Monday, June 26, 2017



Over the years I have submitted my proof to more than 50 professional mathematicians. The mathematicians who have  rejected my 1965 proof have done so primarily because of the belief that there cannot be a proof of FLT using simple mathematical concepts.The few who have actually tried to refute FLT65, have attempted to support this belief with the fact that the division algorithm may or may not apply to integer constants obtained by substituting specific integer values into integer polynomials and reducing them to single integers. But there is no proof that it is true for the integer polynomials of the Fermat equation, and so three of them have resorted to demonstrations that have no relevance to actual solutions of the Fermat equation, to try to make the point that FLT65 may not be valid.

Their idea that the division algorithm might not apply to the integer polynomials of the Fermat equation factor f(Z) = Zp-1 + Zp-2X + Zp-3X2 + ••• + Xp-1 arises from the fact that, for given integer values of X and Z, f(Z) can be reduced to a single integer (a constant), and if that single integer is not prime, in general, one of its integer factors may or may not contain the integer equal to the integer value of Z1 – a.

FLT65 provides a way to determine whether or not any specific single integer value of f(Z) (a polynomial factor of the Fermat equation) can contain the specific single integer value of Z –a (a polynomial factor of Y in the Fermat equation) as a factor, using the division algorithm and its three corollaries. For an integer solution of the Fermat equation, f(Z1) must not only contain Z1 – a, it must be equal to (Z1 – a)p.

The division algorithm and its corollaries, by definition, apply to all polynomials with real number variables, so they apply to polynomials of integer variables in the same way they apply to all polynomials of real numbers because integers are real numbers which, along with non-integer rational and irrational numbers, comprise the field of real numbers. Finally, an integer solution of the Fermat equation, if there is one, is simply one of the infinite number of solutions to one of the Fermat equations, and the three numbers of any solution are a set of three numbers existing in the field of real numbers. FLT65 demonstrates the fact that for the Fermat integer polynomials f(Z) and Z – a, where both polynomials must be factors of Yp, there are no integer values of a, X and Z for which Z – a divides f(Z), because the remainder will always be non-zero.

After more than 40 years, I still have hope that more mainstream mathematicians will join the small, but growing number of mathematicians who agree that there are no fatal flaws in the logic of FLT65.

Edward R. Close, June 26, 2017

For those who are not familiar with Fermat's Last Theorem, I've pasted in a previous discussion and some relevant links below.

The Basic FLT65 Proof
The following steps summarize the logic and mathematics of FLT65. For brevity, I will not present proofs of the steps here because they are so easily proved that they can be proved by a bright high school algebra student. If these steps aren’t obvious go to

STEP #1: The first step in FLT65 was to provide a rigorous proof of the division algorithm and its three corollaries. The reason I provided this proof first, even though it was well known to mathematicians, was to show that it applies to all polynomials across the field of real numbers, including integers, and to highlight the fact that the uniqueness of the dividend and remainder allows the all-inclusive “if and only if” of Corollary III. These points were pointed out in FLT65.

STEP #2: If there is an integer solution for Fermat’s equation: xn + yn = zn, to prove or disprove it, we need only consider n as prime numbers, p >2, and x, y, and z as relatively prime positive integers. Proof of this is included in FLT65 allowing us to proceed to Step 3.

STEP #3: Fermat’s equation can be rewritten as zp – xp = yp, and since all prime numbers >2 are odd, factored as follows: zp – xp = (z-x)( zp-1 + zp-2x + zp-3x2 +•••+ xp-1) = yp

Similarly, zp – yp = (z-y)( zp-1 + zp-2y + zp-3y2 +•••+ yp-1) = xp.

For the next step, and throughout this discussion, keep in mind that we have assumed that there are integer solutions to Fermat’s equation, so the approach is to determine whether this assumption leads to a contradiction. If it leads to a contradiction, FLT is proved.

STEP #4: It is easy to show by simple algebraic division that the only common factor that may be shared between the factors of the Fermat equation is the integer p, and since x, y, and z are relatively prime integers, if either x or y contains p as a factor, the other cannot. See the proofs of this in the original proof in the link above. So we can let y represent the one that does not contain p. It then follows that the two factors of the left hand side of the first equation of step 3 are relatively prime and thus are perfect p-powers of integers. Thus, by inspection of

(z-x)(zp-1 + zp-2x + zp-3x2 +•••+ xp-1) = yp, we see that we can write
(z-x)= Bp, and (zp-1 + zp-2x + zp-3x2 +•••+ xp-1) = Ap, where A and B are positive integers.

STEP #5: If there is an integer solution, then x and y are specific integers X1 and Y1, and the p-1 polynomial in z, f(z)=(zp-1 + zp-2X1 + zp-3X12 +•••+ X1 p-1) = Ap, and BpAp = Y1p. That is, the two factors must be perfect p-powers, integers raised to the pth power.

STEP #6: In a positive integer solution, z >Y1 >A, and by closure of integers, there is a positive integer a, such that A= (z – a), and by corollary II of the division algorithm, when f(z)=(zp-1+ zp-2X1 + zp-3X12 +•••+ X1 p-1) is divided by Z – a, the remainder is equal to f(a) = ap-1+ ap-2X1 + ap-3X12 +•••+ X1 p-1.

