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 when a statement p implies a contrary statement q, which implies p, which implies q, which implies p, … etc ad inf. It is anathema to Platonists, because it is based on a recognition that maths is a human activity  —pursued with rigorously defined, but referencefree, symbol systems— which happens in real time.

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 nth 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 nth powers adding up to a third nth 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:

 

Fermat’s Enigma,

Dissecting the problem which flummoxed the world of maths.

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 [NRn]. They may be found at the end or by clicking on tabs.

 

Fermat’s Enigma,

a logical dissection based on divisibility

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

aⁿ + b ⁿ = cⁿ , where a, b, c and n are natural numbers when n>2. (‘Smallest’ here means with the minimum value of a+b and the earliest n.)

The new dissection, however, starts with a radical change of unknowns. Instead of couching the Fermat Equation in terms of a, b, c and n, as above, we switch to  (a+h)^p + (b+h)^p = (a+b+h)^p ,……………………………………….………………….……………………………………………………………………………………….(1) where h = a + b – c and the two key unknowns a and bare defined as     a = c – b    and b = c – a, where a + b takes the minimum value for which (1 ) is true and  p >2 is the smallest prime for which it is true.. This is the ‘New Fermat Equation’. The important point is that the following reasoning is about three specific unknowns, a, b and p. [NR1]

It is fairly clear that only cases where the original unknowns a and b have no common factor need be considered –because common factors can be cleared immediately from the Original Fermat equation, and also only prime values, p, of n (p>2) are needed. The latter because any example of the Fermat Equation being satisfied by n would also automatically be an example of the Fermat Equation using any prime factor of n.   If n = quy (say) where q, u and y are prime numbers, each of the binomials on the LHS of (1) can be regarded as being qth powers, uth powers or yth powers.  Similarly the bracket on the RHS.              [NR2]

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 p.   (‘Inner’ here means after the first, and before the last.) These coefficients have p in their numerators and various natural number factors less than p in their denominators: so the numerator factor p remains uncancellable. [NR3]

Now (a + b)^p  = a ^p + various pab terms + b ^p.   Also both (a+b) ^p and a ^p + b ^p have the factor (a+b).  So (a+b) ^p – a ^p – b ^p has the factor (a+b).  Hence the rump of the expression (after removing a ^p and b ^p) also has the factor a + b. So this expression “various pab terms” has the factor (a+b) too. It follows that this rump will take the form pab(a+b) times something.

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 = a^p + pab(a+b)T_ab + b^p………………………………….(2)

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 = ((a+b)^p – a ^p  – b^p)/pab(a+b)…………………..(3)

 

Every binomial with a prime exponent has its Inner Tray.

EXAMPLES: The inner tray of (a+b)⁵ is a² + ab + b².  The inner tray when p=7 is a⁴ + 2a³b + 3a²b² + 2ab³ + b⁴.   The inner tray when p=11 is a⁸ + 4a⁷b + 11a⁶b² + 19a⁵b³ +23a⁴b⁴ + 19a³b⁵ + 11a²b⁶ + 4ab⁷ + b⁸. [NR4]

NEW TERM: SHELLING It is convenient to describe the process of removing the outer terms aⁿ and bⁿ from the binomial expansion of (a+b)ⁿ as shelling the binomial expansion.

EXAMPLES (a+b) ³ after shelling becomes 3ab(a+b),  (a+b) ⁵ after shelling becomes 5ab(a+b){a²+ab+b²}.

Note that T_ab begins with a ^p-3, continues with various ab terms, and ends with b ^p-3. The coefficients of T_ab are symmetric, in the sense that the second is the same as the penultimate, the third is the same as the pre-penultimate, etc.

If ((a+b) + h)^p on the RHS of (1) is expanded binomially, treating (a+b) as c, and the two LHS binomials are similarly expanded, the result is:

 a ^p + pa ^{p -1}h +…+pah ^{p -1} + h ^p            +       b ^p +pb ^{p -1}h +…+pbh ^{p -1} + h ^p

= c ^p + pc ^{p -1}h +…+ pch ^{p -1} + h^p……………………………..…………………(4)

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 = pab(a+b)T_ab       from (2)] [NR5]

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)h²a ^{p -2}/2! and p(p-1)h²b ^{p -2}/2! terms on the LHS effectively  shell the      p(p-1)h²c ^{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)a²h ^{p -2}/2! and the p(p-1)b ²h ^{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 pch ^p-1 term on the RHS.
One h ^p remains on the LHS.

