*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 exclusively 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?*

*The initial topic is *Fermat’s Enigma*, on which the author has been working for more than 29 years using only the most basic methods. This is a landmark issue, because the failure to find an elementary demonstration of Fermat’s conjecture hangs like a dark shadow over the elementary subject. *

*The first Blog laid out the preliminary reasoning steps needed to tackle Fermat’s Enigma. Readers are advised to read Blog No. 1 carefully and to work the Exercises appended at the end, before attempting to work the current Blog. (The Answers may be found at the end of this Blog.) *

(*a+h*)* ^{p}* + (

implies, by reasoning and elementary algebra, that *h = pABZV*………………..(21)

Here *A* and *B* are specific unknown natural numbers, while *Z* and *V* are also postulated unknown but *specific* natural numbers defined in terms of *A* and *B*.

The main question is now <<Which unknown is most likely to harbour a contradiction?>>

**THE SUB-STRUCTURE’s POTENTIAL ACHILLES’ HEEL**

The Agenda is to show that the hypothesis that there exists a specific set of natural numbers *a, b, h* satisfying equation (1) —and hence (18), (20), and (21)— leads to contradiction.

Which part of the remarkable sub-structure is most odd, and therefore most likely to be inconsistent? The answer is quite clear: *V* stands out. There is no reason why the initial unknowns *a* and *b* should not be *p*th powers or *p*th powers times *p ^{p}*

*v*, which consists of T* _{ab}* + ascending powers of

Also if *v* is the *p*th power of *V* (with or without a supernumerary factor *p*^{p}^{-1}) then, because all the terms of the polynomial in *h* have the factor *V*, . T* _{ab}* has the factor

Recapping, an explicit statement about the shape of *v* is:

* v *= T* _{ab}* +

[Incidentally when *p*=3 T* _{ab}* = 1 and the Ns disappear. So here

In the early example (see (9) where *p* = 5) T* _{ab}* was

The general formula for N_{1} is

N_{1} = (*p*-2)T* _{ab}*/

Now as T* _{ab}* has a factor

The *V*^{2} term thus produced could, in theory, merge with the *N*_{2}*h*^{2} term in (19) to produce a *V*^{3} term. This is not very likely, but it is possible. Altogether *p*-3 similar unlikely results of merging are needed until the *V*^{p}^{-3} term is reached which could in theory combine with the final *N _{p}*

So a ladder of *p*-3 consecutive, unlikely mathematical *mergers* need to occur if *v* is to be the *p*th power of *V,* possibly times *p*^{p-1}

Each of the mergers considered above is unlikely. That a sequence of *p*-3 of them should obtain in an unbroken chain is evidently very unlikely indeed. So this is a potential Achilles’ Heel: but it needs to be shown that it leads to an explicit, clear-cut contradiction.

There are two distinctly different cases: (1) when the supernumerary *p*^{p}^{-1} factor is *not* a factor of *v*, (and it *is* a factor of *a*, *b*, or *z*)*, *and (2) when the supernumerary factor *is* a factor of *v*.

These two very different cases will be taken separately. In case (2) T* _{ab}* has the factor

CAN *v* BE A *p*th POWER?

Here we address case (1) when *v* does not have the supernumerary factor and *is* a plain *p*th power. So in this section *v *is *V** ^{p}* and one of

Let the exception (the unknown with the supernumerary factor) be *z*. There is a Note at the end of this Part showing how the reasoning laid out below generalises to the other cases, when *a* or *b* happen to have the supernumerary factor.

Now equation (1) and its derivative (19) can only be satisfied by unique natural number unknowns *a, b, *and* h* if, among other things, T* _{ab}* is divisible by

The main question is whether a *V* meeting this condition… plus the full set of requirements (‘mergers’) mentioned above, is logically possible.

Consider the possible structure of *V*:

Now the general form of T* _{ab}* is symmetric: it begins with

Recapping (19), *V ^{p}* = {T

So *V* will have the general form *A*^{p}^{-3} + *B*^{p}^{-3} + *y*, where *y* is whatever natural number is needed to make the total up to *V*..……………………………………(24)

Raising this expression for *V* to the *p*th power, and treating (*A*^{p}^{-3}* + B*^{p}^{-3}) as a unit, using the Prime Binomial Lemma,

*V** ^{p}* =

So, from the reasoning above, the last two *y* terms combined are expected to have the factor *AB*. **[NR18]**

There are two possibilities, *either *that *y=kAB* for some *k*, **[NR19]** *or* *y *does not have the factor *AB*. [*k* here is an unknown integer, not having a factor *A* or *B*. Note that in the first case *k* itself has no factor *A *or* B* because *V** ^{p}* =

[The second case is the “less obviously suitable form of *y*”.]

