*The purpose of this website is to put a body of new thinking and analysis into the public domain. It is an emphasis which can bring reasoning back into mathematics, and not just as a minor item, but as the central core of the discipline. There was, during the 20 ^{th} century, a tendency in maths to distrust reasoning, and with good reason, because the adoption of restrictive set theory in the 1920s came about as a result of reasoning failing, after much effort, to find a valid explanation of Russell’s Paradox.*

*Without reasoning, maths becomes an activity concerned with technical symbolic manipulation something, incidentally, which was automated when Mathematica was devised over 40 years ago. If maths reduces to this, how can we rebut the charge that we are mere handle turners?*

*What has changed? Well there was a gradual drift away from Platonism in interpreting maths in the 20 ^{th} century, and in 1993 a monograph was published by the present author which showed how Russell’s Paradox can finally be explained, by adopting the new concept of dynamic contradiction. This is a contradiction which occurs *sequentially

The author is an older philosopher of mathematics, Christopher Ormell. When he was an undergraduate at Oxford he shook hands with Alan Turing. His first academic paper introduced the concept of superparadox, in which a simple question leads to *n* distinct contradictions, and the number of these contradictions increases exponentially as *n *increases. During a long career, he taught maths at all levels from infant to university. He made discoveries such as the first non-trigonometric formula for the *n*th prime number, and an algebraic formula for [*x*]. He was for ten years Director of a school project which showed that, in teaching, applications of maths are not a minor afterthought. By devising special scenarios in which the use of maths “can enable people in trouble to find a way out”, this project *Maths Applicable*, got disillusioned older students to regain interest and become enthusiastic.

This website will address a series of areas in maths in which reasoning has encountered conceptual confusion. Some of these are zones of difficulty which can only be resolved by thinking outside the box, i.e. which require a background of deep thinking about problems beyond mathematics. Others are simply dense conceptual mazes which require the kind of sustained reasoning which has fallen out of fashion.

Prof. Mike Askew has written a leading article in the *Mathematical Gazette* (104.559. 2-11 (2020)) in which he argues that reasoning needs to become an established habit of mind in mathematics.

He says that we must

<<…shift from asking “How do I teach students to answer this problem?”

to

“What mathematical reasoning do I expect them to engage in as a result of working on this problem?”>> (p. 10)

The first zone of dense, confused reasoning to be addressed on this website will be **Fermat’s Last Theorem**, probably the most contentious topic in maths. New light was thrown onto the problem in 1995 by Andrew Wiles’ unexpected, much celebrated, proof. We now know that Fermat’s conjecture —that no cases occur of two *n*th powers adding up to a third *n*th power (of natural numbers when *n* too is a natural number > 2) — is valid. Wiles’ proof, however, could not be that found by Fermat, and posthumously lost, because Wiles used an advanced modern approach to obtain his result. This is a triumph, but ironically it leaves the underlying problem in an even more unacceptable state:

*Why has no one been able to find an elementary proof of this elementary result after such a long time?*

It is a question which can hardly be asked —with anything much to discuss— until an elementary proof has been found. It was possible to brush this aside when it was unclear whether Fermat’s conjecture was valid. But now we know that the conjecture is *valid*, it has a presence which casts a dark shadow over elementary maths’ capacity to resolve elementary problems.

*The good news is that an elementary proof has recently been found.*

It is the end product of more than 28 years’ tenacious, sustained analysis during which the present author devoted, on average, about an hour a day to the project. His main qualification for embarking on this arduous task was that he was familiar with the kind of 17^{th} century mathematical knowledge available to Fermat, and, having given up higher maths after leaving Oxford, he was not likely to be distracted by unconsciously reverting to more modern abstract ideas. It may be described as a “bare hands” approach, because the author began in January 1991 with no knowledge whatever of modern number theory.

In October 2019 the author sent a version of the present proof to the *Mathematical Gazette* as a contribution to their journal. The first referee seemed to be convinced that there ought to be a disabling mistake, but failed to find one… and resigned. The second referee was equally unable to find such a mistake, and he also resigned. So in May 2020 the editor decided not to publish it in his journal. This is hardly a satisfactory situation. Must a project, which took more than 28 years to complete, be dismissed so casually on unsupported say-so when it offers the resolution of such a hoary enigma?

The proof will be set out on this website in two instalments, because the conclusions of the early reasoning need to be thoroughly digested by the reader before she/he embarks on the later stages. If you are impatient to know how it can lead to closure, you may purchase the specialist monograph which is being published by Ingleside-Ashby Press alongside this exposition.

