pier francesco roggero, michele nardelli, francesco di noto - proof that the primes of wieferich are...

7/29/2019 Pier Francesco Roggero, Michele Nardelli, Francesco Di Noto - PROOF THAT THE PRIMES OF WIEFERICH ARE FINITE

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1

PROOF THAT THE PRIMES OF WIEFERICH ARE FINITE IN NUMBER AS THEY ARE

FINISHED IN THE OTHER GROUPS ap

b

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

Abstract

In this paper we show that the primes of Wieferich are finite in number as they are finished in the other

groups ap

b.


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Index:

1 Groupsp

a b...................................................................................................................................... 3

1.1 Improvement of the Fermat's little theorem and rule to find the divisors ofp

a b ....... 4

1.2 Divisors

n

p of the groups

p

a b........................................................................................ 91.3 Relation (

pa - 1)*( pa + 1) = ( 2pa - 1) ................................................................................122 Wieferich primes................................................................................................................................13

2.1 Existence of divisors 2p of Wieferich and non-existence of divisors p in the group p2 -

1 .................................................................................................................................................16

2.2 Other cases of prime numbers Wieferich in the groupspa b......................................17

2.3 Divisor419 in the group

1-p2819 ..................................................................................19

2.4 Proof that the primes of Wieferich are finite in number as they are finished in the other

groupspa b ............................................................................................................................20

2.5 Exponents special p in the groupspa b similar to Wieferich primes ..........................21


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1 Groupsp

a b

pa b

p 1 integer

a 2 integer also not prime

b integer, we must choose for (p

a b) give, at the beginning for p = 1, a prime number.

It is therefore not necessary to have a and p prime, as it was for Mersenne numbers.

Obviously, no one group can give more consecutive prime numbers.

Let's see what are the dividersp

a b

As a first divisor d for p = 1 we have:

d = a b

It is divisible by all the exponents that are equal to:

p = k(a+b-1)+1

Chosen any s 2, the other divisors are:

d =s

a b

Now if this number is prime its periodicity is given by:

p = k(s

a b 1) + s

If instead the number is composed of the prime factors f1, f2, f3, ... fn, have given periodicity by:

p = k(fn - 1) + s


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1.1Improvement of the Fermat's little theorem and rule to find the divisors of pa b

Given a generic groupp

a b to find its divisors is performed as follows::

- We choose a p, not even prime- We breaks down the number pa b into its prime factors and so are all divisors of pa b: d1,

d2, ... dn

Now the periodicity forp

a b, instead, is given by:

chosen for any s 1 fors

a b

p = k(s

a b 1) + s

and if it is true that dn is a divisor of p-1 we have:

p = kdn + s

with

p divisor ofskdn

a + b; with dn divisor of p - 1

This is more powerful than the Fermat's little theorem which states that1-p

a b is divisible by p,

because also dividers of p-1, as dn, may give rise todn

a b divisible by p.

When dn = p-1 coincides with the Fermat's little theorem.


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Examples:

p12 + 5

For p = 1 is divisible by 17 and for all the exponents equal to:

p = 16k+1

or:

12 + 5 = 171712 + 5 = 22186111067404369973312 + 5 = 4,101862702460022253364261035935e+354912 + 5 = 7,5836984583351248111063210627855e+52

And so on

Are all numbers divisible by 17.

We note that 16 is a multiple of 8 and that this number is a Fibonaccis number and it is connectedwith the modes that correspond to the physical vibrations of a superstring by the following

Ramanujan function:

( )

++

+

=

4

2710

4

21110

log

'

142

'

cosh

'cos

log4

3

18

2

'

'4

0

'

2

2

wtitwe

dxex

txw

anti

w

wt

wx

. (1)

Now consider for p = 4 the number:

412 + 5 = 20741 = 7*2963

In this case all the exponents:


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p = 6k + 4 are divisible by 7p = 1481k + 4 are divisible by 2962

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

p3 + 2


p = 4k+1

or:

3 + 2 = 55

3 + 2 = 2459

3 + 2 = 1968513

3 + 2 = 1594325

And so on


Consider now for p = 5 the number:

53 + 2 = 245 = 5*7*7

All the exponents:

p = 6k + 5 are divisible by 7

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

p2 + 1


p = 2k+1, or all for all odd exponents


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then:

2 + 132 + 152 + 172 + 1

And so on


Consider now for p = 5 the number:

