Pier Francesco Roggero, Michele Nardelli, Francesco Di Noto - "A possible proof of Fermat's Last Theorem"

7/29/2019 Pier Francesco Roggero, Michele Nardelli, Francesco Di Noto - "A possible proof of Fermat's Last Theorem"

1/46

A POSSIBLE PROOF OF FERMATS LAST THEOREM

THROUGH THE ABC RADICAL

Pier Francesco Roggero, Michele Nardelli1,2

, Francesco Di Noto

1 Dipartimento di Scienze della Terra

Universit degli Studi di Napoli Federico II, Largo S. Marcellino, 10

80138 Napoli, Italy

2 Dipartimento di Matematica ed Applicazioni R. Caccioppoli

Universit degli Studi di Napoli Federico II Polo delle Scienze e delle Tecnologie

Monte S. Angelo, Via Cintia (Fuorigrotta), 80126 Napoli, Italy

Abstract

In this paper we show a possible proof of Fermats Last Theorem through the abc

radical. Furthermore, in the various Sections, we have described also some

mathematical connections with , , thence with some sectors of string theory.


2/46

Versione 1.0

12/3/2013

Pagina 2 di 46

Index:

1.PROOF OF FERMATS LAST THEOREM.......................................................................................................... 31.1Radical of an integer ............................................................................................................ 3

2. RULE (THEOREM) OF THE SUM IN A GROUP .................................................................................... 82.1 Connection Fermats last theorem with the abc conjecture ...........................................10

3. UPPER LIMIT FOR ABC CONJECTURE AND ITS PROOF...................................................................133.1 Curious connection with the Golden ratio.........................................................................18

4.EXTENTED RADICAL AB...Z .......................................................................................................................22

4.1 The new abc conjecture extended ....................................................................................234.2 Examples of sums of powers .............................................................................................264.3 Prediction of Computing Minimal Equal Sums Of Like Powers.......................................314.4 Super-general trinomial equation......................................................................................34

5. GOLDBACHS CONJECTURE AND ITS PROOF..................................................................................356 WARING'S PROBLEM .........................................................................................................................42


3/46

Versione 1.0

12/3/2013

Pagina 3 di 46

1.PROOF OF FERMATS LAST THEOREM

Fermat's Last Theorem states that no three positive integers a, b, and c can satisfy the equation an

+

bn

= cnfor any integer value ofn greater than two.

To prove the theorem we first introduce the concept of radical abc.

1.1Radical of an integer

The radical of a positive integer n is defined as the product of distinct (not repeated, ie without

consider the exponent) prime factors of n.

rad (n) = p

Examples

Radical numbers for the first few positive integers are:

1, 2, 3, 2, 5, 6, 7, 2, 3, 10,....

For example,

504 = 23 * 32 *7

and therefore

rad (504) = 2*3*7 = 42

Properties

The function rad is not completely multiplicative.


4/46

Versione 1.0

12/3/2013

Pagina 4 di 46

This means that there isnt, always, the following rule

rad (abc) = rad(a) * rad(b) * rad (c)

This applies only if the three numbers a, b and c are all primes.

Besides the radical of any integer n is the largest square-free divisor of n


5/46

Versione 1.0

12/3/2013

Pagina 5 di 46

Proof of Fermat's theorem

an

+ bn

= cn

for n 3

has no solution in N+

Let s assume that there are positive integers, a and b coprime so are coprime also any powers ak

and

bl. that satisfy the equationa

n+b

n=c

nand lets see how it behaves the radical rad.

rad (an bn cn) = rad (abc) abc < cn

always true for every n 3 because a < c and b < c

We always have that the sign of the radical inequality is < and then

rad (anb

nc

n) < c

nfor every n 3

Only with the Pythagorean triples or with the trivial case a + b = c (hence the abc conjecture) the

radical inequality can be < o > or also = (if we consider a=b)

These two cases correspond to Fermat's theorem with exponents n=2 and n=1 respectively

But we are sure that there is always the radical inequality >

The cases in which the sign of the radical inequality is < are isolated cases, much more rare, are

exceptions that prove the general rule of sign >

Fot the Pythagorean triples as well as a + b = c (conjecture abc) usually the sign ofthe radical

inequality is almost always >


6/46

Versione 1.0

12/3/2013

Pagina 6 di 46

In fact for Pythagorean triples we have

a2

+b2

=c2

rad (a2b

2c

2) = rad (abc) < or >c

2

In Pythagorean triples we can find solutions as well as three numbers a, b and c where no one is a

prime number (example: 16, 63, 65).

One thing is known, that one of the short sides (a or b) is always a multiple of 4.

