=X/K etc...

I'm working on....

Thanks
Ciao
Stefano

Thanks
Ciao
Stefano

 January 21st, 2016, 02:23 AM #12 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Sorry in the previous message there are several typing problems: the rights are of course: If $\displaystyle A\in N$ $\displaystyle A^3 = \sum_{X= 1}^{A} (3X^2-3X+1) = \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \lim_{K\to\infty} \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \int_{0}^{A} 3x^2 dx$ If $\displaystyle A\in Q$ nothing change again (you can't use the standard Sum, so just use the Step Sum): $\displaystyle A^3 = \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \lim_{K\to\infty} \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \int_{0}^{A} 3x^2 dx$ If $\displaystyle C\in R ; but ; C\notin N ; C\notin Q$ you can't use the standard Sum and the Step Sum, because is: $\displaystyle C^3 > \sum_{x= 1/K}^{C'} (3x^2/K-3x/K^2+1/K^3)$ where C' < C is the closest integer or rational to the Real Irrational C that you can rise step 1/K without rest (sorry there is a latex sign but like that is clearer to every reader). So you have to go to the limit (to reach the upper limit): $\displaystyle C^3 = \lim_{K\to\infty} \sum_{x= 1/K}^{C} (3x^2/K-3x/K^2+1/K^3) = \int_{0}^{C} 3x^2 dx$ (To CRGreathouse: will be good think if message can be revised for a longer time....) Last edited by skipjack; January 21st, 2016 at 07:20 AM.
 January 21st, 2016, 06:55 AM #13 Senior Member   Joined: Apr 2015 From: Barto PA Posts: 170 Thanks: 18 complicatemodulus wrote: (To CRGreathouse: will be good [...]....) Where is CRG?
 January 21st, 2016, 10:28 PM #14 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 To Azzjaz: there is of course a "math" solution via equation that I kindly ask to check since I always make typing and stupid error in the develope: With the Sum (if A,B,C are in N) or the Step Sum (if A,B,C are in Q) we can reduce FLT to A^3=? C^3-B^3 to: $\displaystyle \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) =? \sum_{x= B+1/K}^{C} (3x^2/K-3x/K^2+1/K^3)$ So the easy technique we can apply on the sum is to return the lower limit to 1/K (arranging the terms), and reduce what rest to a new little power (is it a cube else) plus some other terms we can prove are bigger than 1: Lowering the B+1/K lower limit to 1/K, without changin the result is equal to put instead of x the x+B in each x dependent term so: $\displaystyle \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \sum_{x= 1/K}^{C-B} (3(x+B)^2/K-3(x+B)/K^2+1/K^3)$ Now we can develope the term and extractly rebuild what we know is again a (more little) cube: $\displaystyle \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \sum_{x= 1/K}^{C-B} (3(x^2+2Bx+B^2)/K-(3x+3B)/K^2+1/K^3)$ Or $\displaystyle \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \sum_{x= 1/K}^{C-B} (3x^2/K-3x/K^2+1/K^3) + \sum_{x= 1/K}^{C-B} (6Bx+3B^2)/K +3B/K^2$ or subtractinge the littlest cube: $\displaystyle \sum_{x= 1/K}^{A+B-C} (3x^2/K-3x/K^2+1/K^3) = \sum_{x= 1/K}^{C-B} (6Bx+3B^2)/K +3B/K^2$ Now we know that the Sum of (2x/K-1/K^2) terms is a square so since we do not have the -1/K^2 we have just to add it in the first sum (square) and subtract in the second remembering the 3B out of the sum: $\displaystyle (A+B-C)^3 = 3B \sum_{x= 1/K}^{C-B} (2x/K-1/K^2) + \sum_{x= 1/K}^{C-B} 3B^2/K+ 3B/K + 3B/K^2$ or $\displaystyle (A+B-C)^3 - 3B (C-B)^2 = \sum_{x= 1/K}^{C-B} 3B^2/K+ 3B/K + 3B/K^2$ or $\displaystyle (A+B-C)^3 - 3B (C-B)^2 = K(C-B)*( 3B^2/K+ 3B/K + 3B/K^2)$ or $\displaystyle (A+B-C)^3 - 3B (C-B)^2 = (C-B)*( 3B^2+ 3B + 3B/K)$ or $\displaystyle (A+B-C)^3 - 3B (C^2+B^2-2BC) = (C-B)*( 3B^2+ 3B + 3B/K)$ or $\displaystyle A^3+B^3-C^3+3A^2B-3A^2C+3AB^2-3B^2C+3AC^2+3BC^2 - 3BC^2 -3B^3+6B^2C = 3B^2C+ 3BC-3B^3 -3B^2 + 3B(C-B)/K)$ or REMEMBERING THAT: A^3+B^3-C^3 =0 $\displaystyle 3A^2B-3A^2C+3AB^2-3B^2C+3AC^2+3BC^2 - 3BC^2 -3B^3+6B^2C = 3B^2C+ 3BC-3B^3 -3B^2 + 3B(C-B)/K)$ or $\displaystyle 3A^2B-3A^2C+3AB^2+3AC^2 - 3BC +3B^2 = 3B(C-B)/K)$ or $\displaystyle K = 3B(C-B)/ (3A^2B-3A^2C+3AB^2+3AC^2 - 3BC +3B^2 )$ or $\displaystyle K = 3B(C-B)/ (3A(B^2+C^2) - 3B(C-B)-3A^2(C-B))$ ...sorry I've to post before loose all....
