However, you have no right to tell anything wrong regarding my proof. From where we get the wrong proof proposed by Lame in ? They are numbers but by no means integers. I challenge that no one in the world can not find integers that satisfy the condition of FLT. Is it interesting? Email: thaotrangtvt3 gmail. And refraiming the problem can also mean refraising the formula that could ultimately proof FLT.

Then you run quickly out of options for c being an integer, I would say. Sorry, the correct spelling is of course 'reframing' and 'rephrasing'. Hey Bert I'm back!!

Incidentally Jeffrey's kicked me off his blog cause my simple math's is giving him a headache!! So what are the other two? Well the 'mathmagicians' can't be bothered to tell us so let's call them 'pseudo Pythagorean Triples' where only two legs are integer and one is irrational. When the integer generated is a square we get a 'PPT' and all the rest are therefore 'pseudoPT' by my definition. Why did he say that? As I've commented elsewhere before 'you can't push a square peg into a round hole' and likewise for the integers, you can't push a cubic or higher power solution into what is after all a Pythagorean equation.

So Pythagoras reigns supreme. I have to agree with Alastair on the Pythagorean influence. In fact, when you generalize the law of cosines you find Fermat as a special case of right triangles. You can show how the isosceles triangle is a trivial support to FLT due to Pythagoras' constant and how the generalized law of cosines holds for equilateral triangles.

The proof however takes a backseat to an amazing pattern that forms the Yin Yang of mathematics. The simplicity is stunning. Then you will find we generate 3 related pythagorean quadruples, 2 with all 4 terms positive and the other with 3 positive and 1 negative term. I have named named them Fermat - Bateman quadruples.

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## Fermat’s Last Theorem: The Curious Incident of the Boasting Frenchman |

So every term of a Fermat triple if one existed would have a simple solution in terms of squares after squaring once or infinitely many times so as Pierre de Fermat said years ago triples above the second power cannot exist. So for it to be repeatedly stated and dogmatically defended that the tools for him to have proved his theorem did not exist in his day is at the least erroneous and at worst a lie.

It is for the reasons given that no quadruples above the 4th power are known simply because they do not and cannot exist. It is worth stating that cubic quadruples also contain infinitely many related trios but none are algebraically of the construction of Fermat-Bateman quadruples. Is there any activity on this blog now? But I need some vistors first.

If you are interested, post a message under "Fermats original solution from ". Once I see the visitors are many enough, I will start to post. Why there? Why not in mathematical publishing magazine? Because I am not a professor,and those magazines will not under any circumstances treat the proof seriously. I already experienced that with Springer. But remember, Fermat himself was an amateur, and not very liked among contemporary so called professionals. So if YOU are interested, Hi all, I have a question for the statistical test: If we assume that the event "Fermat really had the proof" is a random variable, what probability measure would you be able to propose for the acceptance of this null hypothesis "Fermat really had the proof": 0.

This is the first step of how Fermat proved his last theorem.

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I have written out an informal proof proving FLT on the unit circle. It was motivated by trying to find the simplest geometric solution I could find to an even more fundamental proof developed in "Computable Universe". It is hard enough to get people to read a proof of P vs NP based on a new non-standard model of set theory much less use a solution of FLT as a example application. Via my former student Stijn de Vos, I just heard that Colin McLarty , a professor of philosophy and logician at Case Western Reserve University who I remember briefly meeting at a conference in Uppsala a few years ago , claims that it should be possible to formulated a much simpler proof of Fermat's Last Theorem, when compared to Wiles' page monster.

As is well known, Wiles' proof relies heavily on the theoretical machinery introduced by Grothendieck. You don't need sets of sets of numbers, which Grothendieck used in his toolkit and Andrew Wiles used to prove the theorem in the 90s. McLarty retorts: "Fermat's Last Theorem is just about numbers, so it seems like we ought to be able to prove it by just talking about numbers," McLarty said.

It will be very hard. Harvey sees my work as a preliminary step to that, and I agree it is. Any M-Phi'ers around with insider's knowledge on the whole thing? Calculus was gestating in the womb of European scientific culture, and the problems that led to the invention of calculus were the sort of problems that excited people. In contrast, Fermat had a hard time convincing people that problems like his Last Theorem had much value.

Fermat was urging his contemporaries to impose constraints of rationality, and integrality, for no reason other than the pure challenge of it. Equations that are easy to solve in real numbers often become extremely subtle when one adds as a side-condition that the numbers must be rational or must be whole numbers.

