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 August 6th, 2017, 02:16 PM #191 Senior Member   Joined: Mar 2017 From: . Posts: 274 Thanks: 5 Math Focus: Number theory There may not only be different values of a and b but also of n. So, the summation results to polynomials with different coefficients and also different lengths (for different values of a, b and n. All of which are integers. The only way of expressing those polynomials is through that summation notation. Let's not debate on that point. Alright, I agree they result to integers and for any example of such an integer we can denote it with any letter. Still, we want to show how the integers relate to each other, for example if we have integer $p$, we don't denote the other integer as $q$ or $qx$. Instead, we just show how they relate to each other in the summations form. We can easily deduce that each equation above is correct. We cannot factor out any integer from the summation and therefore if we have integers $p$ and $q$ as in the other method, then we know that either of them divides the other to give 1. Hence p=q. Also I hope this discussion is not about who is right and who is wrong but what is right and what isn't. Let's let it be about the math itself and provide insight, modifications or counterexamples where necessary. Also, kindly quit the condescending tone. Last edited by skipjack; August 7th, 2017 at 10:59 AM.
August 7th, 2017, 09:07 AM   #192
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 Originally Posted by Mariga There may not only be different values of a and b but also of n. So, the summation results to polynomials with different coefficients and also different lengths (for different values of a, b and n. All of which are integers. The only way of expressing those polynomials is through that summation notation.
What about $$\sum_{k=0}^n (a^{k+1} - a^k) = \sum_{k=0}^1 a^{k(n+1)}$$
It sure seems like we can express a formula with all kind of different sumations, no?

August 7th, 2017, 09:33 PM   #193
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 Originally Posted by Micrm@ss What about $$\sum_{k=0}^n (a^{k+1} - a^k) = \sum_{k=0}^1 a^{k(n+1)}$$ It sure seems like we can express a formula with all kind of different sumations, no?
But it can be factored out
$$\sum_{k=0}^n (a^{k+1} - a^k) = \sum_{k=0}^n (a^k(a-1) )= (a-1) \sum_{k=0}^n (a^k)$$

So, there are already different ways of expressing the summation.

August 8th, 2017, 05:45 AM   #194
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 Originally Posted by Mariga But it can be factored out $$\sum_{k=0}^n (a^{k+1} - a^k) = \sum_{k=0}^n (a^k(a-1) )= (a-1) \sum_{k=0}^n (a^k)$$ So, there are already different ways of expressing the summation.
Yes, so who says your expressions don't have different ways of expressing themselves? What are the rules your expressions must obey?

 August 8th, 2017, 10:47 PM #195 Senior Member   Joined: Mar 2017 From: . Posts: 274 Thanks: 5 Math Focus: Number theory $\sum \limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}$ cannot be factored out. Neither can we manipulate what's in the summation to give us a different expression of the summation.
August 8th, 2017, 11:26 PM   #196
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 Originally Posted by Mariga $\sum \limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}$ cannot be factored out. Neither can we manipulate what's in the summation to give us a different expression of the summation.
Prove it!

 August 9th, 2017, 03:03 AM #197 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Infact.... $$\sum_{i=1}^n(\sum_{x=1}^{a}(2x-1))^{n-i}(\sum_{x=1}^{b}(2x-1))^{i-1}$$ or: $$\sum_{i=1}^n(\sum_{x=1}^{a}(2x-1))^{n-i}(\sum_{x=1}^{a}(2x-1)+\sum_{x=a+1}^{b}(2x-1))^{i-1}$$ ...so how can you say it cannot be factored out ? Because Wiles prove it ? You are in a cul de sac, and no exit from there using "linearized" terms in this way. you must use the classic develope: $$A^n=\sum_{x=1}^{A}(x^n-(x-1)^n)$$ than with lot of tricks prove that you are forced to an infinite descent (so use the integral) in the case $n>2$ where there is no other way to computate an area bellow a derivate curve (of $Y=2X^n$) with an irrational (right) limit starting from two Integer Areas (A,B we know are integers) having integer (right) limits bellow the same curve. This is a dedekind section plus a concerning on the area over it.... (3th time) Last edited by complicatemodulus; August 9th, 2017 at 03:08 AM.
August 9th, 2017, 08:01 AM   #198
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 Originally Posted by Micrm@ss Prove it!
He can't prove it. He cannot prove it about sets because he has not even defined what a ratio of sets is. And he cannot prove it about numbers because he is relying on a theorem about numbers that is untrue.

The whole thing is pathetic. He asserts that he has a simple proof based on algebra, but it involves denying that sums of products of unspecified numbers are numbers and implicitly asserting that he has a special type of algebra involving summations with rules of inference that he has not specified. And of course he is trying to prove something that already has been proved.

I suggest we stop feeding the troll.

 August 11th, 2017, 01:20 AM #199 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 I re-post there how a Two Hand Clock works: Here the .Gif film where you can see both $n=2$ and $n=3$ counter: Some interesting behaviour are immediately visible: - the hours hand in case $n=2$ moves forward, while the $n=3$ after reach $2$ moves backward. - the Sum of 2 equal numbers, one in base $2$ the other in base $3$ produce a nice behaviour on the Rest ...etc...
 September 2nd, 2017, 10:24 AM #200 Senior Member   Joined: Mar 2017 From: . Posts: 274 Thanks: 5 Math Focus: Number theory I am back after a bit of a break. I can now agree the proof isn't really complete and I couldn't be productive any more due to exhaustion. I will see what can come out of it and share.

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