52 + 1 = 33 = 3*11

All the exponents:


Now consider for p = 4 the number:

42 + 1 = 17

All the exponents:


But beware! For p = 12 we rediscover the number 17:

122 + 1 = 4097 = 17*241

All the exponents:



Then by combining the periodicity of two numbers 17, which has its effective periodicity is given by:




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and we have:

p = 8k + 4 because it repeats for p = 4, 12, 20, 28,

This means that:

48k2 + + 1 is divisible by 17 for the exponents 4, 12, 20, 28,

Also here we note that there is the Fibonaccis number 8 that is also connected with the modes that

correspond to the physical vibrations of a superstring 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

.


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1.2Divisors np of the groups pa b

We have seen that the separate divisors of a group ofp

a b are given by the following formula:

p divisor ofsk(dn)

a + b; with dn divisor of p - 1

If p is a divisor of1-p

a b then 2p is a divisor of1)-p(p

a b.

In general we have that np is a divisor of1)-(pp 1-n

a b.

If there exists dn a divisor of p - 1, the above formula can be simplified in this way:

np is a divisor of)(p 1-na

dn b

In the specific case of 2p we have:

2

p is a divisor of

)p(

a

dn

b

The periodicityp

a b, instead, is given by:

chosen for any s fors

a b, as the divisor np we have

p = k(s

a b)n-1

(s

a b 1) + s

and if exists dn a divisor of p -1 we have:

p = k(p)n-1

(dn) + s

with

np divisor ofsdn +)(k(p)

1-n

a b; with dn divisor ofi p - 1

In the specific case of 2p we have:


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2p divisor ofsdn +)(kpa b

Observations.

Each groupp

a b does not contain NEVER all prime factors taken separately or higher at a power.

This is because each group is only a subset of the set of natural numbers N.

Its for this reason that, for example, in the group of Mersenne p2 - 1 never exist distinct prime factors

in 1093 and 3511 but only their squares and respectively for p = 364k and for p = 1755k

Even in the group of Fibonacci never exists factor 22 , but all other powers n2 with n = 1, 3, 4, 5, 6 ...

are present including so obviously 2.


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Example:

For the groupp

3 - 2 we have that is divisible by 5 with periodicity p = 4k + 3 (in this case dn = p -1 =

4 and s = 3)

In fact:

skdna + b

With k=0 we have

33 - 2 = 25 divisible by 5 but also for

25 .

Here is why:

It is divisible by2

5 with periodicity p = 20k + 3 because

p = k(dn+1)n-1

dn + s = 5*4k + 3

In fact:

2p divisor ofsdn ++1)k(dna b

With k=1 we have

233 - 2 = 94143178825 divisible by 25

It is noted, therefore, that in the group p3 - 2 exists before the divisor 25 for p = 3 and then the divider

5 for p = 7.


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1.3 Relation ( pa - 1)*(p

a + 1) = (2p

a - 1)

A curious relationship, but very useful in the calculations is as follows:

(p

a - 1)*(p

a + 1) = (2p

a - 1)

This means that to find, for example, the factorization of (p

a + 1) is enough to knowthe factorization

of(2p

a - 1) and to removethat of(p

a - 1).

It can, of course, also find the factorization of(2p

a - 1) knowing that of(p

a - 1) and add tothat of(p

a

+ 1)

Example:

( 182 - 1) = 262143 = 33

7 19 73

( 182 + 1) = 262145 = 5 13 37 109

And then you get free factorization of:

( 362 - 1) = 68719476735 = 33

5 7 13 19 37 73 109


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2 Wieferich primes

A prime of Wieferich is a prime number p such that 2p divides 1-p2 - 1.

The only Wieferich primes are known 1093 e 3511.

Apply the rule of paragraph 1.1 and we have:

Consider the first prime 1093.

1092 = 22

3 7 13

We take the divisor 364

The periodicity of the exponents is given considering p2 - 1 by p = 364k

364k2 - 1 is divisible by2

1093

Or for all the exponents 364, 728, 1092, 1456, 1820 is divisible by per2

1093

since exists 1092, 1093 is a Wieferich prime:

10922 - 1 is divisible by2

1093

Mind you but already with exponent 364 we have that 3642 - 1 is divisible by2

1093

Also in paragraph 1.3 will have:

( 1822 - 1)*( 1822 + 1) = ( 3642 - 1)

Since 21093 is a divisor of( 3642 - 1) it cannot be even of ( 1822 - 1) because it is not on the list ofexponents of 364k2 - 1.