 January 21st, 2016, 10:30 PM #15 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 pls check it and asap when correct the conclusion (if possible) Thanks Ciao Stefano
 January 22nd, 2016, 10:11 AM #16 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 There is an error from here: 3B/K must be 3B/K^2 to tap the -1/K^2 we create for the square $\displaystyle (A+B-C)^3 - 3B (C-B)^2 = \sum_{x= 1/K}^{C-B} 3B^2/K+ 6B/K^2$ or $\displaystyle (A+B-C)^3 - 3B (C-B)^2 = K(C-B)*( 3B^2/K+ 6B/K^2)$ or $\displaystyle (A+B-C)^3 - 3B (C-B)^2 = (C-B)*( 3B^2+ 6B/K)$ or $\displaystyle (A+B-C)^3 - 3B (C^2+B^2-2BC) = (C-B)*( 3B^2+ 6B/K)$ or $\displaystyle A^3+B^3-C^3+3A^2B-3A^2C+3AB^2-3B^2C+3AC^2+3BC^2 - 3BC^2 -3B^3+6B^2C = 3B^2C -3B^3 + 6B(C-B)/K$ or REMEMBERING THAT: A^3+B^3-C^3 =0 $\displaystyle A^2B-A^2C+AB^2+AC^2 = 2B(C-B)/K$ or $\displaystyle k= = 2B(C-B)/(A^2B-A^2C+AB^2+AC^2)$ that I let you say if it can be $\displaystyle K\in N$ and K>=1 Last edited by skipjack; January 22nd, 2016 at 11:23 AM.
 January 23rd, 2016, 06:42 AM #17 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 I try to fix all bugs here : PART 1) We all know Fermat Theorem $\displaystyle C^n<>A^n+B^n$ c1- for A,B,C integers, c2- for A,B,C coprimes, c3- for n>=3. The interesting case are for n=odd (for n=even there is a known simple proof) Knowing Mr. Wiles give us a proof (that many people can't understand) The goal is to find a simple proof that Fermat’s (can) wrote with his "simple" sum / infimus instruments. I hope I found a proof using a special Sum (I call Step Sum) on what we can show a "finite" descent in few step: - we strat the problem from A^n=... - than we reduce using sum properties to A^+B^n-C^n + mixed product = f(K). Since we state A^+B^n-C^n = 0 what rest are just n-1 terms that can define K. If we can show that after that K<1 that inply we can cut the salame with slices of rising thickness starting with a column that has an integer width > 1 so we have a contraddiction so we prove the FLT. - that works in the same way for all n>=3, since we can prove for n=3 than using the Newton develope properties and the induction proving for n=m, than prove is true also for n=m+1 so for all n. So be: $\displaystyle A = a*K$ $\displaystyle B = b*K$ C = unknown we suppose $\displaystyle C = c*K$ with: $\displaystyle A \in N$ $\displaystyle B \in N$ C = unknown we wanna check if will exist a $\displaystyle C \in N$ and: $\displaystyle a\in Q$ $\displaystyle b\in Q$ c= unknown we suppose $\displaystyle c\in Q$ $\displaystyle K\in N; K>=1$ We now rewrite FLT as: (1) $\displaystyle (a*K)^n <> (c*K)^n- (b*K)^n$ I use a property (known as the "telescopic sum") of the derivate of the powers curve $\displaystyle y=x^n$ that is $\displaystyle y'= n x^{(n-1)}$ : If $\displaystyle x\in Q : x=A/K$ then the area below the curve $\displaystyle y'= n x^{(n-1)}$ till x, we know is $\displaystyle x^n = A^n/K^n$ can be "squared" using a sum of colums base $\displaystyle 1/K$ and height equal to: $\displaystyle (X^n - (X-1)^n)/K^3$ or in math (here as example n=3): If $\displaystyle A\in N$ $\displaystyle A^3 = \sum_{X= 1}^{A} (3X^2-3X+1) = \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \lim_{K\to\infty} \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \int_{0}^{A} 3x^2 dx$ If $\displaystyle A\in Q$ nothing change again (but you can't use the standard Sum, so just use the Step Sum step 1/k, 2/k 3/K etc...): $\displaystyle A^3 = \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \lim_{K\to\infty} \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \int_{0}^{A} 