Proof for other values of

Comments on Andrew Wiles’ proof

Andrew Wiles’ career

In May 2018, Wiles was appointed Regius Professor of Mathematics at Oxford.

The British Mathematician Andrew Wiles was born on 11th April 1953; he is the son of Maurice Frank Wiles and Patricia Wiles. He received his Bachelor’s degree in Mathematics at Merton College (Oxford). He also carried out research at Clare College, Cambridge and finally got his PhD in 1980.His mathematical research was on the methods of the Iwasawa Theory to work on the arithmetic of elliptic curves with complex multiplication.

He continued the idea of proving the Fermat Theorem in 1986, when Ribet had already proven Serre’s -conjecture and as a result had formulated a connection between Fermat’s Last theorem and the Taniyama-Shimura conjecture. Andrew Wiles presented this idea in 1993 publicly in a conference at Cambridge, but there was a shortcoming within the proof. Consequently, with the help of his ex-student, Richard Taylor, in 1994 he was able to fully support the proof which was published in Annals of Mathematics.

The structure of the proof

It is required to prove the modularity theorem (MT) for elliptic curves. If this is proved, then the Frey curve (FC), given by the equation is impossible as it is not modular, and all elliptic curves are modular if MT is true. If this is so, there is no counterexample for FLT.

If MT is true, then FC is not true (modus tollens).

If the FC is true, then a counterexample, CE, exists for FLT (modus ponens).

Wiles proved MT, and consequently there is no counterexample for FLT

The modularity theorem

Modularity: The modularity theorem, also known as the Taniyama—Shimura conjecture, asserts that every elliptic curve defined over the rational numbers is modular. This means that it has no cusps. A semistable elliptic curve may be described as an elliptic curve that has bad reduction only of a multiplicative type. This is the case where the reduced curve has one singular point with two distinct tangents. We see this in Figure 2 below.

Curiously, Fermat proposed the following equation and the solutions.

Solutions are , . We do not have Fermat’s proof.

Here is Wiles’ introduction to his paper proving the Modular Theorem and consequently implicitly proving FLT. However this was made possible by the findings of other scientists.

“An elliptic curve over is said to be modular if it has a finite covering by a modular curve of the form . Any such elliptic curve has the property that its Hasse-Weil zeta function has an analytic continuation and satisfies a functional equation of the standard type. If an elliptic curve over with a given -invariant is modular then it is easy to see that all elliptic curves with the same -invariant are modular (in which case we say that the -invariant is modular). A well-known conjecture which grew out of the work of Shimura and Taniyama in the 1950’s and 1960’s asserts that every elliptic curve over is modular. However, it only became widely known through its publication in a paper of Weil in 1967 (as an exercise for the interested reader!), in which, moreover, Weil gave conceptual evidence for the conjecture. Although it had been numerically verified in many cases, prior to the results described in this paper it had only been known that finitely many -invariants were modular. In 1985 Frey made the remarkable observation that this conjecture should imply Fermat’s Last Theorem. The precise mechanism relating the two was formulated by Serre as the -conjecture and this was then proved by Ribet in the summer of 1986. Ribet’s result only requires one to prove the conjecture for semistable elliptic curves in order to deduce Fermat’s Last Theorem.”

We shall now examine the case of the Frey Curve which leads us to the proof of FLT.

The Frey Curves

In 1975, Yves Hellegouarch , a French mathematician from the Ecole Normale (ULM) wrote a paper on elliptic curves in Acta Arithmetica. Those curves were associated with Fermat’s last theorem. Hellegouarch associated solutions with FLT. The curve consists of all points in the plane whose coordinates satisfy the relation: . Such an elliptic curve would have very special properties. But it would not be modular, and as Wiles proved all elliptic curves are modular, this curve would not exist, and there would therefore be no counterexample for FLT. This curve has become known as a Frey curve.