### Review of Fermat’s Last Theorem

Later generations came to see this subtlety as a good thing; the fact that these problems were hard but still doable was a sign that these were problems worth doing. But you need to know that for Fermat, making a statement without giving a proof was the norm, not the exception. I think Fermat really came up with an argument that he believed was a valid proof.

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But historians of mathematics think Fermat must have possessed, and believed, an argument of some kind. But can you come up with an infinite sequences of positive integers in which each term is smaller than the one before? If, say, the first term is a million, then the second must be at most ,, and the third must be at most ,, and so on; after a million terms if not sooner , the sequence finds itself boxed into a corner, since the terms of this decreasing sequence are required to be positive integers.

If the first term of the sequence is not a million after all but a larger number, say a billion, then the sequence is still doomed to terminate, even though it can postpone its inevitable termination considerably longer. The upshot is that there is no infinitely long descending sequence of positive integers; regardless of how big its first term is, sooner or later any descending sequence of positive integers must stop. Thus the large rectangle is what the Greeks called a golden rectangle: it consists of a square plus a smaller copy of itself. Golden rectangles all the way down.

Courtesy of Curtis McMullen. What if you want to draw a golden rectangle from scratch, and what if you insist that your golden rectangle should have integer sides, relative to some unit of length nanometers, say? But first, we need to make an innocent-seeming observation related to Figure 1: If the big rectangle has integer sides, then the smaller rectangle does too. To see what this leads to, look at the successively smaller golden rectangles in Figure 2.

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If the largest one has integer sides, then by the observation I made in the preceding paragraph so does the next, and so does the next. We get an infinite sequence of ever-smaller integer-sided rectangles. See the problem? Taking the short side of each of these rectangles, we get an infinite sequence of ever-smaller positive integers.

So there cannot be an integer-sided golden rectangle. The proof we just saw is an example of an indirect proof: to show that something is mathematically impossible, we show that its existence is self-inconsistent or inconsistent with what is already known. For instance, to show that there cannot be an integer-sided golden rectangle, we showed that the existence of an integer-sided golden rectangle would imply the existence of an infinite descending sequence of positive integers, which the principle of infinite descent assures us cannot exist.

But in math, precisely because we are dealing with mental constructs that are-what-they-are by definition rather than empirical observation, proof by contradiction is a legitimate way to reason.

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By looking in all the places where it might be and failing to find it in any of them? Think about the deductive process as a way of navigating around in some space whose locations correspond to states of knowledge. So this style of proof often gives you more elbow room than a direct proof affords. If you liked the proof that there is no integer-sided golden rectangle, you might try to prove, by similar means, that there is no regular pentagon whose five equal sides and five equal diagonals all have integer length; see the Endnotes.

Most historians incline toward the view that Fermat made a mistake. This hypothesis would be strengthened if historians could reconstruct some plausibly Fermat-ish false proofs of FLT. But even without a specific candidate in mind, we can see that Fermat was working on a problem in which mistakes would have been easy to make. This makes it very easy to accidentally construct a bogus indirect proof: just make an algebra error along the way, resulting in a contradiction that unbeknownst to you was not inherent in the collection of premises you started with but rather resulted from an error of reasoning.

We may never know the truth, but unless and until more evidence arrives, that seems the most plausible solution to this mathematical mystery. I am neither a number theorist nor a mathematical historian, so there may be mistakes in this essay; if there are, a they are my sole responsibility, and b please let me know about them! This was sometimes a point of confusion between Fermat and his correspondents. Fortunately, in the case of FLT, the point is moot.

If there were a counterexample to FLT in nonzero whole numbers, then since every whole number is rational there would be a counterexample to FLT in nonzero rational numbers.

The reverse is also true, since if we had a fractional solution, clearing out denominators would give us a whole number solution. The question was: does there exist a set of three rational numbers whose sum is 0 and whose product is 1? There does exist a smaller set of rational numbers having sum 0 and product 1, namely the empty set! Someone with some visual verve could make a great video illustrating this proof, dynamically highlighting the equality of various lengths by sliding and swinging line segments; if you make such a video, please let me know! Are you listening, 3Blue1Brown?

A variant of the principle of infinite descent is the least element principle, also called the well-ordering principle ; another variant is the principle of mathematical induction.