RELATIONS BETWEEN THE GAUSS-EISENSTEIN PRIME NUMBERS AND THEIR CORRELATION WITH SOPHIE GERMAIN PRIMES
Pier Francesco Roggero, Michele Nardelli 1,2 , Francesco Di Noto
1 Dipartimento di Scienze della Terra Università degli Studi di Napoli Federico II, Largo S. Marcellino, 10 80138 Napoli, Italy
2 Dipartimento di Matematica ed Applicazioni “R. Caccioppoli” Università 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 examine the relations between the Gauss prime numbers and the Eisenstein prime numbers and their correlation with Sophie Germain primes. Furthermore, we have described also various mathematical connections with some equations concerning the string theory.
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Index:
1. GAUSSIAN INTEGER ...... 3 1.1 REMARKS ...... 6 1.2 ON SOME EQUATIONS REGARDING LEMMAS AND THEOREMS OF GAUSSIAN PRIMES ………………………………………………………………………………………………...7 2. EISENSTEIN INTEGER ...... 28 2.1 EISENSTEIN PRIME ...... 29 3. RELATIONS BEETWEEN GAUSSIAN PRIMES AND EISENSTEIN PRIMES ...... 30 3.1 CORRELATION WITH SOPHIE GERMAIN PRIMES ...... 32 3.2 PROOF THAT THERE ARE INFINITELY MANY SOPHIE GERMAIN PRIMES ...... 34 4. REFERENCES ...... ………35
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1. GAUSSIAN INTEGER
A Gaussian integer is a complex number whose real and imaginary part are both integers. The Gaussian integers, with ordinary addition and multiplication of complex numbers, form an integral domain, usually written as Z[i].
Definition
Formally, Gaussian integers are the set
Z[i] = {a + bi a,b ∈ Z}.
Norm The norm of a Gaussian integer is the natural number defined as
The norm is multiplicative:
N(zw ) = N(z)N(w).
The units of Z[i] are therefore precisely those elements with norm 1, i.e. the elements :
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Gaussian primes in the complex plane are marked in red :
A Gaussian integer is a Gaussian prime if and only if either:
• a is prime number and b is zero, so we have a prime number of the form 4k+3 2 2 • both are nonzero and p = a + b = (a+bi) (a-bi) is a prime number (which will not be of the form 4k+3).
The prime number 2 is a special case because is the only ramified prime in Z[i]).
The integer 2 factors as 2 = (1+ i)(1− i) = i(1− i)2 as a Gaussian integer, the second factorization (in which i is a unit) showing that 2 is divisible by the square of a Gaussian prime; it is the unique prime number with this property.
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Instead, the prime number 5 is a unique factorization
5 = (2+ i)(2−i).
Gaussian primes are infinite, because they are infinite primes.
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1.1 REMARKS
The prime numbers that are congruent to 3 (mod.4) are also Gaussian primes.
Since there are infinitely many rational primes of the form 4k +3, there are infinitely many necessarily Gaussian primes.
The prime numbers congruent to 1 (mod.4) are the product of two distinct Gaussian primes. In fact, the primes of the form 4k +1 can always be written as the sum of two squares (Fermat's theorem on the sum of two squares) so we have:
p = a 2 +b 2 = (a+bi)( a-bi)
Numerical form.
Since the form 4k + 3 corresponds to the form 6k '+1 with k'
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1.2 On some equations regarding Lemmas and Theorems of Gaussian integers, mod-gaussian convergence of a sum over primes and Gaussian primes .
Now we see some Lemmas and Theorems concerning the ring of Gaussian integers Z[i], N(a + bi ) = a2 + b2 . We note that the letter p denote the Gaussian prime.
Lemma 1
( ) s = ( ) s = res Fk s x / s Ck x , res Fk∗ s x / s Ck∗x , (1.1) s=1 s=1 where π ∞ τ ( )−τ ( − ) π ∞ ( )− ( − ) k a k a 1 tk a tk a 1 C = ∏1+ , (1.2) C ∗ = ∏1+ , (1.3) k ∑ a () k ∑ a () 4 p a=2 N p 4 p a=2 N p
As a consequence of the following representation,
( ) = ( ) k −1( ) (k −k 2 ) 2/ ( ) (−k 3 +6k 2 −5k ) 6/ ( )× (k 3 −4k 2 +3k ) 2/ ( ) (3k 4 −26 k 3 +57 k 2 −34 k ) 24/ ( ) ( ) Fk s Z s Z 2s Z 5s Z 6s Z 7s Z 8s fk s
(1.4) we have ∞ ∞ F (s) τ (a) − τ (a)−τ (a −1) k = ∏1+ k ()1− p 1 = ∏1+ k k (1.5) () ∑ as () ∑ as () Z s p a=1 N p p a=2 N p
( ) ( ) = and so function Fk s / Z s is regular in the neighbourhood of s 1. At the same time we have
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π ()()()= χ ζ = res Z s L ,1 4 res s , (1.6) s=1 s=1 4
which implies (1.2). Numerical values of Ck and Ck∗ can be, for example: ≈ ≈ C2 ,1156101 , C2∗ ,1 524172 . We note that the value 1,156101 is very near to the value 1,15522604 and that the value 1,524172 is very near to the value 1,52313377 ( ,11561 ≈ ,11552 ; ,1 5241 ≈ ,1 5231 ). Values concerning the new universal musical system based on fractional powers of Phi and Pigreco.