Thus the sign of the radical inequality can be > (most often) or < (much more rarely), so we have to

calculate:

For example we have:

25 + 212 =

213

rad (25*144*169) = 2*3*5*13 = 390 > 169

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

27 + 224 =

225

rad (49*576*625) = 2*3*5*7 = 210 < 625

And we have an exception.


7/46

Versione 1.0

12/3/2013

Pagina 7 di 46

rad (abc) < c2

So we have both the sign > that < for the sign of the radical inequality.

Also for the sum a + b = c we can have the sign of the radical inequality > (most often) or , or

rad (abc) > c

This is a necessary and sufficient condition and in this case we have infinite solutions.

When the sign of the radical inequality is > there are also the exceptions (with the sign of the

radical inequality < ) , but the converse is not true (be careful!).

That is, if we have only radical sign with < and NEVER signed > the sum is NOT POSSIBLE.

If, instead, we add two identical elements is the same as thesign of the radical inequality is

In fact

a + a = 2a rad (a a 2a) = rad (2a3) =rad (2a) 2a

if a is a prime we have the sign =, however, if a is a composite number the sign is cn

;

for n = 3: a2

b2c

2> c

3or a

2b

2> c

always true


12/46

Versione 1.0

12/3/2013

Pagina 12 di 46

for n = 4: a2 b2c2 > c4 or a2 b2 > c2 always true because ab > c

for n = 5: a2

b2c

2> c

5or a

2b

2< or > c

3,it depends on a, b and c

for n = 6: a2

b2c

2> c

6or a

2b

2> c

4NEVER TRUE

Thus we have shown that Fermat'last theorem has no solutions for n 6, provided that the abc

weak conjecture is true!

But it makes no sense to choose values of > 0, first because obviously it should be proved that the

abc conjecture or the weak abc conjecture is valid.

In our case itisfair to say thatwhen= 0, theabcconjecturedoes not stand. In factonlyfor> 0,

there exists a constant K which gives a lower limit to have a finite number of solutions.K0 {rad

(abc)} > c

K0 rad (an

bn

cn) = K0 rad (abc) K0 abc < c

n

always true for every n 4 because a < c and b < c and K0 < c

for n = 3 we have

K0 abc > c3

only if

K0 >ab

c2

There is a limit K0 , but as must be

K= K0 < c

K0 >ab

c2

c because we first replaced rad (abc) with abc, more precisely

rad (abc) abc


13/46

Versione 1.0

12/3/2013

Pagina 13 di 46

and so, at fortiori

K0 > )(

2

abrad

c c

We have an absurd and even the case n = 3 has no solution!

So even applying the abc conjecture we have a rebuttal which we always have that the sign of the

radical inequality is < and then

rad (anb

nc

n) < c

nfor every n 3

and

an

+ bn

= cn

for n 3

have no solutions in N+

3. UPPER LIMIT FOR ABC CONJECTURE AND ITS PROOF

The abc conjecture implies that c can be bounded above by a near-linear function of the radical of abc.

But what is that limit?

Recall that the inequality says:

ABC Conjecture: For every > 0, there exists a constant K such that for all triples (a, b,

c) of coprime positive integers, with a + b = c, the inequality

K {rad (abc)}1+

> c

always holds.


14/46

Versione 1.0

12/3/2013

Pagina 14 di 46

We first have to know when given two randomly chosen integers a and b, how likely it is that aand b are coprime. In this determination, it is convenient to use the characterization that a and b

are coprime if and only ifno prime number divides both of them.

Informally, the probability that any number is divisible by a prime (or in fact any integer) P isp

1; for

example, every 7th integer is divisible by 7. Hence the probability that two numbers are both divisible

by this prime is 21

p, and the probability that at least one of them is not 1 - . 2

1

p. For distinct primes,

these divisibility events are mutually independent. For example, in the case of two events, a number is

divisible by p and q if and only if it is divisible by pq; the latter event has probability 1/pq.(Independence does not hold for numbers in general, but holds for prime numbers).

Thus the probability that two numbers are coprime is given by a product over all primes:

( )

==

=

p p pp61607927102.0

6

2

1

1

111

2

1

22 %

)2(

1

= 2

6

= 0,6079....

Here refers to the Riemann zeta function, the identity relating the product over primes to (2) is an

example of an Euler product, and the evaluation of (2) as 2/6 is the famous Basel problem. In

general, the probability ofkrandomly chosen integers being coprime is 1/(k), did not take pairwise

relatively prime, if every pair in the set of integers is relatively prime.

The notion of a "randomly chosen integer" in the preceding paragraphs is not rigorous. One rigorous

formalization is the notion of natural density: choose the integers a and b randomly between 1 and an

integerN. Then, for each upper boundN, there is a probability PNthat two randomly chosen numbers

are coprime. This will never be exactly 26

, but in the limit as N , the probability PN approaches

to 26

.