So the final result of these operations is:

h^p =pab(a+b)T_ab+ph{a ^{p -2}b + …+ ab ^{p -2}}+p(p-1)h²{a ^{p -3}b …+ab ^{p -3}}/2! + … + p(p-1)abh ^p-2………………………(5)

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

So h^p  = pabY, for some Y, a polynomial in p, a, b, h………………………(6)

EXAMPLES When p=3  h³  = 3ab(a+b+2h)………………………(7)
When p=5, we have  h⁵ = 5ab{a³ + 2a²b +2ab² + b³ +h(4a² +6ab + 4b²)+6h²(a+b)+4h³}.(8)

The { } bracket on the RHS of (8) factorises to (a+b+2h){a²+ab+b²+2(a+b)h + 2h²}….(9)  . [NR6]       

 

So h^p = 5ab(a+b+2h){a²+ab+b²+2(a+b)h + 2h²}………………………(10)

Let z =(a+b+2h)………………………(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 p.

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 {a+h+b+h}, as a sum of two odd (pth) powers. But {a+h+b+h} is {a+b+2h} or z. So the LHS of (12) has the factor z and the RHS, on being expanded by the prime binomial lemma, has two terms which have the factor z plus h^p. It follows that h^p has the factor z.  [NR7]

 

This establishes that z is a factor of h^p for all values of p > 2…………………………….(13)

Now because h ^p has the factor z, the following obtains, provided that z has no common factors with a or b:

         h^p = pabzv where v is a polynomial in p, a, b, h………………………………………(14)

 

Here v is effectively “what is left” of the expression for h^p derived from (1) after taking out the factors a, b, z, p. From this point onwards v becomes the fourth unknown alongside a, b, z.

 

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 +2h.

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 h would have it too: so both a+h and b+h would have it as a common factor, contrary to the stated initial condition.

FURTHER:  If a had a common prime factor x with (a+b+2h), it would have the same prime factor x in common with b+2h. Now h^p contains all the prime factors of a, so 2h would have it too.  So b would also have x as a factor —a common factor with a. This would contradict Lemma II. The same reasoning applies to b. So neither a nor b has a common factor with a+b+2h.

This leaves the question whether v has a common factor with a, b or (a+b+2h).  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 + (b+h)^p =  (z – h)^p

or as             (-z+h)^p  + (b+h)^p    =  (-z + b + h)^p ………….……………………………(15)

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 = pab(a+b)T_ab +  pabhW, where W is some natural number………………………..(16)

Now let W = 2T_ab + W where W is a natural number. (Here W is whatever needs to be added to 2T_ab to turn it into W.)

The RHS of (16) becomes  pab(a+b)T_ab + pab2hT_ab + pabhW.

Hence      h ^p =  pab(a+b+2h)T_ab + pabhW……………………………………………..(17)

But h ^p has the factor z, so hW has the factor z. Let hW =W’z where W’ is a natural number.

      So         h^p = pab(a+b +2h){T_ab + W’}…………………………………………….(18)

However we know from (5) that h ^p is equal to pab(a+b)T_ab plus a polynomial in h starting with a pabh term.

So W’ has the general form N₁h + N₂h²+ N₃h³ +…N_p-3h^p-3 , where the Ns are unknown natural numbers —except that N₁ might be zero if the coefficient of the h term in (5) happened to be 2pabT_ab. However the h term in (5) is the shelled result of c ^p-1. This is clearly not the same polynomial as 2pabT_ab.

 

So (18) becomes:

h^p = pab(a+b+2h){T_ab + N₁h + N₂h ² + N₃h ³ + …  …+N_p-3h ^p-3}………………………………………………………………….….(19)

for some natural number coefficients N₁, N₂…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 N₁, N₂, N₃,… 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 a^p-3 and ends with b^p-3 : in-between these end terms all the middle terms have the factor ab. All the powers of h in the { } bracket clearly contain the prime factors of a and b.  So v, as shown in the { } bracket in (19), has no factor in common with a or b. It has been prevented from having any of the prime factors of a or b by the presence of a^p-3 + b^p-3 in T_ab.

