My Math Forum Easy proof of Fermat's last Theorem

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 April 30th, 2019, 11:07 AM #21 Member   Joined: Mar 2019 From: california Posts: 40 Thanks: 0 You will have to show how that qapplies.
 April 30th, 2019, 01:35 PM #22 Member   Joined: Mar 2019 From: california Posts: 40 Thanks: 0 Thank you for waiting for a reply. I felt it was more important to reply to Masche about the other Fermat's theorem. Now that he has agreed that f/l is greater then 1, the proof is very straightforward, being a simple manipulation of an equation. Then when you substitute the ranges in for the different ratios I am reasonably certain that he has already done every check that he can think of to come up with a counter example. If it was me, with his connections, I would be contacting other mathematicians to see what they could come up with. It's all over except for the shouting. If you looked at Masche's reply below, he stated exactly what the the answer should be to your first question. The second part of your reply is whether X^n/2 will get smaller than X^2? And the answer is that it won't, not that it can't but that it would violate the definition of what the Pythagorean theorem means. To do the next iteration and keeping the X^n/2 leg the same would mean that The end point of the X leg would extend past the endpoint of the Z leg. So yes, there is an end to the iterations that can be achieved. The idea of infinite number of them is only due to the fact that the value of n is unknown..As soon as we know n, we can determine how many iterations to make.
April 30th, 2019, 03:13 PM   #23
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Quote:
 Originally Posted by michaelcweir Thank you for waiting for a reply. I felt it was more important to reply to Masche about the other Fermat's theorem.
Let me jump in here if I may.

Quote:
 Originally Posted by michaelcweir Now that he has agreed that f/l is greater then 1, the proof is very straightforward, being a simple manipulation of an equation. Then when you substitute the ranges in for the different ratios
I've tentatively come to that conclusion, though I haven't completed my analysis of the inequality.

Still, I asked you to rewrite your clearly and with whatever modifications are needed now that you realize you had your inequality backward. That's a fair request.

Also, aren't you curious to go through your own argument to see how you arrived at the wrong inequality? When I went through your argument I found that it didn't even prove the WRONG inequality. Your argument didn't prove what you said it did. You should go back through your work and see where you went wrong.

Quote:
 Originally Posted by michaelcweir I am reasonably certain that he has already done every check that he can think of to come up with a counter example.
No. I haven't spent any time on it at all. I tried to indicate that in my previous post to you. In general I almost never get involved in discussions of alleged proofs of FLT. For one thing it's not something I know much about ... either the relevant number theory or the study of "amateur" proofs. For another thing some people who post these kinds of proofs are very sure of themselves, completely out of proportion to their mathematical ability and the likelihood that they're correct. So getting involved with these kinds of threads is something I generally avoid.

In this case I did spend some time on it. I'm not saying I wouldn't in the future. I've just run out of interest. I don't feel like going through the algebra. Also as I mentioned, someone else early on pointed out an error in your algebra which you neither acknowledged nor corrected.

Anyway no, I haven't looked for counterexamples and I also haven't thoroughly analyzed the inequality. I just ran out of steam for this little project. But a clear re-write, line by line, and taking into account the errors that have already been found, would be helpful to me and to others.

Quote:
 Originally Posted by michaelcweir If it was me, with his connections, I would be contacting other mathematicians to see what they could come up with.
LOL. Thanks. I'm not a professional mathematician nor an academic. Just someone with a math and computer background who hangs out on discussion forums.

Quote:
 Originally Posted by michaelcweir It's all over except for the shouting.
You haven't got a proof. Your certainty is unwarranted.

Quote:
 Originally Posted by michaelcweir If you looked at Masche's reply below, he stated exactly what the the answer should be to your first question.
The post you're pointing to is unclear. Can you rephrase what you are talking about?

Quote:
 Originally Posted by michaelcweir The second part of your reply is whether X^n/2 will get smaller than X^2? And the answer is that it won't, not that it can't but that it would violate the definition of what the Pythagorean theorem means. To do the next iteration and keeping the X^n/2 leg the same would mean that The end point of the X leg would extend past the endpoint of the Z leg. So yes, there is an end to the iterations that can be achieved. The idea of infinite number of them is only due to the fact that the value of n is unknown..As soon as we know n, we can determine how many iterations to make.
This is the point in your triangle argument where I ask, "Even if this is true, how does it prove FLT?"

Quote:
 Originally Posted by michaelcweir You will have to show how that qapplies.
This refers to SDK's remark that the p-adic numbers show that your proof can't be right. Here's what he was referring to.

Mathematicians study all kinds of different number systems. One class of them is the p-adic numbers. We don't need the details here but they are important in number theory. In fact Wiles's proof of FLT heavily uses them

It turns out that in the p-adic numbers there are FLT triples, thus making FLT false. However, the p-adic numbers have a lot of the same algebraic properties as the regular old integers. So that if you have an alleged proof of FLT based on elementary algebraic manipulation, it must be wrong, since it would equally well apply to the p-adics.

Theres a Quora thread on this:

https://www.quora.com/What-if-I-foun...s-last-theorem

I'll quote the relevant part.

