The following article is an edited version of the proof in the book Fermat's last theorem proved by Nico de Jong. Pretoria : (c)1992. ISBN 0-620-16639-8 (listed in the South African National Bibliography, Pretoria State Library, 1992, 92-2617)
 

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PREAMBLE AND ABSTRACT

After 350 years of unsuccessful attempts, a mathematically highly advanced proof of Fermat's Last Theorem (FLT) by A. Wiles was accepted and published in Annals of mathematics, May 1995. However, it cannot be Fermat's own elementary demonstration. In the opinion of the present author the following proof is the one Fermat had in mind.

FLT holds that the equation z^w =  x^w + y^w can have a positive integer solution if and only if w = 2. As is well known, a proof for w being any prime suffices. Therefore w is considered a prime number throughout.

Suppose z is a composite positive integer. If for only one of its prime number factors, say p,  p^w  =  some a^w + b^w with w = 2, the existence of at least one Pythagorean triple would be commonplace. If, on the contrary, with w > 2,  p^w = some a^w + b^w , a counterexample to FLT has been found and FLT would already stand refuted.

But in section VI below, a proof is presented of FLT for the case that   p^w  =  any a^w + b^w (p prime, w > 2) has no solution in positive integers, which was approved by the late Prof. Hennie Schutte of the University of Stellenbosch, South Africa. However, he criticised its verbosity and suggested a more concise version which is presented in section VIII. From the proof in section VI follows that FLT holds for any non-primitive equation z^w = n^wp^w = n^wa^w + n^wb^w, namely, that it has no positive integer solution if w > 2. It now only remains to be proved that if z contains no prime factor p for which p^w = some a^w + b^w, the equation z^w = x^w + y^w (z composite; x,y,z relatively prime) has no positive integer solution either. This is easily proved using a binomial model of z^w = ((z-r) + r)^w with all its binomial expansions.

It is shown that a presumed "Fermat equation", emanating from any fixed z^w, is in fact simplified by the descent until we reach the w^th  power of the last prime factor in z for which the equation is no longer tenable. Therefore we have been simplifying what was an inequality all along. At the same time we have thereby proved that the existence of at least one prime factor in z is essential for which p^w =  a^w +  b^w in order to produce a counterexample to FLT.

PROOF OF FERMAT'S LAST THEOREM

 I Introduction

 II The condition for solubility of a non-primitive equation z^w = (na)^w + (nb)^w

 III The mechanism for descent in the demonstration of Fermat's Last Theorem

 IV Intermezzo with w = 2

 V Brain wave

 VI Proof that p^w = a^w + b^w (p, w prime and w > 2) has no positive integer solution

 VII Conclusion

 VIII Remarks on the author's naive approach to the solution of FLT

I    INTRODUCTION:

Fermat's Last Theorem is usually put as : x^w + y^w = z^w has no positive integer solutions for w a natural number > 2. Fermat himself formulated it as: z^w< > [ i.e. not equal to ] x^w + y^w (w > 2) in these words:

It is impossible to separate a cube into two cubes, a fourth power into two fourth powers, or in general any power higher than the second into powers of like degree. I have discovered a truly marvellous proof which this margin is too small to contain.

(Note by Fermat in his copy of Diophantus' Arithmetica.)

• Natural numbers are relatively prime if they do not have a common divisor > 1. Unity (= 1) is not considered a prime.

• An equation in three natural number terms is presented as the greatest term being equated to the sum of the two other terms. The expression involving these two other terms is here called a twoterm.

• An equation of a term to a twoterm is called primitive if any two of the three terms involved do not have a divisor > 1 in common. Examples: 35² = 21² + 28² is not primitive: two of the three terms (therefore all three terms) have a factor 7² in common. 5² = 4² + 3² is a primitive equation, as is 65² = 56² + 33². x,y,z in primitive equations z² = x² + y² are called Pythagorean triples.

• A composite number consists of the product of at least two primes.

• Alphabetical symbols in the algebra used here denote natural numbers.