Since the radical rad (abc) is defined as the product of distinct (not repeated, ie without considering the

exponent) prime factors of abc, where a, b are coprime positive integers, with a + b = c.


15/46

Versione 1.0

12/3/2013

Pagina 15 di 46

If a and b are coprime also c is coprime, because c is the sum of a and b.

The inequality

K {rad (abc)}1+

> c

becomes

{rad (abc)} 62

> c

where6

2= 1,644934...

So to be precise we have that

K= 1 =

6

2-1 = 0,644934..

An assessment of this inequality follows immediately from the third formulation of the conjecture

involves the quality, q(a, b, c), of the triple (a, b, c), defined by:

q(a, b, c) = [ ])(lnln

abcrad

c

For example,

q(4, 127, 131) = log(131) / log(rad(4127131)) = log(131) / log(2127131) = 0.46820... q(3, 125, 128) = log(128) / log(rad(3125128)) = log(128) / log(30) = 1.426565...

A typical triple (a, b, c) of coprime positive integers with a + b = c will have c < rad(abc), i.e. q(a, b,

c) < 1. Triples with q > 1 such as in the second example are rather special, they consist of numbers

divisible by high powers of small prime numbers.

So we have that ABC Conjecture: For every > 0, there exist only finitely many triples

(a, b, c) of coprime positive integers with a + b = c such that q(a, b, c) > 1 + .


16/46

Versione 1.0

12/3/2013

Pagina 16 di 46

Whereas it is known that there are infinitely many triples (a, b, c) of coprime positive integers with a +b = c such that q(a, b, c) > 1, the conjecture predicts that only finitely many of those have q > 1.01 or q

> 1.001 or even q > 1.0001, etc.

The Highest quality triples

q a b c Discovered by

1 1.6299 2 310109 235 Eric Reyssat

2 1.626011

23

25

67

32

2123

Benne de Weger

3 1.6235 191307 7292318283225

4Jerzy Browkin, Juliusz Brzezinski

4 1.5808 283 511132283817

3

Jerzy Browkin, Juliusz Brzezinski, Abderrahmane

Nitaj

5 1.5679 1 237 547 Benne de Weger

So if we rewrite

q(a, b, c) = [ ])(lnln

abcrad

c

q ln rad(abc) = ln c

ln (rad(abc))q

= ln c

(rad (abc))q

= c

and since the MAX q = 1,6299, so far found, the inequality IS EFFECTIVELY TRUE.

{rad (abc)} 62

> c

q 6

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


17/46

Versione 1.0

12/3/2013

Pagina 17 di 46

This inequality is a proof of abc conjecture because it holds for ALL triples (a,b,c) of coprimepositive integers, witha +b =c

CVD

There are exponential bounds that are known.

Specifically, the following bounds have been proven:

( )( )151exp abcradKc < (1986)

( )

c

All these bounds can be replaced and are no longer valid!


18/46

Versione 1.0

12/3/2013

Pagina 18 di 46

3.1 Curious connection with the Golden ratio

A curious observation follows from

{rad (abc)} 62

> c

Another inequality that holds is the following

{rad (abc)} > c

Where is the golden ratio (the value 1,61803398 is very near to the value 1,6449)

In this case

K=

; =

-1 = 0,618..

This is an approximation to the exact formula

(rad (abc))1,6449

In fact, for rad (abc) > 58753630

does not work anymore.

However, at this time also with a hat-trick record that q = 1.6299 works! And there arent

counterexamples though in the future may find the triples with q 1.6449.

We note that the values 1,6299 and 1,6449 are very near to the following values: 1,62108926 and

1,64809064. They are values concerning the aurea frequencies system based on fractional powers

of Phi and Pigreco.


19/46

Versione 1.0

12/3/2013

Pagina 19 di 46

With regard the rational numbers and , the diagram below is a brief excerpt of rational integers. In it

is shown how the integers (and also the prime numbers) can be represented only by powers of. Itmust be assumed that it is possible for any integer, although this seems to be impractical. However,

this form of philosophical investigation is of great importance when we consider the main formula for

the golden ratio:

11111... ++++= = 1,618033989...

number Calculation of

1 1

2 21

+

3 212

+

4 313

5 4113

++

6 4114

++

7 414

+

8 4124

++

9 42114

+++

10 421124

+++

11 515

12 5125

+

13 52115

++

14 52 1125 ++

15 632111125

+++++

16 62116

+

17 6126

++

18 616

+

19 6126

++

20 741116

++


20/46

Versione 1.0

12/3/2013

Pagina 20 di 46

... ...

15127 20120

+

... ...

3.2 Application of the new abc conjecture to prove Fermat's last theorem for allsufficiently large exponents

Let's now try to apply this inequality to the Fermat's theorem.

