January 18th, 2016, 11:15 PM  #1 
Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24  FLT (hope the end of the story...)
I post lot on, but I hope this time all thinks will be at the right place: 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. Knowing Mr. Wiles give us a proof (that many people can't understand) The goal is to find a simple proof that he probably wrote with his "simple" instruments. I hope I found a proof using a special Sum (I call Step Sum) on what we can show an infinite descent that works in the same way for all n>=3, since we can prove for n=3 than using the induction for all following n. 1) If we can prove there is no solution in the integers (c1) for A,B,C having a COMMON FACTOR $\displaystyle K\in N $ (so still if we remove the condition c2) we have proven FLT. So be: $\displaystyle A = a*K $ $\displaystyle B = b*K $ $\displaystyle C = c*K $ We now rewrite FLT as: (1) $\displaystyle (a*K)^n <> (c*K)^n (b*K)^n$ I use a property (known as the "telecopic sum") of the derivate of the powers curve $\displaystyle y=x^n$ that is $\displaystyle y'= \frac {x^{(n1)}} n $ : If $\displaystyle x\in Q : x=A/K $ then the area below the curve $\displaystyle y'= \frac {x^{(n1)}} n $ till x, that we know is $\displaystyle x^n = A^n/K^n $ can be "squared" using a sum of colums base $\displaystyle 1/K$ height ((x^nX1)^n) or in math: $\displaystyle A^3/K^3 = \sum_{X= 1}^{A} (3X^23X+1)/K^3 $ This property rest treu for any K from 1 (and more under certain conditions) to $\displaystyle \infty$ Infact we can pull $\displaystyle K\to\infty$ showing that this easy limit is true: (2) $\displaystyle A^3 = \sum_{x= 1}^{a*K} (3x^2/K^33x/K^3+1/K^3) = \lim_{K\to\infty} \sum_{x= 1}^{a*K} (3x^2/K^33x/K^3+1/K^3) = A^3 $ And 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 how and way the integral works: We show that is we cut the derivate with thinner and thinner colums of base 1/K at the limit for $\displaystyle K\to\infty$ we have the Riemann integral Putting as new variable in the sum $\displaystyle x=X/K$ : $\displaystyle A^3 = \sum_{X= 1}^{a*K} (3X^2/K^33X/K^3+1/K^3) = \sum_{x= 1/K}^{A} (3x^2/K3x/K^2+1/K^3) $ and now pulling $\displaystyle K\to\infty$ $\displaystyle A^3 = \sum_{x= 1/k}^{A} (3x^2/K3x/K^2+1/K^3) = \lim_{K\to\infty} \sum_{x=1/K}^{A} (3x^2/K3x/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 calling N=x*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 MATH] B^3 = \sum_{x= 1}^{b*K} (3x^2/K^33x/K^3+1/K^3)[/MATH] MATH] C^3 = \sum_{x= 1}^{c*K} (3x^2/K^33x/K^3+1/K^3)[/MATH] now we can rewrite (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/K3X/K^2+1/K^3) <> \sum_{x= B+1/K}^{C} (3x^2/K3x/K^2+1/K^3) $ So if you digest the step sum you can see the right hand side of the FLT that JUST if we pull $\displaystyle K\to\infty$ we can REMOVE THE <> sign and write: $\displaystyle \lim_{K\to\infty} \sum_{x= B+1/K}^{C} (3x^2/K3x/K^2+1/K^3) = \int_{B}^{C} 3x^2 dx = C^3 B^3 $ In fact is easy to see 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/K3x/K^2+1/K^3) < \lim_{K\to\infty} \sum_{x= B+1/K}^{C} (3x^2/K3x/K^2+1/K^3) = \int_{B}^{C} 3x^2 dx = C^3 B^3 $ So there is no $\displaystyle K\in Q$ that satisfy the FLT equation, so it means that in our example $\displaystyle C\in R$ And using induction as told this is true for any n>=3 I already explain why it works for n=2: the first derivate is a line, so there will exist a solution in Q, and in N for K=1 I lost lot of time turning words on... 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(x1)^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, I use a variable (of a known function) thickness slice. (sorry I to post it before loose it, I'll recheck asap) Last edited by skipjack; January 18th, 2016 at 11:22 PM. 
January 19th, 2016, 08:16 AM  #2 
Senior Member Joined: Sep 2010 Posts: 221 Thanks: 20 
For y=x^n the derivative y'=nx^(n1) isn't it?

January 19th, 2016, 08:30 AM  #3 
Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 
Yessa ! Sorry lot of typing errors...