Theorem 1
( ) = + ( 2/1 3+4(k −1) 3/ ) ( ) = + ( 2/1 3+2(k 2 +k −2) 3/ ) M k x Ck x O x log x , (1.7) M k∗ x Ck∗x O x log x , (1.8)
where Ck and Ck∗ were defined in (1.2) and (1.3). By Perron formula and by the following expressions,
τ (e)( ) << ε (e)(α ) << ε (α ) (e)(α ) << ε (α ) k∗ n n , tk N , tk∗ N , (1.9) for c = 1+ /1 log x , log T ≈ log x we have
s 1+ε c+iT () = 1 () x + x M k x ∫ Fk s ds O , (1.10) 2πi c−iT s T that, for the eq. (1.7), can be rewritten also as follows:
s 1+ε c+iT ( ) = + ( 2/1 3+4(k −1) 3/ ) = 1 () x + x M k x Ck x O x log x ∫ Fk s ds O . (1.10b) 2πi c−iT s T
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Suppose d = 2/1 − /1 log x . Let us shift the interval of integration to [d − iT ,d + iT ]. To do this consider an integral about a closed rectangle path with vertexes in d − iT , d + iT , c + iT and c − iT . There are two poles in s = 1 and s = 2/1 inside the contour. The residue at s = 1 was calculated in (1.1). The residue at s = 2/1 is equal to Dx 2/1 , D = const and will be absorbed by error term (see below). Identity (1.4) implies
( ) = ( ) k −1( ) ( ) Fk s Z s Z 2s fk s , (1.11)
( ) > ε > where fk s is regular for Rs 3/1 , so for each 0 it is uniformly bounded for Rs > 3/1 + ε . Let us estimate the error term. The error term absorbs values of integrals about three sides of the integration’s rectangle. We take into ( ) = ζ ( ) ( χ ) account Z s s L s, 4 . On the horizontal segments we have + s c iT − x − σ − ∫ Z()()s Z k 1 2s ds << max Z()()σ + iT Z k 1 2σ + 2iT x T 1 << d +iT s σ ∈[]d ,c θ − ( − ) − << x 2/1 T 2 1 log 4 k 1 3/ T + xT 1 log 3/4 T . (1.12)
− It is well-known that ζ (s) ≈ (s −1) 1 in the neighbourhood of s = 1. So on [d,d + i] we get
s d +i 1 1 ()()k −1 x << 2/1 ζ k −1()+ << 2/1 dt << 2/1 k −1 ∫Z s Z 2s ds x ∫2d 2it dt x ∫ k −1 x log x , (1.13) d s 0 0 it − /1 log x and for the rest of the vertical segment we have
s 4/1 2/1 d +iT TT T ()− ()()k −1 x << ζ ()+ 4 dt ()+ χ 4 dt ()+ 2 k 1 dt << ∫ Z s Z 2s ds ∫2/1 it ∫ L 2/1 it , 4 ∫ Z 1 2it d +i s 1t 1 t 1 t ()− + 2/1 + ()− << x 2/1 (log 5 T ⋅ log 8 k 1 3/ 1 T ) << x 2/1 log 3 4 k 1 3/ T . (1.14)
The choice T = x 2/1 +ε finishes the proof of (1.7).
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In conclusion, we note that multiplying both the sides of (1.14) for 8, we obtain the equivalent expression:
s 4/1 2/1 d +iT TT T ()− ()()k −1 x << ζ ()+ 4 dt ()+ χ 4 dt ()+ 2 k 1 dt << 8∫ Z s Z 2s ds 8 ∫2/1 it ∫ L 2/1 it , 4 ∫ Z 1 2it d +i s 1t 1 t 1 t ()− + /1 2 + ()− << 8x 2/1 (log 5 T ⋅ log 8 k 1 3/ 1 T ) << 8x 2/1 log 3 4 k 1 3/ T , (1.14b) that we can connect to the modes corresponding to the physical vibrations of the superstrings by the following Ramanujan modular equation :
∞ cos πtxw ' −π 2 e x w'dx ∫0 π 142 4 anti log cosh x ⋅ πt 2 2 − w' t w' 4 φ () 1 e w' itw ' 8 = . (1.14c) 3 10 +11 2 10 + 7 2 log + 4 4
Thence, we obtain the following mathematical connection :
s 4/1 2/1 d +iT TT T ()− ()()k −1 x << ζ ()+ 4 dt ()+ χ 4 dt ()+ 2 k 1 dt << 8∫ Z s Z 2s ds 8 ∫2/1 it ∫ L 2/1 it , 4 ∫ Z 1 2it d +i s 1t 1 t 1 t ()− + /1 2 + ()− << 8x 2/1 (log 5 T ⋅ log 8 k 1 3/ 1 T ) << 8x 2/1 log 3 4 k 1 3/ T ⇒ ∞ cos πtxw ' −π 2 e x w'dx ∫0 π 142 4 anti log cosh x ⋅ πt 2 2 − w' t w' 4 φ () 1 e w' itw ' ⇒ . (1.14d) 3 10 +11 2 10 + 7 2 log + 4 4
Theorem 2
Let x = elog T / N and N such that x and N / log log T → ∞ as T → ∞ . Then
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sin (t log p) iu ∑ 2 ()+γ 1 2T p≤x eu log log x 4/ ∫ e p dt → Φ()u as T → ∞ (1.15) T T locally uniformly for u ∈ R . Here γ denotes Euler’s constant and Φ is the analytic function given by − 2 u / 4 Φ() = − 1 u u ∏1 J0 (1.16) p∈P p p
where J0 denotes the zeroth Bessel function .
Corollary 1
[ ] Assume RH. Let UT be random variables uniformly distributed on T 2, T . ( (( ) )) ζ ( + ) Then the family /1 log log T 2/ Im log 2/1 iU T satisfies the large deviation principle with speed /1 ((log log T ) 2/ ) and rate function I(h) = h2 2/ . For instance ,
1 1 log λ({t ∈[]T 2, T Im: log ζ ()()2/1 + it ≥ h log log T 2/ }) → −h2 2/ as T → ∞ ()log log T 2/ T (1.17) where h > 0 and λ denotes the Lebesgue measure .