We should lower the limit for n

an

+bn

=cn

{rad (anb

nc

n)} 6

2 (abc) 6

2

(abc) 62

> c

for n = 3: a 62

b 62

c 62

> c3

or a 62

b 62

> or < 63

2

c ,it depends on a, b and c

for n = 4: a 62

b 62

c 62

> c4

or a 62

b 62

> or < 64

2

c ,it depends on a, b and c

for n = 5: a 62

b 62

c 62

> c5

or a 62

b 62

> or < 65

2

c ,NEVER TRUE


21/46

Versione 1.0

12/3/2013

Pagina 21 di 46

Thus we have shown that Fermat's last theorem has no solutions for n 5, but for n =3 and for n

=4 we cannot say nothing a priori.

Please note that we have also increased the amount {rad (anb

nc

n)} 6

2

with (abc) 62

and so it's

very likelythat also for n =3 and for n =4 the inequality is NEVER TRUE

However for greater security. since there are proofs for n = 3 and for n =4 we can use these to

show that no solutions exists for n 3


22/46

Versione 1.0

12/3/2013

Pagina 22 di 46

4.EXTENTED RADICAL AB...Z

We canalsoextend the conceptto the sum ofthreeor more terms all positive coprime integers, and

so wehave determined theradical"extended".

The concept of being all positive coprime integers can also be extended to any finite set of integers S

= {a, b, .... n} to mean that the greatest common divisor of the elements of the set is 1. If every pair in

a finite set of integers is relatively prime, then the set is called pairwise relatively prime and the sum is

true and valid.

Every pairwise relatively prime finite set is relatively prime; however, the converse is not true: {6, 10,15} is relatively prime, but not pairwise relative prime (each pair of integers in the set has a non-trivial

common factor).

We are interested in elements (abcd.n) allpairwise relative prime.

a + b + c + d+ n = z

The sum value z is coprime for definition if they are all its pairwise relatively prime.

rad (abcd.nz) > z

so thatthere isenough orthis is a necessary and sufficient condition to verify thatthe sumis

always >.

The exceptions (if holds.


23/46

Versione 1.0

12/3/2013

Pagina 23 di 46

4.1 The new abc conjecture extended

So we have:

Elements (abcd.n) pairwise relative prime.

a + b + c + d+ n = z

{rad (abcd.nz)}q > z

where

qmust be calculated depending on the number of addends

We need to calculate the exponent of the radical. To do this we must consider that the probability of n

addends there is at least one that is coprime with respect to all the others.

If, for example, lets consider three addends a, b and c how many pairs of numbers came out?

3

2

= 3

For example:

8 + 12 +41 = 61

This is the worst case that can happen and then we have to consider (in the case of all coprime we do

not care for the calculation of probabilities because we have to take the case but worse valid).

So we can afford to have 1 pair of integers not coprime to each other (8 and 12) and 2 pairs of coprime(8:and 41, 12 and 41).

Then the probability is

2

2

6

= 0,3695..

because of the 3 couples that we have, two must be coprime.


24/46

Versione 1.0

12/3/2013

Pagina 24 di 46

So the exponent q is the radical and we have:

q = 2,7058....

It is also intuitive that q should increase because if there are two composite numbers the radical

decreases.

a + b +c = d

{rad (abcd)}2,7> d

If lets consider four addends a, b, c and d how many pairs of numbers came out?

4

2

= 6

For example:

8 + 12 +24 +41 = 85

This is the worst case that can happen and then we have to consider (in the case of all coprime we do

not care for the calculation of probabilities because we have to take the case but worse valid).

So we can afford to have 3 pairs of integers not coprime to each other (8 and 12, 12 and 24, 8 and 24)

and 3 pairs of coprime (8 and 41, 12 and 41, 24 and 41).


3

2

6

= 0,2246.

because of the 6 couples that we have, 3 must be coprime.


q = 4,45....


25/46

Versione 1.0

12/3/2013

Pagina 25 di 46

a + b +c + d = e

{rad (abcde)}4,45

> e

We note that 8 and 24 are the numbers that are connected with the modes that correspond to

the physical vibrations of a superstring by the following Ramanujan functions:

( )

++

+

=

4

2710

4

21110log

'

142

'

cosh

'cos

log4

3

18

2

'

'4

0

'

2

2

wtitwe

dxex

txw

anti

w

wt

wx

. (1)

( )

++

+

=

4

2710

4

21110log

'

142

'

cosh

'cos

log4

24

2

'

'4

0

'

2

2

wtitwe

dxex

txw

anti

w

wt

wx

. (2)

Furthermore, we note also that the values 0,2246 and 4,45 are very near to the values 0,2229 and

4,4498 that correspond to the aurea frequencies based on fractional powers of Phi and Pigreco.


