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.) 

 

Fermat’s Enigma, Part 2

We have seen in Part 1 that Fermat’s Equation in its new form

 

      (a+h)^p + (b + h)^p = (a + b + h)^p ……………………(1)

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 pth powers or pth powers times p^p-1.  And in the light of the symmetry demonstrated in result (15’) this status might be expected to extend to z.  But V or v (which we get by raising V to the power p and possibly multiplying the result by p^p-1) is different. 

 

v, which consists of T_ab + ascending powers of h —starting with an h term and ending with an h^p-3 term— (with coefficients N₁, N₂, N₃…formed of functions of p, a, b), does not look anything like an obvious pth power, still less a pth power times p ^p-1. 

Also if v is the pth 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 V…………..…………………(22)

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

                v = T_ab + N₁h + N₂h² + N₃h³ + …   N^p_p-3h^p-3  as (18).

[Incidentally when p=3 T_ab = 1 and the Ns disappear. So here v = 1. This case is not covered by the following reasoning, but it is addressed in a special Postscript at the end.]

In the early example (see (9) where p = 5)  T_ab was a² + ab + b² and the Ns were simple the expressions, N₁= 2(a+b), N₂ = 2. It is evident, however, from (4) that typical Ns will be polynomial expressions involving p, a and b.       

The general formula for N₁ is 

N₁ = (p-2)T_ab/c –(a^p-2 + b^p-2)/c²……………………………..(23) [NR11] 

Now as T_ab has a factor V, it could in theory, when combined with the N₁h term of v, produce a V² term.  This is not very likely, but it is possible.[NR12]  Here we have a case where two terms with the same factor (here V) are, we suppose, implicitly “merging” to form a V² term. In the sequel the term ‘merge’ will be used to signal an increase of one or more in the power of a relevant factor shared by two added items of the same power of that factor. [NR13]          .                                                                                                                  

The V² term thus produced could, in theory, merge with the N₂h² term in (19) to produce a V³ 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-3h^p-3 term of v to give…. plain V^p (possibly times p^p-1). (Here the exponent of p needs to increase by 3 or possibly p-2.  That the final result of p-3 unlikely mergers should lead to a pth power or to a pth power times p^p-1 can only add to an overall sense of extreme unlikelihood.) [NR14]

So a ladder of p-3 consecutive, unlikely mathematical mergers need to occur if v is to be the pth 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 p in addition to the factor V, so the required chain of unlikely mergers noted above becomes even less likely, involving both p and V, not simply V. [NR15]

CAN v BE A pth POWER? 

Here we address case (1) when v does not have the supernumerary factor and is a plain pth power. So in this section v is V^p  and one of a, b, z has the supernumerary factor p^p-1.

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 V.[NR16]

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 a^p-3 and ends with b^p-3. The terms between the powers a^p-3 and b^p-3 are the interior terms. The interior terms all have the factor ab. [NR17]

Recapping (19), V^p = {T_ab + a term in h, a term in h², a term in h³, …. etc. up to h^p-3. Here the terms with a factor h all have the factor AB and all the interior terms of T_ab have the factor ab, which is A^pB^p. So the factor AB runs through all the terms of V^p, apart from the two outer terms of T_ab,   namely a^p-3 and b^p-3. These are the pth powers of A^p-3 and B^p-3 respectively.

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 pth power, and treating (A^p-3 + B^p-3) as a unit, using the Prime Binomial Lemma,

V^p = a^p-3 + b^p-3 + A^p-3B^p-3 terms + py(A^p-3+B^p-3)VT’ + y^p, where T’ is  an inner tray………………………………………..(25)                      

     

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 = v and v, once the T_ab term has been removed,  has only the overall factor AB,  from N₁h.  [NR20] 

[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-3B^p-3 terms +py(A^p-3+B^p-3)VT’+y{q’AB-pV(A^p-3+B^p-3)T’} ………………………………(28)   [NR21]

Here the py(A^p-3 + B^p-3)VT’ brackets cancel, leaving only –yq’AB to take the  place of y in equation (24).           This is, in effect, another version of y = kAB.

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 = {a^p-3 + b^p-3 + terms with a factor ab} + N₁h + N₂h² +… +N_p-3h^p-3, where the terms in the {} brackets are T_ab, and where h is pABZV.   [This is effectively another version of (19).] 