STEP #7: Corollary III of the division algorithm says that f(z) is divisible by z –a if, and only if, f(a) = 0. But, since f(a) is the sum of p positive integers, it can never equal zero. Thus by assuming there is an integer solution of zp – xp = yp, we have produced a contradiction proving Fermat’s Last Theorem.

Note that the case n = 4 is not addressed in this proof. It was overlooked in FLT65, but this was not a problem because there were several known proofs for n = 4, including one by Fermat himself.
So FLT65 is effectively a complete and valid proof of FLT; but approximately 90% of the mathematicians to whom the proof was submitted over the years did not respond at all. This is because, before Sir Andrew Wiles’ proof was accepted, professional mathematicians received hundreds of supposed proofs of Fermat’s last theorem per year.

If you’ve ever taught mathematics and had to evaluate proofs developed by students, you know it can often be very challenging and time consuming, and attempts at proofs by amateur mathematicians are usually filled with all kinds of errors. In addition, because in more than 300 years, so many first-rate mathematicians had tried to prove or disprove FLT and failed, most mathematicians consider reviewing such ‘proofs’ a waste of time.

I’m sure that this was the reason the first mathematician to whom I sent it rejected it. The reason he gave, however, was that, if FLT65 were true, it would also apply to the case n = 2. [When n = 2, we have z2 – x2 = y2, which does have integer solutions known as the Pythagorean triples, e.g. 3,4,5]. Of course by giving this reason for rejecting FLT65, he revealed the fact that he hadn’t read it, because it is clear to anyone with basic math skills reading the first page that the method of proof of FLT65 doesn’t apply to the case n = 2.

Of the 10% who did respond, most gave the opinion that there had to be a mistake somewhere, but failed to point one out, or provide any mathematical argument supporting their opinion. Of the remaining recipients of FLT65, only a few provided any sort of mathematical demonstration supporting their opinions. Those responses are presented in the article accessed by the links provided above. Those arguments were all easily refuted. However, one of those demonstrations, actually offered with different numerical values by three reviewers, is worth mentioning here because it is a classic example of inadvertent misdirection, and it also shows how tricky a proof of FLT can be.

The argument they put forth was that the division algorithm and corollaries certainly apply to algebraic polynomials, but they may not necessarily apply to the integers obtained when, for specific integer values of z, a and X1, the algebraic polynomials z-a, f(z) and f(a) are reduced to single integer values. This is an interesting conjecture, but none of the reviewers attempted to prove or disprove it, instead they offered what they thought were counterexamples to FLT65 for n = 3. They selected integer values of z, a, and X1 that, when substituted into f(z) and f(a), produced an integer value for f(z) that contained the integer z-a as a factor, even though f(a) did not equal zero, appearing to violate corollary III of the division algorithm.

It is worth contemplating this argument a little more deeply for a moment, because by doing so, we expose the fact that such a demonstration is not actually a counterexample, but is in fact, an inadvertent misdirection, shifting attention away from the fact that z must be part of an integer solution to the Fermat equation. It is not hard to find positive integer values for z, a, and X1 such that f(z) is divisible by z-a, and of course f(a)= ap-1+ ap-2X1 + ap-3X12 +•••+ X1 p-1 will still be non-zero because all the terms are positive integers. But because the values of z and a selected have no relation to the Fermat equation, these demonstrations have no bearing on the logic of FLT65.

FLT65 started with the assumption that there is an integer solution for the Fermat equation. This means that for a numerical example to be relevant, z must be part of an integer solution of Fermat’s equation. The issue is not whether you can find integer values for z, a, and X1 that will make f(z) divisible by z-a; the relevant point here is that, if there is an integer solution, the value of z must satisfy Fermat’s equation. Then, because f(z) and z-a are both polynomials in z, the algorithm and corollaries apply, and the remainder must equal zero for f(z) to equal a perfect p-power, Ap, if the assumption of an integer solution is true. But, of course for Fermat’s equation, f(a) cannot equal zero. -- End of story!

I think these reviewers were so intent on trying to find a way to disprove FLT65, which they were convinced from the beginning could not be valid, that they were blinded to the fact that, if their ‘counterexamples’ were valid, they would actually have provided integer solutions for zp – xp = yp, directly disproving FLT, and thereby also disproving Andrew Wiles’ proof.

So after fifty years, FLT65 still has not been refuted. Those who tried have failed, but only a few besides myself have accepted it as valid, and two of them have since passed away. Many of the mathematicians who have reviewed it believe it cannot be valid, and two even claimed to have refuted it, but their arguments were easily disproved. See the details in the links provided above. 

After fifty years, I would like to have closure; so anyone out there who believes the proof is faulty or incomplete is challenged to provide irrefutable mathematical proof that I can understand supporting that belief. If you can prove to me that FLT65 is wrong, I will acknowledge you proof and send you a check for $100.

Unlike Sir Andrew Wiles’ proof of FLT, which is hundreds of pages long, drawing on a very sophisticated knowledge and understanding of elliptic functions and modular algebra, FLT65 is a relatively simple proof relying only on basic mathematical principles. I believe that Pierre de Fermat will rest easier when FLT65 is recognized as valid, because it proves that he could have proved his famous theorem with mathematics available in 1637. If FLT65 is is at last recognized as correct, I, and poor Fermat will have closure.


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