If the second possibility obtains, the expression which needs to be shown to have the factor *AB, * i.e. the last two terms on the RHS of (25), can be replaced by *pV*(*A*^{p}^{-3} + *B*^{p}^{-3})T’ + *y*^{p}^{-1}……………………………..(26)

[Dividing by *y, *which, it is being assumed, lacks the factor* AB*.]

So let *pV*(*A*^{p}^{-3}+*B*^{p}^{-3})T’+*y*^{p}^{-1} = *q’AB* for some natural number *q*’. It follows, taking *p*-1th roots, that *y* = {*q’AB – pV*(*A*^{p}^{-3}+*B*^{p}^{-3})T’}^{1/}^{p}^{-1}……………..……………(27)

This is an expression for the “less obviously suitable” *y* which might be needed to turn *A*^{p}^{-3} + *B*^{p}^{-3} into *V*. It may be noted that, as a (*p*-1)th root, it is fairly unlikely to be a natural number. [If *y* is not a natural number, it can be disregarded.]

But if the *y* shown in (27) *is* a natural number, substituting it into (25) produces the result that: *V*^{p} = . *a ^{p-3}+b^{p-3}+A^{p-3}B^{p-3 }*terms +

Here the *py*(*A ^{p-3} + B^{p-3}*)

So, we have established the important result that

* V*=*A*^{p}^{-3}+*B*^{p}^{-3}+*kAB*, where *k* is an integer which has no factor *A* or *B*………………………………………..…………….(29)

The main question now becomes whether *A*^{p}^{-3} + *B*^{p}^{-3} + *kAB, *when raised to the power *p*, can satisfy the logical conditions for *v, *that is:

*V** ^{p}* = {

Now, if the first two terms of the { } bracket above are removed, the resulting expression (*Q*) in there will have the factor* AB*, because the remaining terms inside the { } bracket have the factor *ab* and the polynomial terms all have the factor *h*. **[NR22]**

But if the *N*_{1}*h *term is *also* removed from *V** ^{p}*, the resulting expression (

Raising the RHS of (29) to the power *p* using the Prime Binomial Lemma,

*V*^{p}* *= (*A*^{p}^{-3}*+B*^{p}^{-3})* ^{p}* +

So the issue becomes whether *pkABV*(*A*^{p}^{-3}*+B*^{p}^{-3})T_{v} -N_{1}*h* has the factor *A*^{2}*B*^{2}. **[NR24]**

Or, after dividing by *pAB*, whether *k*(*A*^{p}^{-3}*+B*^{p}^{-3})T_{v} -N_{1}*ZV *has the factor* AB……………….……………..*(31)

At this point (31) will be treated as a specific structure composed ultimately of the specific unknowns *A,* *B *and* p.* To ascertain whether *A* is a factor of the expression (31) it is sufficient to substitute *A*=0 in the expression, treated as an algebraic configuration, taking into account the implicit role of *A* in the elements T* _{v}*, N

We look at these in turn.

When *A*=0, T* _{v}* becomes (

When* A=*0*, V* becomes B* ^{p-3}*. . The coefficient N

When* A=*0*, k* becomes initially indeterminate (*V-A*^{ p-3}*-B** ^{ p-3}*)/

So the expression named in (31) after substituting *A*=0 has as its first term zero, because *k* is zero. Its second term is -N_{1} or -(*p*-2)*B ^{p-4}* +

This is not zero. So *A* is not a factor of (31). This is a contradiction. It shows that the only possible form *V* could take (as in (29)) is not consistent with the necessary expression for *V ^{p}* (=

The above reasoning was predicated on the temporary assumption that *z* was the unknown with the supernumerary factor. So it needs to be shown that the same logic applies when *a* or *b* is the supernumerary factor. These unknows *a* and *b* are effectively interchangeable, so it is sufficient to show this in the case when *a* is the exception. When *a* is the basic unknown with the supernumerary factor, the symmetry referred-to in the Lemma III result (15) may be used: replacing –*a* with *z* and *z* with –*a* produces valid equations throughout. **[NR29]**

So (31) becomes *k*’(*Z*^{p}^{-3}*+B*^{p}^{-3})T* _{v}* +

So the hypothesis that *a, b, h* exist as natural numbers satisfying (1) leads to contradiction in all the cases when *v* is a *p*th power.

