Pier Francesco Roggero, Michele Nardelli, Francesco Di Noto "Proof of Fermat-catalan Conjecture...

7/30/2019 Pier Francesco Roggero, Michele Nardelli, Francesco Di Noto "Proof of Fermat-catalan Conjecture Through the ABC

1/49

PROOF OF FERMAT-CATALAN CONJECTURE THROUGH

THE NEW ABC CONJECTURE

Ing. Pier Francesco Roggero, Dott. Michele Nardelli, Francesco Di Noto

Abstract

In this paper we show a proof of Fermat-Catalan conjecture through the new abc

conjecture.


2/49

Versione 1.0

22/3/2013

Pagina 2 di 49

Index:

1. PROOF OF FERMAT-CATALAN CONJECTURE............................................................................. 31.1 RANGE OF VALIDITY OF THE EQUATIONS OF FERMAT-CATALAN ...................11

2. PROOF OF THE CATALAN'S CONJECTURE................................................................................173. SUPER-GENERAL TRINOMIAL EQUATION..................................................................................24

3.1 BEAL'S CONJECTURE......................................................................................................283.1.1 ANOTHER PROOF OF FERMATS LAST THEOREM THROUGH BEALSCONJECTURE..........................................................................................................................30

4. PROOF OF PILLAI'S CONJECTURE..............................................................................................315. PROOF OF TIJDEMAN'S CONJECTURE.......................................................................................356. EXAMPLES OF PILLAIS CONJECTURE.......................................................................................42

6.1 PERFECT POWERS LINKED TO PILLAIS CONJECTURE .........................................47


3/49

Versione 1.0

22/3/2013

Pagina 3 di 49

1. PROOF OF FERMAT-CATALAN CONJECTURE

The FermatCatalan conjecture combines ideas of Fermats Great Theorem and the Catalan

conjecture, hence the name. The conjecture states that the equation

am

+bn

=ck

has only finite solutions in N+

has only finitely many solutions (a,b,c,m,n,k); here a, b, c are positive coprime integers and m, n, kare

positive integers satisfying

knm

111++ < 1

As the new abc conjecture is given:

{rad (abc)} 62

> c

6

2= 1,644934... 1,6449 this is a limit.

We note that this value is very near to the following 1,64809064 1,6480 that is a value regardingsome frequencies connected to the musical system based on the aurea ratio (1,61803398)

This inequality holds for ALL triples (a,b,c) of coprime positive integers, witha +b =c

Lets try to apply the new abc conjecture with the following solutions that are known:


4/49

Versione 1.0

22/3/2013

Pagina 4 di 49

1m + 23 = 32

{rad (1*2*3)} 62

> 9

61,645

= 19,05 > 9

knm

111++ = 0,976.(m 7, so 1

7= 1 is OK)

25

+ 72

= 34

{rad (2*7*3)} 62

> 81

421,645

= 467,99> 81

knm

111++ =0,95

132

+ 73

= 29

{rad (13*7*2)} 62

> 512

1821,645

= 5221,76 > 512

knm

111++ = 0,9444.

27

+ 173

= 712

{rad (2*17*71)} 62

> 5041


5/49

Versione 1.0

22/3/2013

Pagina 5 di 49

24141,645 = 366.946,54 > 5041

knm

111++ = 0,976

35 + 114 = 1222

{rad (3*11*2*61)} 62

> 14884

40261,645

= 851.172,04 > 14884

knm

111++ = 0,95

338

+ 15490342

= 156133

{rad (3*11*2*61*12697*13*1201)} 62

> 3805914951397

798.107.238.7861,645

= 37921958271956418245,68 > 380.591.495.1397

knm

111++ =0,95833

14143

+22134592

= 657

{rad (2*7*101*479*4621*5*13)} 62

> 4902227890625

203.439.016.6901,645

= 4002896288091211907,85 > 4.902.227.890.625

knm

111++ = 0,976.


6/49

Versione 1.0

22/3/2013

Pagina 6 di 49

92623

+ 153122832

= 1137

{rad (2*11*421*7*53*149*277*113)} 62

> 235260548044817

16.025.927.261.4981,645

= 5271498777020594764161,13 > 235.260.548.044.817

knm

111++ =0,976.

177

+762713

= 210639282

{rad (17*13*5867*2*79*33329)} 62

> 443689062789184

(that is divisible for 16 and 64)

6.827.909.123.0741,645 = 1295398290051100057437,9 > 443.689.062.789.184

knm

111++ = 0,976.

We note that if 443689062789184 is divisible for 16 and 64, this value can be connected also with the

following Ramanujan function regarding the modes corresponding to the physical vibrations of the

superstrings:

( )

++

+

=

4

2710

4

21110log

'

142

'

cosh

'cos

log4

3

18

2

'

'4

0

'

2

2

wtitwe

dxex

txw

anti

w

wt

wx

.