II THE CONDITION FOR SOLUBILITY OF A NON-PRIMITIVE EQUATION  z^w = (na)^w + (nb)^w

For ease of use in our demonstration we distinguish between p-type primes and q-type primes as follows :

p-type primes are primes for which p^w < > any a^w + b^w for w prime > 1

q-type primes are primes for which q^w = some a^w + b^w for the same w

Examples of p-type primes for the particular w = 2 are 2, 3, 7, 11 ... These primes are not of the form 4k+1. Examples of q-type primes for w = 2 are: 5, 13, 17, 29 ..., in fact, primes which are of the form 4k+1. Witness: 5² = 4² + 3², 13² = 12² + 5², 17² = 15² + 8², 29² = 21² + 20². This was proved long ago and we need not go into this detail here.

We consider the class P of all z which consist of the product of p-type prime factors p₁, p₂, p₃, . . . p_m+1 for each of which the w^th power < > any a^w + b^w at a particular w > 1.

We also consider the class Q of all z which contain among their prime factors only one q-type prime factor for which, at some w > 1, q^w = some a^w + b^w. The other prime factors in this z are all of the p-type.

It is clear that the classes P and Q are mutually exclusive for each particular w > 1. Any specific w^th power of a composite z being a member of class P cannot belong to class Q as well.

For any z in class P: z^w = n^w p^w < > (na)^w + (nb)^w

For any z in class Q: z^w = n^w q^w = (na)^w + (nb)^w

(n is the product of prime factors of z, (even if they occur more than once in z) for which n = z/p (p any prime factor of z) in class P; and for which n = z/q in class Q.)

The presence of more than one q-type prime factor in z would yield more possible non-primitive Pythagorean equations of the form z^w = (na)^w + (nb)^w (with w = 2) or more counterexamples of Fermat's last theorem of the same form (with w > 2). In order to prove that even one non-primitive counter-example to Fermat's Last Theorem is impossible, it is necessary to contrast the two classes P and Q.

Suppose we have a composite z₁ in the class (P+Q) but we wish to test whether this z₁ is a member of class P or class Q (which are mutually exclusive!). Now:

If z₁ occurs in class P we have:

z₁^w = p₁^w · p₂^w · p₃^w · . . . · p_m^w · p_m+1^w < > (n₁a)^w + (n₁b)^w

If z₁ occurs in class Q we have:

z₁^w = p₁^w · p₂^w · p₃^w · . . . · p_m^w · q^w       =     (n₁a)^w + (n₁b)^w

We divide z₁ by p₁ to obtain z₂, divide the latter by p₂ and obtain z₃, etc. until we MUST arrive at z_m+1 = q or z_m+1 = p_m+1, which are both prime, and of which the w^th power may or may not bifurcate as some a^w + b^w.

Merely depending on which of the two mutually exclusive possibilities is applicable, z₁^w, z₂^w, z₃^w . . . z_m^w were or were not the sum of two not relatively prime w^th powers of natural numbers.

If we would know that with w prime > 2 no q-type prime can possibly exist (which we propose to prove) Fermat's Last Theorem may be considered proved for any non-primitive equation of the form z^w= any (na)^w + (nb)^w (n any factor of z).

Then: All which would remain to be proved is that if a non-primitive equation z^w = (na)^w + (nb)^w in natural numbers does not emanate from z^w (z composite), neither will there be a primitive equation

z^w = x^w + y^w (i.e. with x and y having no factor in common) emanating from the same z^w.

We shall demonstrate this with the aid of the general properties of the binomial theorem. The process which we shall follow is a perfect echo of the process which we followed above in respect of the descent from z₁^w to the w^th power of the last prime in z₁. Descent will be effected again by the w^th power of one p-type prime at a time, in order to retain the only q-type prime factor of z in any interim or final quotient, if it is present in z₁ at all.

III THE MECHANISM FOR DESCENT IN THE DEMONSTRATION OF FERMAT'S LAST THEOREM

In wording his theorem Fermat involved the case of w = 2 as well, by stating : "It is impossible to separate a cube . . . . or in general any power higher than the second into powers of like degree". Therefore we feel justified in considering the general case where w may be = 2 as well as a prime > 2, and we attempt to discover why "it is impossible with powers of like degree" if w is "higher than the second".