Let x be a positive real number and denote by p1, p2 ,..., pn the prime numbers not exceeding x . We have
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2k 2 1 2T 1 1 2T k −λ −λ 2/1 +it = ()1 n ∫∑ 2/1 +it dt ∫ ∑ p1 ... pn dt (1.18) T T ≤ p T T λ + +λ = λ ... λ p x 1 ... n k 1 n and applying the mean value theorem of Montgomery and Vaughan this is equal to
2 2 k −λ −λ 6π k 1 n +θ ∑ p1 ... pn ∑ (1.19) λ + +λ = λ ... λ λ + +λ = λ ... λ 1 ... n k 1 n T 1 ... n k 1 n with θ ≤ 1. The absolute value of the remainder is bounded by (6π /T )k!nk . We note that, for the eq. (1.19) and multiplying both the sides for 4, we can rewrite the eq. (1.18) also as follows :
2k 2 4 2T 1 4 2T k −λ −λ 2/1 +it = ()1 n = ∫∑ 2/1 +it dt ∫ ∑ p1 ... pn dt T T ≤ p T T λ + +λ = λ ... λ p x 1 ... n k 1 n 2 2 k −λ −λ 24 π k = 1 n +θ 4 ∑ p1 ... pn ∑ . (1.19b) λ + +λ = λ ... λ λ + +λ = λ ... λ 1 ... n k 1 n T 1 ... n k 1 n
The eq. (1.19b) can be related with the following Ramanujan modular equation concerning the modes corresponding to the physical vibrations of the bosonic strings :
∞ cos πtxw ' −π 2 e x w'dx ∫0 π 142 4 anti log cosh x ⋅ πt 2 2 − w' t w' 4 φ () e w' itw ' 24 = (1.19c) 10 +11 2 10 + 7 2 log + 4 4 and we obtain :
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2k 2 4 2T 1 4 2T k −λ −λ 2/1 +it = ()1 n = ∫∑ 2/1 +it dt ∫ ∑ p1 ... pn dt T T ≤ p T T λ + +λ = λ ... λ p x 1 ... n k 1 n 2 2 k −λ −λ 24 π k = 1 n +θ ⇒ 4 ∑ p1 ... pn ∑ λ + +λ = λ ... λ λ + +λ = λ ... λ 1 ... n k 1 n T 1 ... n k 1 n ∞ cos πtxw ' −π 2 e x w'dx ∫0 π 142 4 anti log cosh x ⋅ πt 2 2 − w' t w' e 4 φ ()itw ' ⇒ w' . (1.19d) 10 +11 2 10 + 7 2 log + 4 4
Applying sin (t log p) = (pit − p−it ) 2/ i and a version of the mean value theorem, we obtain:
2k 2 1 2T sin ()t log p 1 2k k −λ −λ dt = p 1 ... p n + ∫ ∑ 2k ∑ 1 n T T ≤ 2 k λ + +λ = λ ... λ p x p 1 ... n k 1 n
2 2 2k 2k j 2k − j +θ 2D 2k ∑ ∑ ∑ (1.20) = λ + +λ = λ ... λ λ + +λ = − λ ... λ 2 T j 0 j 1 ... n j 1 n 1 ... n 2k j 1 n with θ ≤ 1 and D is a constant, and we have also that:
2k 2 n Im X 1 2k k −λ −λ E i = p 1 ... p n . (1.21) ∑ 2k ∑ λ λ 1 n i=1 p 2 k λ +... +λ =k ... i 1 n 1 n
Now, for eq. (1.21) and multiplying both the sides for 8, we can rewrite the eq. (1.20) also as follows :
2k 2k 2T () n 8 sin t log p = Im X i + ∫ ∑ dt 8E ∑ T T ≤ p = p p x i 1 i
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2 2 2k 2k j 2k − j +θ 16 D 2k ∑ ∑ ∑ . (1.21b) = λ + +λ = λ ... λ λ + +λ = − λ ... λ 2 T j 0 j 1 ... n j 1 n 1 ... n 2k j 1 n
This expression, can be related with the Ramanujan modular equation that concerning the “modes” that correspond to the physical vibrations of a superstring :
∞ cos πtxw ' −π 2 e x w'dx ∫0 π 142 4 anti log cosh x ⋅ πt 2 2 − w' t w' 4 φ () 1 e w' itw ' 8 = , (1.21c) 3 10 +11 2 10 + 7 2 log + 4 4 and so, we obtain the following relationship :
2k 2k 2T () n 8 sin t log p = Im X i + ∫ ∑ dt 8E ∑ T T ≤ p = p p x i 1 i
2 2 16 D 2k 2k j 2k − j +θ ⇒ 2k ∑ ∑ ∑ = λ + +λ = λ ... λ λ + +λ = − λ ... λ 2 T j 0 j 1 ... n j 1 n 1 ... n 2k j 1 n ∞ cos πtxw ' −π 2 e x w'dx ∫0 π 142 4 anti log cosh x ⋅ πt 2 2 − w' t w' 4 φ () 1 e w' itw ' ⇒ . (1.21d) 3 10 +11 2 10 + 7 2 log + 4 4
Now let X1, X 2 ,... be an i.i.d. sequence of random variables uniformly distributed on the unit circle. We have π 1 2 θ [ iu Im X 1 ]= iu sin θ = () E e ∫ e d J0 u . (1.22) 2π 0
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Applying (1.22), Weierstrass’ product formula, and Merten’s formula, we π (x) Im X / p obtain that ∑i=1 i i converges mod-Gaussian, i.e.