It is a monograph in the series of 21^{st} Century Demystified Mathematics Monographs:

* *

Ingleside-Ashby Press ISBN 0 907669 X First published: July 2020

[ACTION to buy NEEDED Send a Sterling cheque or money order for £10 (UK) £14 (EU) £15 (nonEU) to Ingleside Ashby Press, PO Box 16916, London SE3 7WS, United Kingdom. With your postal address. The monograph will be despatched to you postfree.]

Further monographs will, it is hoped, be published in this series later.

*Background notes are signalled **[NR n]**. They may be found at the end or by clicking on tabs.

Well, the impossible has happened, a fairly simple, elementary, proof of Fermat’s Last Theorem (FLT) has finally been found. It is probable that Fermat’s argument ran on similar lines, though of course there is always the possibility that he had a still shorter, unguessed, more dramatic, proof in his locker. The new proof, though, rests on a solid platform of elementary reasoning —about where common factors exist, and where they are absent. Once established, this platform offers a self-evident staging post for moving towards the summit. The apparent symmetries it offers are too good to be true. Focusing closely onto them must, and does, lead to the emergence of contradictions. Readers may find, however, that they need to pause at the half-way point and thoroughly familiarise themselves with the platform. [To this end eight exercises may be found on the final page of the Notes.] The subsequent course of the proof is also based on reasoning: this later reasoning centres round the only function of the initial unknowns which could possibly fit the necessary conditions for a counter-example.

The new proof starts with a technical hypothesis: that certain unknown numbers exist, the numbers being those involved in the “smallest” solution to the Original Fermat Equation

** a^{n} + b^{ n} = c^{n}** , where

The new dissection, however, starts with a radical change of unknowns. Instead of couching the Fermat Equation in terms of ** a, b, c **and

It is fairly clear that only cases where the original unknowns ** a** and

So the new technical hypothesis is that three specific unknowns *a, b, p* do not exist.

Now the bracket on the RHS of (1) becomes a binomial, *if* *a+b* is replaced by *c*. So (1) can be regarded as composed of three binomials to the prime power *p.* To facilitate further analysis, a lemma will be introduced, which applies to binomial expansions when the exponent is prime. This may be called the ‘Prime Binomial Lemma’.

A NEW LEMMA I All the “inner” coefficients of the binomial expansion of (*a+b*)* ^{p}* have the factor

Now (*a + b*)^{p}*= a** ^{ p}* + various

A NEW TERM The “something” may be denoted T* _{ab}*. This will be called the ‘Inner Tray’ of the prime binomial.

* *

Hence (*a + b*)* ^{p}* =

This result offers a neat way to expand prime binomials, one which will be used repeatedly in the sequel. Its advantage is that it reduces an open-ended expansion to three terms.

Making the Inner Tray the subject T* _{ab}* = ((

*Every binomial with a prime exponent has its Inner Tray.*

EXAMPLES: The inner tray of (*a+b*)^{5} is *a*^{2} + *ab* + *b*^{2}. The inner tray when *p*=7 is *a*^{4} + 2*a*^{3}*b* + 3*a*^{2}*b*^{2} + 2*ab*^{3} + *b*^{4}. The inner tray when *p*=11 is *a*^{8} + 4*a*^{7}*b* + 11*a*^{6}*b*^{2} + 19*a*^{5}*b*^{3} +23*a*^{4}*b*^{4} + 19a^{3}*b*^{5} + 11*a*^{2}*b*^{6} + 4*ab*^{7} + *b*^{8}. **[NR4]**

NEW TERM: SHELLING It is convenient to describe the process of removing the outer terms *a** ^{n}* and

EXAMPLES (*a+b*) ^{3} after shelling becomes 3*ab*(*a+b*), (*a+b*)^{ 5} after shelling becomes 5*ab*(*a+b*){*a*^{2}*+ab+b*^{2}}.

Note that T* _{ab}* begins with

If ((*a+b*) + *h*)* ^{p}* on the RHS of (1) is expanded binomially, treating (

* a** ^{ p}* +

= *c** ^{ p}* +

It may be noticed that the terms *a*^{ p} and *b*^{ p} on the LHS of (4) will *shell* the *c*^{ p} term on the RHS**. **[ *c ^{p} – a^{p} – b^{p}*

Also the *pa*^{ p -1}*h* and *pb*^{ p -1}*h *on the LHS effectively *shell* the *pc*^{ p -1}*h* term on the RHS. (Here the binomial being shelled has a pre- coefficient *ph* .)

