PROOF THAT THE PRIMES OF FIBONACCI ARE
INFINITE IN NUMBER
Ing. Pier Francesco Roggero, Dott. Michele Nardelli, Francesco Di Noto
Abstract
In this paper we show that the primes of Fibonacci are infinite in number plus other
topics.
Versione 1.0
29/03/2013
Pagina 2 di 43
Index:
1. PERIODICITY WITHIN THE FIBONACCI’S SEQUENCE ............................................................ 3
1.1 PROOF THAT THE PRIMES OF FIBONACCI ARE INFINITE ....................................... 5
1.2 DIVISORS p n OF THE GROUPS a p ± B ........................................................................... 6
1.3 TABLE OF DECOMPOSITION OF THE FIBONACCI’S NUMBERS .............................. 8
1.3.1 TABLE OF THE FREQUENCIES FOR INCREASING PERIODICITY....................... 10
1.3.2 TABLE OF THE FREQUENCIES FOR P INCREASING ............................................. 13
1.4 DISTRIBUTION OF THE FIBONACCI’S PRIME NUMBERS ....................................... 14
1.5 FIBONACCI NUMBERS AND ABC CONJECTURE ...................................................... 18
2. PYTHAGOREAN TRIPLE AND ABC CONJECTURE .................................................................. 20
3. PYTHAGOREAN QUADRUPLES AND ABC CONJECTURE EXTENDED ............................... 22
3.1 PROOF OF PYTHAGOREAN QUADRUPLES ................................................................ 24
4. TRIBONACCI NUMBERS AND ABC CONJECTURE EXTENDED ............................................ 25
4.1 PROOF OF TRIBONACCI NUMBERS ............................................................................. 27
5. TRIANGULAR NUMBERS AND ABC CONJECTURE EXTENDED .......................................... 28
6. N − 1 OR N NTH POWERS SUMMING TO AN NTH POWER AND ABC CONJECTURE
EXTENDED .......................................................................................................................................... 30
7. PROOF OF GOLDBACH'S CONJECTURE .................................................................................... 34
8. PROOF OF THE WEAK GOLDACH CONJECTURE.................................................................... 37
9. CONCLUSIONS................................................................................................................................ 39
Versione 1.0
29/03/2013
Pagina 3 di 43
1. PERIODICITY WITHIN THE FIBONACCI’S SEQUENCE
The Fibonacci numbers are a sequence in the frequency of natural integers ε N+, where each number
of the sequence is the result of the sum of the previous two.
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …
By definition, the first two Fibonacci numbers are 0 and 1, and each subsequent is the sum of the
previous two. The sequence is defined by assigning the values of the first two terms, F0: = 0 and F1: =
1, and asking that for every next number is:
Fn = Fn-1 + Fn-2 ,
with
F0 = 0, F1 = 1.
Now given a prime factor p its periodicity within the Fibonacci sequence is given by:
If the prime factor ends with the digits 1, 9, the periodicity is f = a divisor of (p-1) or the same (p-1)
If the prime factor ends with the digits 3, 7 the frequency is f = a divisor of (p +1) or the same (p +1)
See tables in section 3.
Only with the prime factor p = 5, which is a prime number special, the only one that ends with the digit
5 the periodicity is given by f = 5k
For p = 2, another prime special, the only even prime, the periodicity is f = 3k
The prime factors p taken separately are all present in the Fibonacci sequence.
Versione 1.0
29/03/2013
Pagina 4 di 43
Since the Fibonacci sequence is a subset of the set of natural numbers N, DO NOT exist all prime
factors with powers greater than 1.
For example, the factor 22 does not appear anywhere in the list.
This does not mean that the Fibonacci numbers are not divisible by 4, but which, however, some Fn
are divisible by powers of 2 larger, i.e. for 8, 16, 32, 64 and consequently are also divisible by 4.
We note that 8, 16, 32 and 64 are all multiples of 8, number that is connected with the “modes” that
correspond to the physical vibrations of a superstring by the following Ramanujan function:
∞ cos πtxw'


− πx 2 w '
∫0 cosh πx e dx  142

4 anti log
⋅ 2
πt 2
t w'
−
w'