January 19th, 2016, 08:57 AM  #4 
Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 
(Sorry some typing error fixed: derivate, index in (3) etc... ): 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, or you can just change the sign at the end from "<" for odds to ">" for even) Knowing Mr. Wiles give us a proof (that many people can't understand) The goal is to find a simple proof that he probably wrote with Fermat’s "simple" instruments. I hope I found a proof using a special Sum (I call Step Sum) on what we can show an infinite descent that works in the same way for all n>=3, since we can prove for n=3 than using the induction for all following n. 1) If we can prove there is no solution in the integers (c1) for A,B,C having a COMMON FACTOR $\displaystyle K\in N $ (so still if we remove the condition c2) we have proven FLT. So be: $\displaystyle A = a*K $ $\displaystyle B = b*K $ $\displaystyle C = c*K $ 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^{(n1)} $ : If $\displaystyle x\in Q : x=A/K $ then the area below the curve $\displaystyle y'= n x^{(n1)} $ 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  (X1)^n)/K^3 $ or in math (here as example n=3): Starting with know sum step 1: (2) $\displaystyle A^3 = \sum_{X= 1}^{A*K} (3X^23X+1)/K^3 $ This property rest true for any K from 1 (or more under certain conditions) to $\displaystyle \infty$ Infact we can pull $\displaystyle K\to\infty$ showing that this easy limit is true: (3a) $\displaystyle a^3 = \sum_{X= 1}^{a*K} (3X^2/K^33X/K^3+1/K^3) = \lim_{K\to\infty} \sum_{X= 1}^{a*K} (3X^2/K^33X/K^3+1/K^3) = a^3 $ or (3b) $\displaystyle A^3 = \sum_{X= 1}^{A*K} (3X^2/K^33X/K^3+1/K^3) = \lim_{K\to\infty} \sum_{X= 1}^{A*K} (3X^2/K^33X/K^3+1/K^3) = A^3 $ And 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(X1)^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^23X+1) = \sum_{x= 1/K}^{A} (3x^2/K3x/K^2+1/K^3) $ and now pulling $\displaystyle K\to\infty$ $\displaystyle A^3 = \sum_{x= 1/k}^{A} (3x^2/K3x/K^2+1/K^3) = \lim_{K\to\infty} \sum_{x=1/K}^{A} (3x^2/K3x/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 N=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 $\displaystyle B^3 = \sum_{x= 1}^{b*K} (3x^2/K^33x/K^3+1/K^3)$ $\displaystyle C^3 = \sum_{x= 1}^{c*K} (3x^2/K^33x/K^3+1/K^3)$ now we can rewrite (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/K3X/K^2+1/K^3) <> \sum_{x= B+1/K}^{C} (3x^2/K3x/K^2+1/K^3) $ So if you digest the step sum you can see the right hand side of the FLT that JUST if we pull $\displaystyle K\to\infty$ we can REMOVE THE <> sign and write: $\displaystyle \lim_{K\to\infty} \sum_{x= B+1/K}^{C} (3x^2/K3x/K^2+1/K^3) = \int_{B}^{C} 3x^2 dx = C^3 B^3 $ It’s easy to see 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/K3x/K^2+1/K^3) < \lim_{K\to\infty} \sum_{x= B+1/K}^{C} (3x^2/K3x/K^2+1/K^3) = \int_{B}^{C} 3x^2 dx = C^3 B^3 $ So there is no $\displaystyle K\in Q$ 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. I lost lot of time turning words on... I hope proferros will just translate in good math... 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(x1)^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, I use a variable (of a known function) thickness slice. Last edited by complicatemodulus; January 19th, 2016 at 09:11 AM. 
January 19th, 2016, 01:15 PM  #5  
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244  Quote:
Nope. Both sums come out to be $C^3  B^3$ exactly. Again, the sum is independent of $K$. Taking the limit does nothing.  