Any composite or prime number, therefore also the composite z from whose w^th power emanates the sum of two w^th powers or not , may be defined as the sum of two natural numbers z-r and r. Let r successively assume all values from 1 to z-1: this does not change the numerical value of z.

We write z^w as ((z-r) + r)^w and run r through all the natural number values from 1 to z-1. z^w will now be expressed by mentally spreading the complete but finite table of the z-1 binomial expansions which will ensue, one for every value of r. The sum of the first w terms of each binomial expansion we define as term 1 and the last term of each binomial expansion (namely r^w ) we define as term 2 in:

z^w = ((z-r) + r)^w = [the first w terms of the binomial expansion] + term 2 =

[ (z-r)^w + w(z-r)^w-1·r + ... + w(z-r)· r^w-1 ] + term 2 = term 1 + term 2 (1)

If z^w = some x^w + y^w, x^w will certainly figure as some term 1 while its partner y^w will be term 2. Therefore (1) is the binomial model which contains all the candidates for x^w and y^w if a counterexample to Fermat's Last Theorem exists.

If we know that there is one successful term 1 which is a w^th power (term 2 is always a w^th power) we affirm that we have an equation emanating from z^w : z^w = x^w + y^w = some term 1 + term 2 .

If we do not yet know whether there is at least one term 1 which is a w^th power in the z-1 binomial expansions we can defer the decision whether or not we are dealing with an equation or an inequality (namely, z^w = some x^w + y^w, or < > any x^w + y^w) using the descent proposed. We assume z a composite number with as many prime factors as we please.

We use the same descent as discussed in section II above, from the original z^w to a new and small enough z^w which can be judged to be the sum of two w^th powers or not. Now, however, we are at the same time considering the descent with regard to the value of each twoterm (term 1 + term 2).

Each of the twoterms in the set of z-1 twoterms will have the same numerical value, namely z^w . Any possible pair of candidates (term 1 + term 2) pertaining to the r^th binomial expansion is then rewritable as any other pair of candidates (some other term 1 + term 2 ) of some other r^th binomial expansion retaining the same numerical value. We call this the principle of rewritability in our binomial model.

Each two terms of the complete set of twoterms (as defined above) emanating from z^w are candidates for x^w as term 1 pertaining to some r^th binomial expansion and y^w as term 2 ( = r^w ) in the same binomial expansion. Because of the mutual symmetry of x towards y in the Fermatian equation , or the interchangeability as one might wish to call it, one may assert that if there is one successful pair of candidates (term 1 being some w^th power representing x^w while its partner is y^w as term2, which as r^w is a w^th power anyway) , there will be another successful pair in which term 1 will be y^w). The twoterm of a primitive equation (if there is any) emanating from z^w as a sum of two w^th powers will also figure as some term 1 + term2 , because all the twoterms possible occur in the full set of z-1 twoterms .

If none of the terms 1 is a w^th power we have no equation z^w = x^w + y^wemanating from z^w, neither primitive nor non-primitive.

The full set of z-1 pairs of candidates : term 1 + term 2, derived from the binomial expansion of z^w in (1) above , we call: the system z^w

Thus, with z₁ composite, we write:

z₁^w = ((z₁-r) + r)^w

= (z₁-r)^w + w(z₁-r)^w-1 r + . . . . + w(z-r)r^w-1 [term 1] + r^w [term 2] (2)

The method of descent for the proof of Fermat's Last Theorem uses successive division of z₁^w (z₁ composite) by w^th powers of prime factors, based on the following considerations:

z₁ has a prime factor m, say. Therefore we may write: z₁ = nm and z₁^w = n^wm^w . If m is a q-type prime with w > 2 it would be a counterexample of Fermat's Last Theorem, as would be z₁^w = some n^wm^w = n^wa^w + n^wb^w. On the contrary, with w = 2, z² = n²m² = some n²a² + n²b² would be a common occurrence with m being a q-type prime.

Therefore, in considering the cases w = 2 and w > 2 together, it is sufficient to contrast z₁ having only one q-type prime factor with z₁ consisting of nothing else but p-type prime factors. In applying Fermat's descent we give precedence to successive divisions by w^th powers of p-type primes with the purpose to leave the w^th power of the q-type prime in each successive quotient, if there were a q-type prime in z₁ at all. This is the same descent as discussed in section II above.