Again the *p*(*p*-1)*h*^{2}*a*^{ p}^{ -2}/2! and *p*(*p*-1)*h*^{2}*b*^{ p -2}/2! terms on the LHS effectively *shell* the *p*(*p*-1)*h*^{2}*c*^{ p -2} /2! term on the RHS. [Here also the binomial has a pre- coefficient *p*(*p*-1)*h** ^{ 2}*/2!.] And so on…

Eventually the last shelling case occurs when the power of *h* is *p*-2:

the* p*(*p*-1)*a*^{2}*h*^{ p}^{ -2}*/*2! and the *p*(*p*-1)*b*^{ 2}*h*^{ p}^{ -2}/2! terms shell the *p*(*p*-1)*c*^{ 2}*h*^{ p}^{ -2}/2! term leaving *p(p-*1*)abh** ^{ p-2}*.

The next terms, *pah*^{p}^{ -1}and *pbh** ^{ p-1}*, cancel the

One

So the final result of these operations is:

*h** ^{p}* =

Each of the { } brackets have the factor *ab,* and each of their pre- coefficients have the factor *p*.

So *h** ^{p}* =

EXAMPLES When *p*=3 *h*^{3} = 3*ab*(*a+b+2h*)………………………(7)

When *p*=5, we have *h*^{5} = 5*ab*{*a*^{3} + 2*a*^{2}*b* +2*ab*^{2} + *b*^{3} +*h*(4*a*^{2} +6*ab* + 4*b*^{2})+6*h*^{2}(*a+b*)+4*h*^{3}}.(8)

The { } bracket on the RHS of (8) factorises to (*a+b+2h*){*a*^{2}*+ab+b** ^{2}*+2(

** **

So *h** ^{p}* = 5

Let *z* =(*a+b*+2*h*)………………………(11)

From this point onwards *z* will be treated as another major “unknown” alongside *a* and *b* the original unknowns. It too, it is being supposed, is a specific unknown natural number. It may be noticed above that when *p*=3 and *p*=5*, z* is a factor of *h** ^{ p}*.

These two examples suggest the conjecture that (*a+b+2h*) might be a factor of *h** ^{p}* for all

PROOF of the CONJECTURE that *z* is a FACTOR of *h* * ^{p}*.

Expressing the RHS of (1) in terms of *z*, (*a+h*)^{ p }+ (*b+h*)^{ p} = (*z-h*) * ^{p}* ……………………(12) Now the LHS of (12) has the factor {

This establishes that *z* is a factor of *h** ^{p}* for all values of

Now because *h** ^{ p}* has the factor

* h** ^{p}* =

Here *v* is effectively “what is left” of the expression for *h ^{p}* derived from (1) after taking out the factors

The next phase of the exposition looks at the question whether the factors *a,b,z,v* revealed in (14) above have any sub-factors in common.

LEMMA II *a* and *b* have no common factor. Further, neither *a* nor *b* has a common factor with *a*+*b* +2*h*.

PROOF: It is given that *a+h* and *b+h* have no common factor. If *a* and *b* had a common prime factor *x,* *h* * ^{p}* would clearly have it from (6). So

FURTHER: If *a* had a common prime factor *x* with (*a+b*+2*h*), it would have the same prime factor *x* in common with *b*+2*h*. Now *h ^{p}* contains all the prime factors of

This leaves the question whether *v* has a common factor with *a, b* or (*a+b*+2*h*). To prove that it does not, requires two things, a symmetry result (see Lemma III below) and an explicit formula for *v *forthcoming.

LEMMA III The original equation (1) can be written as:

(*z-b-h*)* ^{p}* + (

or as (-*z+h*)* ^{p}* + (

Let *a*’=-*z.* Equation (15) becomes identical with equation (2) apart from the dashed version of *a*. Also (11) rewritten becomes -a = -z + b + 2h. Now letting z’ = -a, (15) becomes z’ = a’+b + 2h. So equation (11) too remains the same apart from the dashes attached to *a* and *z*.