e 4 φw' (itw') 
1 
. (1)
8=
3
  10 + 11 2 
 10 + 7 2  

+ 
log  



4
4
 




If we choose a prime number n of Fn then its decomposition gives rise to prime factors ALL news and
ALL DISTINCT (with power equal to 1), which are all prime factors that have not yet appeared in the
Fibonacci sequence for n lower.
Only for n primes we can have Fibonacci numbers Fn primes.
However, if we choose a number n that isn’t prime of Fn, then in the decomposition of it we have
prime factors that are already present in the previous Fibonacci numbers in the sequence for n lower
and there is at least one new prime factor (one or more than one) .
Versione 1.0
29/03/2013
Pagina 5 di 43
1.1 PROOF THAT THE PRIMES OF FIBONACCI ARE INFINITE IN NUMBER
We know that so far have been discovered the following 33 n primes of Fn (except n = 4 that is even)
n = 3, 4, 5, 7, 11, 13, 17, 23, 29, 43, 47, 83, 131, 137, 359, 431, 433, 449, 509, 569, 571, 2971, 4723,
5387, 9311, 9677, 14431, 25561, 30757, 35999, 37511, 50833 e 81839.
The first 14 are the following:
2, 3, 5, 13, 89, 233, 1597, 28657, 514229, 433494437, 2971215073, 99194853094755497,
1066340417491710595814572169, 19134702400093278081449423917,
475420437734698220747368027166749382927701417016557193662268716376935476241
Since all prime factors p taken separately are present in the Fibonacci sequence, then in the
decomposition of the Fibonacci numbers Fn with n prime number we have always NEW prime factors
all distinct or prime factors with power equal to 1.
The number of prime factors m resulting from the decomposition of Fn may be m = 1, 2, 3 or m ≥ 1.
Because the number of the news prime factors is cyclic, with the increase of magnitude of the
Fibonacci numbers also increases the number m of the single prime factors.
This cyclicity is due to the fact that there is a periodicity of the prime factors inside the Fibonacci
sequence.
All the new prime factors of a decomposition of Fn with n prime numbers give rise to a periodicity k*n
or a cyclicity on the number of the new prime factors that are repeated, in fact, with the periodicity
k*n.
It then becomes increasingly rare to find new prime numbers for n increasing of Fibonacci.
However, the number of prime factors m is cyclic and therefore will find always a new prime number
of Fibonacci Fn with a single prime factor m = 1 for some n prime and with periodicity given by k*n.
Versione 1.0
29/03/2013
Pagina 6 di 43
1.2 DIVISORS p n OF THE GROUPS a p ± B
We have seen that the divisors distinct p of a group ap ± b are given by the following formula:
p divisor of ak(dn)+s ± b
with dn divisor of p - 1
Thus if p is a divisor of a p-1 ± b then p2 is a divisor of a p(p-1) ± b.
In general we have that pn is a divisor of a p
n −1
( p −1)
±b
If there is a divisor dn of p – 1 the above formula is simplified in this way:
pn is a divisor of a p
n −1
(dn )
±b
In the specific case of p2 one has:
p2 is a divisor of a p(dn) ±b
The periodicity ap ± b, instead, is given by:
chosen any s for as ± b, as divisor pn we have
p = (as ± b)n-1 (as ± b - 1) + s
and if it is true that there is a divisor dn of p – 1, we have:
p = k (p)n-1 (dn) + s
with pn divisor of
a k(p)
n -1
( dn ) + s
±b
with dn divisor of p - 1
In the specific case of p2 one has: p 2 divisor of
a kp ( dn ) + s
±b
Versione 1.0
29/03/2013
Pagina 7 di 43
Observation.
Each group ap ± b does not contain NEVER all the prime factors taken separately or raised to
the power.
This is because each group is only a subset of the set of natural numbers N.
That is why, for example, in the group of Mersenne 2p - 1 never exist the distinct prime factors
1093 and 3511 but only their squares 10932 and 35112 respectively for p = 364k and p = 1755k
In the group of Fibonacci, for example, never exists the factor 22, but all the other powers of 2n with n
= 1, 3, 4, 5, 6 ... are present including so obviously the 2.
Versione 1.0
29/03/2013
Pagina 8 di 43
1.3 TABLE OF DECOMPOSITION OF THE FIBONACCI’S NUMBERS
1
2
3
4
5
6
7
8
9
10
11
1
1
2
3
5
8
13
21
34
55
89
1
1
2
3
5
2^3
13
3 7
2 17
5 11
89
12
13
14
15
16
17
144
233
377
610
987
1597
24 32
233
13 29
2 5 61
3 7 47
1597
18
19
20
21
22
23
2584
4181
6765
10946
17711
28657
23 17 19
37 113
3 5 11 41
2 13 421
89 199
28657
24