26/46

Versione 1.0

12/3/2013

Pagina 26 di 46

4.2 Examples of sums of powers

(3,1,3): mean exponent 3, the terms on the right of the equal sign = 1, 3 addends

7093

=6313

+ 4613

+ 1933

(they are 4 primes)

7189057 + 97972181 + 251239591= 356400829

rad (7189057*97972181*251239591*356400829 ) = 193*461*631*709 = 39.804.651.767 >

356.400.829

We have the sign > of the radical inequality.

We observe that, as mentioned in section 3.1, if there are links between prime numbers and the

powers of the golden ratio, also the numbers that appear in this example, can be considered asnew vibrations of strings (corresponding, for example, to the exotic particles).

This impliesthatapart from the factthat there are solutions, there may also be exceptions(if


27/46

Versione 1.0

12/3/2013

Pagina 27 di 46

95800 = 23 52 479217519 prime

414560 = 25

5 2591

The integers 95800 and 424560 are not coprime and so the rule of the sum is not applicable!

If we apply the new abc conjecture extended at the case (4, 1, 3)we obtain:

{rad (abcd)}2,7

> d

a4

+b

4+

c

4=

d

4

(3) = 1,202

{rad (a4b

4c

4d

4)}

2,7= rad (abcd)}

2,7 a

2,7b

2,7c

2,7d

2,7

a2,7

b2,7

c2,7

d2,7

>d4

a2,7

b2,7

c2,7

> d1,3

Always true.

So certainly we have a set of finite solutions, as we know.

In fact in our example

958004

+ 2175194

+ 4145604

= 4224814

{rad (95800*217519*414560*422481)}2,7

= {rad (2*3*5*479*2591*140827*217519)}2,7

=

{1140531558873718710}2,7

= 5,6779260050628588354701626563897e+48

422481

4

= 31858749840007945920321

And so

{rad (95800*217519*414560*422481)}2,7

>4224814

CVD.

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


28/46

Versione 1.0

12/3/2013

Pagina 28 di 46

Also in the famous example for exponent n =5 there are several solutions such as

(5, 1, 4)

275

+ 845

+ 1105

+ 1335

= 1445

but we have

27 = 33

84 = 22

3 7 prime

110 = 2 5 11

133 = 7 19

The integers 84 and 110 are not coprime (also 27 and 84 arent also coprime) and so the rule of the

sum is not applicable!

If we apply the new abc conjecture extended at the case (5, 1, 4)we obtain

{rad (abcde)}4,45

> e

a5 +b5+c5 +d5 = e5

rad (a5b

c

5d e

5)}

4,45) = rad (abcde)}

4,45 a

4,45b

4,45c

4,45d

4,45e

4,45

a4,45

b4,45

c4,45

d4,45

e4,45

> e5

a4,45

b4,45

c4,45

d4,45

>e0.55

So certainly we have a set of finite solutions, as we know.

In fact in our example

275

+ 845

+ 1105

+ 1335

= 1445

{rad (27*84*110*133*144)}4,45

= {rad (2*3*5*7*11*19)}4,45

= {43890}4,45

=

455540332822932351085,61..

1445

= 61917364224

And so


29/46

Versione 1.0

12/3/2013

Pagina 29 di 46

{rad (27*84*110*133*144)}4,45 > 1445

CVD

We note that the number 144 is a Fibonaccis number. Furthermore, 144 = 24*6, where 24 is the

number that correspond to the modes of the physical vibrations of the bosonic strings.

A set of four positive integers a, b, c and dsuch that a2

+ b2+ c

2+ d

2= z

2is called a Pythagorean

quadruple. The simplest example is (1, 2, 2, 3), since 12

+ 22

+ 22

= 32. The next simplest (primitive)

example is (2, 3, 6, 7), since 22 + 32 + 62 = 72.

Lets apply also in this case with theabc conjecture extended

a2

+ b2+ c

2= z

2

rad (2

a2

b2

c 2z ) < or >2

z

Then the inequality may be > or


30/46

Versione 1.0

12/3/2013

Pagina 30 di 46

rad (49*196*484*729) = 2*3*7*11 = 462 < 729

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

fourth example:

42

+ 82

+ 192

= 212

rad (16*64*361*441) = 2*3*7*19 = 798 > 441

This derives from the fact that we cannot randomly choose four positive integers a, b, c and z so

always worth the Pythagorean quadruple!

We observe that the numbers 16 and 64 are multiples of 8 that is a Fibonaccis number and is

also the number concerning the modes corresponding to the physical vibrations of the

superstrings.

