Path To Fermat

       

Path to Fermat        Manuel Oliveros Martínez  2009                            © Certified to the Public Domain 

        Dedicated to   

Dorum    whomever he or she is    that at Scribd helped me indicating my use of monic functions  to prove the impossibility of natural roots was flawed    helping one in the error is far more important than showing one is right    The Fermat Theorem proof as I arrived at it   

 

This  work  has  been  done  along  some  other  things  in  the  second  half  of  August  till  today  September 3.   I  have  tried  over  20  approaches  that  allowed  me  familiarity  with  the  problem,  arriving  to  a  successful  approach  at  attempt  21,  then  making  the  single  page  version  as  23.  Dorum's  observation allowed me to get to some final format as version 25.  My  guide  in  the  attempts  has  been  that  the  solution  should  be  simple  and  affordable  to  someone  of  Fermat's  era.  Half  of  the  first  attempts  were  as  represented  here  more  than  anything  numerical  evaluations  and  graphical  representations  of  what  happening.  Then  Newton's (Al‐Karaji's) binomy appeared in my mind as the tool to be used and I made almost  as much attempts refining the idea as before getting acquainted with the problem.  We  can't  know  what  Fermat's  proof  was.  He  had  it,  he  postulated  the  theorem;  and  if  not  Newton's  binomy  he  surely  had  mathemathics  enough  to  appreciate  this  TRICK,  or  other  tantamount to it.   I am yet convinced there must be even more concise proofs maybe even from just some greek  math approach (mainly proportion) and from n‐dimensional math when properly formulated  looking  at  the  functions  of  x^n+y^n  (x+y)^n  seen  as  some  intersection  of  something  with  something (or a subset) at the respective dimensions, what can be some interesting problem  to approach.  I think as interesting as the problem itself is why it doesn't apply for exponents 1 and 2, and  even that some more general rule should be developed to encompass the general behaviour.  The  why  is  surely  proven  in  practical  terms,  maybe  or  not  as  well  in  the  general  theoretical  grounds,  and  neither  the  general  encompassing  behaviour  properly  described;  not  being  a  conversant  mathematician  I  am  ill  prepared  to  know  about  or  proceed  ahead,  but  may  try  someday.  More. This is not as much a mathematical proof as a case of the art of problem solving. I am far  more  good  at  that  than  at  mathematics,  and  of  course  I  am  not  even  the  shadow  of  any  licensed mathematician, how less their most accomplished representatives like Andrew Wiles,  that first proved it in some way I am surely even unable to read, how less understand.   I about 24 years ago had the thought to be  en cierto sentido simétrico de Christopher Wren  in some way symmetrical to Christopher Wren  for  he  was  going  from  mathematics  to  architecture;  I  have  always  been  going  a  bit  the  opposite way.  Furthermore, modest that their abilities in helping me to get math in my head, I acknowledge  this present is a gift from the past masters dead and alive that I had. In fact I was doing graphic  statics with the computer and downloading related old texts and alike just prior to tackle this  thing. My love for their work has been rewarded just simply too much. 

a  1

b  2

b mayor que a

1

c( n b ) 

 a  n   b  n

n

10

c( n b)



b

n

n

1

0.1

0

1

2

3

4

5 n

6

7

8

9

10

100

10 c( n b)



b

n

n 1

0.1

0

10

20 n

b

n

a

lim n∞

an  bn b

n

1

relación cierta para cualquier a y b

1

n  4 a  1  25

C

n

c1 ( a b )  a  b b  1  25



n

n

C

a b

 c1 ( a b )

C

1

n  4 x  1  26

Z

z( x y ) 

 x  n   y  n

y  1  26

Z

x y

n

 z( x  13 y  13)

Z

1

n  200 a  1  25

C

n

c1( a b )  a  b b  1  25



n

n

C

a b

 c1( a b )

C

1

n  27 x  1  26

Z

z( x y ) 

 x  n   y  n

y  1  26

Z

x y

n

 z( x  13 y  13)

Z

1

n  4

r( α) 

c  3

n   c  r1 ( α)    cos( α) n  sin( α) n   

n

r1 ( α) if 0  deg  α  90 deg  180  deg  α  270  deg r1 ( α  90 deg) otherwise

a( α)  r( α)  cos( α)

b ( α)  r( α)  sin( α)

c3 n4

2

b( α) 0

2

2

0 a( α)

2

1

n  9

r( α) 

c  3

n   c  r1 ( α)    cos( α) n  sin( α) n   

n

r1 ( α) if 0  deg  α  90 deg  180  deg  α  270  deg r1 ( α  90 deg) otherwise

a( α)  r( α)  cos( α)

b ( α)  r( α)  sin( α)

c3 n9

2

b( α) 0

2

2

0 a( α)

2

1

n  4

r( α) 

n   c  r1 ( α)    cos( α) n  sin( α) n   

c  5

n

r1 ( α) if 0  deg  α  90 deg  180  deg  α  270  deg r1 ( α  90 deg) otherwise

a( α)  r( α)  cos( α)

c5

b ( α)  r( α)  sin( α)

n4

5

4

b( α) 3

2

1

1

2

3 a( α)

