Pier Francesco Roggero, Michele Nardelli, Francesco Di Noto "Proof of the Diophantine Equation...

7/30/2019 Pier Francesco Roggero, Michele Nardelli, Francesco Di Noto "Proof of the Diophantine Equation Involving Factorial"

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PROOF OF THE DIOPHANTINE EQUATION INVOLVING

FACTORIAL

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

Abstract

In this paper we try to show diophantine equations involving factorial.


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Index:

1. RADICAL OF AN INTEGER ............................................................................................................. 32. UPPER LIMIT FOR ABC CONJECTURE AND ITS PROOF .......................................................... 43. CONNECTION FERMATS LAST THEOREM WITH THE ABC CONJECTURE ........................ 94. THE BROCARD-RAMANUJAN DIOPHANTINE EQUATION x! + 1 = y2 .................................10

4.1 PROOF OF THE GENERAL EQUATION x! + 1 = yk

......................................................124.2 LEMMA ...............................................................................................................................14

5. DABROWSKI DIOPHANTINE EQUATION x! + A = y2

OR A VARIANT OF BROCARDDIOPHANTINE EQUATION x! + A = y

k............................................................................................15

6. ANOTHER VARIANT OF BROCARD DIOPHANTINE EQUATION x! zk =y

k.......................17

6.1 OTHER VARIANT OF BROCARD DIOPHANTINE EQUATION x! zn

= yk..............24

6.2 OTHER VARIANT OF BROCARD DIOPHANTINE EQUATION x! y = yk...............26

6.3 OTHER VARIANT OF BROCARD DIOPHANTINE EQUATION x! z! = yk..............29

7. APPLICATIONS OF FERMAT-CATALAN CONJECTURE.........................................................327.1 CONNECTION WITH FERMATS LAST THEOREM ....................................................337.2 CONNECTION WITH EQUATION INVOLVING FACTORIAL ....................................35

8. CONCLUSIONS................................................................................................................................36


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1. RADICAL OF AN INTEGER

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

considering 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.

This means that there isnt, always, the following rule

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

This applies only if three numbers a and b are coprimes (cconsequently is also coprime).

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


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2. 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.

We first have to know when given two randomly chosen integers a and b, how likely it is that a

and b are coprime. In this determination, it is convenient to use the characterization that a and b

are coprime if and only if no 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 is 1 - . 21

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 pp%61607927102,0

6

2

1

1

111

2

1

22 .


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)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 result of famous Basel problem.

In general, the probability of k randomly 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 rigorousformalization 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 PNapproaches

2

6

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

The inequality

K {rad (abc)}1+

> c

becomes

{rad (abc)} 6

2

> c

where6

2= 1,644934...

So to be precise we have that

K= 1


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=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 + .

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 1.6260 112 325673 22123 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


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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

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

CVD

We know there are exponential bounds are known.

Specifically, the following bounds have been proven:

( )( )( )198615

1exp abcradKc < ,

( )( )1991

3

2

2exp

c

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


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3. CONNECTION FERMATS LAST THEOREM WITH THEABC CONJECTURE

an

+bn

=cn

rad (an

bnc

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

2

6

< c2

18

because ab > c

since the upper limit is given by

{rad (abc)} 62

> c

cn

< c26

218

= c3

so if n 3 we cannot have solutions!

This is the proof ofFermats last theorem.


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4. THE BROCARD-RAMANUJAN DIOPHANTINEEQUATION x! + 1 = y2

It is believed or conjectured that the Brocard-Ramanujan diophantine equation

x! + 1 = y2

has only finitely many integral solutions (x, y).

Unaware of Brocards work, Ramanujan proposed the same problem.

The three known solutions are

(x, y) = (5, 4), (11, 6), (71, 7)

4! + 1 = 52

= 25

5! + 1 = 112

= 121

7! + 1 = 712

= 5041

Using crafty computer programming, Berndt and Galway claim that there are no other solutions for n

109

That is one billion factorial!