It follows that there is a symmetry here, and any result obtained with the unknowns plain, automatically becomes a valid result with the unknows dashed. A similar result obtains if *b*’ is substituted for –*z* and vice versa………………………………………………..….(15’). **[NR8]**

The key expansion (5), simplified, becomes:

* h** ^{ p}* =

Now let *W* = 2T* _{ab}* +

The RHS of (16) becomes *pab(a+b*)T* _{ab} *+

Hence *h** ^{ p}* =

But *h** ^{ p}* has the factor

* *So * h** ^{p}* =

However we know from (5) that *h* * ^{p}* is equal to

So *W*’ has the general form N_{1}*h* + N_{2}*h*^{2}+ N_{3}*h ^{3}* +…N

So (18) becomes:

*h** ^{p}* =

for some natural number coefficients N_{1}, N_{2}…etc. So here we have an explicit expression for *v*, i.e. the curly bracket in (18) above.

This result (19) is a plain, ordinary algebraic consequence of equation (1). It represents and embodies the essence of the problem posed by FLT. If (19) is possible, (1) is possible. If (19) is impossible (1) is impossible.

From (5) it may be clearly seen that the coefficients N_{1}, N_{2}, N_{3},… in (19) represent polynomial functions of *p, a, b*. Note that the *h* in the ( ) bracket implies that the highest power of *h* in the { } bracket is *p*-3.

The { } bracket in (19) is *v*. So now the outline form of *v* introduced in (14) i.e. *h*^{p}*= pabzv,* has been found. [*v* had previously been introduced effectively as *h* ^{p}*/pabz*.]

The postponed question whether *v* has factors in common with *a, b, z* can now be addressed.

T* _{ab}* begins with

This leaves the question whether *v* (or in effect T* _{ab}*) is co-prime to

*h** ^{p}* =

When *a*’ = –*z*, there is clearly no change in *h*, because *z* becomes –*a* and *a* becomes –*z*. (*a’+b*+2*h*) becomes –*z + b* +2*h* or –*a*. So the first four factors on the RHS of (19) become *p*(-*z*)*b*(-*a*) or *pabz* : no change. The { } bracket remans as *v*. The T* _{a’b}* of (19) becomes T-

This completes the demonstration that all possible pairs of *a, b, z, v* are co-prime.

It now follows from (14) *h** ^{p}* =

This is pivotal. Because each one of the four unknowns *a, b, z, v* is co-prime to the other three, and the LHS of (19) is a *p*th power, each of *a, b, z, v* will be a *p*th power too *with the proviso that one and only one of them has the extra factor* *p ^{p}*

(It is convenient to call the special unknown among the set *a, b, z, v* which has the extra factor *p ^{p}*

So, taking *p*th roots:

*h*=*pABZV*………………………………………………….(21) where *A, B, Z, V* denote the *p*th roots of *a, b, z, v* *except that* one of these —the exception— has to be pre-divided by *p*^{p}^{-1} before taking the *p*th root. (Incidentally the exception, being the only unknown with the factor *p,* might have additional hidden factors *p** ^{p}*,

It can be proved that the exception with the extra factor *p* * ^{p-1}* necessarily has the total factor

The results (20) and (21) are pivotal to everything which follows. (21) is the extremely neat, apparently symmetric “platform” from which it is possible to show that the hypothesis equation (1) is impossible.

(21) also provides a simple explicit formula for *h*, the central, crucial unknown whose possibility is, in effect, the main target of the inquiry.

**IMMEDIATE IMPLICATIONS ** .

Result (20) shows incidentally that *a* and *b* cannot be chosen indiscriminately. Both *a* and *b* will only be suitable to be inserted into (1) if they are *p*th powers, with one of them possibly also multiplied by *p** ^{ p-1}*.

(20) shows also that the *a *and *b* chosen initially must be such that *a+b+2h* forms another natural number *p*th power, again possibly multiplied by *p** ^{p-1}* (but only if neither

So what is revealed here is a remarkably neat, apparently symmetric, apparently consistent, sub-structure underlying the Fermat Problem. The next step is to identify the Achilles Heel of this apparently consistent sub-structure. This is the half-way point in the exposition. Readers are advised fully to familiarise themselves with the results as a preparation for Part 2 of this blog.

————————————————————————————————————————————–

Reference Notes

**[NR1]** Any conjecture is, of course, an hypothesis. The form of the proof of a negative conjecture like Fermat’s is that one tries to show that accepting the hypothesis leads to a contradiction. This means that we treat the hypothesis as valid and hope for a contradiction flowing from it.