7/49

Versione 1.0

22/3/2013

Pagina 7 di 49

438 + 962223 = 300429072

{rad (43*2*3*7*29*79*109*275623)} 62

> 902576261010649

124.303.909.686.2221,645

= 153259525699804607160993,02 > 902.576.261.010.649

knm

111++ = 0,95833.

The first of these (1m+2

3=3

2) is the only solution where one of a, b or c is 1; this is the Catalan

conjecture, proven in 2002 by Preda Mihailescu. Technically, this case leads infinitely many solutions

(since we can pick any m for m>6), but for the purposes of the statement of the Fermat-Catalan

conjecture we count all these solutions as one.

PROOF

am

+ bn

= ck

knm

111++ < 1

As the new abc conjecture is given:

{rad (amb

nc

k)} 6

2

> ck

Lets apply

rad (amb

nc

k) abc < c

m

k

cn

k

c


8/49

Versione 1.0

22/3/2013

Pagina 8 di 49

Please not that we cannot write, as is as in the case of Fermat's Great Theorem the inequality

rad (anb

nc

n) < c

3

because

it doesn't apply

a < c and b < c.

The values of a, b and c may be any as long as obviously > 1

ck

< c 62

++ 1

n

k

m

k

This implies that we have no solutions if

k 6

2

++ 1

n

k

m

k

and hence dividing by k

1 6

2

++

knm

111

It depends on values of m, n and k

++

knm

111 2

6

= 0,6079

and hence we have solutions if and only if

2

6

= 0,6079 512

1821,645

= 5221,76 > 512

knm

111++ = 0,9444.

Also here, we note that 512 / 8 = 64 and that 8 is a Fibonaccis number and the number of

the modes corresponding to the physical vibrations of the superstrings by the following

Ramanujan function:

( )

++

+

=

4

2710

4

21110log

'

142

'

cosh

'cos

log4

3

18

2

'

'4

0

'

2

2

wtitwe

dxex

txw

anti

w

wt

wx

.


11/49

Versione 1.0

22/3/2013

Pagina 11 di 49

1.1 RANGE OF VALIDITY OF THE EQUATIONS OF FERMAT-CATALAN

am

+ bn

= ck

we have solutions, according to the new abc conjecture, if and only if

26= 0,6079 2

6

\

x < 6,33

then the only possible values are

2

1,

3

1and

6

1

2)If we choose

5

1+

6

1+

x

1> 2

6

\

x < 4,14



12/49

Versione 1.0

22/3/2013

Pagina 12 di 49

21 ,

31 and

41

3)If we choose

6

1+

7

1+

x

1> 2

6

\

x < 3,35


2

1,

3

1

4)If we choose

7

1+

8

1+

x

1> 2

6

\

x < 2,94

then the only possible value is

2

1

and thus also for

7

1+

k

1+

x

1> 2

6

\

k = 8, 9, 10, 11, 12..

then the only possible value is


13/49

Versione 1.0

22/3/2013

Pagina 13 di 49

2

1

5)If we choose

3

1+

4

1+

x

1> 2

6

\

x < 40,7


5

1,

6

1.

40

1

Please note that the value of

2

1exceeds the upper limit

knm

111++ < 1 and therefore isnt good.

6)If we choose

2

1+

3

1+

x

1< 1

x > 6


7

1,

8

1.

N

1

N


14/49

Versione 1.0

22/3/2013

Pagina 14 di 49

7)If we choose

2

1+

4

1+

x

1< 1

x > 4


51 ,

61 .

N1

N

8)If we choose

2

1

+ 5

1

+ x

1

< 1

x > 3,33


4

1,

6

1.

N

1

N

From the set of solutions above we have only solutions:

in the conjecture of Fermat-Catalan when one of the exponents m or n or k is equal to 2.

In our examples this happens almost always, except in the cases 1, 2, 3 and 5.

In example 1)


15/49

Versione 1.0

22/3/2013

Pagina 15 di 49

4

1+

5

1+

x

1> 2

6

\

x < 6,33


2

1,

3

1and

6

1

So we can have

a4

+ b5

= c3

or any permutation of 3, 4 and 5, as

a3

+ b5

= c4

a3

+ b4

= c5

The last equation cannot give solutions.

In the same way we have

a5

+ b6

= c4

a4

+ b6

= c5

a4

+b

5

=c

6

The last equation cannot give solutions.

So we have the subset

a4

+ b5

= c3

a3

+ b5

= c4


16/49

Versione 1.0

22/3/2013

Pagina 16 di 49

a5 + b6= c4

a4

+ b6

= c5

The solutions for this quartet of equations is very complicated and can only be obtained with the aid of

computers.

Also remember that we can choose any values for a, b and c, not the usual values "small" that we are

accustomed.

So that we have solutions and, therefore, the difficulty lies in the exponential growth of these powers.


17/49

Versione 1.0

22/3/2013

Pagina 17 di 49

2. PROOF OF THE CATALAN'S CONJECTURE

Catalan's conjecture (or Mihilescu's theorem) is now a theorem that was conjectured by themathematician Eugene Charles Catalan in 1844 and proven in 2002 by Preda Mihilescu.

