## A *Non*–*Abstract Algebraic* Proof of Fermat’s Last Theorem

*Non*

By DM Delinferni, Jr. Chief, Pure and Applied Mathematics Branch, National Diabetes Center (*dmd@delinferni.diabetes-mellitus.org*)

See also **https://diabetes-mellitus.org/nonabstractalgflt.pdf **

The *only universal* restriction to the set forming

*Pythagorean Triples*[of the (n=2) second order] can be represented by the following equivalent identities labeled as (#

**1)**and (#

**2)**respectively:-

For any base **x **where **X _{0}^{2}+Y_{0}^{2}=Z_{0} ^{2}**and

**a=Z**

_{0}-Y_{0}**#1) ****(2ax _{0}**

**+**

**a**

^{2})^{2}+ (2ax_{0}^{2}**+**

**2a**

^{2}x_{0}**+**

**(a**

^{3 }– a)/2)^{2}= (2ax_{0}^{2}**+**

**2a**

^{2}x_{0}**+**

**(a**

^{3 }– a)/2)^{2}

^{}Consider two right-triangles, **A** and **B** which are related in that

a) the base of the larger triangle (**B**) is the square of the base of the smaller triangle (**A**)

b) the difference between the hypotenuse and the vertical arms of both triangles is the same and equal to some measure, “**a**”

c) we will be interested in framing the vertical arms, “** x**” and “

**”, of the both triangles, (**

*y***A**) and (

**B**), as functions of the vertical arm, “

**, of the smaller triangle, (**

*d”***A**), as well as of the difference, “

**a**”.

x = (=x^{1/2})^{2 }= (d + **a**)^{2} – d^{2 }= (2**a**d + **a**^{2}) = __2a ∙ a ^{2}_{d}__

x^{2} = (y + **a**)^{2} – y^{2} = (2**a**y + **a**^{2}) = (2**a**d + **a**^{2})^{2} = 4 **a**^{2}d^{2}+ 4**a**^{3}d + __a ^{4}__

2**a**y = 4 **a**^{2}d^{2}+ 4**a**^{3}d + **a**^{4 }– **a**^{2}

y = 2**a**d^{2} + 2**a**^{2}d + (**a**^{3 }– **a**)/2 = __2a ∙ 2a ^{2} ∙ (a^{3 }– a)/2_{d}__

z = y + **a **= __2____a____ ∙ 2____a__^{2}__ ∙ (____a__^{3 }__+ ____a____)/2___{d}

And if **Z _{1 }– Y_{1} =**

**1**, substituting

**k=a**and

^{-1}**x**, then :-

_{1}=x_{0}+(a-1)/2**#2)**** ****k[( 2x _{1}**

**+**

**1)**

^{2 }+ ( 2x_{1}^{2}**+ 2**

**x**

_{1})^{2 }=( 2x_{1}^{2}**+ 2**

**x**

_{1 }+1)^{2}]or where subscripts represent the base variable “**x**_{1}**“ **and “**∙**“ separates its powers,

#**2a)**** ****k[( 2****∙****1 _{x1})^{2 } + ( 2**

**∙**

**2**

**∙**

**0**

_{x1})^{2 }= ( 2**∙**

**2**

**∙**

**1**

_{x1})^{2}]Then, if some **X _{1}^{n }+ Y_{1}^{n }= Z_{1}^{n } **were to represent another true identity, it would of necessity map to any other true identity including, but not limited to,

**X _{0}^{2 }+ Y_{0}^{2 }= Z_{0}^{2}. **So, dividing both sides of

**X**by

_{1}^{n }+ Y_{1}^{n }= Z_{1}^{n }**k=(Z**,

_{1}-Y_{1})and letting **A**=** X _{1}/k**,

**B**=

**Y**, and

_{1}/k**C**=

**Z**, then

_{1}/k**(C-B) = 1,**

*and :-***k(A ^{n} + B^{n} =**

**C**

^{n})Therefore via identity #2a

**k[ ( ^{n}**

**√(**

**2**

**∙**

**1**

_{x1})^{2})^{n }+ (^{n}**√**

**(2**

**∙**

**2**

**∙**

**0**

_{x1})^{2 })^{n}= (^{n}**√**

**(2**

**∙**

**2**

**∙**

**1**

_{x1})^{2})^{n}]substituting **A**=^{n}**√(****2****∙****1 _{x1})^{2 } ** ,

**B**=

^{ n}**√**

**(2**

**∙**

**2**

**∙**

**0**and

_{x1})^{2 }**C**=

^{n}**√**

**(2**

**∙**

**2**

**∙**

**0**

_{x1})^{2},then in order for **A**, **B**, and **C** to all be ** non-zero** and rational with any odd n>2, all three (

**A**,

**B**, and

**C**) would of necessity have to have rational

**2n**roots – which seems on its face to be absurd as B and C only differ by 1. This is confirmed via a very nice proof offered by Professor A. Rosenbloom of the University of Toronto,

^{th}**since B and C differ by 1, and B=Y**

^{2n}, and C=B+1=Z^{2n}, then 1 = Z^{2n }– Y^{2n}and 1=(Z^{n}+Y^{n})( Z^{n}-Y^{n}). Therefore both (Z^{n}+Y^{n}) and ( Z^{n}-Y^{n}) must divide and equal 1 implying either Z or Y =0 and the other =1.**And, therefore, FLT must hold for non-zero rationals with any odd n>2.**

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