Stirlings approximation says

x! ~

x

e

xx

2

So

109! ~

109

9

9

10102

e > 109 (108)

910 > 10 410*89 +

This implies that 109! has at least eight billion and four digits.

Assuming the number of atoms in our universe stays constant, it is suspected the universe contains

1080

atoms. So, take the number of suspected atoms in the universe and raise it to the 100,000,000

power, and we have an approximation that is smaller than 109!.


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The number is unfathomable.

In particular, the diophantine equation

x! + 1 = yk

has no solutions fork > 1, except possibly fork = 2 andk = 4.

But also in the casek = 4, namely for the equationx4 1 =n!, there are actually no solutions.

Then is believed that the equation has no solutions if k 3

However, the seemingly simplest case, of the Brocard-Ramanujan Diophantine Equation

x! + 1 = y2

remained open.

Further evidence that this equation has only finitely many solutions.


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4.1 PROOF OF THE GENERAL EQUATION x! + 1 = yk

x! + 1 = yk

has only finitely many integral solutions (x, y) for k =2.

Let s apply the new abc conjecture

a + b = c

{rad (abc)} 62

> c

This inequality is a proof of abc conjecture because it holds for ALL triples (a,b,c) of coprime

positive integers, witha +b =c

rad (x! yk) = rad (x! y) = rad( n

x

primen

=

* y) nx

primen

=

* y

where

nx

primen

=

is the product of only distinct prime factors or the PRIMORIAL n

(example rad (4!) = rad (2*3) = 6)

{ nx

primen

=

y} 62

> yk

Since


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rad( nx

primen

=

) = n en

(n is the largest prime so that n x)

substituting

{ en

y} 62

> yk

so we have a threshold for k, that depends on nx

primen

=

or x! or better n enand we prove that k is

bounded.

Certainly k = 2 as the solution is fine and does work.

It is clear that more the value of{( nprimen

=

) y} 62

is larger, k can be large.

If

{( nprimen

=

) y} 62

= y3

Then

y3

> yk

k < 3 or k = 2 we have finite solutions


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4.2 LEMMA

Since we have used the new abc conjecture

{rad (abc)} 62

> c

and not the weakabc conjecture

{rad (abc)}2

> c

We can assume for

x! + 1 = yk

{( nprimen

=

) y} 62

> y

k

and not

{( nprimen

=

) y}2

> yk

that would provide no solutions, either k = 2


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5. DABROWSKI DIOPHANTINE EQUATION x! + A = y2 OR AVARIANT OF BROCARD DIOPHANTINE EQUATION x! + A = y

k

Dabrowki, a Polish mathematician, states that

x! + A = y2

where A is an integer which isnt a perfect square, has only finitely many solutions (x, y).

Let s apply the new abc conjecture, but with general case

x! + A = yk

rad (x! Ayk) = rad (x! Ay) = rad( n

x

primen=

*Ay) n

x

primen=

* Ay

where

nx

primen

=

= n

{ nx

primen

=

Ay} 62

> yk

Since

rad( nx

primen

=

) = n en


substituting


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{ en

Ay} 62

> yk


primen

=

or x! or better n en

and depends on A, and

we prove that k is bounded, but in this case since A can take any value, eccept a perfect square,

the exponent k can assume k 3

It is clear that more the value of{( nprimen=

) Ay} 62


Conclusions:

Replacing 1 with an integer constant A does not change anything for the demonstration given before.

k is always boundedand certainly k = 2 or k 3 as the solution is fine and does work.

What is important to note, however, is that, once set the value of A, we have only finitely many

solutions.


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6. ANOTHER VARIANT OF BROCARD DIOPHANTINE

EQUATION x! zk =yk

We can also considered, more generally, the diophantine equation

x! zk

=yk

We know that most of these have only finitely many solutions.

