Path to Fermat

8/14/2019 Path to Fermat

1/24


2/24

Dedicatedto

Dorum

whomeverheorsheis

thatatScribdhelpedmeindicatingmyuseofmonicfunctions

toprovetheimpossibilityofnaturalrootswasflawed

helpingoneintheerrorisfarmoreimportantthanshowingoneisright

TheFermatTheoremproofasIarrivedatit


3/24

Thiswork has been done along some other things in the second half of August till today

September3.

Ihave triedover20approaches thatallowedme familiaritywith theproblem,arriving toa

successful approach at attempt 21, then making the single page version as 23. Dorum's

observationallowed

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 attemptswere as represented heremore than

anything numerical evaluations and graphical representations of what happening. Then

Newton's(AlKaraji's)binomyappearedinmymindasthetooltobeusedandImadealmost

asmuchattemptsrefiningtheideaasbeforegettingacquaintedwiththeproblem.

We can'tknowwhat Fermat'sproofwas.Hehad it,hepostulated the theorem;and ifnot

Newton's binomy he surely had mathemathics enough to appreciate this TRICK, or other

tantamount

to

it.

Iamyetconvincedtheremustbeevenmoreconciseproofsmaybeevenfromjustsomegreek

mathapproach(mainlyproportion)andfromndimensionalmathwhenproperlyformulated

looking at the functions of x^n+y^n (x+y)^n seen as some intersection of something with

something(orasubset)attherespectivedimensions,whatcanbesome interestingproblem

toapproach.

Ithinkas interestingastheproblem itself iswhy itdoesn'tapplyforexponents1and2,and

eventhatsomemoregeneralruleshouldbedevelopedtoencompassthegeneralbehaviour.

Thewhy is surelyproven inpractical terms,maybeornotaswell in thegeneral theoretical

grounds, and neither the general encompassing behaviour properly described; not being a

conversantmathematician Iam illprepared to know aboutorproceed ahead,butmay try

someday.

More.Thisisnotasmuchamathematicalproofasacaseoftheartofproblemsolving.Iamfar

more good at that than atmathematics, and of course I am not even the shadow of any

licensedmathematician,howlesstheirmostaccomplishedrepresentativeslikeAndrewWiles,

thatfirstproveditinsomewayIamsurelyevenunabletoread,howlessunderstand.

Iabout24yearsagohadthethoughttobe

enciertosentidosimtricodeChristopherWreninsomewaysymmetricaltoChristopherWren

for he was going from mathematics to architecture; I have always been going a bit the

oppositeway.

Furthermore,modestthattheirabilitiesinhelpingmetogetmathinmyhead,Iacknowledge

thispresentisagiftfromthepastmastersdeadandalivethatIhad.InfactIwasdoinggraphic

staticswiththecomputeranddownloadingrelatedoldtextsandalikejustpriortotacklethis

thing.My

love

for

their

work

has

been

rewarded

just

simply

too

much.


4/24

a 1 b 2 b mayor que a

c n b( ) a n

b n

1

n

0 1 2 3 4 5 6 7 8 9 100.1

1

10

c n b( )n

b n

n


5/24

0 10 200.1

1

10

100

c n b( )n

b n

bn

an

n

an

bn

bn

lim

1 relacin cierta para cualquier a y b


6/24

n 4 c1

a b( ) an

bn

1

n

a 1 25 b 1 25 Ca b

c1 a b( )

C


7/24

C


8/24

n 4 z x y( ) x n

y n

1

n

x 1 26 y 1 26 Zx y

z x 13 y 13( )

Z


9/24

Z


10/24

n 200 c

1

a b( ) an

bn

1

n

a 1 25 b 1 25 Ca b

c1 a b( )

C


11/24

C


12/24

n 27 z x y( ) x n

y n

1

n

x 1 26 y 1 26 Zx y

z x 13 y 13( )

Z


13/24

Z


14/24

n 4 c 3 r1 ( )c

n

cos ( )n sin ( )n

1

n

r ( ) r1 ( ) 0 deg 90 deg 180 deg 270 degif

r1 90 deg( ) otherwise

a ( ) r ( ) cos ( ) b ( ) r ( ) sin ( )

c 3

n 4

2 0 2

2

0

2

b ( )

a ( )


15/24

n 9 c 3 r1 ( )c

n

cos ( )n sin ( )n

1

n



a ( ) r ( ) cos ( ) b ( ) r ( ) sin ( )

c 3

n 9

2 0 2

2

0

2

b ( )

a ( )


16/24

n 4 c 5 r1 ( )c

n

cos ( )n sin ( )n

1

n



c 5

a ( ) r ( ) cos ( ) b ( ) r ( ) sin ( )n 4

1 2 3 4 51

2

3

4

5

b ( )

a ( )


17/24

n 12 c 5 r1 ( )c

n

cos ( )n sin ( )n

1

n



c 5

a ( ) r ( ) cos ( ) b ( ) r ( ) sin ( )n 12

1 2 3 4 51

2

3

4

5

b ( )

a ( )


18/24

n 3 c 10 r1 ( )c

n

cos ( )n sin ( )n

1

n



a ( ) r ( ) cos ( ) b ( ) r ( ) sin ( ) c 10

0

30

60

90

120

150

180

210

240

270

300

330

0

5

10

r ( )

n 3


19/24

n 7 c 10 r1 ( )c

n

cos ( )n sin ( )n

1

n



a ( ) r ( ) cos ( ) b ( ) r ( ) sin ( ) c 10

0

30

60

90

120

150

180

210

240

270

300

330

0

5

10

r ( )

n 7


20/24

n 100 c 10 r1 ( )c

n

cos ( )n sin ( )n

1

n



a ( ) r ( ) cos ( ) b ( ) r ( ) sin ( ) c 10

0

30

60

90

120

150

180

210

240

270

300

330

0

5

10

r ( )

n 100


21/24

n 3 c 10 r1 ( )c

n

cos ( )n sin ( )n

1

n



c 10

a ( ) r ( ) cos ( ) b ( ) r ( ) sin ( )n 3

2 4 6 8 10

0.2

0.4

0.6

0.8

mod b ( ) 1( )

a ( )


22/24

n 4 c 6

b a( ) cn

an

1

n

1 2 3 4 5 61

2

3

4

5

6

b a( )

a


23/24

ORIGIN 1

n 5 c 7 b a( ) cn an

1

n

i 1 floor c( ) 1

Ai

i Bi

mod b Ai 1 Rmn min B( )

2 4 6

2

4

6

b a( )

a

2 4 60

0.2

0.4

0.6

0.8

B

A

B

1

0.997

0.98

0.913

0.718

0.182

Rmn 0.1822586154


24/24

SINGLE PAGE PROOF OF THE THEOREM OF FERMAT

Manuel Oliveros Martnez, 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

cn

an

bn

cn

say a band 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.

Making c x b x x b( )n

an

bn

x b( )n

0

n

k

combin n k( ) xn k

bk

an

bn

k

eliminating the common b^n term at left and right

0

n 1

k

combin n k( ) xn k

bk

an

k

or establishing a polynomial on x of the nth power at the left

0

n 1

k

combin n k( ) xn k

bk

an

0k

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 cant 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.

entirely pinpoints (as a itself) at the integer fielda

n

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

0

n 1

k

combin n k( ) xn k

bk

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.

Path to Fermat

Documents

Transcript of Path to Fermat