Division by the w^th power of a p-type prime in the p₁^th line of candidates (where r = some prime factor p₁ of z₁) would yield :

z₂^w = z₁^w / p₁^w = ((n₁.p₁ - p₁) + p₁)^w / p₁^w = ((n₁.p₁ - p₁ )^w + w. (n₁.p₁ - p₁)^w-1 . p₁ + . . . + p₁^w) / p₁^w (3)

It will be observed that in the resulting quotient the place of the last term, which in the binomial expansion of z₁^w was filled by p₁^w has been reduced to 1^w, and that each of the other terms of this expansion , after division by p₁^w is still a natural number.

This expression for the quotient z₂^w _, again viewed as a twoterm with 1^w as term 2 is at the same time the first binomial expansion manifestation of a fresh system z₂^w = ((z₂ - r) + r)^w in which r is = 1, but which is now run through all the natural numbers from 1 up to z₂-1. As before, this will not change the numeric value of z₂^w. In other words, the first line of the new binomial expansion generates a full new system z₂ with all the candidates for x₂^w and y₂^w in fresh terms 1 each of which is paired by some term 2, which is always a w^th power .

If only some term 1 among the z₂-1 pairs of candidates would be some w^th power we would have a z₂^w which is the sum of two w^th powers. And if this would be so, z₁^w (from which we have just been descending by a magnitude of the w^th power of a p-type prime) would be the sum of two w^th powers, albeit in at least a non-primitive way.

In this way we continue to divide z₂ by one of its p-type prime factors p₂ and obtain z₃ , in order that if z₃^w is the sum of two w^th powers then z₂^w will prove to have also been the sum of two w^th powers, and so on with z₃, z_{4, . . .} z_k. Of necessity we must reach the last prime in the original z₁. If this last prime is of the p-type there is no non-primitive equation z₁^w = some n^wa^w + n^wb^w , but a non-primitive inequality :

z₁^w  < > n^w · (any a^w + b^w).

If, on the contrary, this last prime is of the q-type we can write z₁^w = n^wa^w + n^wb^w, in which n is the product of the primes in z by whose w^th power the total descent was effected.

IV   INTERMEZZO WITH w = 2

Preliminary discussion:

We note the analogy of Fermat's problem with the problem of finding natural number solutions to the Pythagorean equation z² = x² + y². Let us therefore examine some conditions for solution in the Pythagorean environment before we assign a prime number > 2 to the exponent.

We know that one q-type prime in composite z is sufficient to produce a non-primitive equation

z² = some (na)² + (nb)². But how do we prove that composite z without a q-type prime factor cannot lead to a primitive Pythagorean triple x,y,z ? In other words how do we prove that with such a composite z there would be no z² = x² + y² with x,y relatively prime ? It could be pointed out that in order to have a natural number solution for a primitive equation z² = x² + y² (z composite) we need z consisting solely of primes of the form 4k+1 (which with w =2 are identical with q-type primes).

Put q₁² = some a² + b²

q₂² = some c² + d²

Now z² = q₁²q₂² = q₁²c² + q₁²d²

= q₂²a² + q₂²b²

which are two non-primitive Pythagorean equations emanating from z².

On account of the following so-called Fibonacci identity there are also two primitive equations emanating from the same z²:

z² = (a² + b²) · (c² + d²) = (ac-bd)² + (ad + bc)² and:

= (a² + b²) · (d² + c²) = (ad-bc)² + (ac + bd)²

Thus z² (with z having as only factors two different q-type primes q₁ an q₂) would yield four essentially different sums of two squares, two in a primitive and two in a non-primitive way. We look at an example where q₁ = 5 and q₂ = 13 . z = 5 · 13 = 65. Reverting to the model portrayed by the identity (1) and putting w = 2 we write z² = (z-r)² + 2(z-r)r + r² .