This leaves the question whether v (or in effect T_ab) is co-prime to z. The approach here brings in Lemma III, from which it may be deduced that

h^p = pa’b(a’+b+2h){T_a’b + N’₁h + N’₂h ² + N’₃h ³ + …   N’_p-3h ^p-3}………………….(19’), since adding a dash to all the a_s in the reasoning above changes nothing.  [The Ns have been dashed too, because they are now polynomials based on a’, b and p. ]

When a’ = –z,  there is clearly no change in h, because z becomes –a and a becomes –z.  (a’+b+2h) becomes –z + b +2h 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-_zb which ends with b^p-3, while all the remaining terms have the factor z. Also all the h terms in the { } bracket have the prime factors of z.    Thus everything in the {  } bracket has the prime factors of z except b^p-3. So v has no factor in common with z  [NR9]                                                                                      

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

It now follows from (14) h^p = pabzv  that each of a, b, z, v is a pth power except that one and only one of a, b, z, v also has the extra factor p^p-1.  …………………………….………………………………………………………….(20) [NR10]

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 pth power, each of a, b, z, v will be a pth power too with the proviso that one and only one of them has the extra factor p^p-1 needed to make the p factor of the RHS a pth power of p.

(It is convenient to call the special unknown among the set a, b, z, v  which has the extra factor p^p-1, the exception.)

 

So, taking pth roots:

h=pABZV………………………………………………….(21) where A, B, Z, V denote the pth 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 pth root. (Incidentally the exception, being the only unknown with the factor p, might have additional hidden factors p^p, p^2p, p^3p,… etc.)

It can be proved that the exception with the extra factor p ^p-1 necessarily has the total factor p ^2p-1. [See the exercises at the end.]

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 pth 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 pth power, again possibly multiplied by p^p-1 (but only if neither a nor b has the factor p^p-1.)    (p ^p-1 is the supernumeraryfactor.)                                                    .

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 aⁿ + bⁿ is always smaller than (a + b)ⁿ, when a and b are > 0, and much smaller when a and/or b are both large.

If we add 1 to both a and b and also add 1 to a+b, the same result applies. For example 5⁷+ 9⁷ < 12⁷ and 6⁷ + 10⁷ < 15⁷. 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 2h^h approximately: whereas the single power is only approximately h^h.  So the inequality has switched, and the LHS has become the larger item. This is true whatever positive natural numbers a and b are involved at the beginning. It seems to highlight the essence of the problem summed up in the question: can the LHS (a+h)ⁿ + (b+h)ⁿ ever exactly equal the RHS (a+b+h)ⁿ ? An agenda suggests itself from this:  we should work out what equation (1) tells us about h.

[NR2] EXAMPLE a¹⁵ + b¹⁵ = c¹⁵  is also an example of cubic powers, namely a⁵ , b⁵ and c⁵  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 pa ^p-1h or pha ^p-1.

Similarly the 2^nd term of (b+h) ^p is phb ^p-1.  So these are the two outer terms of phc ^p-1 and we can subtract them from the RHS of (3) to shell phc ^p-1.

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 b ² in each case multiplied by coefficients.  The result is therefore 2ab. The ‘2’ here cancels the 2! in the denominators of the coefficients.

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 h^p term as the subject of the equation on the LHS after all the shelling has been completed.

(5) demonstrates that h ^p has the factors a, b and p and that what is left is a polynomial (Y) consisting of terms in a, b, h, and p.

[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 c³ -a²b -b²a or c³ -abc. This is c(a²+ab+b²) or cT_ab.    The large h( ) bracket is 2hT_ab +2hc². The last two terms are 2ch +2h².  So now we have (c+2h)T_ab+2h{c²+3ch+2h²}. Here the { } bracket factorises into (c+2h)(c+h). So (c+2h) is a common factor, and the expression for h⁵ becomes 5ab times (c+2h){T_ab+2ch+2h²}. This is the RHS of (10).

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

z^p -pzh(z-h)T – h^p, where T is an inner tray.

[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 p, which implies that it has the factor p^p. But one and only one of a, b, z, v can have this factor. So one and only one of them must have the factor p^p-1. Because they have no common factors each of a, b, z, v must be a pth power.

 

SOME EXERCISES

• Find the coefficients of T_ab for (a+b)¹³.
• Show that if a, b or z has the supernumerary factor p^p-1, that the corresponding unknown A, B, Z has the factor p as well.
• Find an explicit formal for N₁ in terms of a, b, and p.
• Find an explicit formula for N₂ in terms of a, b, p and N₁.
• Prove that, when a and b have no common factor, T_ab is always odd.
• Prove that T_ab = T_AB (mod p).
• Prove that pab(a+b)T_ab has the factor h.
• Prove that the T_ab of (a+b)⁵ is not divisible by 5 for all natural numbers a, b.

 

Solutions will be given in the Blog next time.