23

and 32

are two powers of natural numbers, whose values 8 and 9 respectively are consecutive. The

theorem states that this is the only case of two consecutive powers. That is to say, that the only

solution in the natural numbers of

ck bn = 1

for c, k, b, n > 1

1 + bn

= ck

is

1 + 23

= 32

so worth the inequality

knm

111++ < 1

But what value we have to choose for m?

We use the trick of taking m = so that we have he following inequality

1

+ 23

= 32

kn

11+ < 1

Lets apply the new abc conjecture and so let's show this conjecture:

rad (1*bnc

k) bc < c

n

k

c


18/49

Versione 1.0

22/3/2013

Pagina 18 di 49

ck

< c 62

+ 1

n

k


k 6

2

+ 1n

k


1 6

2

+

kn

11

It depends on values of n and k

+

kn11 26

= 0,6079


2

6

= 0,6079 <

+

kn

11< 1

This is a stronger inequalityof the initial statement

kn

11+ < 1

which then is wrong, because for


19/49

Versione 1.0

22/3/2013

Pagina 19 di 49

+

kn

11 2

6

= 0,6079

we cannot have integer solutions.

We can prove this way:

If n = 3 we have

1 + b3

= ck

hence we have solutions if and only if

2

6

= 0,6079 <

+

kn

11< 1

2

6

= 0,6079 <

+

k

1

3

1< 1

k < 2

2

18

3

= 3,6417

k >2

3= 1,5

and hence

1, 5 < k < 3,6417

Having to take only integer values

k = 2 or

k = 3 ,this is impossible otherwise

1 + b3 = c3

Has no sense, the difference between two cubes is never equal to 1


20/49

Versione 1.0

22/3/2013

Pagina 20 di 49

therefore remains only this equation that is valid

1 + b3

= c2

with two unknowns b and c

Is then the solution sought

1 + 23

= 32

But If we choose n = 4 we have

1 + b4

= ck


2

6

= 0,6079 <

+

kn

11< 1

2

6

= 0,6079 <

+

k

1

4

1< 1

k < 2

2

24

4

= 2,7938

k >3

4= 1,3333

and hence

1,3333< k < 2,7938



21/49

Versione 1.0

22/3/2013

Pagina 21 di 49

We note that in the inequality k < 22

24

4

= 2,7938, we have 24, number that is connected with the

modes corresponding to the physical vibrations of the bosonic strings by the following Ramanujan

function:

( )

++

+

=

4

2710

4

21110log

'

142

'

cosh

'cos

log4

24

2

'

'4

0

'

2

2

wt

itwe

dxex

txw

anti

w

wt

wx

.

Furthermore, we note that 1,3333< k < 2,7938 are very near to the following values: 1,33333333

and 2,79256959, (the first is equal) values corresponding to frequencies concerning the musical system

based on the Phi (i.e. the aurea ratio 1,61803398)

k = 2

1 + b4

= c2

with two unknowns b and c

But for what value of n we have that k = 2

1 + bn

= c2


2

6

= 0,6079 <

+

2

11

n< 1


22/49

Versione 1.0

22/3/2013

Pagina 22 di 49

n < 2

2

12

2

= 9,2655.

n > 2

and hence

2 < n < 9,2655.


1 + bn = c2

n = 3, 4, 5, 6, 7, 8 or 9

with two unknowns b and c.


23/49

Versione 1.0

22/3/2013

Pagina 23 di 49

Since the equation is symmetric

1 + bn

= ck

is equivalent to solve the unknown n or k

So, for example, we LIMIT the field equations valid

1 + b2 = c3, 4, 5, 6, 7, 8, 9

1 + b3, 4, 5, 6, 7, 8, 9

= c2

Solving this set of equations of only 14 equations, we find the only possible solution is the

following

1 + 23

= 32

Thus, for example, it isn't possible to have an equation as

1 + 103

= 2,687

Apart from the fact that c is not an integer, what is important to note is the value of the exponent of

the inequality abc that exceeds the maximum value of6

2= 1,6449

(rad (10*2,68))q

= 2,687

q = 2,098 > 1,6449

that's no good and it shows that all the other infinite equations with any n and k aren't possible,

except field equation above.

Notes that the field equations have n exponent or k exponent equal to 2, otherwise we don't have

solutions.

This occurs also in the general case of the conjecture of Fermat-Catalan where one of the

exponents m or n or k is equal to 2.


