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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.
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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
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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.
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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
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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 >
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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.
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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
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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
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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.
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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.
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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 + .
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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
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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!
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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.
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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
++
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... ...
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
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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
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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.
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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.
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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).
Then the probability is
3
2
6
= 0,2246.
because of the 6 couples that we have, 3 must be coprime.
So the exponent q is the radical and we have:
q = 4,45....
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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.
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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
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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.
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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
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{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
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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
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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).
Then the probability is
4
2
6
= 0,1365859.
because of the 10 couples that we have, 4 must be coprime.
So the exponent q is the radical and we have:
q = 7,32....
a + b +c + d + e = f
{rad (abcde)}7,32
> e
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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.
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----------------------------------------------------------------------------------------------------------------------------
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.
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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
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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
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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 >
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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)
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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
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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)
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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
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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
( ) ( )( )
=
+
+
+
+=
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)
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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
.
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--------------------------------------------------------------------------------------------------------------------------
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.
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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.
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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 ...
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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