But then, inevitably,2

1093 is a divisor of( 1822 + 1).

It has, therefore, that for p2 + 1 the exponents p that allow the division by2

1093 are:

p = 364k+182

Or for all the exponents 182, 546, 910, 1274, 1638. 2002


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182364k2 + + 1 is divisible by 21093

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

Let us now consider the second Wieferich prime 3511.

3510= 2 33

5 13


The periodicity of the exponents is given considering p2 - 1 by p = 1755k

1755k2 - 1 is divisible by2

3511

Or for all the exponents 1755, 3510, 5265, 7020, is divisible by 23511

since exists 3510, 3511 is a Wieferich prime:

35102 - 1 is divisible by2

3511

Mind you but already with exponent 1755 we have that 17552 - 1 is divisible by2

3511

Also in paragraph 1.3 will have:

( 17552 - 1)*( 17552 + 1) = ( 35102 - 1)

Since2

3511 is a divisor of( 17552 - 1) it cannot be even of ( 17552 + 1) because it is already of ( 35102 -

1).

In this case for p2 + 1 it doesnt EXIST a divisor as 3511, much less2

3511 , that these two numbers

never appear in the prime factors of that group because the periodicity for 1755k2 - 1 is 1755k, which is

odd because 1755 is not divisible by 2.

Since we have:

Divisors of p2 - 1 of the form ak


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Divisors of p2 + 1 of the form ak +2a

now ifa = 1755,2

acannot give an integer.


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2.1Existence of divisors 2p of Wieferichand non-existence of divisors p in the group p2 - 1

Considering the paragraphs 1.1 and 1.2 we have:

(1) p divisor of k(dn)2 - 1; with dn divisor of p - 1and

(2) 2p divisor of )(kp2 dn - 1; with dn divisor of - 1

For Wieferich primes we DONT HAVE divisor p but only the divisor 2p in the group p2 - 1

and since 2p COINCIDES with p we have that:

2p divisor of k(dn)2 - 1; with dn divisor of p - 1

That is the rule (2) disappears and IS NOT MORE VALID.


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2.2 Other cases of prime numbers Wieferich in the groups pa b

For the group1-p

3 - 1

It has that 2p divides it by p = 11 and p = 1006003

We consider only the first number 11.

10 = 2 5


The periodicity of the exponents is given consideringp

3 - 1 by p = 5k

5k3 - 1 is divisible by 211

Or for all the exponents 5, 10, 15, 20, 25 is divisible by 211

Being present 10, 11 is a prime number of Wieferich of the group1-p

3 - 1

103 - 1 is divisible by 211

Mind you but already with exponent 5 we have that5

3 - 1 that is divisible by 211

In the groupp

3 - 1 there is NEVER a factor 11 taken individually.

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

For the group1-p

19 - 1

It has that 2p divides it by p = 3, 7, 13, 43, 137, 63061489

We consider only the first number 11 3.

2 = 2 1


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We take the only divisor 2


19 - 1 by p = 2k

2k19 - 1 is divisible by

23

Or for all the exponents 2, 4, 6, 8, 10, 12 is divisible by2

3


19 - 1:

219 - 1 is divisible by 23 (indeed 19 * 19 1 = 361 1 = 360; and 360/9 = 40)

In the groupp

19 - 1 there is NEVER a factor 3 taken individually.


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2.3 Divisor4

19 in the group1-p

2819

For the group1-p

2819 - 1

It has that 4p divides it by p = 19

We consider prime number 19.

18 = 2 3^2 (2 * 32)



2819 - 1 by p = 3k

3k2819 - 1 is divisible by 419

Or for all the exponents 3, 6, 9, 12, 15, 18 is divisible by 419


2819 - 1

182819 - 1 is divisible by 419

Mind you but already with exponent 3 we have that3

2819 - 1 that is divisible by 419

In the groupp

2819 - 1 there is NEVER a factor 19 taken individually. Even the factors 219 and319 cannot appear.


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2.4Proof that the primes of Wieferich are finite in number as they are finished in the other groupsp

a b

In all groupsp

a b we have a factorization into prime factors for increasing p, but they are NOT all

individually.