3x^2 dx$ Now we know in our problem for sure A,B are in N, while for sure there is a solution for $\displaystyle C\in R$ but we don't know if it can be in the subset Q or N. So If $\displaystyle C\in R ; but ; C\notin N ; C\notin Q$ we can't use the standard Sum and the Step Sum, because for any $\displaystyle C' \in Q, K\in N$ is: $\displaystyle C^3 > \sum_{x= 1/K}^{C'} (3x^2/K-3x/K^2+1/K^3)$ where C' < C is the closest integer or rational to the Real Irrational C, that you can rise step 1/K without rest (sorry there is a latex sign but like that is clearer to every reader). So in case C is Irrational you have to go to the limit $\displaystyle K\to\infty$ to reach the upper limit with the last infimus step: $\displaystyle C^3 = \lim_{K\to\infty} \sum_{x= 1/K}^{C} (3x^2/K-3x/K^2+1/K^3) = \int_{0}^{C} 3x^2 dx$ As told by induction we can extend this for all n. Now the step will require you a minimum of flexibility since I will use the Sum operator in a new way, that is perfectly equal to the known one, in the result, but in my opinion show in a most easy and clear way how and why the integral works: We show that if we cut the area bellow the derivate with thinner and thinner colums of base 1/K and height $\displaystyle (X^n-(X-1)^n)/K^n$ at the limit for $\displaystyle K\to\infty$ we have the Riemann integral Putting as new variable in the sum $\displaystyle x=X/K$ (in the index dependet terms), to left unchanged the result we must works on the upper and lower limit that becomes: $\displaystyle A^3 = \sum_{X= 1}^{A} (3X^2-3X+1) = \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3)$ and now pulling $\displaystyle K\to\infty$ $\displaystyle A^3 = \sum_{x= 1/k}^{A} (3x^2/K-3x/K^2+1/K^3) = \lim_{K\to\infty} \sum_{x=1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \int_{0}^{A} 3x^2 dx = A^3$ I won't spend more time on this point since you've to digest it as it is. To help, you can make both limit using the (2) calling A=a*K and solving it using the known value of the sum of each single term. Once understood this point, in the same way, we can apply the sum to B and C considering now it's a Rational so we can reach it with a finite K, step 1/K. $\displaystyle B^3 = \sum_{x= 1/K}^{b*K} (3x^2/K-3x/K^2+1/K^3)$ $\displaystyle C^3 = \sum_{x= 1/K}^{c*K} (3x^2/K-3x/K^2+1/K^3)$ now we can rewrite FLT as: (1) $\displaystyle A^3= (a*K)^n =? (c*K)^n- (b*K)^n$ using the sum: $\displaystyle A^3 = \sum_{x= 1}^{A*k} (3x^2/K-3x/K^2+1/K^3) <> \sum_{x= B+1/K}^{C} (3x^2/K-3x/K^2+1/K^3)$ So if you digest the step sum you understand that for sure in the right hand side of the FLT if we pull $\displaystyle K\to\infty$ we can REMOVE THE <> sign and write: $\displaystyle A^3 = \lim_{K\to\infty} \sum_{x= B+1/K}^{C} (3x^2/K-3x/K^2+1/K^3) = \int_{B}^{C} 3x^2 dx = C^3 -B^3$ It’s easy to imagine that on the monotone smooth rising function if the limit exist (and we prove is C^3 -B^3) that it is the only one, infact for any $\displaystyle K\in Q$: $\displaystyle \sum_{x= B+1/K}^{C} (3x^2/K-3x/K^2+1/K^3) < \lim_{K\to\infty} \sum_{x= B+1/K}^{C} (3x^2/K-3x/K^2+1/K^3) = \int_{B}^{C} 3x^2 dx = C^3 -B^3$ So there is no $\displaystyle K\in N ; K>=1$ that satisfy the FLT equation, so it means that in our example the solution can be found just a the limit so, if we want to obtain the equal sign instead of the "<>" we have to immagine that fixing A,B we have to rise K to infinity so make an infinite serie of loop to rise the upper limit C that is $\displaystyle C\in R$ And using induction, as told, this can be easy proved true for any