Now running r through the values 1 to 64 and imagining the complete table of binomial expansions, which we learnt to see as term 1 for the sum of the first w (here: 2) terms of the expansion and term 2 for the values of r² , at various rankings of r² in the system z² we come across all the possible values which x² and y² can take as generalized before:

65² twoterms

((65- 1) + 1)² = 64² + 2.64. 1 + 1²

((65-16) + 16)² = 49² + 2.49.16 + 16² = 63² + 16² primitive

((65-25) + 25)² = 40² + 2.40.25 + 25² = 60² + 25² non-primitive

((65-33) + 33)² = 32² + 2.32.33 + 33² = 56² + 33² primitive

((65-39) + 39)² = 26² + 2.26.39 + 39² = 52² + 39² non-primitive

If we complete the binomial table of the system 65² we will, of course, discover that all the squares in the

16^th, 25^th, 33^rd , and 39^th ranking of r as term 2 recur as some term 1 lower down the table complemented by their partners 63², 60², 56² and 52² respectively as term 2.

If z would exclusively consist of more than the two q-type prime factors displayed here there would be more possibilities to apply the Fibonacci identity and consequently more primitive Pythagorean equations would emanate from such z².

If we would multiply such z by only one p-type factor p, no primitive equation u² + v² would emanate from p²z², but a number of non-primitive equations would still exist, all emanating from p²z².

If there would be one or more p-type prime factors in z but only one q-type prime, no primitive equation z² = x² + y² would emanate from z². However, because the square of a q-type prime would be = some a² + b² we would have: z² = n² (a² + b²) = (na)² + (nb)² by the effect of n² "embracing" a² + b². (n is the product of all p-type primes in z)

If z consists of a product of p-type primes only, we would have a number of non-primitive inequalities:

z² < > n² · (any a² + b²) by the same effect of embracement, this time under the < > sign, and we may also write

z² < > (na)² + (nb)².

We must, however, still prove that if the equation z² = (na)² + (nb)² has no solution in natural numbers ( n being the product of all p-type factors of z but one), neither will there be a primitive equation z² = x² + y² (x,y,z having no common factor) emanating from the same z².

The Fibonacci identity proves that a number of primitive Pythagorean equations are constructible from the squares of the prime factors of z, if all the prime factors of z are of the q-type. Of course a number of non-primitive equations of the form z² = (na)² + (nb)² are constructible as well (n here being the product of all prime factors of z but one).

But the Fibonacci identity z² = (a² + b²) · (c² + d²) = (ac-bd)² + (ad+bc)² = (ad-bc)² + (ac+bd)²

is not very helpful in proving that no primitive equation can possibly emanate from z² if z is the product of p-type primes only. We must remove all suspicion that there may be another mechanism at work which promotes the emanation from z² of a primitive equation z² = x² + y². However, by applying identity (2) in section III above, we provide a simple proof as follows:

THEOREM:

If z is a composite natural number containing no prime factors of the form 4k+1, then z² = x² + y² has no solution in natural numbers.

Proof:

We compile a composite z which consists of p-type primes only, i.e. 2 and those of the form 4k-1.

Say we assume some z² = x² + y². This leads to a contradiction as follows: We draw op the binomial model as in (2) in which x² must figure as some term 1 and y² as its complement term 2, and descend as demonstrated in section II above until we reach the square of the last p-type prime, which is by definition any a² + b². This means that any effort to assign natural number values to a and b, and therefore to the last prime in z descended to, will change its value and with it the value of the original z. Descent from the original z² by successive squares of p-type primes cannot then be distinguished from simplifying an inequality z² < > any term 1 + term2 instead of an equation.

Therefore any z², z having no q-type prime factor, cannot equal some n²a² + n²b² and the assumption with which we started was false. Actually, the truth of the assumption confirms the existence of one q-type prime factor in z and the falsity of the assumption forces the non-existence of one q-type prime factor in the original z by the process of the descent as demonstrated above.

Again, we do not have to prove that if z² < > n²a² + n²b² there cannot also be a primitive Pythagorean triple x,y,z, because if there would be some primitive equation z² = x² + y² , both x² and y² would figure as term 1 and term 2 in some twoterm of the system z² in natural numbers.