We now turn to the second possibility, that *v* has the supernumerary factor.

CAN *v *BE A* pth *POWER times *p** ^{p-1}*?

Now the remaining possibility needs to be considered, namely that it is *v* which is the exception, and *a, b, z* are plain *p*th powers.

At first sight it is difficult to see how the shape established for *v*,

* v* = T* _{ab}* +

could possibly take the form *p*^{p-1}*V*, for any conceivable version of *V*.

(Incidentally *V* itself has the factor* p*, see the remark three lines after (21).)

The first term of (19), T* _{ab}*, has no explicit factor

So T* _{ab}* =

Incidentally T* _{ab}* has

Now T* _{ab}* treated as

T* _{ab}*+

Incidentally *L *here* *has no factor *A* or *B, *otherwise the whole of the RHS of (19) would have a factor *A* or *B*. [*v, a, b* have no common factors.]

Hence *pVY + N*_{1}*pABZV = p*^{2}*V*^{2}*L*.

Hence *Y + N*_{1}*ABZ = pVL*……………………(35)

So *V * = (*Y + N*_{1}*ABZ*)/*pL*. **[NR32]**

Hence *v* = *p*^{p-1}*V** ^{p}* = (

Now *v* – *b** ^{p-3}* is an expression composed entirely of terms with a factor

So it is a logical consequence of the original hypothesis in this case that (*Y *+*N*_{1}*ABZ*)* ^{ p}*/

To check whether this is the case the type of reasoning used previously from (31) onward is needed here again. Putting *A*=0 in (37) should reduce this expression to zero.

The special values of* Y *and* L* will be denoted with underlining when *A*=0. So when *A*=0 the expression (37) becomes (*Y** ^{p}*)/

Hence, multiplying (38) by *p**L*: *Y*^{p}*-p**L*^{p}*b ^{p-3}*=0……………………………….(39)

Can the LHS of (39) be zero? No! Because it is a *p*th power (> 0) minus *p* times a *p*th power (> 0). This is a contradiction. **[NR33]**

Thus a contradiction results both from the assumption that *v* has the factor *p*^{p-1}*,* and that it does not. It follows that there can be no natural number solutions for the Fermat Equation (1) when *p* > 3. QED

POSTSCRIPT The Case when *p*=3.

When *p* = 3 the above reasoning does not apply, because both *v* and *V* equal 1.

Here *z = a + b* + 2*h* as before and (20) becomes *h** ^{3}* = 3

Let *z* be the exception with the factor 9, so *z* = 9*Z ^{3}*.

Now *a+b* has the factor 3 so *A* has the form 3*A** + j* and *B* has the form 3*B** – j *where *A* and *B* are some natural numbers: also *j = *1 or 2)*.* So, cubing and adding, *A*^{3} + *B*^{3} (or *a+b*) will have the factor 9. **[NR35]**

Now 9*Z** ^{3}* =

So* *9*Z*^{3}* – a – b = *6*ABZ*………………………………..………………..(41)

Here the LHS has a factor 9 at least. So 6*ABZ* has the factor 9. It follows that *Z* has the factor 3. **[NR36]** . (The other factors of *Z* —not 3— will be called its “ordinary” factors.)

Let *S = A+B*, then *S*^{3} = *a + b* + 3*ABS* …………………………..……..(42)

(*S* incidentally also has the factor 3 at least, or 3*A* + 3*B* above.) **[NR37]**

So 9*Z*^{3}*-S*^{3}* *= 3*AB*(2*Z-S*)………………………………….(43) Also 8*Z*^{3}* – S*^{3} = (4*Z ^{2}* + 2

Subtracting*, Z** ^{3}* = {3

. Now (46) implies that the { } and the ( ) brackets will each be composed entirely of factors of *Z*^{3}. So *S* in the ( ) bracket must be composed only of factors of *Z*. However the { } bracket cannot now meet this condition, because it contains -3*AB* (*A*, *B *co-prime to *Z*) and other terms which all have the ordinary factors of Z. The only combination of the factors of *Z* in the two brackets which is consistent with (46) is that the curly bracket reduces to 3 and (*S*-2*Z*) equals Z^{3}/3. But because *Z*> *A* and *Z* > *B* and *S *> *A* and *S* > *B* the { } bracket is larger than 4*AB* and hence much larger than 3. This is a contradiction.