n>=3 I already explain why it works for n=2: the first derivate is a line, the second a flat line, so there will exist a solution in Q, and in N for K=1, for any integer A we choose with a couple of integer B and C, and as we know with the parametric equation to generate all the pitagorean triplets. In the Part.2 in the next post I will prove with math on the example n=3 what now are or seems just void words. To help to figure out what happen here some pictures, and what suggest to me to use this property to try to solve the FLT. Here a picture where I show the property of the derivate of the power curves y=x^n, where via integer or rational Columns of height coming from the Partial Binomial Develope (x^n-(x-1)^n), called gnomons, we can see the telescopic sum property rest true for any K till infinity. (pls, sorry, there are errors in the lables I've no time to fix... but I hope is clear enough) Here a Xls table that helps to show how the descent works (going closer and closer to the upper limit C in R): That is the base to understand Beal proof that follows... I just "invent" another way to cut the "salame": the modular algebra use same thickness slice "moduluo", I use a variable (of a known function) thickness slice. End Part 1.
 January 23rd, 2016, 08:58 AM #18 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Part.2 I hope I left no bugs here: The point is now to see and prove with classic "reduction as absurdum" on the example for n=3. Using the Step Sum (I show befor can works also in Q) we can reduce FLT to A^3=? C^3-B^3 to: $\displaystyle \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) =? \sum_{x= B+1/K}^{C} (3x^2/K-3x/K^2+1/K^3)$ So the easy technique we can apply on the sum is: 1) return the lower limit to 1/K (arranging the terms of the sum) 2) reduce what rest (till possible) to a new little power (is it a cube or else) plus some other terms depending just on A,B,C and we will see also depending on K. Newton's develope assure use that till we have a sum contining x^m terms we can reduce them in littlest powers, till the last x^1 term that can be transformed in a square (easy proof unsing known induction). Point 1- Lowering the B+1/K lower limit to 1/K, without changing the result is equal to put instead of x the x+B in each x dependent term so: $\displaystyle A^3= \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \sum_{x= 1/K}^{C-B} (3(x+B)^2/K-3(x+B)/K^2+1/K^3)$ Now we can develope the terms and extractly rebuild what we know is again a (more little) cube, plus a square, plus other non x dependent terms: $\displaystyle \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \sum_{x= 1/K}^{C-B} (3(x^2+2Bx+B^2)/K-(3x+3B)/K^2+1/K^3)$ Or $\displaystyle \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \sum_{x= 1/K}^{C-B} (3x^2/K-3x/K^2+1/K^3) + \sum_{x= 1/K}^{C-B} (6Bx+3B^2)/K +3B/K^2$ or subtractinge the littlest cube: $\displaystyle \sum_{x= 1/K}^{A+B-C} (3x^2/K-3x/K^2+1/K^3) = \sum_{x= 1/K}^{C-B} (6Bx+3B^2)/K +3B/K^2$ Now we know that: 1- the Step Sum from 1/k till A+B-C of 3x^2/K-3x/K^2+1/K^3 is the cube of (A+B-C): 2- the Step Sum from 1/k till C-B of (2x/K-1/K^2) terms is the square of (C-B) so since we do not have the -1/K^2 we have just to add it in the first sum (square) and subtract in the second remembering the 3B out of the sum: $\displaystyle (A+B-C)^3 = 3B \sum_{x= 1/K}^{C-B} (2x/K-1/K^2) + \sum_{x= 1/K}^{C-B} 3B^2/K+ 3B/K^2 + 3B/K^2$ or $\displaystyle (A+B-C)^3 - 3B (C-B)^2 = \sum_{x= 1/K}^{C-B} 3B^2/K+ 6B/K^2$ or $\displaystyle (A+B-C)^3 - 3B (C-B)^2 = K(C-B)*( 3B^2/K+ 