Both twoterms x² and y² could then be rewritten as some n²a² + n²b² if the last prime in z would have been of the q-type but, without a q-type prime factor in z, any natural number substitution for a² and b² would now lead to the impossibility of rewriting n²a² + n²b² as any term 1 + term 2 which would be numerically equivalent to z², regardless of the form which term 1 would take, i.e. whether it is a w^th power or not.

We have to conclude that, with w = 2, the absence of a q-type factor in z rules out the existence of a primitive and non-primitive Pythagorean equation emanating from z². But, strictly speaking , if there is only one q-type prime factor, the reasoning in the above paragraph does not rule out the existence of some other term 1 also being a square and forming a primitive Pythagorean triple together with z² and term 2 (the latter always a square).

However, as we saw in the above diversion on the Fibonacci identity, something more than one q-type prime in z is needed for z² producing a primitive Pythagorean triple, namely: the condition that all primes in z are of the q-type. (Actually if, with composite z , one primitive Pythagorean triple emanates from z², at least another one will also exist: they come in pairs.)

V    BRAIN WAVE

If we could prove (which we are going to do in the next section) that for any prime p, and w prime > 2, there is no p^w = some a^w + b^w we have, in fact, proved Fermat's Last Theorem. For we have a situation identical with the case z² < > x² + y² in which composite z has no q-type prime factors. It will be recalled that the case for z² is based on the general properties of the binomial theorem for all w > 1 and the descent from z^w = ((z-r) + r)^w and the subsequent binomial expansion (1).

With w > 2 we still have term 2 as a w^th power of r which ranges from 1^w to (z-1)^w . Term 1 (consisting of the first w terms of the binomial expansion involved will have more terms of the binomial expansion than is the case with w = 2, but division by successive p₁^w, p ₂^w, p₃^w . . . etc. (all w^th powers of p-type prime factors) will always result in the first binomial expansion of the next z^w (one magnitude p₁^w, p₂^w, . . . etc. smaller) with r^w (term 2) = 1^w, from which the full table of the binomial expansions for the new and smaller system z can be completed, as was argued in the first instance for the general case with w prime and > 1.

The advantage of the demonstration in z² is that it places the comprehension of the reasoning involved more easily within the grasp of one's imagination without loss of generality. The awe we always felt towards the 'intractability' of this major Fermat problem has proved the greatest obstacle in finding the solution for more than 350 years.

VI    PROOF THAT  p^w   =  a^w + b^w   (p, w PRIME AND w > 2)  HAS NO POSITIVE INTEGER SOLUTION

(A)

If p^w = a ^w + b ^w (p,w prime, w > 2) , a and b are relatively prime, otherwise p would not be prime.

p < a + b, because if p = a + b , p^w = (a + b)^w > a^w + b^w .

Also: a < p; b < p; thus a + b < 2p

Therefore ½·(a + b) < p < a + b

(B)

Because the prime w > 2 is always odd, a^w + b^w is factorable as:

(a + b) · (a^w-1 - a^w-2 · b + a^w-3 · b² - . . . . - a · b^w-2 + b^w-1 )

In order to examine whether a+b and what we here call the cyclotome: a^w-1 - a^w-2 ·b + . . . + b^w-1 have any factors in common, we apply the principle on which Euclid's algorithm rests to the two factors a+b and the cyclotome. We subtract a+b from the cyclotome as many times as we please. Any factor which a+b has in common with the cyclotome it would also have in common with the result of a repeated addition or subtraction. Division, which is the same as repeated subtraction, of a+b into the cyclotome yields an integer quotient plus a remainder w·a^w-1 or w·b^w-1 , depending on starting the division from the cyclotome's first term or from its last.

Neither a nor b (which are relatively prime to each other) have a factor in common with either a+b or the cyclotome. Nor is a+b a factor of the cyclotome, for a+b is not a factor of w · a^w-1 or w · b^w-1.

If a+b has a factor w, it follows that this factor occurs in the cyclotome as well, but only once, for w occurs only once in w · a^w-1 and w · b^w-1. It also follows that if w occurs as a factor in a+b repeatedly (namely to some power), it will still occur in the cyclotome only once!