4

5

1

n  12

r( α) 

n   c  r1 ( α)    cos( α) n  sin( α) n   

c  5

n

r1 ( α) if 0  deg  α  90 deg  180  deg  α  270  deg r1 ( α  90 deg) otherwise

a( α)  r( α)  cos( α)

c5

b ( α)  r( α)  sin( α)

n  12

5

4

b( α) 3

2

1

1

2

3 a( α)

4

5

1

n  3

r( α) 

n   c  r1 ( α)    cos( α) n  sin( α) n   

c  10

n

r1 ( α) if 0  deg  α  90 deg  180  deg  α  270  deg r1 ( α  90 deg) otherwise

a( α)  r( α)  cos( α)

b ( α)  r( α)  sin( α)

c  10 n3 90

120

10

60

150

30 5

r( α) 180

0

0

210

330

240

300 270 α

1

n  7

r( α) 

n   c  r1 ( α)    cos( α) n  sin( α) n   

c  10

n

r1 ( α) if 0  deg  α  90 deg  180  deg  α  270  deg r1 ( α  90 deg) otherwise

a( α)  r( α)  cos( α)

b ( α)  r( α)  sin( α)

c  10 n7 90

120

60 10

150

30 5

r( α) 180

0

0

210

330

240

300 270 α

1

n  100

r( α) 

n   c  r1 ( α)    cos( α) n  sin( α) n   

c  10

n

r1 ( α) if 0  deg  α  90 deg  180  deg  α  270  deg r1 ( α  90 deg) otherwise

a( α)  r( α)  cos( α)

b ( α)  r( α)  sin( α)

c  10 n  100 90

120

60

10 150

30 5

r( α) 180

0

0

210

330

240

300 270 α

1

n  3

r( α) 

c  10

n   c  r1 ( α)    cos( α) n  sin( α) n   

n

r1 ( α) if 0  deg  α  90 deg  180  deg  α  270  deg r1 ( α  90 deg) otherwise c  10

a( α)  r( α)  cos( α)

b ( α)  r( α)  sin( α) n3

0.8

0.6 mod( b( α) 1)

0.4

0.2

2

4

6 a( α)

8

10

n  4

c  6 1

n

b ( a)  c  a



n

n

6

5

4 b( a) 3

2

1

1

2

3

4 a

5

6

ORIGIN  1 1

n  5

c  7

A  i

n

b ( a)  c  a

  i 

B  mod b A 1

i

i



n

n

i  1  floor( c)  1 Rmín  min( B)

6

b( a) 4

2

2

4

6

a

B

0.8 0.6 0.4 0.2 0

2

4 A

 1   0.997    0.98   B  0.913   0.718     0.182 

Rmín  0.1822586154

6

SINGLE PAGE PROOF OF THE THEOREM OF FERMAT Manuel Oliveros Martínez, Arquitecto 2009-09-02 (REVISION 4) (C) CERTIFIED TO THE PUBLIC DOMAIN a b c natural integers positive bigger than or equal to 1, n exponent equal or bigger than 3, proof it is not possible n

n

n

say c  a  b ab and so there's not a natural c that satisfies the preceding equations in such conditions. To say that there's no natural c such that c^n=a^n+b^n when a and b naturals and n  3 is to say that the set of the nth power of naturals and the sum of of 2 of such powers is disjoint, so no c^n can become a^n+b^n whatever the a and b. For the sake of demonstration, assume this assertion is false and that exists such c^n. c  x  b

Making

n

n

( x  b )  a  b

n

n k k n n  combin(n k)x b   a  b n

k0

eliminating the common b^n term at left and right n 1

n k k n  combin(n k)x b   a

k0 or establishing a polynomial on x of the nth power at the left n 1

n k k n  combin(n k)x b   a  0

k0

This is a polynomial with positive coefficients on all x^n, x^(n-1) ... x and then -a^n. To negate Fermat's theorem we need to find some integer solution to x, for it might produce a natural c satisfying the equality.  

If any solution to it can´t be integer, Fermat theorem is proven. If contrarily, we meet an integer solution to it (x natural and lesser than a, we have proven in the shorthand article), Fermat's theorem, is false.

Now I will proceed to explain what makes that the above equation impossible to meet for an integer, waiting for better symbolical or whatever more formal proof in what I am also working at. a

n

n 1

entirely pinpoints (as a itself) at the integer field

 combin(n k)x

k0

n k

b

k



this term, even if integer, lacking the b^n term that would reconciliate it with some sum x+b of integers to be elevated to the nth potence, causes the pair x+b originating the summation be ejected from the integer field (not pinpoint on it), and since b integer, x can never be

As a corollary, the single real number x+b in the positive field that meets the equation, is irrational, for it belongs to the class of the root of a natural number rhe root of which is not an integer.