6B/K^2)$ or $\displaystyle (A+B-C)^3 - 3B (C-B)^2 = (C-B)*( 3B^2+ 6B/K)$ or $\displaystyle (A+B-C)^3 - 3B (C^2+B^2-2BC) = (C-B)*( 3B^2+ 6B/K)$ or $\displaystyle A^3+B^3-C^3+3A^2B-3A^2C+3AB^2-3B^2C+3AC^2+3BC^2 - 3BC^2 -3B^3+6B^2C = 3B^2C -3B^3 + 6B(C-B)/K$ and REMEMBERING THAT for FLT statment: A^3+B^3-C^3 =0 $\displaystyle A^2B-A^2C+AB^2+AC^2 = 2B(C-B)/K$ or $\displaystyle K= 2B(C-B)/(A^2B-A^2C+AB^2+AC^2)$ that I let you say if it can be $\displaystyle K\in N$ and K>=1 The point, I remember, is that if what rest from C^n-B^n is a power in N than it must be K independet... while we prove (I hope) it is not If you try as example: A=5, B=6, C=7 you can see what happen still if you don't see more or you don't use wolframalpha for the graph. So the conclusion is: till we rest in the integers, or rationals while all X terms will be re.transfomed in a non dependet K terms, the rest will alway depend by K... so.... (Beal will follows !) In other terms: from n>=2 the derivate is a curve so there are minimum 3 terms in what I call "the rational derivate". the x dependent can be transformed as seen to have a little cube and a little square, but a rest will always be present, so we can conclude: first derivate = curve ? Than no way... DINNER TIME...KIDS & WIFE AGAIN OUT OF CONTROLL... really hope no bugs... Last edited by complicatemodulus; January 23rd, 2016 at 09:28 AM.
 January 23rd, 2016, 11:33 PM #19 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Still little bug in: when you see "C" as upper limit you've to read: - $\displaystyle C\in N$ if the sum is step 1 (classic sum) ($\displaystyle C_N$) - $\displaystyle C\in Q$ if the sum is step 1/K (K>1 Step Sum) ($\displaystyle C_Q$) - $\displaystyle C\in R$ if the sum is pulled at the limit for $\displaystyle K\to\infty$ or in the integral. ($\displaystyle C_R$) Probably will be better to call them with different index, for example: $\displaystyle C_N < C_Q < C_R$ well knowing that we consider them in this order only Thanks
 January 25th, 2016, 09:56 PM #20 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Of course the first think you can do is try with n=2 K=1 starting from: $\displaystyle \sum_{x=1}^{A}(2x-1) = \sum_{x=1}^{C}(2x-1) - \sum_{x=1}^{B}(2x-1)$ or: $\displaystyle \sum_{x=1}^{A}(2x-1) = \sum_{x=B+1}^{C}(2x-1)$ Than apply the subtraction of the littlest square from the bigger and the "lowering" technique: $\displaystyle \sum_{x=1}^{A}(2x-1) = \sum_{x=1}^{C-B}(2(x+B)-1)$ or $\displaystyle \sum_{x=1}^{A}(2x-1) = \sum_{x=1}^{C-B}(2x-1)+ (C-B)*2B$ or $\displaystyle \sum_{x=C-B+1}^{A}(2x-1) = (C-B)*2B$ or $\displaystyle \sum_{x=1}^{A+B-C}(2(x+C-B)-1) = 2BC-2B^2$ or $\displaystyle \sum_{x=1}^{A+B-C}(2x-1) + (A+B-C)*2(C-B)= 2BC-2B^2$ or $\displaystyle (A+B-C)^2 + (A+B-C)*2(C-B)= 2BC-2B^2$ or $\displaystyle (A+B-C)^2 + 2AC-2AB+2BC-2B^2-2C^2+2BC-2BC+2B^2=0$ or $\displaystyle A^2+B^2+C^2+2AB-2AC-2BC + 2AC-2AB+2BC-2B^2-2C^2+2BC-2BC+2B^2=0$ or $\displaystyle A^2+B^2-C^2 =0$ So, and this is the point, once we again apply the FLT rule, we return to the identity, so we know it can works with some integer A,B,C called Pitagorean tryplets. If what we have in the right side cannot be transfromed in a power of an integer (as happen from n>=3) after the lowering technique we find that when we apply again FLT equation A^n+B^n-C^n=0 the REST will vanish just at one K that is $\displaystyle K<1 : K\notin N$ And seems not another void turn... Here on the graph is more easy to see and understand: or: Thanks Ciao Stefano

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