We are now in a position to argue whether the product of a+b and the cyclotome could equal p^w if p is a prime. If w occurs in a+b as a factor it must occur there to the power w-1, because one single prime factor w is accounted for in the cyclotome. With the exception of w, no factors of a+b can possibly occur in the cyclotome and vice versa. Therefore all factors other than w must be groupable as w^th powers inside a+b as well as inside the cyclotome, otherwise p^w would not be a w^th power.

If a+b contains no factor w, p^wwill therefore consist of at least two relatively prime w ^th powers, which means that p is not prime. The only possibility left for p^w = a^w + b^w to be true lies in p being = w, obtaining: w^w = a^w + b^w = ( a + b) · the cyclotome.

However, from (A) above we obtained ½(a + b) < w < a + b

Because p (which, in this only possibility, = w) < a + b, a + b must be some power of w. The cyclotome being > 1 must be = w and a + b = w^w-1. This means that a+b is to be exactly w^w-2 times larger than p. For the smallest case, where w = 3, this works out at a+b being 3 times larger than p. This contradicts our findings under (A) above. Therefore p^w =  a^w + b^w (p,w prime, w > 2) has no positive integer solution.

This completes the proof of Fermat's "Last" Theorem.

VII    CONCLUSION:

With w = 2 we have a z^w microcosm in which can occur squares of both p-type and q-type primes as factors. The z^w macrocosm with w > 1 includes the microcosm with w = 2.

In the z^w macrocosm with w > 1 the minimum requirement of a q-type prime in z for the existence of a Pythagorean equation and for a counterexample of Fermat's Last Theorem is the same. That the requirement cannot be met with w > 2 does not imply that the requirement does not exist. The rules of the proof are exclusively directed by those of the binomial theorem in concert with the fundamental theorem of arithmetic.

VIII    REMARKS ON THE AUTHOR'S NAIVE APPROACH TO THE SOLUTION OF FLT
 

". . . I will relate what has been done with Fermat's problem without using any sophisticated methods. Let me say, that these attempts should not be looked down on. On the contrary, they show much ingenuity, and they have helped to understand the intrinsic difficulties of the problem. I'll point out, in various cases, how these attempts have brought to light quite a number of other interesting, perhaps more difficult problems than Fermat's.
If I have decided to group these various results under the heading of 'the naive approach', it is only because Fermat's problem has proved itself to be at another level. In fact, it is possible that all other approaches tried as yet may someday be considered naive. Who knows?"
Paolo Ribenboim, 13 lectures on Fermat's last theorem, p.51. New York: Springer, c1979
 

Comprehension of what the present author thinks is Fermat's own proof can be classed at different levels. The proof as presented is obviously aimed at the high school level of the educated general public. If the proof's level is to be raised to make it more acceptable for inclusion in future editions of elementary textbooks aimed at the freshman's level, the following suggestive remarks may apply, especially with reference to section II: The condition for the solubility of a non-primitive equation: z^w = (na)^w + (nb)^w.

1. It is argued that Fermat's descent is nothing else but using the principle of putting an equation in lowest terms. The first occasion to illustrate this principle in elementary textbooks arises when the subject of the incommensurability of rational and irrational numbers is broached. The classic example presented in textbooks is a discussion of the solubility of the example z² = 2q² (z,q at first assumed natural numbers) with the remark that we suppose the fraction z/q already in lowest terms, i.e. that z and q are relatively prime. A remark that this equation may be put in lowest terms only if none of the squares of prime factors of z be = 2 (in this case) is omitted. Such a stipulation appears to be considered superfluous. However, if we define z a natural composite number, q as the product of all prime factors in z but one, raise both z and q to the wth power (w > 2), and thus obtain z^w = 2q^w we are still dealing with the incommensurability of natural versus irrational numbers, but suddenly we find ourselves in the middle of FLT's domain.
Suppose that a lemma:  " p^w = a^w + b^w (p,w any prime, w > 2) has no positive integer solution " has already been proved. Then we can, for the purpose of determining the solubility of z^w = (np_k)^w = n^w(a^w + b^w) [p_k any prime factor of z; n = z/p_k; a and b any positive integer] immediately re-obtain the result of our lemma by the simple process of putting this supposed equation z^w = (na)^w + (nb)^w [z,n,a,b natural numbers] in lowest terms and conclude that in this case the process amounts to simplification of an inequality instead of an equation.