π ()x Im X − 2 iu ∑ i u / 4 2 ()+γ i=1 1 u u log log x 4/ pi → − e E e ∏1 J0 (1.23) ∈ p p P p as x → ∞ , locally uniformly for u ∈C , where π (x) denotes the number of primes not exceeding x . By means of (1.20), (1.21), and the analogous results for odd integers, we can apply for fixed x the method of moments and obtain the following weak convergence
sin (t log p) iu ∑ ≤ 1 2T p x p → u → ∞ e dt ∏ J0 as T . (1.24) ∫T T p≤ x p
The improvement of Theorem 2 follows from combining (1.23) with the following proposition.
Proposition 1
Let c > 1 be a constant. Define x = elog T / N with N = (c'ec 2 4/ )log log T , c'> 1 another constant. Then for T sufficiently large independent of c and c' ,
sin (t log p) iu 2T ∑ ≤ 1 p x p = u + (()N + ()2 N ) e dt ∏ J0 O /1 c' 2c / log x (1.25) ∫T T p≤ x p uniformly for u ≤ c , u ∈ R . The remainder is o(exp (− c2 (log log T ) 4/ )), if c'log c'> /1 e .
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eiu = (iu )k / k!+θu2N ' /(2N )!' u ∈ R θ ≤ 1 From the Taylor expansion ∑k ≤2N '−1 , , with and N'= N , we obtain
sin (t log p) k 2N ' iu ∑ k 2N ' 1 2T p≤x ()iu 1 2T sin ()t log p u 1 2T sin ()t log p e p dt = ∑ ∑ dt +θ ∑ dt ∫T ∫T () ∫T T k ≤2N '−1 k! T p≤ x p 2N !' T p≤ x p
(1.26) with θ ≤ 1. The remainder is by (1.20)
N ' 2N ' 2 N ' c 1 1 (c n) O + 2N ' ∑ . (1.27) N !' 2 p≤ x p T
Using the bound (N )!' ≥ (N /' e)N ' , elementary results in the theory of primes, /1 p = log log x + c + O( /1 log x) n = π (x) ≤ 2x / log x namely the formulas ∑ p≤ x 1 and , and finally N'= N , this is
N N N ec 2 log log T 2c2 1 N 2c2 O + = O + (1.28) 4N log x c' log x for sufficiently large T independent of c,c' (since c,c'> 1).
Now let X1, X 2 ,... be an i.d.d. sequence of random variables uniformly distributed on the unit circle. The moments in (1.26) are by (1.20) and (1.21) equal to those of the stochastic model plus a remainder which is bounded by ()2D /T nk k! . The resulting remainders in (1.26), k ≤ 2N'−1, add up to N N O((c2n) /T )= O((2c2 / log x) ). Hence, (1.26) is equal to
k ()k n iu Im X i + ()()N + ()2 N ∑E ∑ O /1 c' 2c / log x (1.29) k ≤2N '−1 k! i =1 pi
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for sufficiently large T . Applying the above Taylor expansion again, we obtain
n Im X k 2N ' iu ∑ i k n 2N ' n u i=1 ()iu Im X u Im X = pi = i +θ i ∏ J0 E e ∑E∑ E ∑ (1.30) ≤ p ≤ − k! = p ()2N !' = p p x k 2N ' 1 i 1 i i 1 i with θ ≤ 1. The remainder already appeared in (1.26) and is O(( /1 c')N ) for sufficiently large T . Now, we note that multiplying in the eqs. (1.26) and (1.30) both the sides for 8, we obtain the following relationship:
sin (t log p) k 2N ' iu ∑ k 2N ' 8 2T p≤x ()iu 8 2T sin ()t log p u 8 2T sin ()t log p e p dt = ∑ ∑ dt +θ ∑ dt ⇒ ∫T ∫T () ∫T T k ≤2N '−1 k! T p≤ x p 2N !' T p≤ x p k 2N ' 8()iu k n Im X 8u 2N ' n Im X ⇒ i +θ i ∑E∑ () E ∑ . (1.30b) k≤2N '−1 k! i=1 pi 2N !' i=1 pi
Also this expression, can be related with the Ramanujam modular equation that concerning the “modes” that correspond to the physical vibrations of a superstring :
∞ cos πtxw ' −π 2 e x w'dx ∫0 π 142 4 anti log cosh x ⋅ πt 2 2 − w' t w' 4 φ () 1 e w' itw ' 8 = , (1.30c) 3 10 +11 2 10 + 7 2 log + 4 4 thence, we obtain the following mathematical connection:
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sin (t log p) k 2N ' iu ∑ k 2N ' 8 2T p≤x ()iu 8 2T sin ()t log p u 8 2T sin ()t log p e p dt = ∑ ∑ dt +θ ∑ dt ⇒ ∫T ∫T () ∫T T k ≤2N '−1 k! T p≤ x p 2N !' T p≤ x p k 2N ' 8()iu k n Im X 8u 2N ' n Im X ⇒ i +θ i ⇒ ∑E∑ () E ∑ k≤2N '−1 k! i=1 pi 2N !' i=1 pi ∞ cos πtxw ' −π 2 e x w'dx ∫0 π 142 4 anti log cosh x ⋅ πt 2 2 − w' t w' 4 φ () 1 e w' itw ' ⇒ . (1.30d) 3 10 +11 2 10 + 7 2 log + 4 4
Now we briefly discuss Selberg’s result about the rate of convergence in the central limit theorem of Im log ζ ( 2/1 + it ). From Theorem 2 we obtain
Lemma
Let x = elog T / N and N such that x and N / log log T → ∞ as T → ∞ . Assume further than N = O(log log T ) . Then