46368
25 32 7 23
25
26
27
28
29
75025
121393
196418
317811
514229
52 3001
233 521
2 17 53 109
3 13 29 281
514229
30
31
32
33
34
35
832040
1346269
2178309
3524578
5702887
9227465
23 5 11 31 61
557 2417
3 7 47 2207
2 89 19801
1597 3571
5 13 141961
36
14930352
24 33 17 19 107
Versione 1.0
29/03/2013
Pagina 9 di 43
37
38
39
40
41
24157817
39088169
63245986
102334155
165580141
73 149 2221
37 113 9349
2 233 135721
3 5 7 11 41 2161
2789 59369
42
43
44
45
46
47
267914296
433494437
701408733
1134903170
1836311903
2971215073
23 13 29 211 421
433494437
3 43 89 199 307
2 5 17 61 109441
139 461 28657
2971215073
48
49
4807526976
7778742049
26 32 7 23 47 1103
13 97 6168709
50
51
52
53
12586269025
20365011074
32951280099
53316291173
52 11 101 151 3001
2 1597 6376021
3 233 521 90481
953 55945741
54
55
86267571272
1,39583862445000000E+11
23 17 19 53 109 5779
5 89 661 474541
56
57
58
59
2,25851433717000000E+11
3,65435296162000000E+11
5,91286729879000000E+11
9,56722026041000000E+11
3 72 13 29 281 14503
2 37 113 797 54833
59 19489 514229
353 2710260697
60
61
62
63
64
65
1,54800875592000000E+12
2,50473078196100000E+12
4,05273953788100000E+12
6,55747031984200000E+12
1,06102098577230000E+13
1,71676801775650000E+13
23 32 11 31 41 61 2521
4513 555003497
557 2417 3010349
2 13 17 421 35239681
3 7 47 1087 2207 4481
5 233 14736206161
66
67
68
69
70
71
2,77778900352880000E+13
4,49455702128530000E+13
7,27234602481410000E+13
1,17669030460994000E+14
1,90392490709135000E+14
3,08061521170129000E+14
23 89 199 9901 19801
269 116849 1429913
3 67 1597 3571 63443
2 137 829 18077 28657
5 11 13 29 71 911 141961
6673 46165371073
72
73
74
4,98454011879264000E+14
8,06515533049393000E+14
, 1304969544928657
25 33 7 17 19 23 107 103681
9375829 86020717
73 149 2221 54018521
Versione 1.0
29/03/2013
Pagina 10 di 43
1.3.1 TABLE OF THE FREQUENCIES FOR INCREASING PERIODICITY
2
3
3^2
5
2^3
13
7
7^2
17
11
89
3k
4k
12k
5k
6k
7k
8k
56k
9k
10k
11k
2^4
233
29
61
47
1597
19
37, 113
41
421
199
28657
12k
13k
14k
15k
16k
17k
18k
19k
20k
21k
22k
23k
23, 2^5
24k
5^2, 3001
521
53, 109
281
514229
25k
26k
27k
28k
29k
31
557, 2417
2207
19801
30k
31k
32k
33k
Versione 1.0
29/03/2013
Pagina 11 di 43
3571
141961
34k
35k
3^3, 107
73, 149, 2221
9349
135721
2162
2789, 59369
36k
37k
38k
39k
40k
41k
211
433494437
43, 307
109441
139, 461
2971215073
42k
43k
44k
45k
46k
47k
2^6, 1103
97, 6168709
48k
49k
101, 151
6376021
90481
953, 55945741
50k
51k
52k
53k
5779
661, 474541
54k
55k
14503
797, 54833
59, 19489
353, 2710260697
56k
57k
58k
59k
2521
4513, 555003497
3010349
35239681
1087, 4481
14736206161
60k
61k
62k
63k
64k
65k
9901
269, 116849, 1429913
67, 63443
137, 829, 18077
71, 911
6673, 46165371073
66k
67k
68k
69k
70k
71k
103681
72k
Versione 1.0
29/03/2013
Pagina 12 di 43
9375829, 86020717
54018521
73k
74k
Versione 1.0
29/03/2013
Pagina 13 di 43
1.3.2 TABLE OF THE FREQUENCIES FOR P INCREASING
2
3
5
7
11
13
17
19
23
29
31
3k
4k
5k
8k
10k
7k
9k
18k
24k
14k
30k
37, 113
41
43, 307
47
53, 109
59, 19489
19k
20k
44k
16k
27k
58k
61
67, 63443
71, 911
73, 149, 2221
79
83
15k
68k
70k
37k
78k
84k
89
11k
97, 6168709
49k
Versione 1.0
29/03/2013
Pagina 14 di 43
1.4 DISTRIBUTION OF THE FIBONACCI’S PRIME NUMBERS
Distribution of the Fibonacci’s numbers and of the Fibonacci’s primes up to 10^n
Table with the initial values:
n
10^n
a
Number of F(n)
≈ 5n
1
2
3
4
5
6
…
10
100
1000
10 000
100 000
1 000 000
….
6
11
16
20
25
30
…
b
Number of F(n)
primes (indexes)
≈7n
4
12
21
26
33
43
…
b/a
Ratio between
the two numbers
0,66
1,09
1,31
1,3
1,32
1,43
…
Indexes = 3, 4, 5, 7, 11, 13, 17, 23, 29, 43, 47, 83, 131, 137, 359, 431, 433, 449, 509, 569, 571, 2971,
4723, 5387, 9311, 9677, 14431, 25561, 30757, 35999, 37511, 50833 e 81839.
Fibonacci’s prime numbers
2, 3, 5, 13, 89, 233, 1597, 28657, 514229, 433494437, 2971215073, 99194853094755497,
1066340417491710595814572169,
19134702400093278081449423917,
475420437734698220747368027166749382927701417016557193662268716376935476241
n
10^n
1
2
3
4
5
6
10
100
1000
10 000
100 000
1 000 000
a
Number of F(n)
≈ 5n
6
11
16
20
25
30
b
Number of F(n)
primes ≈ 1,5 n
3
5
6
7
8
9
a/b
Ratio between
the two numbers
2
2,2
2,6
2,8
3.1
3,3
Versione 1.0
29/03/2013
Pagina 15 di 43
…
….
…
Here the graph with the two curves F(n) and F(n) primes