24/49

Versione 1.0

22/3/2013

Pagina 24 di 49

3. SUPER-GENERAL TRINOMIAL EQUATION

We can alsoprove the caseofsuper-general trinomialequation

Aam

+ Bbq

= Cck

rad (A*B*C*am

bn

ck

) ABCabc < ABC cm

k

cn

k

c

Cck

< (ABC) 62

c 62

++ 1

n

k

m

k

ck

< (A 62

B 62

C1

6

2

) c 62

++ 1

n

k

m

k

Now to solve this inequality we assume that

(A 62

B 62

C1

6

2

) = ce

6

2

with

e 0

so

ck

< ce

6

2

c 62

++ 1

n

k

m

k

ck

< c 62

+++ e

n

k

m

k1


25/49

Versione 1.0

22/3/2013

Pagina 25 di 49


k6

2

+++ e

n

k

m

k1


1 6

2

+++ k

e

knm

111

It depends on values of n, k and e

+++

k

e

knm

111 2

6

= 0,6079


2

6

= 0,6079 <

+++

k

e

knm

111< 1

with

e 0

2

6

= 0,6079 <

++

knm

111< 1

(A 62

B 62

C 162

) = ce

6

2

and we have the case of Fermat-Catalan conjecture with A = 1, B = 1 and C =1, in fact:


26/49

Versione 1.0

22/3/2013

Pagina 26 di 49

(1 62

1 62

1 162

) = c0

= 1

instead if e=1 we have:

2

6

= 0,6079 <

++ knm

211

< 1

a special case with

(A 62

B 62

C 162

) = c 62

Of course it must always apply the inequality

knm

111++ < 1

so e is limited by n and k

This shows by the new abc conjecture that we have only finitely many solutions (a,b,c,m,n,k) where

(m or n or k) = 2, one of the three, otherwise the inequality doesn't hold.

--------------------------------------------------------------------------------------------------------------------------

Famous example of Ramanujan

a3

+ b3

= 1729

rad (a3

* b3

* 1729) > 1729

In fact there are 2 solutions

123

+ 13

= 1729


27/49

Versione 1.0

22/3/2013

Pagina 27 di 49

rad (123 *13 * 1729) = 2*3*7*13*19 = 10374 > 1729

or even better (for the quantity or quality of the radical)

103

+ 93

= 1729

rad (103

* 93

* 1729) = 2*3*5*7*13*19 = 51870 > 1729


28/49

Versione 1.0

22/3/2013

Pagina 28 di 49

3.1 BEAL'S CONJECTURE

Beal's conjecture is a conjecture that states

If

am

+ bq

= ck

where a, b, c, m, n, and kare positive integers with m, n, and k 3 then a, b, c have a common factor.

Examples

To illustrate, the solution 33

+ 63

= 35

has bases with a common factor of 3, and the solution 76

+ 77

=

983

has bases with a common factor of 7. Indeed the equation has infinitely many solutions, including

for example

( )[ ] ( )[ ] ( ) 1++=+++ mmmmmmmmm bababbaa

for any 3,, >mba . But no such solution of the equation is a counterexample to the conjecture, since

the bases all have the factor mm ba + in common.

The example 73

+ 132

= 29

shows that the conjecture is false if one of the exponents is allowed to be 2.

Properties

A variation of the conjecture where m, n, and k(instead ofa, b, c) must have a common prime factor is

not true. See, for example 274

+ 1623

= 97

(3, 4 and 7 have no common factor)

PROOF

By the caseofsuper-general trinomialequation


29/49

Versione 1.0

22/3/2013

Pagina 29 di 49

Aam

+ Bbq

= Cck

We have Beals conjecture replacing

A = B = C =1

And so we return to the FermatCatalan conjecture BUT with exponentm,n, andk 3 (2 is not

allowed).

This shows by the new abc conjecture that we have only finitely many solutions (a,b,c,m,n,k) where(m or n or k) = 2, one of the three, otherwise the inequality doesn't hold (remember that a, b, c must

be coprime).

In fact we have trivial solutions with a common factor only with m,n, andk 3.


30/49

Versione 1.0

22/3/2013

Pagina 30 di 49

3.1.1 ANOTHER PROOF OF FERMATS LAST THEOREMTHROUGH

BEALS CONJECTURE

Beal's conjecture is a generalization of Fermat's last theorem, which corresponds to the case

zyx == . If xxx cba =+ with 3x , then either the bases are coprime or share a common factor. If

they share a common factor, it can be divided out of each to yield an equation with smaller, coprime

bases.

Now since Beal's conjecture is true, Fermats last theorem can be proven by contradiction:

Let there be positive integers n 3 anda, b, c such that

an

+ bq

= cn

where c is the smallest possible.

Then, by the Beal's conjecture with m = n = k = n it is shown there exists a base p dividing each ofa,

b and c. However, then

n

p

a

+

n

p

b

=

n

p

c

which contradicts C being the smallest possible, BECAUSE C DOESNT HAVE TO BE

DIVISIBLE BYp.

This shows the positive integers n, a, b, c cannot exist, thus proving Fermat's last theorem.

CVD


31/49

Versione 1.0

22/3/2013

Pagina 31 di 49

4. PROOF OF PILLAI'S CONJECTURE

Pillai's conjecture concerns a general difference of perfect powers .