January 19th, 2016, 09:29 PM  #6  
Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24  Quote:
Quote:
So that rising K we always stop the sum at the same upper limit if it's not fixed but it's the variable of the problem ? You probably miss the fact that the upper limit C is not fixed in the problem, but is the variable that is "limited" just by the formula you choose: in the left hand side of the rational step sum you MUST use JUST $\displaystyle K \in N$ so you can just rise JUST to $\displaystyle C \in Q$ Viceversa in the right hand you can make infimus step, so you can rise to $\displaystyle C \in R$ So since C is variable, if it is in N or in Q , rising K, we have to find all same results as happen for n=2 due to flat derivate, or if n>=3 another K that must be equal to the limit we already prove we reach when $\displaystyle K\to\infty $ Since we prove (numerically or remembering that the first derivate is a curve, or just seeing what happen to the limits/terms of the sums) that rising K we can just go closer and closer to the "real" unknown C: the sign "<" rest always true... Again: since we can prove (viapushing the sum to the limit) that one limit exist, since we can see and prove we are on a monotone rising function Only one limit can exist So if the right hand can give higer result, that is the right limit we have to consider. I left the "Step Sum" formulation because it is more clear what happen rising K:  both the lower limits, of both the sum (left and right hands), fells,  both the upper limits, of both the sum, can rise, but just going to the limit for $\displaystyle K\to\infty $ we can have an integral where:  the lower limit is exactly zero, and  the upper limit can be an irrational (so bigger than the one in the left hand side), and  All the bigger order infimus terms will vanish, so just the first survive  so right hand (once proven true) is always bigger that the left hand side (as told remembering that the upper limit C is the only variable in the equation) And, sorry, the point is you don't understand that this is the "right" process that take the sum to the integral (the one can be used to teach what really connect the sum with vanishing terms to the integral...) I post the picture, you've just to figure out, there, what happen rising K... You make thinner and thinner slice till you cover the area bellow the derivate so you make the integral (it's exactly a Riemann integration process, the only difference is that you've to understand that we always cover the same area, indipendently by the K we choose, JUST if we stop at the same x, while here we do not since C is not fixed "a priori" ). I know this is the point to close the proof, I hope some professor can better explain... I really appreciate concernings on, but just with proof of what you said is really right (while now it seems viceversa). Thanks Ciao Stefano Last edited by complicatemodulus; January 19th, 2016 at 09:38 PM.  
January 19th, 2016, 10:07 PM  #7 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244  Okay. On rereading your post, I found that the situation is much worse. We already know that $$\sum_{x = 1/k}^{ak}\left(\dfrac{3x^2}{k}  \dfrac{3x}{k^2} + \dfrac{1}{k^3}\right) = a^3k^3 = \int_0^{ak}3x^2\,dx.$$ I think we can agree on that. Now, let's look at the limit as $k\to\infty$. It's not so clear what happens on the lefthandside, so let's ignore it for now. The expression in the middle approaches $\infty$... Okay... The integral on the right approaches $\displaystyle \int_0^\infty3x^2\,dx$ which diverges... Hmmm... That gives us $\infty = \infty$. Unfortunately, that's meaningless. In fact, the statement $$A^3 = \sum_{x = 1/k}^A\left(\dfrac{3x^2}{k}  \dfrac{3x}{k^2} + \dfrac{1}{k^3}\right) = \lim_{k\to\infty}\sum_{x = 1/k}^A\left(\dfrac{3x^2}{k}  \dfrac{3x}{k^2} + \dfrac{1}{k^3}\right)$$ is wrong for that reason. $A^3$ is (I presume) finite, whereas the limit of the last sum is infinite. The only time this is correct is when $A$ does not depend on $k$, but you've already stated that it does. Then what's the significance of $314^{1/3}$ in your spreadsheet? 
January 19th, 2016, 10:42 PM  #8  
Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24  Quote:
So we, for sure, can't go to infinity with the upper limit, or we prove the solution of our problem is wrong, that is what I did going infimus 8that is more clear and easy to be understood). The UPPER LIMIT EXIST and is bounded; what means $\displaystyle C\in R$ is that C (one we prove is not in N or in Q) is an irrational An irrational is a bounded number since we can prove that there exist a bigger one ( C+1, for example...) but we cannot write it on a piece of paper... Quote:
When we are talking of a limit for $\displaystyle K= \to\infty$ we are not saying: Since A is finite $\displaystyle K= \infty$, so must be $\displaystyle a = A/\infty$ ... You agree ? Quote:
If it has an integer or a rational root you'll find a finite K (1 or bigger) that will exactly stop the process with rest zero, while if the root is an irrational you can continue to rise K and you'll never stop the process, but it do not means that your number is infinity, just that it is an irrational... and you always have a proof of that since at the next step you go over what you expect... but time by time less over... So we are are walking on a curve me from left and you from right. The tangent to the point where we are is the same just when we meet on the same point... Last edited by complicatemodulus; January 19th, 2016 at 11:01 PM.  
January 20th, 2016, 07:30 AM  #9 
Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 
Just another note: Since we work base 10 when we extract the root (here the cubic of 341 as example) if we keep K=10^m, we have the result of the root with m digits. If you try to rise m from 1 to p, each number you reach when the next x let the sum bigger than 341, is the rigth one. This is different from other faster method that gives lot of wrong digits after the significant one you decide when you decide to stop the approximation. Last edited by complicatemodulus; January 20th, 2016 at 07:34 AM. 
January 20th, 2016, 02:43 PM  #10 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 
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, $KA$. 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 $KA$. This is the last post I'll make on this subject, since it seems hopelessly full of contradictions and errors. 

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