…
Versione 1.0
29/03/2013
Pagina 16 di 43
As we can see, the ratio between the two types of numbers grows more and more up to 3,3 which is
obviously also the mean ratio of the respective frequencies: 5/1,5 = 3.33 ... until
10^6 = 1000000; then the Fibonacci primes are becoming increasingly rare, up to 4*10^78 there are in
fact just 15 and the initial mean distribution n*1,5 no longer applies, since up to 10^76 there should be
76*1,5 = 114 Fibonacci primes, and instead there are only 15, about one-eighth of the mean estimate,
in fact 114/15 = 7,6 about 8.
One could introduce a correction constant equal to about n/8, for n numbers higher than 76 in such a
way that (n) primes ≈ 1,5 n*n/8 that for n = 79 we will have (1,5*76 )/8 = 114*1,5/8 = 21,375 about 15
real value. The exact ratio is instead 175/15 = 11,4, slightly greater than 8 (not to be confused n of F
(n) with n of 10^n)
For example, for n = 100, we will have 100*1,5 = 150; 150/8 = 18,75, and we would have a real
value of number of Fibonacci primes next to 19, probably very near to the real value, still unknown.
About the curves, the curve of prime numbers of Fibonacci (of course under the curve of Fibonacci’s
numbers up to 10^n) will flatten immediately after the first values, since up to 10^76 there are only 15
primes of Fibonacci, an average of one every five powers of 10, because 76/15 = 5,06, while up to
10^6 there are 9, about one in every 9/6 = 1,5 powers of 10
A mean statistical estimate in this sense could be given by the formula
n*1,5/2√n, in fact, for n = 6, we have 6*1,5/2*2,44 = 9/4,89 = 1,8, very near to 1,5 powers of 10.
We see now with n = 76
76*1,5/2*8,7 = 114/17,43 = 6,55 very near, for excess, to the mean 5,06 of powers of 10
which contain average a Fibonacci’s prime number
Since up to n = 76 there are 11,6 groups of 6,55 powers of 10, 11,6 is a rough estimate, by defect, of
the number of the primes up to 10^76, with the real value 15.
For n = 100, we would have 100*1,5/2*10 = 300/20 = 15, then on average a Fibonacci’s prime number
every 15 powers of 10. Since in 100 there are 100/15 = 6,6 groups of 15 powers of 10 we would have
100/6,66 = 15,01 Fibonacci primes, always estimated by defect, but neat to 19 provided with the
previous estimate, i.e. 18,75 about 19 Fibonacci primes up to 10^100.
A better estimate might be the value obtained by this estimate, plus its square root:
n/mean+2√n (1)
Examples
For n = 76, 11,6 + √11.6 = 11,6 + 3,40 = 15, fully centered on the real value, 15
For n = 100, 15,01 + √15,01 = 15,01 + 3,87 = 18,88, very near to the previous estimate 18,75
For n = 6, we have instead 6/1,5 +2√6 = 4 +2*2,44 = 4+4,89 = 8,89 ≈ 9 real value of Fibonacci’s
Versione 1.0
29/03/2013
Pagina 17 di 43
prime numbers up to 10^6
As n increases also increases the mean of powers of 10 which contain a prime number of Fibonacci,
and also increases the square root of n and its double, the estimated values with the (1) grow too, and
then the their curve, and this is a great indication of the infinitude of the Fibonacci’s prime numbers,
indirect support to the demonstration according to their cyclic factorization.
Versione 1.0
29/03/2013
Pagina 18 di 43
1.5 FIBONACCI NUMBERS AND ABC CONJECTURE
If we consider the set of Fibonacci numbers and let’s try adding two Fibonacci numbers we see how it
applies in the abc conjecture
8 + 144 = 152
rad (8*144*152) = rad (175.104) = rad (210 * 32 * 19) = 114 (2 * 3 * 9)
114 < 152
55 + 89 = 144
rad (55*89*144) = rad (704.880) = rad (24 * 32 * 5 * 11 * 89) = 29.370
29.370 > 144
We have that then the inequality can be > or <
This statement was easily predictable because chosen any two Fibonacci numbers their sum can never
take any integer or also more limited if we choose an even as a sum of two odd Fibonacci numbers.
In fact, for example the first even integer that cannot derive by the sum of two odd numbers of
Fibonacci, but only by three Fibonacci numbers:
12 = 8 + 2 + 2
We note that 152 is divisible for 8 and 144 is divisible for 24 and 8 and 24 are connected with the
“modes” that correspond to the physical vibrations of a superstring and of a bosonic string by the
following Ramanujan functions:
Versione 1.0
29/03/2013
Pagina 19 di 43
∞ cos πtxw'