Pillai conjectured that the gaps in the sequence of perfect powers tend to infinity.

This is equivalent to saying that each positive integer occurs only finitely many times as a difference

of perfect powers: more generally, in 1931 Pillai conjectured that for fixed positive integersA,B, Cthe

equation

A + Bbn

= Cck

has only finitely many solutions (b,c,n,k) with (n,k) (2,2).

Lets apply the new abc conjecture

rad (A*B*C*bn

ck

) ABCbc < ABCcn

k

c

Cck

< (ABC) 62

c 62

+ 1

n

k

ck

< (A 62

B 62

C1

6

2

) c 62

+ 1

n

k


(A 62

B 62

C1

6

2

) = ce

6

2

with

e 0


32/49

Versione 1.0

22/3/2013

Pagina 32 di 49

so

ck

< ce

6

2

c 62

+ 1

n

k

ck

< c 62

++ e

n

k1


k6

2

++ e

n

k1


1 6

2

++

k

e

kn

11

It depends on values of n, k and e

++

k

e

kn

11 2

6

= 0,6079


2

6

= 0,6079 <

++

k

e

kn

11< 1

with

e 0


33/49

Versione 1.0

22/3/2013

Pagina 33 di 49

if e=0 we have:

2

6

= 0,6079 <

+

kn

11< 1

(A 62

B 62

C 162

) = ce

6

2

and we have the case of Catalan conjecture with A = 1, B = 1 and C =1, in fact:

(1 62

1 62

1 162

) = c0

= 1

instead if e=1 we have:

2

6

= 0,6079 <

+

kn

21< 1

a special case with

(A 62

B 62

C 162

) = c 62

Of course it must always apply the inequality

knm

111++ < 1

so e is limited by n and k


34/49

Versione 1.0

22/3/2013

Pagina 34 di 49

26

= 0,6079 <

++

k

e

kn

11< 1

However. all the cases are a stronger inequalityof the initial statement

knm

111++ < 1

This shows by the new abc conjecture that we have only finitely many solutions (b,c,n,k) with (n,k)

(2,2), otherwise the inequality doesn't hold.


35/49

Versione 1.0

22/3/2013

Pagina 35 di 49

5. PROOF OF TIJDEMAN'S CONJECTURE

Tijdeman's conjecture states that there are at most a finite number of consecutive powers.

Stated another way, the set of solutions in integers b, c, n, kof the exponential diophantine equation

A +bn

=ck

for exponents n and k 2, is finite.

If A =1 the theorem was proven by Mihailescu and it's the Catalan's conjecture, as we have seen before

and is also known asTijdeman's theorem.But with A 2, n and k 2 we have an unsolved problem, called the generalized Tijdeman problem. It

is conjectured that this set also will be finite. This would follow from a yet stronger conjecture of Pillai

stating that the equation

A + Bbn

= Cck

only has a finite number of solutions, that we have shown above.

In fact

Lets apply the new abc conjecture

rad (A*bnc

k) Abc < Ac

n

k

c

ck

< (A) 62

c 62

+ 1

n

k


A 62

= ce

6

2


36/49

Versione 1.0

22/3/2013

Pagina 36 di 49

with

e 0

so

ck

< ce

6

2

c 62

+ 1

n

k

and then follow all the calculation as before


2

6

= 0,6079 <

++

k

e

kn

11< 1

with

e 0

if e=0 we have:

2

6

= 0,6079 , ( a , constant), multiplying, both sides of the

previous relation, for xe , and integrating, with respect to x , between the limits zero and infinity, we

obtain another divergent series, defined by:

( )

=0

00!n

xnnx dxexn

Adxxfe . (5.1)

Now, we have the following relation:

= 0 !1 k

k

kx k

xB

e

x. (5.2)

Applying to this relation, the integration of eq. (5.1), we have that:

+===0 0 0 0 1

22

11

!

1

1 k k kkk

xk

kx

x

BBdxexk

Be

dxxe. (5.3)

Now we compute the integral that is to the left-hand side of the (5.3). We obtain:

( )

( )

+

=+

==

=0 0 0 0 0

2

2

2 162

1

11 k k

kx

x

xx

x

x

kdxxe

e

dxexe

e

dxxe . (5.4)

Thence, from the (5.3), we obtain:

=1

2

22

3

6kkB

. (5.5)


39/49

Versione 1.0

22/3/2013

Pagina 39 di 49

The left-hand side of the (5.5) is just a divergent series, that is represented from the value of the right-hand side of this expression.

Substituting in the (5.2), x to x , we have:

( )

=

0 !1 k

k

kx k

xB

e

x. (5.6)

For 2x , the precedent relation is divergent, thence, applying to the same relation, the integration

of eq. (5.1), we have:

( )

( )

+

=

+==

=

0 00

00

2

2

1

61

1

11 k k

kx

x

x

x

x

kdxxedx

e

xedx

e

xe

; (5.7)

( )( )

++==

00

0 1

22

111

!

