*NOTE 1 The author’s bare hands (elementary) proof of Fermat has now been on-line for four years —and during these 48 months no hint of a counter-argument has emerged. No challenge has been posted. It is a scandal that no-one in the mathematic establishment has taken the trouble to even *try* to understand the new proof. They have been lazy to the *n*th degree, luxuriating in their (now no longer widely recognised) self-proclaimed superiority.*

*If they had looked at the new bare hands proof, they would have seen almost immediately that it is securely based on simple reasoning (binomial and prime factoring), and is most probably roughly equivalent to what Fermat claimed he had found more than 350 years ago. *

*Making this known would be a plus. It would raise the status of elementary reasoning and bare hands (first principles) maths —something which evidently does not appeal to the chiefs, priests or gurus of the subject. There is evidently little stomach among these gurus for revealing Fermat’s posthumous triumph, even after four-centuries of studied neglect. There is equally little welcome, among the subject’s elite, for the palpable confidence boost such a disclosure could make for the over-worked, under-appreciated, teachers of advanced maths (years 11-12) in schools… not to mention their bright students. The gurus of the subject seem to be currently lying low (a kind of hiding), and many of them may be quite unaware of the dire crisis in the subject, which is growing by the month in school classrooms. *

* *

**NOTE 2** The argument given in the previous blog (49) overlooked the possible case, namely when y-x = 1.

*The section involved is:*

* *

<< 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 + 2×2/(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. >>

**CORRECTION** Yes, this is OK but there is an overlooked case , namely in (5) when y=x + 1, because in this case 2x^{2 }*can* be divided by 1.

**COMMENT** However this cannot happen, because, if so, the LHS of (5) becomes even and the RHS of (5) is odd.

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*MAIN TOPIC The way to colour networks*

*We now return to look at the steps involved in four colouring a typical fully triangulated, fully weeded network.*

*Diagram I shows a fully triangulated, fully weeded network of 27 nodes. An initial node O is surrounded by a circuit of five nodes, followed by two circuits of ten nodes and finally an end node W. *

*We could choose any order-5 node for removal. In this exposition the Node marked X has been chosen to be removed. This is illustrated in Diagram II. We know from the new proof that that any 3-cluster which refills the space opened-up after removing X will be colourable with three colours. *

*(There are actually five different ways in which the circuit of X can be re-filled with diagonal pairs, thus forming different 3-clusters. One such 3-cluster will have A and B having the same colour.)*

*The two nodes A and B are now amalgamated… shown as AB in Diagram III. The node D is now of order 4 and can be weeded…*

*Proceeding in this fashion this will eventually lead to a trivial network with four nodes and an obvious 4-colouring.** *

*Working backwards, we now systematically identify the colours needed at each stage of the deconstruction process. This will be a 4-colouring needed for the given 27-node network.*

**CHRISTOPHER ORMELL 1 ^{st} July 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 Also comments on the reasoning in this or earlier blogs in this series can be submitted by email to the same address. This includes any counter-argument submitted as a bid for the prize offered in blog 49.**