*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 x^{128} + y^{128} = z^{128} can be reduced to X^{4} + Y^{4} = Z^{4}, where X=x^{32}, Y = y^{32} and Z = z^{32}.

So the challenge is to show that X^{4} + Y^{4} = Z^{4} isn’t possible. This, once proved, takes care of itself and all higher powers of 2.

Well if X^{4} + Y^{4} = Z^{4 }is possible we have (X^{2})^{ 2} + (Y^{2})^{ 2} = (Z^{2})^{ 2}, so {X^{2}, Y^{2}, and Z^{2}} 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 X^{4} + Y^{4} =Z^{4}.

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) m^{2} – n^{2} = a = X^{2} (a square) (ii) 2mn = b = Y^{2} (a square) and (iii) m^{2} + n^{2} = c = Z^{2} (a square).

Now we have m^{2} = X^{2} + n^{2} and m^{2} = Z^{2} – n^{2}……………………………(1)

Or *both* m^{2} + n^{2} and m^{2} – n^{2} are squares.

[A SIDE-SHOW If we subtract the two equations in (1) we get 0 = X^{2} – Z ^{2 }+2n^{2} So 2n^{2} = Z^{2} – X^{2}…..(2)

Lets call this a “dualled Pythagorean triple”.

To find them let Z= 2p^{2} + q^{2} X = 2p^{2} – q^{2} n = 2pq.

This generates these variant dualled triples if 2p^{2} > q^{2}.]

The main question, though, is really whether m^{2} + n^{2} and m^{2} – n^{2} can both be squares.

We have m^{2} – n^{2} = X^{2} and m^{2} + n^{2} = Z^{2}, where X and Z have no common factor.

So, adding 2m^{2} = X^{2} + Z^{2}. ………………………………………………………………….(2)

Is this possible?

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

If so (1) becomes 2m^{2} = 2m^{2} + 2m(x-y) + x^{2} + y^{2}…………..(3)

So 2m = (x^{2} + y^{2})/(y-x)………………………………………(4)

Now if x and y did have a common factor t, m would have it too, because x^{2} + y^{2}has the factor t^{2}. So X and Z would have a common factor ….which is impossible. So x and y don’t have a common factor.

Also 2m = (y^{2} – x^{2})/(y-x) + 2x^{2}/(y-x)

or 2m = y+x + 2x^{2}/(y-x)…………………………………………………….(5)

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

So 2m isn’t a natural number.

So 2m^{2} = X^{2} + Z^{2} is impossible.

It follows that X^{4} = Y^{4} + Z^{4} 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 ^{th}Dec 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**