1

k k k

k

k

k

xk

k

k BBdxexk

B . (5.8)

Equalling the results of the two last relations, we obtain:

=

1

2

2 2

3

6k kB

, (5.9)

that is identical to the (5.5).

With regard the string theory, we can to obtain mathematical connections with some equations

concerning a two-parameter family of Wilson loop operators in N = 4 supersymmetric Yang-Mills

theory which interpolates smoothly between the 1/2 BPS line or circle, principally some equations

concerning the one-loop determinants

With regard the calculation of the 2-loop graphs for the Wilson loop with a cusp in the case of non-

zero , the resulting expression can be written as a sum of the contribution of ladder graphs and the

interacting graphs ( )( ) ( )( ) ( )( ) ,,, 2int22

VVV lad +=

( )( )( )

+

+

+

+=

02

2

2 log1

1log

sin

coscos,

i

i

i

i

ladez

ez

ze

ze

z

dzV

( )( ) ( ) ( ) ++=1

0

222

int 1,1cos2,coscos4, zzzdzYV . (5.10)

The integrand in the last expression is the scalar triangle graph the Feynman diagram arising at

one-loop order from the cubic interaction between three scalars separated by distances given by the

arguments


40/49

Versione 1.0

22/3/2013

Pagina 40 di 49

( )

=2

3

2

2

2

1

4

2

2

13

2

23

2

12

11,,

wxwxwxwdxxxY

,

22

jiij xxx = . (5.11)

This integral is known in closed form. For2

13

2

23

2

12 , xxx < it is equal to

( ) +

+

+=

2

12

2

12

3

1,, 22

2

2

13

2

13

2

23

2

12

AtsLi

AtsLi

AxxxxY

+

++ 2

1

ln2

1

ln2lnln

AtsAts

ts

2

13

2

12

x

xs = ,

2

13

2

23

x

xt= , ( ) sttsA 41 2 = . (5.12)

Thence, we have that:

( ) =

= 23

2

2

2

1

4

2

2

13

2

23

2

12

11,,

wxwxwxwdxxxY

+

+

+=

2

12

2

12

3

122

2

2

13

AtsLi

AtsLi

Ax

+

++

2

1ln

2

1ln2lnln

AtsAtsts . (5.12b)

This expression is valid for 1, , then for 1z the first

two arguments ofY in (5.10) are less than unity and in that regime Y evaluates to

( ) ( ) ( ) ( ) ( ) ( )

++++=

iiiii

zeezezzeLizeLiz

iY 1loglog1loglog1

6sin22

2

. (5.13)

The integration then gives

( )( )

+=

1

0 sin3

dzY . (5.14)

With the prefactor we find the final expression (valid by analytical continuation for all


41/49

Versione 1.0

22/3/2013

Pagina 41 di 49

( )( ) ( )( )

+

=

sin

coscos

3

4,2intV . (5.15)

The first integral in (5.10) can also be done analytically. Again one should take care in choosing

branch cuts for the logarithms, where the principle branch is for small . The result is

( )( )( ) ( ) ( ) ( )

+

+

=

32

2

2

2

32

2

2

363

sin

coscos4,

ieLiieLiV

ii

lad . (5.16)

Thence, form (5.10), we have the following expression:

( )( )( )

=

+

+

+

+=

02

2

2 log1

1log

sin

coscos,

i

i

i

i

ladez

ez

ze

ze

z

dzV

( ) ( ) ( ) ( )

+

+

=

32

2

2

2

32

2

363

sin

coscos4

ieLiieLi

ii . (5.16b)

We have the following mathematical connection:

( )

( )=

+==

=

+

0 00

00

2

2

1

61

1

11 k k

kx

x

x

x

x

kdxxedx

e

xedx

e

xe

( ) ( )( )

=

+

+

+

+=

02

2

2 log1

1log

sin

coscos,

i

i

i

i

ladez

ez

ze

ze

z

dzV

( ) ( ) ( ) ( )

+

+

= 3

2

22232

2

363

sincoscos4

i

eLiieLi ii . (5.17)


42/49

Versione 1.0

22/3/2013

Pagina 42 di 49

6. EXAMPLES OF PILLAIS CONJECTURE

We see a table with x and y consecutive numbers with exponents 2 and 3, and then a similar table with

exponents 3 and 4 for Pillai

x^2 y^3 Algebraic

difference

y^3-x^2

Square root of

the difference

Observations

final digits

1, 7, 9 of the

differences

3^2 = 9 2^3=8 1 1 14^2=16 3^3=27 11 3,31 1

5^2= 25 4^3= 64 39 6,24 9

6^2=36 5^3=125 89 9,43 9

7^2=49 6^3= 216 167 12,92 7

8^2=64 7^3=343 279 16,70 9

9^2=81 8^3=512 431 20,70 1

10^2=100 9^3=729 629 25,07 9

11^2=121 10^3=1000 879 29,64 9

12^2=144 11^3=1331 1187 34,45 7

13^2=169 12^3=1728 1559 39,48 914^2=196 13^3=2197 2001 44,73 1

15^2=225 14^3=2744 2519 50,18 9

As we can see, the successive differences grow more and tend to infinity (their square roots grow

about 4 or 5 at least up to x = 15), and therefore the only possible solution is the first,

with one difference, as Catalan's conjecture. Furthermore, we note that the final digits of the

differences are always 1, 7 and 9, are identified two successive groups of 9, 9, 7, 9, 1 after the pair 1, 7

Initial.