A Frey curve is given by the equation related to an triple. In other words, elliptic curves may provide solutions to equations. It was Yves Hellegouarch, who in 1975 put forward the idea and its relationship to Fermat’s equation. “Given a prime number, , we know there exists a constant such that no abelian curve, defined under , allows for any rational point of order for ; nevertheless, the numerical value of , for any is not known.” (Acta Arithmetica, XXVI, 1975).

Gerhard Frey (1982) found that the curve had unusual properties. Fermat’s conjecture was now related to the Taniyama-Shimura-Weil conjecture and essentially if there were a counterexample to Fermat’s Last Theorem, there would exist a curve that would not be modular. Frey in 1986 suggested that the Taniyama-Shimura-Weil conjecture implies Fermat’s Last Theorem. In 1985, Jean-Pierre Serre proposed that a Frey curve could not be modular. This showed that a proof of the semi-stable case of the Taniyama-Shimura conjecture would imply Fermat’s Last Theorem. This was the challenge given to Wiles: prove the Taniyama-Shimura-Weil conjecture.

A study of Fermat’s remarks on Diophantus

We find his remarks in Bassine’s Précis des Oeuvres Mathématiques et de l’Arithmetique de Diophante, translated from Latin into French. and published in 1621 by Claude Bachet. The important question was finding the sum of two odd squares equal to 16. Now we know that the difference between two odd squares equals 16 or but not the sum of two odd squares in terms of natural whole numbers. It was the study of this problem that allowed Fermat to put forward his conjecture without giving a proof. Bassine . Fermat was studying Arithmetica, which seemingly inspired Fermat’s conjecture. Fermat was working on Problem II.8 of Arithmetica which raises the question: how can a given square number be split into two other squares; in other words, for a given rational number , find rational numbers u and v such that . Now Diophantus chose an even square 16 and shows how to solve this sum-ofsquares problem for (the solutions being and . However, However, the difference between and , , has a solution in natural numbers: , and corresponds to the first Pythagorean triple known. Thus, we can reformulate the problem in these terms: if we have , where is even, and and are odd, then if there is an integer solution for , there will not be one for , and this observation leads as to the case of , which splits up into: . We can therefore split into two factors, , which has integer solutions, and , which does not.

Proof by Diophantus: To divide a given square into a sum of two squares:

We first let the square and divide 16 into a sum of two squares.

Let the first summand , and the second , which is to be a square. We make the square of the difference of an arbitrary multiple of which is diminished by the root of 16 which is .

We find the square of , which is . Let this expression be equal to . Thus (artifact in OCR). Actually .

Now add to both sides then subtract 16. Thus, we obtain , hence .

We now have 256/25 and 144/25. The sum of these numbers is 16 and each summand is a square. Thus .

One possible proof for

Splitting the equation:

Let with and relatively prime in and .

We begin with .

Let and . We need only prove that is not an integer square if is a square.

Let and according to the Euclid formula.

Therefore , the square root of which is .

. So, is a square, but not .

Can we conclude that so not equal to .

Statement for FLT with

FLT can be stated thus: With and , and , the equation has no integer solutions. We chose and to be odd numbers and integers, and we must show why is not a whole number but an irrational one.

We shall now endeavour to find a method allowing us to prove FLT. We need to show why is not an integer, and we shall use the two factors of to prove this.

When , both and must be integers, otherwise cannot be an integer, and we have a possible proof of Fermat’s conjecture.

Let . Now is even because , being two odd numbers is even and must be an term to divide , in the present case .

If is divisible by which itself is an even cube, we are left with which we must prove, cannot divide (1) (which is equal to ) with an integer cube root, at the same time as , when . If both and divide (1), and are both integer cubes, then we have a counterexample, since the third term , would also be a cube. It is easy to find values of and which allow to divide (1) and leave a residual since , and is the even factor of equation (1). Let us show happens. Let , then, . We can now take out of the equations and see what might be the result.

We begin, by substituting for , with .

We now divide by and consequently, we find .

Let us examine this result. If , . Surprisingly we have the sum of 2 cubes. In this case, however, , so, is not an integer cube root when . If , , then .