2. The promise of a proof that there is no natural number solution for p^w = a^w + b^w (p,w prime, w > 2) is not made good until section VI is reached. That proof is also at the level of the educated general public. Criticism was levelled at its verbosity by the late Professor Hennie Schutte of the University of Stellenbosch, South Africa. He rewrote it, perhaps because his version requires more deftness in handling formulae, and mailed it to the present author in 1989. After translation into English it would run as follows:
Suppose a^w + b^w = p^w (a,b,p,w natural numbers; p,w prime, w > or = 3; a and b relatively prime -- otherwise p would not be prime).
p^w = a^w + b^w = (a+b) . (a^w-1 - a^w-2b + a^w-3b² - . . . + b^w-1)  = (a+b) . (m(a+b) + wb^w-1)  (1) [m natural number; remainder theorem]
[Note that Professor Schutte makes no reference to the motivation for the long division of a+b into a^w-1 - a^w-2b + a^w-3b² - . . . + b^w-1, which is to determine a common factor, if there would be any, by Euclid's algorithm. -- NdJ ]
(i) Because p is prime it follows that a + b = p^k for some natural number k between 0 and w. From this follows:
p^w = a^w + b^w = p^k . (mp^k + wb^w-1) [ 0 < k < w ].
Therefore: mp^k + wb^w-1 = p^w-k . . . . . . . . . . . (2)
If k = 1 or k = w-1 it follows that p is a factor of wb^w-1
(i-i)
If p is a factor of b^w-1 then p is a factor of b and from (1) above we obtain that p is also a factor of a, contradicting our assumption that a and b are relatively prime. Therefore p is not a factor of b^w-1.
If p = w it follows that a^w + b^w = w^w. But (a + b)^w > a^w + b^w for w > 1, yielding another contradiction.
(i-ii)
If 1 < k < w-1 it follows from (2) above that at least p² is a factor of wb^w-1, and because p and w are both prime, p must be a factor of b. As before, see (1) above, this leads us to conclude that p is a factor of a, which again is contradictory to the assumptions made.
(ii) All cases lead to contradictions of assumptions made. Therefore the equation p^w = a^w + b^w has no natural number solution.
The level of handling symbols and formulae here imposes distinctly higher demands on the reader than the present author's original version dealt with in section VI. It is left to the reader to decide which version is more forceful or beautiful.

3. Re: The impossibility of a primitive equation: z^w = x^w + y^w (x,y,z relatively prime; z composite)
Because the identity z^w = ((z-r) + r)^w = ((z-r) + r)^w + w(z-r)^w-1r + . . . + r^w, seen as a twoterm as explained, yields all possibilities for z^w = x^w + y^w, whether x,y,z are relatively prime or not, the descent (or putting the identity in lowest terms) can be performed, prime by prime, until the last prime factor in z is reached.
A quicker way of writing z^w = (np)^w as an assumed Fermat equation, and putting it in lowest terms, is to find any z/p = n (n the product of all prime factors in z but one). In z^w = (np)^w = ((z-r) + r)^w and the concomitant general binomial expansion the factor n^w will be discerned among the values r^w will take (because n < z-1) and the writing in lowest terms can be effected in one swoop.
The realization that the binomial table with z-1 sets of twoterms as explained is the only locus of all possible candidates for x^w + y^w (as a primitive or non-primitive emanation from z^w) settles FLT immediately for all values of z^w. After putting  z^w in lowest terms, we are under an obligation to appoint natural number values for the supposed equation obtained, viz: p^w = a^w + b^w (p any prime factor of z to which we happen to descend), which obligation is impossible to fulfil.
Note that without recourse to this binomial model we would have great difficulty in proving the primitive case.
The same mathematical construct (namely writing the binomial table for z^w in lowest terms) with w = 2 and some prime factor (of the form 4k+1) of z being the sum of two squares, allows us to descend to the square of this prime. All z-1 sets of term 1 + term 2 would then be equal to z² and to each other, preparing the ground for the occurrence of at least two primitive "Fermat/Pythagoras" equations if all prime factors of z would be of the form 4k+1.