With regard the following case:

82

+ 122

+ 242

= 282

(144 + 64 + 576 = 784)

rad (8*12*24*28) = 2*3*7 = 42


31/46

Versione 1.0

12/3/2013

Pagina 31 di 46

4.3 Prediction of Computing Minimal Equal Sums Of Like Powers

We can also predict how many addends we want to have any solution if we increase the exponent n

If n = 6 how many addends we want?

Let's try with 5 addends

If lets consider five addends a, b, c, d and e how many pairs of numbers came out?

5

2

= 10

For example:

8 + 12 +24 + 36 + 41 = 121

This is the worst case that can happen and then we have to consider (in the case of all coprime we donot care for the calculation of probabilities because we have to take the case but worse valid).

So we can afford to have 6 pairs of integers not coprime to each other (8 and 12, 12 and 24, 8 and 24, 8

and 36, 12 and 36, 24 and 36) and 4 pairs of coprime (8 and 41, 12 and 41, 24 and 41, 36 and 41).


4

2

6

= 0,1365859.

because of the 10 couples that we have, 4 must be coprime.


q = 7,32....

a + b +c + d + e = f

{rad (abcde)}7,32

> e


32/46

Versione 1.0

12/3/2013

Pagina 32 di 46

Also here we note that 8 and 24 (and also 12 that is a sub-multiple of 24) are the numbers thatcorrespond to the modes of the physical vibrations of the superstrings and of the bosonic strings.

Furthermore, the values 0,13658 and 7,32 are very near to the values 0,13677 and 7,31104 that

correspond to the aurea frequencies based on fractional powers of Phi and Pigreco.

If we apply the new abc conjecture extended at the case (6, 1, 5) we obtain

{rad (abcde)}7,32

> e

a6

+b

6+

c

6+

d

6+ e

6= f

6

{rad (a6b

c

6d

6e

6f6)}

7,32= { rad (abcdef)}

7,32 a

7,32b

7,32c

7,32d

7,32e

7,32f

7,32

a7,32

b7,32

c7,32

d7,32

e7,32

>f -1,32

Always true.

But also for exponent n = 7 and n = 8 and five addends the inequality holds, in fact

n = 7 a7,32

b7,32

c7,32

d7,32

e7,32

>f -0,32

n = 8 a7,32

b7,32

c7,32

d7,32

e7,32

>f -0,68

So it takes 5 addends for the exponents n = 6, 7 or 8

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

The same procedure will lead to the conclusion that 6 addends are enough to representatives

n = 9, 10, 11, 12

Definitely get these addends is very difficult, because it is to manipulate huge numbers, if we take the

power of only 100012

= 1,e+36 and it is certainly likely that the numbers are above 1000.


33/46

Versione 1.0

12/3/2013

Pagina 33 di 46

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

It therefore seemsthat the conjecture of Lander, Parkin and Selfridge in 1966 that:for every n > 3, k >= n - 1

Where n is the exponent and k the number of addends is NOT VALID and WRONG.


34/46

Versione 1.0

12/3/2013

Pagina 34 di 46

4.4 Super-general trinomial equation

We can also prove the caseofsuper-general trinomialequation

Aap

+ Bbq

= Ccr

that certainly will have solutions because

rad (Aap

* Bbq

* Ccr) > Cc

r

for someA, B, C, p, q, rhavethe valuesa,b and cthatsatisfy this equation

Famous example of Ramanujan:

a3

+ b3

= 1729

rad (a3

* b3

* 1729) > 1729

In fact there are 2 solutions

123

+ 13

= 1729

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

We note that 10374 is divisible for 13 and 21 and that 51870 is divisible for 5, 13 and 21. They are

all Fibonaccis numbers.

Applying the new abc conjecture a fortiori we have

{rad (abc)} 62

> c

{rad (Aap

* Bbq

* Ccr

} 62

> Cc

r


35/46

Versione 1.0

12/3/2013

Pagina 35 di 46

always true.

5. GOLDBACHS CONJECTURE AND ITS PROOF

At this point we can also proveeasilythe Goldbach conjecture.

Given 2prime numbersa and bthe sumcoversthe entire set ofnatural numberspositive integersEVENbecause

a + b = c

rad (abc) = ab rad (c) > ab

ab > c

that is always > because a and b are the addends of the sum and therefore this is true and valid!