Let s apply the new abc conjecture, it can also be applied to the difference and then + or -

rad (x!zk

yk) = rad (x! zy) = rad( n

x

primen

=

*zy) nx

primen

=

* zy

where

nx

primen

=

= n

{ nx

primen

=

zy} 62

> yk

Since

rad( nx

primen

=

) = n en


substituting


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{ en

zy} 62

> yk

so we have a threshold for k, that depends on or x! or better n enand depends on z, and we prove

that k is bounded.

It is clear that more the value of{( nx

primen

=

) zy} 62


Since

z < y

from par. 3 we have

rad (z) = y 62

{ en

zy} 6

2

{ en

y

2

61

+

} 6

2

If

{ en

y2

61

+

} 62

= y3

Then

y3 > yk

k < 3 or k = 2 we have finite solutions.

Conclusions:

k is always boundedand certainly k = 2 as the solution is fine and does work.

In fact we have the following solutions with k =2


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4! + 52

= 72

= 49

5! + 72

= 132

= 169

5! + 132

= 172

= 189

5! + 292

= 312

= 961

and with a difference

6! - 122

= 242

= 576

We note that 576 = 242

and that 24 is related to 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

wtitwe

dxex

txw

anti

w

wt

wx

.

We have said, above, that

rad (z) = y 62

; { en

zy} 62

{ en

y2

61

+

} 62

(A)

if

{ e

n

y

2

61

+

} 6

2

= y

3

(B)

then

y3

> yk



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With regard the value6

2 , (see eqs. A and B) we have also some interesting mathematical connections

with some sectors of string theory.

If a series, ( )

=0 !n

n

n xfn

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

0

0

!n

xnnx dxex

n

Adxxfe . (6.1)

Now, we have the following relation:

= 0 !1 k

k

kx k

xB

e

x. (6.2)

Applying to this relation, the integration of eq. (6.1), we have that:

+===0 0 0 0 1

22

11

!

1

1 k k kkk

xk

kx

x

BBdxexk

Be

dxxe. (6.3)

Now we compute the integral that is to the left-hand side of the (6.3). 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 . (6.4)

Thence, from the (6.3), we obtain:

=1

2

22

3

6kkB

. (6.5)

The left-hand side of the (6.5) is just a divergent series, that is represented from the value of the right-

hand side of this expression.

Substituting in the (6.2), x to x , we have:

( )

=

0 !1 k

k

kx k

xB

e

x. (6.6)

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

of eq. (6.1), we have:


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( )

( )

+

=

+==

=

0 00

00

2

2

1

61

1

11 k k

kx

x

x

x

x

kdxxedx

e

xedx

e

xe

; (6.7)

( )( )

++==

00

0 1

22

111

!

1

k k k

k

k

k

xk

k

k BBdxexk

B . (6.8)

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

=1

2

223

6kkB , (6.9)

that is identical to the (6.5).

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 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 . (6.10)

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

( )

=2

3

2

2

2

1

4

2

2

13

2

23

2

12

11,,

wxwxwxwdxxxY

,

22

jiij xxx = . (6.11)

This integral is known in closed form. For 2132

23

2

12 , xxx < it is equal to


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( ) +

+

+=

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 = . (6.12)

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 . (6.12b)

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

two arguments ofY in (6.10) are less than unity and in that regime Y evaluates to

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

++++=

iiiii zeezezzeLizeLi

z

iY 1loglog1loglog1

6sin22

2

. (6.13)

The integration then gives

( )( )+

=1

0 sin3

dzY . (6.14)

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


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The first integral in (6.10) can also be done analytically. Again one should take care in choosingbranch 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 . (6.16)

Thence, form (6.10), we have the following expression:

( )( )( )

=

+

+

+

+=

02

2

2 log

1

1log

sin

coscos,

i

i

i

i

lad

ez

ez

ze

ze

z

dzV

( ) ( ) ( ) ( )

+

+

=

32

2

2

2

32

2

363

sin

coscos4

ieLiieLi

ii . (6.16b)

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

( ) ( ) ( ) ( )

+

+

=

32

2

2

2

32

2

363

sin

coscos4

ieLiieLi

ii . (6.17)


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6.1 OTHER VARIANT OF BROCARD DIOPHANTINE EQUATION x! zn

= yk

x! zn =yk

We know that most of these have only finitely many solutions.