For example, in the group

1-p

2 - 1 of Wieferich we have 2 divisor

2

p that we know to be p = 1093 e3511.

There arent the prime factors p = 1093 e 3511 taken individually in the group 1-p2 - 1.

This is because the group p2 - 1 results in increasing numbers p for which, of course, only a subset of

the entire set of natural numbers.

It is true that the group 1-p2 - 1 or p2 - 1 contains all the factors and prime numbers but are NOT all

individually.

This rule also applies to all other groups pa - 1 and more generally for the groups pa b, as we have

seen in the previous examples.Rather, there are groups, such as

1-p2819 - 1, which has as a divisor 419 and there are never any 19,

219 e 319 .

Only when we consider all groups pa b we have the whole set of natural numbers and then we have

all the factors and prime numbers individually arising from the factorization.

Each grouppa b therefore contains "anomalies" or numbers 2p , 3p , 4p . but inside they

CANNOT EVER be present all the factors or primes individually.

In addition, each groupp

a b contains a finite number of dividers2

p ,3

p ,4

p because, beinginfinite groups pa b, we can find all prime factors p of the set of natural numbers.

If the group 1-p2 - 1 contains two primes divisor 2p , and the group1-p

3 - 1 contains two other 2p

and the group1-p

19 - 1 contains six divisor 2p , then in the totality of all groups pa - 1 or better pa

b means that having available ALL the natural numbers is also has the infinity of prime numbers or

factors resulting.

Its for this reason that within each group pa - 1 or pa b has only a finite number, for example, of2p and that lack primes p taken individually but that we find in other groups pa - 1 or pa b to

complete the entire panorama of infinite factors or primes of all natural.


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If the divisors, for example, 2p were infinite in the group of Wieferich p2 - 1 we wouldnt find2p in the other groups pa b and this would be an absurd.

2.5Exponents special p in the groupsp

a bsimilar to Wieferich primes

Consider the group:

p3 - 2

and

2p-3 - 2

Is divisible by2

5 with p = 5

In fact:

2-53 - 2 = 25 is divisible by

25

The periodicity is given, considering the groupp

3 - 2, by p = 20k+3:

Or for all the exponents 3, 23, 43, is divisible by2

5 .

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

Consider the group:

p3 + 2

and

2p-3 + 2

Is divisible by2

7 with p =7


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In fact:

2-73 + 2 = 245 is divisible by

27

The periodicity is given, considering the groupp

3 + 2, by p = 42k+5:

Or for all the exponents 5, 47, 89, is divisible by2

7 .

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

Consider the group:

p2 + 1

Is divisible by 23 with p = 3

In fact:

32 + 1 = 9 is divisible by 23

The periodicity is given, considering the group p2 + 1, by p = 6k+3.

Or for all the exponents 3, 9, 15, 21 is divisible by 23 .

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

Consider also the group:

p2 + 1

And ask if it is divisible by 3p

Is divisible by 33 with p = 23

In fact:

232 + 1 = 513 is divisible by 33

The periodicity is given, considering the group p2 + 1, by p = 18k+9.


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Or for all the exponents 9, 27, 45, 63 is divisible by 33 .

Each group ap

b therefore contains the "anomalies" or numbers divisors of the group p2, p

3, p

4, .... but

inside they CAN NOT EVER be present all the factors or prime numbers p individually. This happens

and it is intrinsic to all groups ap

b.

If in a group like the most famous 2^(p-1)-1 there are anomalies or irregularities, or there are two

prime numbers that divide it only for p^2 and never for p, this could mean that in a string may also

exist vibration modes that nobody imagined, for example with frequency quadratic but NOT as

frequency SINGLE. These may give rise to strange particles.

References:

1) PROOF THAT THE FERMAT PRIME NUMBERS ARE ONLY THE FIRST FIVE AND

ALL THE OTHER NUMBERS ARE COMPOSITE.Pier Francesco Roggero, Michele

Nardelli1,2, Francesco Di Noto

2) 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 stringaFrancesco Di Noto, Michele Nardelli

3) Sul Problema di Brocard (versione estesa n!+k = m^2 quadrato perfetto)

Gruppo B. RiemanFrancesco Di Noto, Michele Nardelli

Tutti sul sito http://nardelli.xoom.it/virgiliowizard/

pier francesco roggero, michele nardelli, francesco di noto - proof that the primes of wieferich are...

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