1 1 sin ()t log p b − 2 dt sup λt ∈[]T 2, T : ∑ ∈[]a,b − ∫ e t 2/ = O()/1 log log T . ()+ γ a π T log log x 2/ p≤ x p 2 a (1.31) In fact, the right hand side can be replaced by o( /1 log log T ). φ ( ) We denote by n u the left hand side of (1.15). We can bound the right hand side by (log log x / log log T →1) c log log x 2 − 1 e u 2/ ()φ ()u / ()log log x + γ 2/ −1 /u du + O π ∫ n . (1.32) −c log log x c log log T 18 Versione 1.0 18/06/2013 Pagina 19 di 35 > ( ) ≤ We choose c 0 such that J0 u has no zeros for u c . An inspection of the φ ( ) = φ( )( + ( )) ≤ proof of Proposition 1 shows that n u u 1 O u / log x , u c . On the other hand, we have φ(u / (log log x + γ ) 2/ )= 1+ O(u / log log T ), u ≤ c log log x . Plugging in these estimates gives the lemma. This lemma combined with the following bound { ∈[ ] ( ) ≥ } = ( ) t T 2, T : r ,1 x t clog log log T O /1 log log T (1.33) where c > 0 is a constant, yields Selberg’s result 1 Im log ζ ()2/1 + it b − 2 dt log log log T sup λt ∈[]T 2, T : ∈[]a,b − e t 2/ = O . (1.34) < () ∫a π a b T log log T 2/ 2 log log T Let x and y be positive real number, ap and bp be complex numbers with ≤ ≤ ap 1 and bp log p / log x , and k be a positive integer. By repeating the arguments in (1.18) and (1.19), we obtain 2k k 2T a 1 p ≤ 1 + π ()π () k + dt k! 6 k! x /T , (1.35) ∫T ∑ 1 2it ∑ 2 T p≤ x p p≤ x p 2k k 2T b 1 p ≤ 1 log p + π ()π () k ∑ + dt k! ∑ 6 k! x /T , (1.36) ∫T 2/1 it ()k T p≤ x p log x p≤ x p 2k k 2T a 1 p ≤ 1 + π ()π ()()− π k + dt k! 6 k! x y /T . (1.37) ∫T ∑ 2/1 it ∑ T y< p≤ x p y< p≤ x p If x ≤ T /1 k , the first two terms are bounded by (Ak )k and the third by (k(log log x − log log y + A))k , A > 0 some constant. 19 Versione 1.0 18/06/2013 Pagina 20 di 35 Now, if we multiply for 4 both the sides of the eq. (1.36), we obtain the equivalent expression: 2k k 2T b 4 p ≤ 1 log p + π ()π () k ∑ + dt 4k! ∑ 24 k! x /T , (1.37b) ∫T /1 2 it ()k T p≤x p log x p≤x p and this expression, can be related with the following Ramanujan modular equation concerning the modes corresponding to the physical vibrations of the bosonic strings, i.e. ∞ cos πtxw ' −π 2 e x w'dx ∫0 π 142 4 anti log cosh x ⋅ πt 2 2 − w' t w' 4 φ () e w' itw ' 24 = , 10 +11 2 10 + 7 2 log + 4 4 and thence, we obtain the following mathematical connection : 2k k 2T b 4 p ≤ 1 log p + π ()π () k ∑ + dt 4k! ∑ 24 k! x /T ⇒ ∫T 2/1 it ()k T p≤ x p log x p≤ x p ∞ cos πtxw ' −π 2 e x w'dx ∫0 π 142 4 anti log cosh x ⋅ πt 2 2 − w' t w' e 4 φ ()itw ' ⇒ w' . (1.37c) 10 +11 2 10 + 7 2 log + 4 4 Returning to the eqs. (1.35) – (1.37), for example, we obtain for a function g(u) ≤ 1 20 Versione 1.0 18/06/2013 Pagina 21 di 35 2V 2V 2T Λ() 2T b a 1 1 n log n = 1 p + p + () ≤ + g dt + + O 1 dt ∫T /1 V ∑ 2/1 it /1 V ∫T ∑2/1 it ∑ 1 2it T log T n≤T /1 V n log T T p≤T /1VVp p 2 ≤ T /1 p ≤ 32V ((AV )V + (AV )V + O(1)V ). (1.38) Also here, multiplying both the sides for 8, we obtain the following similar expression: 2V 2V 2T Λ() 2T b a 8 1 n log n = 8 p + p + () ≤ + g dt + + O 1 dt ∫T /1 V ∑ 2/1 it /1 V ∫T ∑2/1 it ∑ 1 2it T log T n≤T /1 V n log T T p≤T /1VVp p 2 ≤ T /1 p ≤ 24 2V ((AV )V + (AV )V + O(1)V ), (1.38b) and this expression can be related with the Ramanujan’s modular equations concerning the modes corresponding to the physical vibrations of the superstrings and the bosonic strings, i.e. ∞ cos πtxw ' −π 2 e x w'dx ∫0 π 142 4 anti log cosh x ⋅ πt 2 2 − w' t w' 4 φ () 1 e w' itw ' 8 = , 3 10 +11 2 10 + 7 2 log + 4 4 ∞ cos πtxw ' −π 2 e x w'dx ∫0 π 142 4 anti log cosh x ⋅ πt 2 2 − w' t w' 4 φ () e w' itw ' 24 = . 