The idea of switching unknowns is not new. It forms the basis, for example, of Cardan’s cubic equation solution. In the present case the reasoning behind the switch begins with the observation that *a ^{n} + b^{n}* is always smaller than (

If we add 1 to both *a* and *b* and also add 1 to *a+b*, the same result applies. For example 5^{7}+ 9^{7} < 12^{7} and 6^{7} + 10^{7} < 15^{7}. It still holds if we add 2 to *a, *and 2 to *b,* and 2 to *a+b*. And so on… But if we add a larger number, here denoted *h,* the sum of the two powers gradually closes the gap on the single power as *h* increases. Eventually, when *h* becomes much larger than both *a* and *b*, the two powers add up to 2*h ^{h}* approximately: whereas the single power is only approximately

**[NR2]** EXAMPLE a^{15} + b^{15} = c^{15} is also an example of cubic powers, namely a^{5} , b^{5} and **c**^{5} each cubed.

**[NR3]** INNER TRAY COEFFICIENTS This can be seen in Pascal’s Triangle. The inner numbers of line 3 are 3 3, those of line 5 are 5 10 10 5, and of line 7 they are 7 21 35 35 21 7, … and so on.

**[NR4]**The coefficients of the inner trays when *p*= 5, *p*=7 and *p*=11 may be noted, respectively:

1 1 1

1 2 3 3 2 1

1 4 11 19 23 19 11 4 1

If we displace each of these sets one place, and add it to the original we get:

1 1 1

1 1 1 Add & multiply by *p*:

__5 10 10 5__ (Pascal numbers)

Also:

1 2 3 3 2 1

1 2 3 3 2 1 Hence:

__7 21 35 42 35 21 7__ (Pascal numbers)

Also:

1 4 11 19 23 19 11 4 1

1 4 11 19 23 19 11 4 1

__11 55 165 330 462 462 330 165 55 11__ (Pascal numbers)

The underlined sets of numbers are familiar lines of Pascal’s Triangle with the end 1s missing.

By performing these steps in reverse order an algorithm for the Inner Tray coefficients can be found. [See exercises.]

NOTE: Binomials to non-prime powers do not have Inner Trays.

**[NR5]** SHELLING RESULT (3)

The 2^{nd} term in the expansion of (*a+h*)* ^{p}* is

Similarly the 2^{nd} term of (*b+h*)* ^{ p}* is

Next we do the 3^{rd} term of the two expansions on the LHS …and so on.

THE LAST SHELLING CASE Here we are, in effect, shelling *c* or (*a+b) *^{2} with *a ^{ 2}* and

This result (5) is significant. There were two *h ^{ p}* terms on the LHS of (4) and only one on the RHS. Thus we are left with one

(5) demonstrates that *h ^{ p}* has the factors

**[NR6]**This factorisation of the { } bracket in (8) was discovered by chance. To see how it arises we put *c* in place of *a + b*. The first four terms are *c*^{3} *-a ^{2}b -b^{2}a* or

**[NR7]** The RHS of (12) expanded by the Prime Binomial Lemma is

*z ^{p} -pzh*(

**[NR8] **Exactly the same reasoning and algebra is involved, except that the initial symbol *a *is replaced by *a*’ throughout and also the initial symbol *z* is replaced with *z*’ throughout. The final result is then interpreted by replacing *a*’ with *z *and replacing *z*’ with *a*.

**[NR9] **And the “something” left after taking out the separate factors *a,b,z*, will be alongside *a, b, z* if these unknowns have no common factor. *p *is the initial factor, which has been there from the beginning.

**[NR10]** The *p *factor on the RHS of (20) implies that *h ^{p}^{ }*has the factor

**SOME EXERCISES**

- Find the coefficients of T
for (_{ab}*a+b*)^{1}^{3}. - Show that if
*a, b*or*z*has the supernumerary factor*p*, that the corresponding unknown^{p-1}*A, B, Z*has the factor*p*as well. - Find an explicit formal for N
_{1}in terms of*a, b*, and*p*. - Find an explicit formula for N
_{2}in terms of*a, b, p*and N_{1}. - Prove that, when
*a*and*b*have no common factor, Tis always odd._{ab} - Prove that T
_{ab}__=__T(mod_{AB}*p*). - Prove that
*pab*(*a+b*)Thas the factor_{ab }*h*. - Prove that the T
of (_{ab}*a+b*)^{5}is not divisible by 5 for all natural numbers*a, b*.

Solutions will be given in the Blog next time.