− πx 2 w '
∫0 cosh πx e dx  142

4 anti log
⋅ 2
πt 2
t w'
−
w'

e 4 φw' (itw') 
1 
8=
,
3
  10 + 11 2 
 10 + 7 2  
+ 

log  



4
4
 




∞ cos πtxw'


− πx 2 w '
∫0 cosh πx e dx  142

4 anti log
⋅ 2
πt 2
−
w'

 t w'
4
(
)
'
e
φ
itw
w
'

24 = 
.
  10 + 11 2 
 10 + 7 2  
+ 

log  



4
4
 




Versione 1.0
29/03/2013
Pagina 20 di 43
2. PYTHAGOREAN TRIPLE AND ABC CONJECTURE
A Pythagorean triple consists of three positive integers a, b, and c, such that a2 + b2 = c2.
Let’s apply also in this case with the abc conjecture
rad ( a
2
b 2 c 2 ) < or > c 2
In Pythagorean triples we can also find that none of the three numbers a, b and c are prime (example:
16, 63, 65) and that and that a cathetus (a or b) is always a multiple of 4.
Then the inequality may be > or <, we do not know before we calculate it.
For example:
5 2 + 12 2 = 13 2
rad (25*144*169) = 2*3*5*13 = 390 > 169
-------------------------------------------------------------------------------------------------------------------------
7 2 + 24 2 = 25 2
rad (49*576*625) = 2*3*5*7 = 210 < 625
But in this case we have
rad (abc) < c
This derives from the fact that we cannot randomly choose three positive integers a, b and c so always
worth the Pythagorean theorem!
Versione 1.0
29/03/2013
Pagina 21 di 43
In fact for generating a Pythagorean triple we must use Euclid's formula.
Given an arbitrary pair of positive integers m and n with m > n. The formula states that the integers
a = m 2 − n 2 , b = 2mn , c = m 2 + n 2
form a Pythagorean triple.
Versione 1.0
29/03/2013
Pagina 22 di 43
3. PYTHAGOREAN QUADRUPLES AND ABC
CONJECTURE EXTENDED
A set of four positive integers a, b, c and d such that a2 + b2+ c2 + d2 = z2 is 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.
Let’s apply also in this case with the abc conjecture extended
a2 + b2+ c2 = z2
rad ( a
2
b 2 c 2 z 2 ) < or > z 2
Then the inequality may be > or <, we do not know before we calculate it.
The first three examples give the sign of the inequality of “<”, that are exceptions and the fourth
example gives as a sign of inequality of “>”
12 + 22 + 22 = 32
rad (1*4*4*9) = 2*3 = 6 < 9
------------------------------------------------------------------------------------------------------------------------22 + 32 + 62 = 72
rad (4*9*36*49) = 2*3*7 =42 < 49
But in this case we have
rad (abc) < c
------------------------------------------------------------------------------------------------------------------------72 + 142 + 222 = 272
Versione 1.0
29/03/2013
Pagina 23 di 43
rad (49*196*484*729) = 2*3*7*11 = 462 < 729
------------------------------------------------------------------------------------------------------------------------fourth example:
42 + 82 + 192 = 21
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!
Versione 1.0
29/03/2013
Pagina 24 di 43
3.1 PROOF OF PYTHAGOREAN QUADRUPLES
In fact for generating a Pythagorean quadruple we can use this formula.
Thus, all primitive Pythagorean quadruples are characterized by the Lebesgue Identity
(m
2
+ n2 + p2 + q2
) = (2mq + 2np ) + (2nq − 2mp ) + (m
2
2
2
2
)
2
+ n2 − p2 − q2 .
From this formula we see that two sides are always even
b = 2(mq + np ) ,
c = 2(nq − mp ) ,
while the other side and the longer side of all (for that Pythagoras would be the hypotenuse) may be
prime numbers or the one or the other or both.
a2 + b2+ c2 = z2
It follows that the radical
rad ( a
2
b 2 c 2 z 2 ) = rad (abcz)
cannot be simplified because a, b and c can never be all prime numbers.
Consequently the sign of the inequality may be > or <, we do not know before we calculate it.
This shows that in the set of Pythagorean quadruple, we cannot cover the entire set entire set of natural
numbers N + because the sign of the radical inequality can be < or >.
Versione 1.0
29/03/2013
Pagina 25 di 43
4. TRIBONACCI NUMBERS AND ABC CONJECTURE
EXTENDED
The tribonacci numbers are like the Fibonacci numbers, but instead of starting with two
predetermined terms, the sequence starts with three predetermined terms and each term afterwards is
the sum of the preceding three terms. The first few tribonacci numbers are:
0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, 10609, 19513, 35890,
66012, ….
The tribonacci constant
1 + 3 19 + 3 33 + 3 19 − 3 33
= 1,8939286
3
is the ratio toward which adjacent tribonacci numbers tend. It is a root of the polynomial
x3 − x2 − x − 1, approximately 1.83929 and also satisfies the equation x + x−3 = 2. It is important in the
study of the snub cube, that’s an Archimedean solid with 38 faces: 6 squares and 32 equilateral
triangles. It has 60 edges and 24 vertices.
When calculating its volume appears the tribonacci constant
Let’s apply also in this case with the abc conjecture extended
4 + 24 + 44 = 72
rad (4*24*44*72) = 2*3*11 = 66 < 72
Also here, we note that have the two numbers 24 and 72 = 24 * 3, thence the number related to the
modes corresponding to the physical vibrations of the bosonic strings by the following Ramanujan
function:
Versione 1.0
29/03/2013
Pagina 26 di 43
∞ cos πtxw'


− πx 2 w '
∫0 cosh πx e dx  142

4 anti log
⋅ 2
πt 2
t w'
−
w'

e 4 φw' (itw') 

24 =
.
  10 + 11 2 
 10 + 7 2  
+ 

log  



4
4
 




------------------------------------------------------------------------------------------------------------------------2 + 7 + 13 = 22
rad (2*7*13*22) = 2*7*11*13 = 2002 > 22
Versione 1.0
29/03/2013
Pagina 27 di 43
4.1 PROOF OF TRIBONACCI NUMBERS
We can also choose as the sum of three prime numbers that give an odd number or a prime number
a+b+c=z
Here's an example with 4 primes
7 + 13 + 149 = 163
rad (7*13*149*163) = 2210117 > 163
In this case we have with a, b and c primes
rad (abcz) = abc rad(z) > abc
and always
abc > z
But we have seen that the sign of the inequality may be > or <, we do not know before we calculate it.
This shows that in the set of Tribonacci numbers, we cannot cover the entire set entire set of natural
numbers N + because the sign of the radical inequality can be < or >.
Versione 1.0
29/03/2013
Pagina 28 di 43
5. TRIANGULAR NUMBERS AND ABC CONJECTURE
EXTENDED
A triangular number or triangle number counts the objects that can form an equilateral triangle. The
nth triangle number is the number of dots composing a triangle with n dots on a side, and is equal to
the sum of the n natural numbers from 1 to n. The sequence of triangular starting at the 0th triangular
triangle, is:
0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120 ....
The triangle numbers are given by the following explicit formulas:
n
Tn = ∑ k = 1 + 2 + 3 + ... + n =
k =1
n(n + 1)  n + 1

