Fermat, the last frontier

The 20-page bare hands proof of Fermat’s conjecture which I posted on-line in 2020 in the first two of these blogs has now survived intact for almost four years —with no sign whatever of a demolishing counter-argument.

I have since that time posted essays in the New English Review which are likely to infuriate deniers that Fermat’s line of reasoning will ever be found.

In my opinion it is these diehards who should be ashamed of their acceptance of a situation which tolerates a complete disregard for the theoretic power of simple reasoning in maths.  These 359 years during which the best brains of the subject have failed to re-discover Fermat’s reasoning are surely deeply embarrassing, because they show a “couldn’t care less” attitude to the situation. A proud subject would have made sure it did not put-up-with such a classic loss of lucidity.  How can anyone be proud of a subject which tolerates such acceptance of its own toothlessness for so long?

It is only in the last two months, though, that I have realised that my bare hands posts of 2020 were incomplete. They did not cover the case when n= 4 or any higher power of 2.

Fermat himself used a meta-argument (‘method of descent’) to cover this case: this is the only bit of Fermat’s work on the solution which we know.  It is a valid argument, but not as satisfying as a direct reductio ad absurdum would be.  So can we find a direct line of reasoning which sews-up this final case?

Yes!  

Anything like x¹²⁸ + y¹²⁸  = z¹²⁸  can be reduced to X⁴ + Y⁴ = Z⁴, where X=x³², Y = y³²  and Z = z³².

So the challenge is to show that X⁴ + Y⁴ = Z⁴ isn’t possible.  This, once proved, takes care of itself and all higher powers of 2.

Well  if X⁴ + Y⁴ = Z⁴  is possible we have (X²) ² + (Y²) ² = (Z²) ², so {X², Y², and Z²} has to be a Pythagorean triple {a, b, c}.  It will have its unique m, n.  a and b are also coprime, because we can remove any common factor from X⁴ + Y⁴ =Z⁴.

For this to happen three conditions must be met.   There is an unique m and an unique n (with no common factor) such that                                                                                                                                (i) m² – n²  = a = X² (a square)   (ii) 2mn  = b = Y² (a square)   and    (iii) m² + n² = c = Z² (a square).

Now we have  m² = X² + n²          and           m²  = Z² – n²……………………………(1)

Or  both m² + n² and m² – n² are squares.

[A SIDE-SHOW If we subtract the two equations in (1) we get  0  = X² –  Z ² +2n²   So 2n² = Z² – X²…..(2)

 Lets call this a “dualled Pythagorean triple”.

To find them let      Z= 2p² + q²    X =  2p² – q²   n     =  2pq.

This generates these variant dualled triples if 2p² > q².]

The main question, though, is really whether m² + n² and m² – n² can both be squares.

We have m² – n² = X²      and   m² + n²  = Z², where X and Z have no common factor.

So, adding 2m² =  X² + Z². ………………………………………………………………….(2)

Is this possible?

Let         X = m + x    and    Z = m – y where x, y are natural numbers.

If so (1) becomes     2m² =  2m² + 2m(x-y) + x² + y²…………..(3)

So                      2m =  (x² + y²)/(y-x)………………………………………(4)

Now if x and y did have a common factor t, m would have it too, because     x² + y²has the factor t².  So X and Z would have a common factor ….which is impossible. So x and y don’t have a common factor.

Also              2m =  (y²  – x²)/(y-x)  +  2x²/(y-x)

or    2m = y+x   +  2x²/(y-x)…………………………………………………….(5)

There is no way in which the denominator  (y-x) can be cancelled by any factor of 2x², because y has none of the prime factors of x.

So 2m isn’t a natural number.

So   2m² = X² + Z² is impossible.

It follows that X⁴ = Y⁴ + Z⁴ is impossible and the same result follows for any higher power of 2.    QED      

A prize of £500 will be given to the first person to find a disabling glitch in this reasoning before Christmas 2024.  Send your proposed glitch to per4group@gmail.com on or before 24^thDec 3024.

NOTE: the next blog in this series (50) will show in detail how to four-colour a fully triangulated, fully weeded network of n=26 nodes.    

CHRISTOPHER ORMELL 3^rd June 2024 If you would like a free online copy of the P E R Narrative Maths Manifesto, send an email requesting this to per4group@gmail.com