They are absent then the final digits 3 and 5.

The fact that he chose:

x3

- (x + 1)2

= k

x3

- x2

- 2x -k - 1 = 0

The solutions of this equation gives the values of x, being the divisors of the number (k 1), by the

rule of Ruffini.


43/49

Versione 1.0

22/3/2013

Pagina 43 di 49

We can observe that the trinomial

x3

- x2

- 2x - 1

for any value of x results in an odd integer with the final digit that can only be 1, 7 or 9.

This proves precisely for induction.

Since Pillais conjecture applies to the equation

A + Bbn = Cck

Cck - Bbn = A

the difference A is an integer odd or even that can also be repeated a finite number of times.

In fact the sequence of the powers perfect is the following.

1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100, 121, 125, 128, 144, 169, 196, 216, 225, 243, 256, 289, 324, 343,

361, 400, 441, 484, 512, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1000, 1024, 1089, 1156, 1225, 1296,1331, 1369, 1444, 1521, 1600, 1681, 1728, 1764

Examples:

25 33 = 5

122

27

= 16


44/49

Versione 1.0

22/3/2013

Pagina 44 di 49

Now we see the same table for the conjecture Pillai, limited to the exponents 3 and 4 in place of 2 and

3.

x^3 y^4 Algebraic

difference

y^4- x^3

Square root of

the difference

Observations

final digits

1, 3, 7 and 9 of

the differences

3^3 = 27 2^4 = 16 11 3,31 1

4^3=64 3^4=81 17 4,12 7

5^3=125 4^4= 256 131 11,44 1

6^3= 216 5^4=625 409 20,22 9

7^3=343 6^4=1296 953 30,87 3

8^3=512 7^4=2401 1889 43,46 9

9^3= 729 8^4=4096 3367 58,02 7

10^3=1000 9^4=6561 5561 74,57 1

11^3=1331 10^4= 10000 8669 93,10 9

12^3=1728 11^4= 14641 12913 113,63 3

13^3=2197 12^4=20736 18539 136,15 9

14^3=2744 13^4=28561 25817 160,67 7

15^3=3375 14^4=38416 35041 187,19 1

Now, however, we note that the differences are still growing faster than in the table above, and

therefore also their square roots.

Now the differences between the final digits there is also the number 3, but there is a little more order

in the sequences of the final digits: we identify two identical successive groups: 7 1 9 3 9:07 1 9 3 9

after initial sequence 1; but also the third group begins with 7 and 1.

Further calculations with successive powers and their differences (from Pillai's conjecture) show some

interesting regularities, which we quote:

Consecutive powers (for conjecture Pillai)

1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81,100,121,125,128,144, 169,196,216, 225,243, 256, 289, 324,

343, 361, 400, 441, 484, 512, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1000,1024, 1089, 1156,

1225, 1296, 1331, 1369, 1444, 1521, 1600, 1681, 1728, 1764


45/49

Versione 1.0

22/3/2013

Pagina 45 di 49

consecutive differences

3 Fibonacci

4 =2^2

1 only solution for the catalans conjecture (proved)

7 prime

9 =3^22 Fibonacci

5 Fibonacci

4 =2^2

13 Fibonacci

15 =3*517 prime

19 prime

21 Fibonacci4 = 2^2

3 Fibonacci

16 =4^2

25 =5^2

27 =3^320 =2*2*5

9 =3^2

18 =2*3*3

33 =3*11

35 =5*7

19 prime18 =2*3*3

39 =3*13

41 prime

43 prime28 =2*2*7

17 prime

47 prime49 = 7^2

51 prime

53 prime

55 Fibonacci

57 =3*19

59 = prime

61 =prime39 =3*13

24 =2*2*2*3

65 =5*13


46/49

Versione 1.0

22/3/2013

Pagina 46 di 49

67 prime69 = 3*23

71 prime35 =5*7

38 = 2*19

75 = 3*5*5

77 =7*7

79 prime

81 =9^2= 3^4

47 prime36 =6^2

We note that 4 occurs three times, twice on 3, 9 twice, 7 and 13 once, 6, 8, 10 and 12 zero times.

The odd numbers in blue differ by 2 to groups of two, three or four consecutive numbers, with a group

of eight consecutive odd numbers. After this cycle falls to about half last number in blue (ex. 71/35 =

2.02, 81/47 = 1.7) and after two or three sizes too small recurs again, obviously with consecutive

numbers gradually larger.