We have only two exceptions, but with a=b and not coprime:

1 case:

2 + 2 = 4

2*2 = 4

onlyin this onecase we have touse thecomplete formulawith rad (abc)

rad (abc) > c

rad(2*2*4) = rad(24 ) = 2 < 42 < 4

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

2 case:

3 + 3 = 6

3*3 = 9 that is > 6


36/46

Versione 1.0

12/3/2013

Pagina 36 di 46

Also in this onecase we have touse thecomplete formulawith rad (abc)

rad(abc) > c

rad(3*3*6) = rad(33

* 2) = 6 = 6

6 = 6

In all other cases, with a and b different primes the inequality with the sign > is always valid and

true.

ab > 2n for n > 3

Examples:

113 + 127 = 240

rad(113*127*24

3 5) > 240

113*127*2*3*5 = 430.530 > 240

or also

113*127 > 240

14351 > 240

We note that 240 is a multiple of 24 that is the number that represent the physical vibrations of

bosonic strings

--------------------------------------------------------------------------------------------------------------------------.

10.001.009 + 100.313 = 10.101.322

rad (10.001.009*100.313*2 7 112

67 89) > 10.101.322

10.001.009*100.313*2 7 11 67 89 > 10.101.322

10.001.009*100.313 > 10.101.322

We have,in this case, only two exceptions (iethe sign oftheradicalinequality< , but where the two

chosen numbers are equal a = b),but whatinterests us is thatthere is asign >


37/46

Versione 1.0

12/3/2013

Pagina 37 di 46

The solutionsare endless and infiniteand cover theentire setof natural numbers N+becausethesign oftheradicalinequalityis always and only >.

Obviously we have also with the new abc conjecture

{rad (abc)} 62

> c

{rad (abc)} 62

= a 6

2

b 62

( rad (c)) 62

> ab

ab > c

always TRUE.

Now we want remember, as written above, that:

( )

==

=

p p pp61607927102.0

6

2

1

1

111

2

1

22 %.

)2(1

= 2

6

= 0,6079....

where refers to the Riemann zeta function, the identity relating the product over primes to (2) is an

example of an Euler product, and the evaluation of(2) as 2/6 is the famous Basel problem.

With regard the value6

2, we have some interesting mathematical connections with some sectors of

string theory.

If a series, ( )

=

0 !n

n

n xf

n

xA , is divergent, for ax > , ( 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 . (3)

Now, we have the following relation:

= 0 !1 k

k

kx k

xB

e

x. (4)


38/46

Versione 1.0

12/3/2013

Pagina 38 di 46

Applying to this relation, the integration of eq. (3), we have that:

+===0 0 0 0 1

22

11

!

1

1 k k kkk

xk

kx

x

BBdxexk

Be

dxxe. (5)

Now we compute the integral that is to the left-hand side of the (5). 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 . (6)

Thence, from the (5), we obtain:

=1

2

22

3

6kkB

. (7)

The left-hand side of the (7) is just a divergent series, that is represented from the value of the right-

hand side of this expression.

Substituting in the (4), x to x , we have:

( )

=

0 !1 k

k

kx k

xBe

x. (8)

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

of eq. (3), we have:

( )

( )

+

=

+==

=

0 00

00

2

2

1

61

1

11 k k

kx

x

x

x

x

kdxxedx

e

xedx

e

xe

; (9)

( )( )

++==

00

0 1

22

111!

1

k k k

k

k

k

xk

k

k BBdxexkB . (10)

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

=1

2

22

3

6kkB

, (11)

that is identical to the (7).

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


39/46

Versione 1.0

12/3/2013

Pagina 39 di 46

theory which interpolates smoothly between the 1/2 BPS line or circle, principally some equationsconcerning 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 theinteracting 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 . (12)

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

( )

=2

3

2

2

2

1

4

2

2

13

2

23

2

12

11,,

wxwxwxwdxxxY

,

22

jiij xxx = . (13)

This integral is known in closed form. For 2132

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

1ln

2

1ln2lnln

AtsAtsts

2

13

2

12

x

xs = ,

2

13

2

23

x

xt= , ( ) sttsA 41 2 = . (14)

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 . (15)


40/46

Versione 1.0

12/3/2013

Pagina 40 di 46

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

two arguments ofY in (12) are less than unity and in that regime Y evaluates to

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

++++=

iiiii

zeezezzeLizeLiz

iY 1loglog1loglog1

6sin22

2

. (16)

The integration then gives ( )( )

+=

1

0 sin3

dzY . (17)

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


41/46

Versione 1.0

12/3/2013

Pagina 41 di 46

( )

( )=

+==

=

+

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

( ) ( ) ( ) ( )

+

+

=

32

2

2

2

32

2

363

sin

coscos4

ieLiieLi

ii . (21)


42/46

Versione 1.0

12/3/2013

Pagina 42 di 46

6 WARING'S PROBLEM

In the problem of Waring anypositive integercanbewritten as a sumof spowersk-th of integers. Itis

important to find whichisthe maximum numbersthatgivesthe possibilityto write everypositive

integer.

For eachdegree k, theminimumnumber sthatoccursthe problemofWaringisdenoted byg(k).