Quote:
 As a final note, depending on exactly what you mean by elementary, it might well be impossible to find such a proof. As in, such a proof doesn't exist. In particular, the answer by Nguyen Quang Do at Fermat's Last Theorem simple proof shows that no proof relying only on addition, multiplication and unique prime factorization can possibly exist for Fermat’s Last Theorem as any such argument would work for the p-adic integers ... it can be shown that Fermat’s Last Theorem is false for the p-adic integers (that is, there are p-adic integers $x, y, z$ such that $x^p + y$^p = z^p for $p>2$)
In other words: If you have an FLT "proof" based on elementary algebra; it must be wrong, since it applies equally to the p-adics, in which FLT is false.

Last edited by Maschke; April 30th, 2019 at 03:26 PM.

April 30th, 2019, 06:37 PM   #24
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Quote:
 Originally Posted by Maschke Let me jump in here if I may. This refers to SDK's remark that the p-adic numbers show that your proof can't be right. Here's what he was referring to. Mathematicians study all kinds of different number systems. One class of them is the p-adic numbers. We don't need the details here but they are important in number theory. In fact Wiles's proof of FLT heavily uses them It turns out that in the p-adic numbers there are FLT triples, thus making FLT false. However, the p-adic numbers have a lot of the same algebraic properties as the regular old integers. So that if you have an alleged proof of FLT based on elementary algebraic manipulation, it must be wrong, since it would equally well apply to the p-adics. Theres a Quora thread on this: https://www.quora.com/What-if-I-foun...s-last-theorem I'll quote the relevant part. In other words: If you have an FLT "proof" based on elementary algebra; it must be wrong, since it applies equally to the p-adics, in which FLT is false.
This link is great. The quora link is not bad, but the answer on quora links to a stackexchange post which is perfect. There is a comment there that clearly explains the reasoning much more clearly than I would have, even if I was inclined to try and explain it, which I'm not. Thanks for sharing this. I'll quote it to save people some time clicking.

Quote:
 There is a "trick", due to Marc Krasner, which prevents you from wasting time in examining "elementary" arithmetic proofs of Fermat's Last Theorem. "Elementary" means precisely that the proof uses only addition and multiplication (operations in a ring), and perhaps also the existence and unicity of decomposition into prime factors (so the ring in question is factorial). I suppose this is the case here, although not all details are given. Then, without checking anything, you can be assured that the reasoning is certainly wrong. This is because all such "elementary" arguments can be repeated word for word in the ring $Z_p$ of p-adic integers, which is factorial (and a lot more !), but in which FLT is false, because in the field $Q_p$ of p-adic numbers, the equation $x^p + y^p = 1$ always has non trivial solutions (if you take $y$ to be a high power of $p$, then p-adic analysis tells you that $1 - y^p$ has a p-th root in $Q_p$).

 May 6th, 2019, 05:32 PM #25 Member   Joined: Mar 2019 From: california Posts: 40 Thanks: 0 i posted handwritten analysis, but I had to upload 3 times before it was readablea s pdf.
 May 6th, 2019, 05:34 PM #26 Member   Joined: Mar 2019 From: california Posts: 40 Thanks: 0 It's in the other proof.
May 6th, 2019, 05:34 PM   #27
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Quote:
 Originally Posted by michaelcweir i posted handwritten analysis, but I had to upload 3 times before it was readablea s pdf.
IMO even crude ASCII math typed in here is preferable to handwritten. You can copy/paste symbols from https://math.typeit.org/.

Or you can pick up a bit of Mathjax. Just hit the Quote button to see how I did this:

$x^n + y^n = z^n$

But even without that, just straightforward ASCII is fine.

x^n + y^n = z^n

That's a million times better than trying to decode a handwritten exposition.

Last edited by Maschke; May 6th, 2019 at 05:37 PM.

May 19th, 2019, 01:19 PM   #28
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Hi Masche
I'm doing my best to comply with your request. I decided the best way to do that is to follow Fermat's example. Which is to give you the bare minimum to solve the proof.

I have posted 2 diagrams the use of the Nautilus process to determine that Z, for any n greater than 2, must be less than y+1. Since Z must be greater than Y, Y<Z<Y+1. Thus Z must be irrational. This proves FLT..

from fig. 2 Z^2 +X^2 < X^2 +(Y+1)^ 2

This implies Z^2 < (Y+1)^2

OR z < y+1
Attached Files
 fermat nautalis20190519_14010157.pdf (6.6 KB, 5 views)

 May 19th, 2019, 02:27 PM #29 Senior Member   Joined: Mar 2019 From: iran Posts: 150 Thanks: 9 z can be greater than y+1 for example 28^10 + 30^10 = (31.24)^10 31.24 is greater than 30+1 you are begging the question in fig 2 you assume side z is smaller than side y+1 then you get z is smaller than y+1 Thanks from Maschke
 May 20th, 2019, 01:06 PM #30 Member   Joined: Mar 2019 From: california Posts: 40 Thanks: 0 Yes that is certainly true. But it also is true that 31.24 is not an integer. If you say Z^n is an integer and Z^n = X^n + Y^n, then Z^n

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