They are present Fibonacci numbers (3, 5, 13, 21), square (4, 9, 16, 25, 36, 49, 81 and even a cube

(27). We assume that this trend continues as the semiregular differences growth of consecutive powers,

and could contribute to another definitive proof of Pillai's conjecture, according to which these

differences are repeated in a finite number of times.

This means that the smaller numbers of this estimate will never be more later, and therefore their

presence exists in a finite number of times, just as says the conjecture, so that proves true even if only

by empirical means these numerical evidence, but not analytical.

Conclusions

Remember that Pillai's conjecture states that for any given positive integer A there are only a finite

number of pairs of perfect powers whose difference is A, but from the proof above that we have that

for fixed positive integers A, B, C the equation

Cck

- Bbn

= A

has only finitely many solutions (b,c,n,k) with (n,k) (2,2).

This is equivalent to saying that each positive integer A is valid and occurs only finitely many times

as a difference of perfect powers.


47/49

Versione 1.0

22/3/2013

Pagina 47 di 49

6.1 PERFECT POWERS LINKED TO PILLAIS CONJECTURE

A sequence of perfect powers can be generated by iterating through the possible values for m and k.

The first few ascending perfect powers in numerical order (showing duplicate powers) are the

following:

22

= 4, 23

= 8, 32

= 9, 24

= 16, 42

= 16, 52

= 25, 33

= 27, 25

= 32, 62

= 36, 72

= 49, 26

= 64, 43

=

64, 82

= 64,

The sum of the reciprocals of the perfect powers (including duplicates) is 1:

=

=

=2 2

11

m kk

m.

The first perfect powers without duplicates are:

(sometimes 1), 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81,100,121,125,128,144, 169,196,216,

225,243, 256, 289, 324, 343, 361, 400, 441, 484, 512, 529, 576, 625, 676, 729, 784, 841, 900,

961, 1000,1024, 1089, 1156, 1225, 1296, 1331, 1369, 1444, 1521, 1600, 1681, 1728, 1764

The sum of the reciprocals of the perfect powersp without duplicates is:

( ) ( )( )

=

=p k

kkp 2

...874464365.011

where (k) is the Moebius function function and (k) is the Riemann zeta function.

Instead the sum of 1/(p1) over the set of perfect powersp, excluding 1 and excluding duplicates, is 1:

=+++++++=p p

.1...31

1

26

1

24

1

15

1

8

1

7

1

3

1

1

1


48/49

Versione 1.0

22/3/2013

Pagina 48 di 49

This is sometimes known as the Goldbach-Euler theorem.

But what happens if we add the differences ofthe reciprocals of the perfect powers?

= + 2 1

1

k kk pp=

4

1+ 1 +

7

1+

9

1+

2

1+

5

1+

4

1+

13

1+.=

Since it must appear all natural numbers (with also duplicates) we have a sum a bit more of the series

of theHarmonic series, thats a divergent infinite series.

This proves also that the gaps in the sequence of perfect powers tend to infinity.


49/49

Versione 1.0

22/3/2013

Pagina 49 di 49

References all on this web site http://nardelli.xoom.it/virgiliowizard/

1) IL CONCETTO MATEMATICO DI ABBONDANZA E IL RELATIVO

GRAFICO PER LA RH1Francesco Di Noto, Michele Nardelli (Gruppo B. Riemann)

2) NOVITA SULLA CONGETTURA DEBOLE DI GOLDBACH

Gruppo B.Riemann

Francesco Di Noto,Michele Nardelli3)Appunti sulla congettura abc

Gruppo B. Riemann*

*Gruppo amatoriale per la ricerca matematica sui numeri primi, sulle loro congetture e sulle loro

connessioni con le teorie di stringa

Francesco Di Noto, Michele Nardelli

4) I numeri primoriali p# alla base della dimostrazione definitiva della congettura di Goldbach

(nuove evidenze numeriche)Francesco Di Noto, Michele Nardelli

5) ESTENSIONI DELLE CONGETTURE,FORTE E DEBOLE, DI GOLDBACH

(a k= primi , conNe kentrambi pari o dispari)

Gruppo B. Riemann*

Francesco Di Noto, Michele Nardelli

*Gruppo amatoriale per la ricerca matematica sui numeri primi, sulle loro congetture e sulle loro

connessioni con le teorie di stringa.

6) IPOTESI SULLA VERITA DELLE CONGETTURE SUI NUMERI PRIMI CON

GRAFICI COMET E CONTRO ESEMPI NULLI

(Legendre, Goldbach, Riemann)

Michele Nardelli ,Francesco Di Noto,

Pier Francesco Roggero, Michele Nardelli, Francesco Di Noto "Proof of Fermat-catalan Conjecture...

Documents

Transcript of Pier Francesco Roggero, Michele Nardelli, Francesco Di Noto "Proof of Fermat-catalan Conjecture...