= 
2
 2 
Even if we choose the set of triangular numbers and apply the abc conjecture we obtain the same
results as the Pythagorean triples and Fibonacci numbers.
Moreover, for the triangular numbers we already know from German mathematician Carl Friedrich
Gauss discovered that every positive integer is representable as a sum of at most three triangular
numbers, writing in his diary his famous words, " num = ∆ + ∆ + ∆".
Note that this theorem does not imply that the triangular numbers are different (as in the case of
20=10+10), nor that a solution with three nonzero triangular numbers must exist.
Let’s apply the abc conjecture extended.
We have, for example, choosing the triangular numbers 36 and 300
36 + 300 = 336
rad (36*300*336) = rad (3.628.800) rad ((28 * 34 * 52 * 7) = 210
210 < 336
In almost all other cases, we have instead
Versione 1.0
29/03/2013
Pagina 29 di 43
rad (abc) > c
We have that then the inequality can be > or <
This statement was also easily predictable because chosen any two triangular numbers their sum can
never take any integer or also more limited if we choose an even number as a sum of two odd
triangular numbers.
In fact, for example the first even integer that cannot derive by the sum of two odd triangular numbers,
but only by three triangular numbers:
8=6+1+1
This shows that in the set of triangular numbers, we cannot cover the entire set entire set of natural
numbers N + because the sign of the radical inequality can be < or >.
Versione 1.0
29/03/2013
Pagina 30 di 43
6. N − 1 OR N NTH POWERS SUMMING TO AN NTH
POWER AND ABC CONJECTURE EXTENDED
Another generalization is searching for sequences of n + 1 positive integers for which the nth power of
the last is the sum of the nth powers of the previous terms. The smallest sequences for known values of
n are:
•
•
•
•
•
n = 3: {3, 4, 5; 6}.
n = 4: {30, 120, 272, 315; 353}
n = 5: {19, 43, 46, 47, 67; 72}
n = 7: {127, 258, 266, 413, 430, 439, 525; 568}
n = 8: {90, 223, 478, 524, 748
There can also exist n − 1 positive integers whose nth powers sum to an nth power (though, by
Fermat’s last theorem, not for n = 3); these are counterexamples to Euler’s sum of powers conjecture.
The smallest known counterexamples are
•
•
n = 4: (95800, 217519, 414560; 422481)
n = 5: (27, 84, 110, 133; 144)
Let’s apply the abc conjecture extended.
We have, for example, by n =3 and 2 primes
•
n = 3: {3, 4, 5; 6}.
27 + 64 +125 = 216
rad (27*64*125*216) = 2*3*5 = 30 < 216
This is an exception
------------------------------------------------------------------------------------------------------------------------Example with 4 primes:
(3,1,3): 709=631+461+193
Versione 1.0
29/03/2013
Pagina 31 di 43
7189057 + 97972181 + 251239591= 356400829
rad (7189057*97972181*251239591*356400829) = 193*461*631*709 = 39.804.651.767
356.400.829
>
So the sign of the inequality may be > or <, we do not know before we calculate it.
This shows that in the set of sequences of 4 positive integers for which the 3th power of the last is the
sum of the 3th powers of the previous terms, we cannot cover the entire set entire set of natural
numbers N + because the sign of the radical inequality can be < or >.
------------------------------------------------------------------------------------------------------------------------Another sequence with no primes:
•
n = 5: (27, 84, 110, 133; 144)
14348907 + 4182119424 +16105100000 + 41615795893 = 61917364224
rad (1434890*4182119424*16105100000*41615795893*61917364224) = 2*3*5*7*11*19 = 43890 <
61.917.364.224
This is an exception
------------------------------------------------------------------------------------------------------------------------Another sequence with no primes:
(5,1,4): 85359=85282+28969+3183+55
503284375 + 326725621763716143 + 20401754572881571915849 + 4511146005966249681574432
= 4531548087264753520490799
rad(503284375*326725621763716143*20401754572881571915849*4511146005966249681574432*
4531548087264753520490799)
=
2*3*5*11*37*59*491*769*1061*42641
=
12.306.038.584.733.736.810 < 4.531.548.087.264.753.520.490.799
This is an exception
Versione 1.0
29/03/2013
Pagina 32 di 43
-------------------------------------------------------------------------------------------------------------------------
•
n = 4: (95800, 217519, 414560; 422481) with one prime 217519
84229075969600000000 +
31858749840007945920321
2238663363846304960321
+
29535857400192040960000
=
rad(84229075969600000000*2238663363846304960321*29535857400192040960000*31858749840
007945920321)
=
2*3*5*479*2591*140827*217519
=
1.140.531.558.873.718.710
<
31.858.749.840.007.945.920.321
This is an exception
------------------------------------------------------------------------------------------------------------------------(4,1,3) 145087793=122055375+121952168+1841160
11491219604030165199360000
+
2,2118623769909863494842339525018e+32
2,2193594025582085161084008789063e+32 = 25488497887618394389866690
+
Rad
(11491219604030165199360000*2,2118623769909863494842339525018e+32*2,219359402558208
5161084008789063e+32*4,431221894461390905894286825008e+32
=
2*3*5*13*67*229*271*4327*25037*145087793
=
25488497887618394389866690
<
25488497887618394389866690
This is an exception
Versione 1.0
29/03/2013
Pagina 33 di 43
By induction we can NEVER know a priori the sign of the radical inequality for any sequences of k
positive integers for which the nth power of the last is the sum of the nth powers of the previous terms.
Versione 1.0
29/03/2013
Pagina 34 di 43
7. PROOF OF GOLDBACH'S CONJECTURE
Goldbach's conjecture is a conjecture that states:
Every even integer greater than 2 can be expressed as the sum of two primes (that may be also
equal)
Goldbach was a German mathematician and his problem resists well as 271 years and is also part of
the eighth Hilbert problem still unresolved!
For example,
