I said I might do this so this is part 1 of 2

I use some techniques that are different than I have yet seen, but I haven't read everything so I could be wrong. Casting the relationships of numbers as the progress to ∞ is pretty much big O. and so if I say that (x^2)/(x^2+2x+1) goes to 1 as an asymptote then it is similar to that notation or concept.

It is possible to assign properties to number in conditions and thus properties of the results. This will be come clear in the steps.

x^2 + y^2=z^2 x^2 + (x+a)^2 = (x+a+b)^2 a<a+b obviously a and b are integers the result is : x-b=2ab x/2b -1/2 = a x must be odd, if b is odd 2(x/2b-1/2)=2a x/b -1=2a=even integers can be assigned a 'character' of any modulo beside 2 the possible combinations of x and b are limited by big O and the need to be an odd product x/b is an odd integer x and b have a integer number set relationship (a=b=n) where n is an integer x,y,z (3,4,5)*n

There are more steps to the squared function, but these are the types of sets that would be used in the x^n methods where n>2 The answer hinges on the fact that certain sets of integer numbers cannot be multiplied or divided to produce products in some sets. 2*x is never =1 . a^2 is > a unless a=1. (x/n -1/3 =integer) requires that n have a factor of 3 and x has a set of possible values like 1 4 7 ...

It is my opinion that Fermat assumed the solution to be trivial, otherwise it would have been a big deal and he would have spouted about it. If he is to be believed then it was my opinion that the solution must be trivial and what might have seemed very obvious to him, so much so, that it was not worth wasting a new sheet of paper.

A non trivial solution would seem to make Fermat mistaken or a liar and so the enigma would have no meaning and even if solved would just be some random numerical relationship. That is my opinion anyway.

The many different number sets would have been second nature to them as it is me as I play with number table sets all the time and it would be no more difficult than me deMorganing a page of gates. You could hardly get a gold plaque at Intel for deMorganing a 100 gates. A prime has no divisors that are not 1 which yield an integer by definition. A proof that depends on that fact would have been considered trivial. IMHO Now I can sleep as these equations have been rolling through my head for an hour while I was trying to sleep and now they are out and gone I hope.

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