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 N₁h term is also removed from V^p, the resulting expression (Q’) will have the factor A²B².   [NR23]     So the current task is to see if the result (29) can meet this condition.

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

              

V^p = (A^p-3+B^p-3)^p + k^pA^pB^p + pkABV(A^p-3+B^p-3)T_v, where T_v is an inner tray……………………………(30) 

So the issue becomes whether pkABV(A^p-3+B^p-3)T_v -N₁h has the factor A²B².  [NR24]

Or, after dividing by pAB, whether    k(A^p-3+B^p-3)T_v -N₁ZV 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, N₁, Z and V.  [NR25]

We look at these in turn.

When A=0, T_v becomes (B^p-3)^p-3    [NR26] and Z becomes B/(p)^(p-1)/p, because z=a+b+2h which reduces in this case to z = b or Z^pp^p-1 = B^p. 

When A=0, V becomes B^p-3.                                                                  .   The coefficient N₁ becomes (p-2)B^p-3/B – B^p(p-2)-2 (see (23)). [NR27] 

When A=0, k becomes initially indeterminate (V-A ^p-3-B ^p-3)/AB. However cancelling V with –B ^p-3 in the numerator leaves  A^p-3  divided  by  AB,   or A^p-4/B i.e. zero.)  So k becomes 0.

     So the expression named in (31) after substituting A=0 has as its first term zero, because k is zero.  Its second term is -N₁ or -(p-2)B^p-4 + B^p(p-2)-2  times ZV [which = B/(p)^(p-1)/p times B^p-3].

    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 (=v) as set out in (19). [NR28]                

   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 + N’₁AV …………………….………(32) and the question is now whether this has a factor ZB, where k’ is a natural number and N’₁ is the coefficient of h in the expansion of v expressed in terms of –z and b.    Putting Z=0 cannot reduce the new expression to zero, because k’ is zero as before and N’₁ reduces to exactly the same expression in B as shown in the last note on p.15.   

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

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

                   v = T_ab + N₁h + N₂h² + N₃h³ + …   N^p_p-3h^p-3 ………………..(19)

 could possibly take the form p^p-1V, 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 V or p. But since the RHS of (19) as a whole has the factors V and p, and the h terms all have the factor pV, it may be deduced that T_ab has the factor V and p too. [NR30]

So                   T_ab = pVY for some unknown natural number Y………….(33)                                  

Incidentally T_ab has only the factor pV, not higher powers of p or V. It is only by merging T_ab and N₁h to form a new p²V² term that the final result can (improbably) emerge. (This then needs to merged with N₂h² to form a new (improbable)  p³V³ term to be merged with… and so on.) So if T_ab had higher powers of p or V as factors this process could not begin.

Now T_ab treated as pVY and N₁h treated as N₁pABZV have explicit factors p and V, but the pair need to merge to produce a factor p²V²: otherwise the RHS of (23) would only have an overall factor pV. So [NR31]

                  T_ab+N₁h = p²V²L for some natural number L………….…..(34)

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 + N₁pABZV = p²V²L.

Hence                  Y + N₁ABZ        = pVL……………………(35)

So                                           V     = (Y + N₁ABZ)/pL.    [NR32]

Hence                         v = p^p-1V^p    =  (Y + N₁ABZ) ^p/pL^p……………….(36)

Now v – b^p-3 is an expression composed entirely of terms with a factor A,  because T_ab has the form a^p-3 + ab terms + b^p-3 and all the powers of h  in (19) have the factor A..

So it is a logical consequence of the original hypothesis in this case that              (Y +N₁ABZ) ^p/pL^p -b^p-3 should have the factor A………………………..(37) 

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)/pL^p -b^p-3) and this should be zero………………………………(38)

Hence, multiplying (38) by pL:    Y^p -pL^pb^p-3=0……………………………….(39) 

  

Can the LHS of (39) be zero? No! Because it is a pth power (> 0) minus p times a pth 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 + 2h as before and (20) becomes h³ = 3abz, where a, b, z are cubes (of A, B, Z respectively) except that one of them also has the supernumerary factor 9 and h=3ABZ…….…………………………..…(40)

Let z be the exception with the factor 9, so z = 9Z³.   [NR34]

Now a+b has the factor 3 so A has the form 3A + j and B has the form    3B – j where A and B are some natural numbers: also  j = 1 or 2). So, cubing and adding,  A³ + B³ (or a+b) will have the factor 9.  [NR35]

Now  9Z³ = a + b + 6ABZ.  From this  a+b has the factor Z.