Let s apply the new abc conjecture, it can also be applied to the difference and then + or -

rad (x!zn

yk) = rad (x! zy) = rad( n

x

primen

=

*zy) nx

primen

=

* zy

and we the same as in par. 6

{ en zy} 62

> yk

so we have a threshold for k, that depends on or x! or better n enand depends on z, and we prove

that k is bounded.


primen

=

) zy} 62


If

{ en

zy} 62

= y3

Then

y3

> yk


But for the symmetry of the equation we can have also n = 2.


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since Stirlings approximation is

x! ~

x

e

xx

2 ex

x

e

x

From the proof of Fermat-Catalan we know that

2

6

= 0,6079 y

k

Since

rad( nx

primen

=

) = n en


substituting


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{ en y} 62

> yk


primen

=

or x! or better n en, and we prove that k is

bounded.


primen=

) y} 62


If

{ en

y} 62

= y4

Then

y4

> yk

k < 4 or k = 2 or k = 3 we have finite solutions

Conclusions:

Replacing 1 with an unknown variable y does not change anything for the demonstration given before.

k is always boundedand certainly k = 2 and k = 3 as the solution is fine and does work.

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

Example:

with

x! - y = y2

2! 1 = 12


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3! 2 = 22

2! + 2 = 22

3! + 3 = 32

with

x! + y = y3

2! 1 = 13

3! + 2 = 23

4! + 3 = 33

5! + 5 = 53

6! + 9 = 93


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6.3 OTHER VARIANT OF BROCARD DIOPHANTINE EQUATIONx! z! = yk

Another related equation

x! z! =yk

We know that we can have only finitely many solutions.

Let s apply the new abc conjecture, doesnt matter.

x! - y = yk

rad (x! z! yk) = rad (x! z! y) = rad( n

x

primen

=

nz

primen

=

*y) MAX( nx

primen

=

nz

primen

=

) * y

where

MAX( nx

primen

=

nz

primen

=

) is the maximum among the values of x and z

n

{ nxorz

primen

=

y} 62

> yk

Since

rad( nxorz

primen

=

= n en

(n is the largest prime so that n x or z)

substituting


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{ en

y} 62

> yk

so we have a threshold for k, that depends on MAX( nx

primen

=

nz

primen

=

) or better en

(n is the largest prime

so that n x or z), and we prove that k is bounded.

It is clear that more the value of{ nxorz

primen

=

y} 62


If

{ nxorz

primen

=

y} 62

= y3

Then

y3

> yk

k < 3 or k = 2 we have finite solutions

Conclusions:

Replacing 1 with a factorial z! does not change anything for the demonstration given before.

k is always boundedand certainly k = 2 as the solution is fine and does work.

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

Examples:

Apart from the solutions already known

1! + 4! = 52

1! + 5! = 112

1! + 5! = 712


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we have

4! + 5! = 122

Alos here, we note that 122

= 144 = 24 * 6, where 24 is related to 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

wtitwe

dxe

x

txw

anti

w

wt

wx

.


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7. APPLICATIONS OF FERMAT-CATALAN CONJECTURE

From the FermatCatalan equation

am

+ bn

= ck

1)If

knm

111++ > 1

We have infinite solutions inside each of these (2, 2, k), (2, 3, 3), (2, 3, 4), (2, 3, 5).

2)If

knm

111++ = 1

We have finitely many solutions inside each of these (2, 4, 4), (2, 3, 6).

From the proofobtained from the new abc conjecture we have

3) 26

= 0,6079

Pier Francesco Roggero, Michele Nardelli, Francesco Di Noto "Proof of the Diophantine Equation...

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Transcript of Pier Francesco Roggero, Michele Nardelli, Francesco Di Noto "Proof of the Diophantine Equation...