10 +11 2 10 + 7 2 log + 4 4 21 Versione 1.0 18/06/2013 Pagina 22 di 35 Thence, we can obtain, for example, the following mathematical connection: 2V 2V 2T Λ() 2T b a 8 1 n log n = 8 p + p + () ≤ + g dt + + O 1 dt ∫T /1 V ∑ 2/1 it /1 V ∫T ∑2/1 it ∑ 1 2it T log T n≤T /1 V n log T T p≤T /1VVp p 2 ≤ T /1 p ≤ 24 2V ((AV )V + (AV )V + O(1)V )⇒ ∞ cos πtxw ' −π 2 e x w'dx ∫0 π 142 4 anti log cosh x ⋅ πt 2 2 − w' t w' 4 φ () 1 e w' itw ' ⇒ . (1.38c) 3 10 +11 2 10 + 7 2 log + 4 4 Now, we consider the following function: −1 F()z = (ez +1) , (1.39) Developing in series the (1.39), we obtain: −1 − − ()+ F()z = (ez +1) = e z ∑()()−1 k ezk = ∑ −1 k e z 1 k . (1.40) k≥0 k ≥ 0 ≥ The equation (1.40) is defined in the complex half-plane Re z 0 with the exception of points in which ez = −1. Now, we examine the following function − p (ez −1) , p > 0 (1.41) − p performing the series expansion type (1.40) of the function (ez −1) , we find: − − + − z p − pz p k − zk p k 1 − z()p+k ()e −1 = e ∑ ()−1 e = ∑ e . (1.42) k≥0 k k ≥ 0 k 22 Versione 1.0 18/06/2013 Pagina 23 di 35 the series on the right hand-side of equation (1.42) is defined in the half- ≥ z = plane Re z 0 with the exception of points in which e 1. Integrating the equation (1.42) between the limits 0 and infinity, we find: ∞ + − ∞ + − − p p k 1 − ()+ p k 1 1 ()ez −1 dz = ∑ e z p k dz = ∑ ; (1.43) ∫0 ∫0 + k ≥0 k k ≥0 k p k ∞ ∞ − p − dz 0 − − − (ez −1) dz = e pz = − t p 1()()()1− t p dt = Γ p Γ 1− p e z = t ∫ ∫− p ∫ with . (1.44) 0 0 ()1− e z 1 Thence, we can rewrite all also as follows: ∞ + − ∞ + − ∞ − p p k 1 − ()+ p k 1 1 − dz ()ez −1 dz = ∑ e z p k dz = ∑ = e pz = ∫0 ∫0 + ∫0 − z p k ≥0 k k ≥0 k p k ()1− e 0 − = − t p−1()()()1− t p dt = Γ p Γ 1− p . (1.45) ∫1 Therefore: p + k −1 1 π = Γ()()p Γ 1− p = (0 < p <1) ∑ + ()π , (1.46) k ≥0 k p k sin p 1 For p = we have that 2 1 + − π k 1 1 = ∑ 2 . (1.47) ≥ 1+ 2k 2 k 0 k 22z −1 1 Remembering that: Γ()2z = Γ()z Γ z + , we obtain: π 2 23 Versione 1.0 18/06/2013 Pagina 24 di 35 1 Γ + k Γ()k +1 1 + − Γ()+ π 2k k 1 = 2 = 2k 1 = 1 2 2k , (1.48) 1 2 1 2 k k k!Γ Γ()k +1 22k ()k! Γ 2 2 and thence: 2k −2k π 2 = ∑ + . (1.49) k ≥0 k 1 2k 2 1 Thence, for p = the eq. (1.45), for the eqs (1.47) – (1.49), can be rewritten 2 also as follows: 1 ∞ − p p + k −1 ∞ p + k −1 + − π ()z − = − z()p+k = 1 = k 1 1 = ∫ e 1 dz ∑ ∫ e dz ∑ ∑ 2 , (1.50) 0 ≥ k 0 ≥ k p + k ≥ 1+ 2k 2 k 0 k 0 k 0 k or also as follows: 1 ∞ − p p + k −1 ∞ p + k −1 + − ()z − = − z()p+k = 1 = k 1 1 = ∫ e 1 dz ∑ ∫ e dz ∑ ∑ 2 0 ≥ k 0 ≥ k p + k ≥ 1+ 2k k 0 k 0 k 0 k 1 1 Γ + k Γ()k +1 + k −1 1 2 1 Γ()2k +1 π 1 1 2k = = = = = ∑ 2 2k ≥ 1+ 2k 1 1+ 2k 2 1 1+ 2k 2 k k 0 k k!Γ Γ()k +1 22k ()k! Γ 2 2 2k −2k π = 2 = ∑ + . (1.50b) k ≥0 k 1 2k 2 Multiplying both the sides of (1.42), for e−qz , and integrating with respect to z, between the limits 0 and infinity, we find: ∞ + − ∞ + − − p − p k 1 − ()+ + p k 1 1 ()ez −1 e qz dz = ∑ e z q p k dz = ∑ ⇒ ∫0 ∫0 + + k ≥0 k k ≥0 k q p k 24 Versione 1.0 18/06/2013 Pagina 25 di 35 ∞ ∞ − p − − − dz 0 + − − Γ(q + p)Γ(1− p) − ⇒ ( z − ) qz = pz qz = − q p 1()− p = z = ∫e 1 e dz ∫e e − z ∫ t 1 t dt for e t 0 0 ()1− e 1 Γ()q +1 (1.51) Therefore: p + k −1 1 Γ(q + p)Γ(1− p) = (0 < p < ,1 q > 0) ∑ + + Γ()+ , (1.52) k ≥0 k q p k q 1 For p = q = 2/1 , we find: 2k 2−2k = 2 ∑ + . (1.53) k ≥0 k 1 k Thence, in conclusion, for p = q = 2/1 , we find: ∞ + − ∞ + − − p − p k 1 − ()+ + p k 1 1 Γ(q + p)Γ(1− p) ()ez −1 e qz dz = ∑ e z q p k dz = ∑ = ⇒ ∫0 ∫0 + + Γ()+ k ≥0 k k ≥0 k q p k q 1 2k 2−2k ⇒ = 2 ∑ + . (1.53b) k ≥0 k 1 k Now, we note that the eqs. (1.50b) and (1.53b) can be related with the eqs. (1.15) and (1.32) for obtain the following mathematical relationships: 1 For example, multiplying the eq. (1.53b) for π , we obtain: sin (t log p) iu ∑ 2 ()+γ 1 2T p≤x eu log log x 4/ ∫ e p dt ⇒ T T c log log x 2 − 1 e u 2/ ()φ ()u / ()log log x + γ 2/ −1 /u du + O ⇒ π ∫ n −c log log x c log log T 25 Versione 1.0 18/06/2013 Pagina 26 di 35 ∞ + − ∞ + − 1 − p − 1 p k 1 − ()+ + 1 p k 1 1 Γ(q + p)Γ(1− p) ⇒ ()ez −1 e qz dz = ∑ e z q p k dz = ∑ = ⇒ π ∫0 π ∫0 π + + πΓ()+ k ≥0 k k ≥0 k q p k q 1 − 1 2k 2 2k 2 1 ⇒ = π ∑ + π . (1.54) k ≥0 k 1 k 1 Indeed, we have that the following relation: sin (t log p) iu ∑ 2 ()+γ 1 2T p≤x eu log log x 4/ ∫ e p dt ⇒ T T c log log x 2 − 1 e u 2/ ()φ ()u / ()log log x + γ 2/ −1 /u du + O π ∫ n . (1) −c log log x c log log T 2 Form this, we can to obtain π : sin (t log p) iu ∑ 2 2 ()+γ 1 2T p≤ x = eu log log x 4/ ∫ e p dt / π T T c log log x −u 2/ ()φ ()()+ γ − + 1 ∫ e n u / log log x 2/ 1 /u du O . (2) −c log log x c log log T 2 Also the following relation yields to π . Indeed: ∞ + − ∞ + − 1 − p − 1 p k 1 − ()+ + 1 p k 1 1 Γ(q + p)Γ(1− p) ()e z −1 e qz dz = ∑ e z q p k dz = ∑ = ⇒ π ∫0 π ∫0 π + + πΓ()+ k≥0 k k≥0 k q p k q 1 1 2k 2−2k 2 ⇒ = π ∑ + π . (3) k ≥0 k 1 k 2 The eqs. (2) and (3), thence, are both equal to π . 