So 9Z³ – a – b = 6ABZ………………………………..………………..(41)

Here the LHS has a factor 9 at least. So 6ABZ 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 S³ = a + b + 3ABS …………………………..……..(42) 

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

            So         9Z³-S³ = 3AB(2Z-S)………………………………….(43)                                         Also 8Z³ – S³ = (4Z² + 2ZS + S²)(2Z–S)   [NR38]

 Subtracting, Z³ = {3AB -4Z² -2ZS – S²}(2Z-S)……………………(45)                                           Or                Z³ = {4Z² + 2ZS + S² – 3AB}(S-2Z). ………………………(46)                       

                                                             .                                                                                                   Now (46) implies that the { } and the ( ) brackets will each be composed entirely of factors of Z³.  So S in the ( ) bracket must be composed only of factors of Z. However the { } bracket cannot now meet this condition, because it contains -3AB (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-2Z) equals Z³/3. But because Z> A and Z > B and S > A and S > B the { } bracket is larger than 4AB 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 pabcT_ab had a factor h follows from result (5) (Part 1). But some factors of h, namely, p, A, B, Z are already present in pabc.  [In the case of c establishing this was No 8 in the Exercises.] So it follows that T_ab has the factor V.

[NR13] Now two terms each with the factor V can of course produce a sum with the factor V². A simple example is V² – 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 AB raised to a high power.

[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 N₁h 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 n > 1, and those without the factor (AB).But y cannot have the factor AB to a higher power n, where n > 1, so we end up with the two cases discussed in the main text.

[NR20] We have reverted here to (25) leaving the py(A^p-3+B^p-3)VT’ term in place (now a natural number) and treating y^p as yy^p-1.

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

y{pXVT’ +y^p-1}where X is         A^p-3 + B^p-3. This is then an irrational number y, times a natural number, which disqualifies it as part of an expression for V.

[NR21] We are treating the y^p in (25) again as y y^p-1.

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

[NR23] In other words, after removing N₁h,  the lowest power of (AB) present is (AB)² implicit in h².

[NR24] The inner terms of the (  )^p bracket and k^pA^pB^p have high powers of (AB) as factors anyway, so they are not under scrutiny.

[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,N₁, 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 (A^p-3 + B^p-3)^p, so all its terms have the factor A except the last, which is B^p-3 raised to the power p-3.

[NR27] Here the T_ab in (23) reduces to b^p-3 and the (   )/c² term reduces to       b^p-2/c². 

[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 p² because V also has the factor p. Thus T_ab has the factor pV necessary to start the cascade of improbable mergers off which will result in the RHS of (19) becoming p^p-1V^p.

[NR31] In which case the improbable cascade of p-3 mergers needed to turn the RHS of (19) into p^p-1V^p could not begin.

[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-1.  Most of the p factors in the denominator are cancelled by p^p-1, leaving a single p factor in the denominator.

[NR33] The only way in which a pth power minus p times a pth power (of natural numbers) could be zero is if both of the pth 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  b^p-3, so Y cannot reduce to zero. The same reasoning applies to L in (34). 

[NR34] That h^p = 3abz follows from (1) when p=3. (a+h)³ + (b+h)³ = (c+h)³. So one of a, b, z has the factor 9.

[NR35] The j³ terms cancel and the all the other terms have the factor 9 or 27.

[NR36] The 9Z³ 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 -2h. z is Z^p or p^p-1Z^p and h is pABZV. So both z and 2h have the factor Z.

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 – 2h, z = b (mod p). So Z = B (mod p) . So z – b has the factor p².  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 N₁ = (p-2)T_ab/c –(a^p-2 + b^p-2)/c².

5 N₂ = {½(shelled c^p-2)/pab  -2N₁}/c.  [Equate the h² term from (5) with the h² term from (19).]

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 a, b are odd put a=1 and b=1. So 2^p = 1 +2pT_ab.+1.  If T_ab were even, this would only have the factor 2. So  T_ab is odd when a = b = 1. In the general case the coefficients are the same so the result will be the same.

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 a² + ab + b².  The table of possible non-zero residues of T_ab mod 5 is as follows:  a = 1,2,3,4 across   b= 1,2,3,4 down

3   2   2   1

2   2   4   3

2   4   2   2

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