Generalizing, for any integer value of :

We can show this numerically using a computer program.

We can conclude that the second term of the multiples of equation (1), , and we let . However, must be an integer to find the counterexample.

Let us attempt to show why does not have an integer cube root.

Now by squaring and subtracting .

Let . We can see that is not a cube with an integer.

We can test this. If , then , when . For , , then . We can see that, in cases where , .

When , .

For , we have the expansion . As .

Let , . We now have .

For , . If , . So, .

The factors of are .

So . Proving that is not an integer when would be difficult, but when we divide the equation by , we find , providing .

Generalizing for

There is an easy way to prove that if is an integer, and we chose 2 for our examples, , the root of is not an integer but an irrational number.

, and .

Now , so . If or or both are not natural numbers, then is not a whole number and the theory is proved. We can make and we now have .

We are now required to prove that is an irrational number. If for example factor and not , the theorem would also be proved without looking for the value of .

In general, for we can factorise in the following way:

The even factor always divides the Fermat equation when we choose .

So, we choose . Therefore . We can easily find through division of .

Let the first factor be , even, the second one be odd. With this choice of values of and , divides and the quotient is with . We have seen that when we divide by , the quotient is an odd natural number. It is convenient to begin with the division by , and we find without having to calculate it from the formula. The question now is how can we prove that has no integer root.

We must choose and with . Otherwise would not be an integer.

What we need to prove is that factor does not have an integer root.

The following equation is the one we need to solve to prove that there is no counterexample:

Theorem

Let us try this demonstration.

First, we let . So, has an root 2, which divides equation (1), and leaves the odd number . It would be difficult if not impossible to prove directly from the expansion of , with , an integer, that could divide the equation and leave .

Let . We know that , so, we have in terms of . We substitute and we only keep .

Proof of FLT

Object: To prove that and divide the equation , with a natural number, and an irrational number, thus is irrational since .

We shall now try the following method to prove this theorem. The same applies for and if they are the unknown values using a similar equation.

Let and .

We wish to show that is not an integer. We divide the equation by .

We now have .

We divide and separately by .

Thus .

And changing sides for , we now have the key to our proof.

Thus .

We find the root of both sides.

and therefore, equation (2), substituting for . The important element is , which will always be smaller than 1. is a positive decimal and smaller than 1. is also a rational number and a fraction.

Why is ? The odd factor is in fact much larger than . If we look at the factors for , we find and . On one hand, we have a subtraction of and , on the other the addition of two squares and the product . This is true whatever the value of . Consequently, and so .

Let . Now must be an integer and must divide (or ) for to be an integer. However, divides leaving . Moreover, lies between and , as , so (or ) lies between , so and . In fact, .

Consequently, is a fraction, and so is not an integer and cannot divide leaving a whole number. This is the key point.

Thus, there is no counterexample for FLT, and cannot be an integer, since and we have shown to be irrational.

Consequently, as cannot be an integer, and , is irrational.

We could choose , but the result for would be the same. For instance, would divide the equation and leave but would not be an integer as shown above.

Testing the results numerically

Let us look at some examples. Here are some numerical examples that confirm the results of our equation.

1. Let .

   If . Consequently . To find we divide by 8, and we have 163. Using the formula, .

   Now , so .

   Now .

   Let us check with the cube root of .

   Therefore, our equation allows us to find the value of , which is irrational and indicates why it is so.

   Note that .

2. Let us take the example of , and .

   Now . .

   . Both are identical to five decimal places.

3. Let . We have made , and . Now . So, . Let . . . . Now . Cube root of . So, when is equal to an root of an even number, then we can see that is not an integer.

Example shows that when or become greater and greater, the value of which we call , becomes smaller and smaller and probably tends to 0.

Direct demonstration of the irrationality of

The following demonstration shows directly why is irrational whatever the value of .