26 Versione 1.0 18/06/2013 Pagina 27 di 35 The same can be make with the eq. (1.50b). As we note, the eq. (1.54) can be + related with π (and thence with φ = 5 1 ) by the following Ramanujan 2 expression regarding just π : 24 10 +11 2 10 + 7 2 π = log + . (1.55) 142 4 4 We observe that the eq. (1.55) contain the number 24, that is related to the physical vibrations of the bosonic strings. Thence, for the (1.54) for p = q = 2/1 , we obtain the following mathematical connection: sin (t log p) iu ∑ 2 ()+γ 1 2T p≤x eu log log x 4/ ∫ e p dt ⇒ T T c log log x ∞ − 1 ()z − p −qz −u 2/ ()φ ()()+ γ − + 1 ⇒ e 1 e dz e n u / log log x 2/ 1 /u du O ⇒ π ∫0 ∫ −c log log x c log log T 2 ⇒ × 24 10 +11 2 10 + 7 2 log + 4 4 142 c log log x −u 2/ ()φ ()()+ γ − + 1 ∫ e n u / log log x 2/ 1 /u du O . (1.56) −c log log x c log log T A general relationship that link, π , prime numbers, Riemann Hypothesis and physical vibrations of bosonic strings. 27 Versione 1.0 18/06/2013 Pagina 28 di 35 2. EISENSTEIN INTEGER Eisenstein integers are the intersection points of a triangular lattice in the complex plane: Eisenstein integers (named after Gotthold Eisenstein), are complex numbers of the form: z = a + bω . where a and b are integers and 1 π ω = (−1+ i 3)= e 2 i 3/ 2 is a primitive (non-real) cube root of unity. The Eisenstein integers form a triangular lattice in the complex plane, in contrast with the Gaussian integers which form a square lattice in the complex plane, (see the two figures above ). 28 Versione 1.0 18/06/2013 Pagina 29 di 35 2.1 EISENSTEIN PRIME The Eisenstein primes which are equal to a natural number 3n -1 are 2, 5, 11, 17, 23, 29, 41, 47, 53, 59, 71, 83, 89 (with differences (after 5) equal to 6 (sexy numbers) or 12, for example. 41-29 = 12, 71-59 = 12, 83 = 12 -71, etc..) or even more. For larger numbers we have in fact the biggest differences, but always of the form 6k, for example 419-401 = 18 = 6 * 3 All, less the initial prime number 2 are of the form 6k - 1 (since, for n even, we have the form 6n -1) and therefore are not Gaussian integers, which are of the form 6k +1 as already seen before. Eisenstein primes are congruent to 2 mod 3 and Mersenne primes (except the smallest, 3) are congruent to 1 mod 3; thus no Mersenne prime is an Eisenstein prime. 29 Versione 1.0 18/06/2013 Pagina 30 di 35 3. RELATIONS BEETWEEN GAUSSIAN PRIMES AND EISENSTEIN PRIMES We have seen that the Gaussian primes "g" are of the form: g = 4k+3 with k = 0, 1, 2,… The sequence is the following: 3, 7, 11, 19, 23, 31, 43, 47, 59, 67, 71, 79, 83, 103, 107, 127, 131, 139, 151, 163, 167, 179, 191, 199, 211, 223, 227, 239, 251, 263, 271, 283, 307, 311, 331, 347, 359, 367, 379, 383, 419, 431, 439, 443, 463, 467, 479, 487, 491, 499, 503, 523, 547, 563, 571,… Eisenstein primes “e” are of the form : e = 3k+2 with k = 0, 1, 2,… The sequence is the following: 2, 5, 11, 17, 23, 29, 41, 47, 53, 59, 71, 83, 89, 101, 107, 113, 131, 137, 149, 167, 173, 179, 191, 197, 227, 233, 239, 251, 257, 263, 269, 281, 293, 311, 317, 347, 353, 359, 383, 389, 401, 419, 431, 443, 449, 461, 467, 479, 491, 503, 509, 521, 557, 563, 569, 587, … 30 Versione 1.0 18/06/2013 Pagina 31 di 35 From the intersection of the two sequences we obtain prime numbers of Gauss-Eisenstein. From the two preceding formulas 3k +2 and 4k +3, since the least common multiple between 3 and 4 is 12 we have the following formula for primes common Gauss-Eisenstein ρ: ρ = 12k + 11 with k = 0, 1, 2,… The sequence is the following: 11, 23, 47, 59, 71, 83, 107, 131, 167, 