A similar contradiction is obtained if *A* or *B* has the supernumerary factor 9. It follows that the Fermat Equation cannot be satisfied when *p*=3. QED

**Conclusion**

What difference does such a proof make? Well, it shows that a tacit axiom in pure mathematics which has been accepted since the time of Gauss is not valid. For over 200 years it has been assumed that the best hope of making progress in pure maths is by *generalising problems*, so that a given challenge on a lower level of generality becomes a more abstract challenge on a higher level. This has had the unfortunate long-term effect of moving the research focus of the subject away from anything even the moderately mathematical public could understand. It is now “in the clouds”, and the disconnect which this has produced, is not doing the subject’s reputation with the educated public any good. It is in danger of turning sour.

This proof uses only concepts and methods taught to young people aged 17-18. Working it, and hence seeing the power of elementary reasoning at first hand, is something which can give school maths at the advanced level a much needed boost. CPO.

**Monographs by the Author:**

1* New Thinking about the Nature of Mathematics* (1992). A collection of essays by Philip Davis, Ray Monk, Warwick Sawyer, Paul Ernest, David Henley, Eric Blaire and Chris Ormell.

2 *Some Varieties of Superparadox* (1993)

3* Some Criteria for Sets in Mathematics* (1996)

4* The Peircean Applicability of Mathematics* (1997)

5* After Descartes* (2000)

6 *Exotic Infinity and a long decline of confidence in mathematics* (2008)

These volumes are available from Ingleside-Ashby, PO Box 16916, London SE3 7WS, UK. Prices: £10 except nos 3, 4, 5 which are £15. Postfree inside the UK. For EU add £5 postage, for non-EU add £5. Cheques and money orders should be payable in Sterling to “Ingleside-Ashby”.

**Some comments:**

[Of 1] “It engagingly raises some issues in the philosophy of mathematics” Prof. Robert Thomas, Editor *Philosophia Mathematica*. [University of Manitoba]

[Of 2] “I would certainly recommend the monograph as an essential read for those with any interest in the foundations of the subject”Stephen Jones in *The Mathematical Gazette*.

[Of 3] “Ormell quite reasonably insists on knowing what mathematical objects are, before addressing the question of what a set of mathematical objects is. His notion of the nature of mathematical objects is that they are ‘honorific existents’. This seems to work pretty well. I suspect that any comprehensive description of the nature of mathematical objects will have to incorporate some such notion”. Fred Richman in *Philosophia Mathematica*.

[Of 4] “The author’s impassioned plea is to recognise honestly that the main reason for supporting mathematics at every level is its applicability: and that the possibility of this applicability is not a mystery, but comes directly from Peirce’s <<*Mathematics is the study of what is true of hypothetical states of things*>.” Prof. Clive Kilmister, formerly London University.

**NOTES AND REFERENCES **

**[NR11] **The challenge to establish this was No. 4 in the set of Exercises. It may be noticed that the terms here are of order *p-*4.

**[NR12] **That *pabc*T* _{ab}* had a factor

**[NR13] **Now two terms each with the factor *V *can of course produce a sum with the factor *V*^{2}. A simple example is *V*^{2} –* V* and *V*. When two terms, after adding, thus produce a factor with a higher power of their previous common factor, the process may be called ‘merging’.

**[NR14] **That this is so extremely unlikely might serve as evidence that the current approach is on the right lines. It is not, of course, any kind of proof.

**[NR15] **If we multiply together the two notional very small probabilities that these sets of mergers will happen, the probability that both conditions are met, is, of course, a whole level of magnitude smaller.

**[NR16] **See [NR8] (Part 1).

**[NR17] **We do not need to know anything about these interior terms of T* _{ab }*except that they have the factor

**[NR18] **That is to say, once the *a*^{(}^{p}^{-3)} and *b*^{(}^{p}^{-3)} terms have been removed, what is left must have the factor *AB* only. [Not (*AB*)^{2} because the term N_{1}*h* in (19) only contains the factor *AB*.]