Obviouslyg(1) =1.

To write, for example, the number 7how many squaresare needed?

It takes 4, in fact:

7 =22 +1+1+1

In this casewe havethat g(2) = 4

Of course, towriteall the othernatural numberscan servea maximum of 4squares.

Thismeans thatto writeall othernatural numbersg(2) 4, or 2, 3 or4 squares.

Let's apply the extended radical

a2

+b2

+c2

+d2

= e

rad (a2

b2

c2d

2e) = rad (abcde) > e

always trueNote that inthis casethe sumisalwayscertainlydefined, theresultvalue, andin our case e, does not

belongto the groupof squaresn2

.

Thus, there arepositive integers"critical"that requirea maximum number ofk-th powers.

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

How many cubes does it take to write 23?

23 =3

2 +3

2 + 1 + 1 + 1 + 1 + 1 + 1 + 1

We observe thatwe will have9 terms and g (3) = 9

Of course, towriteall the othernatural numbers can servea maximum of 9cubes.

Let's apply the extended radical

a12

+a22

+a32

+a42

+a52

+ a62

+a7

2+ a8

2+ a9

2 =e

rad (a12

a22

a32

a42

a52

a62

a72

a82

a92

e) = rad (a1a2a3

a4a5 a6

a7a8 a9

e) > e

always trueAlso inthis casethe sumisalwayscertainlydefined, theresultvalue, andin our case e, does not

belongto the groupof squaresn2

.


43/46

Versione 1.0

12/3/2013

Pagina 43 di 46

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

Insteadto write79as the sum ofthe fourthpowersare neededmore than 19fourth powers:

79 =42 +

42 +42 +

42 + 1*15 (fifteen 1)

Sowe haveg (4) = 19

Of course, towriteall the othernatural numbers can servea maximum of19fourth powers.

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

The rule to find g (k) is the following:

g(k) =k2 + integer part

k

2

3- 2

It is clear thatapplyingtheextended radical we have alwayssolutions.

In fact, themaximumnumber oftermsisgiven by the formulag(k) and the Warings problem is thussolved.

Insteadthe ruleto find theintegers "critical" can be easily deduced.

Areall betweenk2 and

k3 , more preciselythecritical numberforthefifthpowerisgiven by:

integer part ( 5

5

2

3) 1 = integer part (

32

243) 1 7 1 = 6

223 = (52 )*6 + 1*31 = 2

5+ 2

5+ 2

5+ 2

5+ 2

5+ 2

5+ 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 +

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 223

It takesthen37fifthpowersandg (5) = 37

Of course, towriteall the othernatural numbers can servea maximum of37fifthpowers.


44/46

Versione 1.0

12/3/2013

Pagina 44 di 46

The general ruleto find the numbersCritical"x for eachdegree kis as follows:

x = integer part (k

)2

3( 1))*

k2 +k2 - 1

Example

For thesixthpower, by applying the rule, we have:

703 = ( 62 )*10 + 1*63

Or73sixthpowers and g(6) = 73

Of course, towriteall the othernatural numbers can servea maximum of73sixthpowers.

Thetables1 and 2 belowgivesthecritical numbersaccording to the degree.


45/46

Versione 1.0

12/3/2013

Pagina 45 di 46

TAB. 1:

Degree kCritical

number x

1

2 7

3 23

4 795 223

6 703

7 2175

8 6399

9 19455

10 58367

11 176127

12 528383

13 1589247

14 476774315 14319615

16 42991615

17 129105919

18 387186687

19 1161822207

20 3486515199

TAB. 2

x(k) 1 7 23 79 223 703 2175 6399 19455 58367 ...

g(k) 1 4 9 19 37 73 143 279 548 1079 ...

k 1 2 3 4 5 6 7 8 9 10 ...


46/46

Versione 1.0

12/3/2013

Pagina 46 di 46

References

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

1) IL CONCETTO MATEMATICO DI ABBONDANZA E IL RELATIVO

GRAFICO PER LA RH1 - Francesco Di Noto, Michele Nardelli (Gruppo B. Riemann)

2) NOVITA SULLA CONGETTURA DEBOLE DI GOLDBACH - Gruppo B.Riemann

Francesco Di Noto,Michele Nardelli

3)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,

7) Una nota sulle serie divergenti e loro utilizzazione Pasquale Cutolo -

http://www.maecla.it/Matematica/PASCUT3.pdf (see also: La Comunicazione - numero unico

2008-2009)

8) Generalized quark-antiquark potential at weak and strong coupling Nedav Drukker and

Valentina Forini arXiv:1105.5144v1 - 25/05/2011

Documents

Pier Francesco Roggero, Michele Nardelli, Francesco Di Noto - "A possible proof of Fermat's Last Theorem"