4=2+2
6=3+3
8=3+5
10 = 3 + 7 = 5 + 5
12 = 5 + 7
14 = 3 + 11 = 7 + 7
and so on
Let’s now consider the conjecture abc and apply to the sum of 2 primes that give an even number.
The abc conjecture states in terms of three positive integers, a, b and c (hence the name), which have
no common factor and satisfy a + b = c. If d denotes the product of the distinct prime factors of abc,
the conjecture essentially states that d is usually not much smaller than c. In other words: if a and b are
divisible by large powers of primes, then c is usually not divisible by large powers of primes.
So we need to introduce the concept of radical for d
For a positive integer n, the radical of n, denoted rad(n), is the product of the distinct prime factors of
n. For example:
•
•
•
rad(16) = rad(24) = 2
rad(17) = 1
rad(18) = rad(2·32) = 2·3 = 6
Versione 1.0
29/03/2013
Pagina 35 di 43
Let’s apply it to the case of Goldbach:
If we take two “different” primes a and b we have:
rad (abc) > c
if we write
a + b = 2n
we can write
rad (abc) = ab rad(c) > c
for n > 3
PROOF:
In our case the inequality is even stronger.
In fact we have
rad (abc) = ab rad (c) > ab
ab > c
we can even neglect rad (c) for n > 3 and so we have
ab > 2n
ab > 2
h
p 1j p 2k …. p nz
This inequality is always true because the product of two “different” prime numbers a * b is
always greater than the sum of two prime numbers a + b
a*b > a + b = c
Versione 1.0
29/03/2013
Pagina 36 di 43
Only for n = 2 and n = 3 we have that the sign of the inequality is < for n = 2 and = for n = 3 and we
must consider the full radical rad (abc), see the examples that follows.
This happens simply because when we add 2 prime numbers that are equal to each other the sign of
inequality is =, except the case 2 +2 = 4 where the sign of inequality is <
If a=b
a + a = c = 2a
rad (a*a*2a) = rad (2a3 ) =2a
We must consider the full radical rad (abc) and not the simplified rad (abc) = ab rad(c) > c, where a
and b are different primes.
In other words the inequality is always true and is > if the two primes a and b are different and
is = if the two primes are equal a = b, except the case 2 + 2.
rad (abc) ≥ c
CVD
Versione 1.0
29/03/2013
Pagina 37 di 43
8. PROOF OF THE WEAK GOLDACH CONJECTURE
The conjecture seen before is called strong -to distinguish it from a weaker corollary. The strong
Goldbach conjecture that implies the weak conjecture that all odd numbers greater than 7 are the
sum of three odd primes.
Since we have shown that the strong conjecture is true, it follows that the weak Goldbach conjecture
will be true by implication.
In fact any odd number can be written as a sum.
u = (u – 3) +3
The term (u – 3) , which is even, can be written as a sum of two primes (a and b) after the strong
Goldbach's conjecture.
Accordingly then just add 3 and we get all the odd numbers as the sum of 3 odd primes for u > 7
In fact if we apply the abc “extended” conjecture as
a + b +c = d
with a, b and c primes
rad (abcd) > d
PROOF:
rad (abcd) = abc rad (d) > abc
if a, b and c are different primes
we can even neglect rad (d) and so we have
abc > d
This inequality is always true because the product of three prime numbers a * b * c is always
greater than the sum of three prime numbers a + b + c
Versione 1.0
29/03/2013
Pagina 38 di 43
abc > a + b +c = d
If we choose 3 primes all equal to each other a = b =c the sign of the inequality is =
a + a + a = 3a
rad (a*a*a*3*a) = 3a
and only for a =3 we have the sign of inequality <
In other words the inequality of abc “extended” conjecture is always true and is > if the three
primes a b and c are different and is = if the three primes are equal a = b = c, except the case 3 +
3 + 3.
rad (abcd) ≥ d
CVD
Example:
The first odd number is 9 (u > 7)
3+3+3=9
rad (3*3*3*9) = rad (35) = 3 < 9
---------------------------------------------------------------------------------------------------------------------------The second odd number is 11
3 +3 +5 = 11
rad (3*3*5*11) = rad (32 *5*11) = 165 > 11
In all other cases, with a, b and c different primes the inequality of abc “extended” with the sign
> is always valid and true.
Versione 1.0
29/03/2013
Pagina 39 di 43
9. CONCLUSIONS
Randomly selected three positive integers a, b and c for Pythagorean triple or we choose any two
Fibonacci numbers or we choose any two triangular numbers we don’t cover the full set of the natural
numbers N+ and so the abc conjecture does not guarantee if the inequality.
rad (abc) > c
or
rad (abc) = c
and we don’t know which is the sign of inequality before calculating.
Only in the case of the sum of 2 primes we are guaranteed that
rad (abc) = ab rad(c) > c
or better, as we have seen
ab > 2n
for n > 2
This shows that any chosen two prime numbers their sum covering the entire set of numbers of
even natural numbers 2n
Similarly, in other words, we can say that any chosen set of numbers with certain rules, the abc
conjecture gives a sign for inequality rad (abc) <, > or = c
Instead, if we choose the set of prime numbers that has no rule prediction of the next prime and
are therefore completely random and not subject to any rule, we have that the sum of two primes
in the abc conjecture is always
rad (abc) > c
always for any a and b primes
CVD
With regard the Fibonacci’s numbers (the Fibonacci sequence is:
Versione 1.0
29/03/2013
Pagina 40 di 43
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, ....)
we remember that if a Fibonacci number is divided by its immediate predecessor in the sequence, the
quotient approximates φ (the golden ratio); for example., 987/610 ≈ 1.6180327868852. These
approximations are alternately lower and higher than φ, and converge on φ as the Fibonacci numbers
increase. With regard the mathematical connection with the string theory, we recall the following
Ramanujan functions:
∞ cos πtxw'