**[NR19] **The *y* term here covers all possibilities. These are then split into three, those with the plain factor *AB,* those with the factor (*AB*)* ^{n}* where

**[NR20] **We have reverted here to (25) leaving the *py*(*A ^{p}*

If *y* is irrational, the two *y* terms can again be treated as

*y*{*pXVT*’ +*y ^{p}*

**[NR21] **We are treating the *y ^{p}* in (25) again as

**[NR22] ***h* has the factor *AB* from (21).

**[NR23] **In other words, after removing N_{1}*h, * the lowest power of (*AB*) present is (*AB*)^{2} implicit in *h*^{2}.

**[NR24] **The inner terms of the ( )* ^{p}* bracket and

**[NR25] **If this structure reduces to zero when *A* has been equated to zero in the ( ) bracket of (31) and in the expressions for respectively *k*, T_{v},N_{1}, *Z*, and *V*, this will be an unequivocal sign that the expression (31) has the factor *A*. If not, it will be an equally unequivocal sign that (31) does *not *have the factor *A.*

**[NR26] **T* _{v}* relates to (

**[NR27] **Here the T* _{ab}* in (23) reduces to

**[NR28] **It has been shown both that *V* must have the form of (24) and that that a contradiction flows from this.

**[NR29] **The expression (32) is the part which has to be shown to have the factor *ZB*. The full expression equivalent to (30) also contains many terms which clearly have the factor *ZB*.** **

**[NR30] **Notice that *h* from (21) is *pABZV*, so *h* now has the factor *p*^{2} because *V* also has the factor *p*. Thus T* _{ab}* has the factor

**[NR31] **In which case the improbable cascade of *p*-3 mergers needed to turn the RHS of (19) into *p ^{p}*

**[NR32] **Making *V* the subject. On the next line the expression for *V* is raised to the power *p* and then multiplied by *p ^{p}*

**[NR33] **The only way in which a *p*th power minus *p* times a *p*th power (of natural numbers) could be zero is if both of the *p*th powers are individually zero. Now the version of *Y* which remains when *A* is replaced by 0 is determinable from (33). But T* _{ab}* reduces to a positive

**[NR34] **That *h ^{p}* = 3

**[NR35] **The j^{3} terms cancel and the all the other terms have the factor 9 or 27.

**[NR36] **The 9Z^{3} term has the factor 243, but –*a -b* has only *necessarily* got the factor 9. Of course *Z* might have a higher power of 3 as factor, in which case all the powers in the subsequent reasoning would be increased correspondingly. *S* would also have the factor 9 but the { } bracket in (46) would only, at most, have 3 among the factors of Z. In this way the form of contradiction introduced in the text becomes even more clear.

**[NR37] **See [NR36] above.

**[NR38] **By standard factorisation.

**ANSWERS**

1 c = *a + b = z* -2*h*. *z* is *Z ^{p}* or

2 The Pascal numbers divided by 13 are 1 6 22 55 99 132 132 99 55 … The Inner Tray coefficients: 1 5 17 38 61 71 61 38 17 5 1. Here 1+5 = 6 5 + 17 = 22 17+38 = 55 38+61 = 99 61+71 = 132, etc. To get these figures, subtract each known inner tray number from the next Pascal number-divided-by-13. Start with the first inner tray number, 1.

3 If *a* has the supernumerary factor where *a = z – b* – 2*h*, *z **= **b* (mod *p*). So *Z **=** B* (mod *p*) . So *z – b* has the factor *p*^{2}. It follows that h has the factor p2. Only *a* can have the factor *p*, so *A* has the factor *p*. Similarly, b, z.

4 N_{1} = (*p*-2)T* _{ab}*/

5 N_{2} = {½(shelled *c ^{p}*

6 When one of the pair *a, b* is odd T* _{ab}* has an odd term at one end and all the other terms are even. When both

7 Each term in both expressions has the same residue mod *p*.

8 This follows from (5): every term in (5) has the factor *h* except *pab*(*a+b*)T* _{ab}*.

9 The inner tray of (*a+b*)^{5} is *a*^{2}* + ab + b*^{2}. The table of possible non-zero residues of T* _{ab}* mod 5 is as follows:

3 2 2 1

2 2 4 3

2 4 2 2

1 3 2 3 There are no 0s in this table.