179, 191, 227, 239, 251, 263, 311, 347, 359, 383, 419, 431, 443, 467, 479, 491, 503, 563, 587, 599, 647, 659, 683, 719, 743, 827, 839, 863, 887, 911, 947, 971, 983, 1019, 1031, … It should be noted that all the prime numbers that are derived from this formula 12k +11 are all prime numbers that are simultaneously Gauss and Eisenstein primes. These numbers are then congruent to 11 modulo 12. We note that is 12 = 24/2, where 24 is the number connected to the modes corresponding to the physical vibrations of the bosonic strings by the following Ramanujan’s modular equation: ∞ cos πtxw ' −π 2 e x w'dx ∫0 π 142 4 anti log cosh x ⋅ πt 2 2 − w' t w' 4 φ () e w' itw ' 24 = . 10 +11 2 10 + 7 2 log + 4 4 31 Versione 1.0 18/06/2013 Pagina 32 di 35 3.1 CORRELATION WITH SOPHIE GERMAIN PRIMES We observe that some of the previous sequence numbers are also Sophie Germain primes. A Sophie Germain prime is a prime p such that 2 p+1 is also prime. The number 2 p + 1 is instead called safe prime . The Sophie Germain prime numbers up to 1000 are: 2, 3, 5, 11 , 23 , 29, 41, 53, 83 , 89, 113, 131, 173, 179 , 191 , 233, 239 , 251 , 281, 293, 359 , 419, 431, 443, 491, 509, 593, 641, 653, 659 , 683 , 719 , 743 , 761, 809, 911 , 953, 1013, 1019 , 1031 , … Let’s recall the list of numbers ρ = 12k+11 with k = 0, 1, 2,… 11 , 23 , 47, 59, 71, 83 , 107, 131 , 167, 179 , 191 , 227, 239 , 251 , 263, 311, 347, 359, 383, 419 , 431 , 443 , 467, 479, 491 , 503, 563, 587, 599, 647, 659 , 683 , 719 , 743 , 827, 839, 863, 887, 911 , 947, 971, 983, 1019 , 1031 , … Out of 40 numbers of Sophie Germain, as many as 20, are also Gaussian- Eisenstein (marked in bold in the two sequences). So we have a new connection, with the numbers of Sophie Germain, of which ρ is a subset and is exactly the half. We can say that 50% of the Sophie Germain primes are also Gauss- Eisenstein primes This can be demonstrated with a simple consideration: 32 Versione 1.0 18/06/2013 Pagina 33 di 35 In the list of Sophie Germain primes, if we want to know if its any number N is a number ρ of the form 12k +11, obviously we must subtract 11 and divide by 12, if the result is integer, then N = ρ A few examples: 3779 = Sophie German prime = prime ρ since (3779-11)/12 = 3768/12 = 314 (integer) Instead 4793 (Sophie German prime) it is not because (4793-11) / 12 = 4782/12 = 398,5 is not an integer, and then 4793 is not a prime number of Gauss-Eisenstein. Now chosen any prime number N of Sophie Germain and applying the rule above, we have (N -11) / 12 = integer or half-integer The division produces only 2 possible results, or an integer, and then N is also a prime number of Gauss-Eisenstein or is a half-integer of the form x+1/2 and N isn’t a Gauss-Eisenstein prime. As a result 50% of the Sophie Germain primes are also Gauss-Eisenstein primes. Also here, we note that is present the number 12 that, as we have described above, is connected to the modes corresponding to the physical vibrations of the bosonic strings. 33 Versione 1.0 18/06/2013 Pagina 34 di 35 3.2 PROOF THAT THERE ARE INFINITELY MANY SOPHIE GERMAIN PRIMES It is conjectured that there are infinitely many Sophie Germain primes. Since Gauss-Eisenstein primes are infinite and there is a relationship with Sophie Germain primes, we can by induction affirm that Sophie Germain primes are infinite. This is because the set ρ of Gauss-Eisenstein of the form 12k +11 is infinite and these numbers are an infinitely subset so that produce endless Sophie Germain primes. . 34 Versione 1.0 18/06/2013 Pagina 35 di 35 4. REFERENCES 1. Andrew V. Lelechenko – “Multidimensional exponential divisor function over Gaussian integers” – arXiv:1211.0724v1 [math.NT] – 4.11.2012 2. Martin Wahl – “On the Mod-Gaussian convergence of a sum over primes” – arXiv:1201.5295v2 [math.NT] – 08.07.2012 3. Pasquale Cutolo – “A note on the Study of some particular functions. Considerations and observations” – www.matematicamente.it/approfondimenti . in Italian 35