− πx 2 w '
e
dx 
∫0 cosh πx

142
4 anti log
⋅ 2
πt 2
t w'
−
w'

e 4 φw' (itw') 
1 
8=
, (1)
3
  10 + 11 2 
 10 + 7 2  

+ 
log  



4
4
 




∞ cos πtxw'


− πx 2 w '
∫0 cosh πx e dx  142

4 anti log
⋅ 2
πt 2
−
w'

 t w'
4
(
)
e
φ
itw
'
w'

24 = 
.
  10 + 11 2 

 10 + 7 2 

+ 
log  



4
4
 




(2)
Palumbo and Nardelli (2005) have compared a simple model of the birth and of the evolution of the
Universe with the theory of the strings, and translated it in terms of the latter obtaining:
R
1
1


− ∫ d 26 x g −
− g µρ g νσ Tr (Gµν Gρσ ) f (φ ) − g µν ∂ µ φ∂ν φ  =
2
 16πG 8

∞
2

1
1 ~ 2 κ
2 
1/ 2
= ∫ 2 ∫ d 10 x(− G ) e −2Φ  R + 4∂ µ Φ∂ µ Φ − H 3 − 102 Trν F2  ,
2κ 10
2
g10


0
( )
(3)
A general relationship that links bosonic and fermionic strings acting in all natural systems.
It is well-known that the series of Fibonacci’s numbers exhibits a fractal character, where the forms
5 −1
repeat their similarity starting from the reduction factor 1 / φ = 0,618033 =
(Peitgen et al.
2
1986). Such a factor appears also in the famous fractal Ramanujan identity (Hardy 1927):
Versione 1.0
29/03/2013
Pagina 41 di 43
0,618033 = 1 / φ =
5 −1
= R(q) +
2
5
 1 q f 5 (−t ) dt 
3+ 5

1+
exp
1/ 5
4/5 
∫
2
 5 0 f ( −t ) t 




3 
5
,
R(q) +
π = 2Φ −
20 
 1 q f 5 (−t ) dt  
3+ 5

1+
exp

1/ 5
4/5  
∫
2
 5 0 f (−t ) t  

and
Φ=
where
,
(4)
(5)
5 +1
.
2
Furthermore, we remember that π arises also from the following identities (Ramanujan’s paper:
“Modular equations and approximations to π” Quarterly Journal of Mathematics, 45 (1914), 350-372.):
π=
(
)(
)

 2 + 5 3 + 13 
12
24
log 
log 
 , (5a) and π =
130
2
142




 10 + 11 2 
 10 + 7 2  

+ 
  . (5b)




4
4




From (5b), we have that
24 =
π 142

log 


 10 + 11 2 

+


4


 10 + 7 2  




4


.
(5c)
The introduction of (4) and (5) in (3) provides:





1
1
 R
26
− ∫ d x g −
⋅
− g µρ g νσ Tr (Gµν Gρσ ) f (φ ) +
16G

 8





3 
5
2Φ −  R(q ) +


20 
 1 q f 5 (−t ) dt  
3+ 5


1+
exp
1/ 5
∫


t 4 / 5  
2
 5 0 f −t


(
)
Versione 1.0
29/03/2013
Pagina 42 di 43




∞ R
1 µν
3
5
⋅
− g ∂ µ φ∂ν φ ] = ∫ 2 ⋅ 2Φ −
R (q ) +
5
0

q
2
20
κ 11
 1
3+ 5
f (−t ) dt  

1+
exp

1/ 5
4/5  
∫
2

 5 0 f (−t ) t  
κ112
1 ~ 2
1 / 2 −2Φ
10
µ
−
+
4
∂
Φ
∂
Φ
−
−
d
x
(
G
)
e
R
H
Trν
3
µ
∫
2





3 
5
2Φ −  R ( q ) +
2 Rg102

5
20 
 1 q f (−t ) dt  
3+ 5

1+
exp
1/ 5
4/5  
∫

2
 5 0 f (−t ) t  

[
( F ) ],
2
2
(6)
which is the translation of (3) in the terms of the Theory of the Numbers, specifically the possible
connection between the Ramanujan identity and the relationship concerning the Palumbo-Nardelli
model.
Versione 1.0
29/03/2013
Pagina 43 di 43
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 , con N e k entrambi 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,