We use the same algorithm we used to find : express the equation, divide by the RHS. Take the second term to the RHS. Find the root of both sides, and express the result equal to . We can clearly see why is irrational, and cannot therefore be a natural number, consequently, there can be no counterexample for FLT.

Let , where and are odd, is even. We wish to show that is not an integer and is in fact irrational. We will use the same method and equation above to prove is irrational.

Now .

Therefore,

We now take the root of both sides, and we have , where .

Therefore , equation (3), or . Where or

Now we know that is a natural number since it is the difference between two odd natural and , but what about ?

We now have and as is irrational, and is getting smaller as becomes greater and greater with respect to and tending to zero.

We can verify the exactitude of the above equation.

The data. Let , , , and .

We find and .

Now . Consequently, our equation holds.

We assume that our equation is exact for all natural numbers, and , odd and for all values of .

We also noticed that if or is the number we are looking for, and the other two terms are natural numbers, then, using the same method as explained above:

We find that , and . and , or , or are irrational numbers.

One point remains, is a natural number when .

Let , , and from the Euclid equation.

We now substitute in equation (3), and for , , and as indicated above.

Thus, is a natural number when we use and to find Pythagorean triples, otherwise is an irrational number.

Discussion

The key point in giving a proof of FLT is in fact the relationship between the non-modular semi-stable elliptic curve or Frey curve and the equation , where , and are nontrivial natural numbers. Indeed corresponds to the curve given by

If all semi-stable elliptic curves are modular, then the Frey curve could not exist because it is non-modular, so FLT is proved if the Modularity Theorem is proved. This was accomplished by Wiles. Without the suggestion given by Hellegouarch, and the proof that the Frey curve could not be modular, by Jean-Pierre Serre, FLT would not have been proved. Consequently, a group of mathematicians made it possible to prove FLT along with many other theorems: the logic is as follows, if the Frey curve is related to FLT, and if it is not modular, whereas all semi-stable elliptic curves are modular, then Frey curves do not exist and FLT is proved, there are no integer solutions for x, y, and z..

As Gödel had proved: Every logically consistent system of arithmetic contains at least one proposition that cannot be proven or disproven within the system itself. This means that no formal system can be complete in the sense of containing all true statements (First Theorem). This seems to be the case of FLT. It was necessary to find the proof outside the domain of classical algebra, using modus tollens and with reference to many recent discoveries in number theory.

Conclusion

We believe we developed a way of proving the theorem with intuition, a heuristic, and a little serendipity.

If , we have two factors, the first, , is even, and the second is odd. Consequently, to find a counterexample of FLT, and must be cubes, and their cube roots must be integers. All we had to do is set . Thus, we could divide out the even numbers, and we were left with the odd factor. We knew then that divided equation (1); we had to show that did not have an integer root, . This could be applied to . Our proof was given by the equation:

Fermat’s intuition began with the study of Pythagorean triples which we explored in different ways in our preparatory research and especially the problem raised by Diophantus, concerning the possibility of an even square being the sum of two odd squares, which probably lead to Fermat’s proof for .

I was also interested by the way Wiles, with the discoveries of Hellegouarch (who found a relationship between elliptic curves and FLT), Serre, Frey, and Ribet who proved that such a relationship was exact, had found the missing link that would allow him to prove FLT.

We hope our attempt can be verified and show that for FLT, we remained inside the scope of century mathematics.

We also found a simple direct way of showing why was not an integer, but an irrational number using our equation: .

I would like to thank Pr. Matthieu Manant and Pr. Antoine Auberger for their help and advice in my research. And especially my thesis supervisor Jean-Marc Labat, Sorbonne professor in IT, and my alma mater, Ecole National Supérieur des Telecoms (Telecom Paris), and of course the University of Oxford where I still study (Calculus, Astrophysics, and Ancient Greek). Also, Dr. Myriam Wisdom for her encouragements and patience. And I must not forget my ex-aequo at the Agrégation, Pr. Florent Gusdorf, former professor at the Ecole Polytechnique and the director of Languages and International communication, for his long-term friendship and his encouragement.