Goldbach’s 2 Conjectures with Proof
Mantzakouras Nikos
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Mantzakouras Nikos. Goldbach’s 2 Conjectures with Proof. 2021, 10.13140/RG.2.2.32893.69600.
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The 2 Goldbach's Conjectures with Proof
Research Proposal · June 2020
DOI: 10.13140/RG.2.2.32893.69600
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Goldbach’s 2 Conjectures with Proof
Mantzakouras Nikos
M.Sc in Applied Mathematics && Physics
e-mail: nikmatza@gmail.com
Abstract
The 1741 Goldbach [1] made his most famous contribution to mathematics with the conjecture
that all even numbers can be expressed as the sum of the two primes (currently Conjecture
referred to as ”all even numbers that is greater than 2 can be expressed as the sum- two primes).
Yet, no proof of Goldbach’s Conjecture has been found. This assumption seems to be right for a
large multitude of numbers using numerical calculations. Some examples are 10 = 3 + 7, 18 =
7 + 11, 100 = 97 + 3, and so on. But there is a multitude of primes of even numbers which,
i would say that it is irregular, but increases with the order of the even numbers. The same
happens and with an odd number as example 25, ie. 25 = 3 + 3 + 19 = 3 + 5 + 17 = 3 + 11 + 11 =
5 + 7 + 13 = 7 + 7 + 11. We say only 5 cases and only those that meet the constant sum of 25.
First turns in Theorem 3 that there may be at least a pair of primes, such that the sum is equal to
every even number. But at the same time reveals that the method of finding all pairs satisfy this
condition. According to Theorem 4, the second guess interrupted to form the first guess and this
is primarily elementary, to prove the truth of 1. The alternative and the side stream assistant,
for the prove Goldbach’s Conjecture in this research, was the Mathematica program which is the
main tool for data collection, but also for finding all the steps in order to demonstrate each part
of the proof. We must mention that Vinogradov is proving to 1937, that for every sufficiently
large number can be expressed, that is the sum of the three primes. And finally, the Chinese
mathematician Chen Jing as demonstrated for a big prime and with a constant number that is
the sum of the tree first in 1966 [2]. Finally, the investigation of Goldbach’s Conjecture has
acted as a catalyst for the creation and development of many methods that are useful, with many
theorems that help and other areas of mathematics.
Definition.
A natural number p > 1 is called a prime number if and only if the only divisors of is ±1 and ±p. A natural
number n > 1 which is not prime, will be called composite. Otherwise, we could say that are relative primes
if they do not have another shared divisor other than 1. The 2 (is the first of primes) but outside of 2 is odd
numbers. The 25 first primes numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 61, 67, 71, 73, 79,
83, 89, 97. We note that the number 1 is considered neither prime nor composite.
Lemma 1.
Every prime number greater than 1, it is odd and is written in the form 2k + 1 and therefore the sum
of 2 primes is an even number.
Proof
Let p, prime in N. Suppose pk , pk+λ by κ, λ ∈ N, pk+λ > pk then we will have pk + pk+λ = 2k + 1 +
2(k + λ) + 1 = 2(2κ + λ + 1) to be even as easily seen.
1
Lemma 2.
The sum of 3 odd numbers it is an odd number.
Proof
If p, prime in N, natural numbers. Assume, pk , pk+λ , pk+ρ κ, λ, ρ ∈ N, pk+ρ > pk+λ > pk then we’ll
have pk + pk+λ + pk+ρ = 2k + 1 + 2k + 2λ + 1 + 2k + 2ρ + 1 = 2(3k + λ + ρ + 1) + 1 that’s odd as easy it
seems.
Theorem 1.
If we have a integer n = 2h, h ∈ N even number, asking to find the upper and lower limit of the sum
of two odd, which equal with n.
Proof
From Lemma 1 we have:
i) n = 2(2h − λ + 1) ⇒ 2h − λ = n/2 − 1 ⇒ 2h > n/2 − 1 ⇒ h >
n−2
4 ,
λ > 0, h ∈ N.
n−4
ii) The largest odd should be less than n − 3. Suppose therefore that 2h + 1 ≤ n − 3 ⇒ h ≤
· So
2
n−2
n−4
n−2
<h≤
the integer part of a
. Were
the limits of the number of all cases, it will be
4
2
4
rational number.
I.The Existence Theorems 2 or 3 primes, with constant sum.
Theorem 2.
For n > 2, n = 2h, h, w ∈ N and n = p1 + p2 , which p1 , p2 are prime numbers, and there is a positive integer j with j = 2w if n/2 is odd and j = 2w + 1 if n/2 is even, then n/2 − j and n/2 + j is odd
and possibly also prime numbers [8].
Proof
i) We assume that we have two odd and possibly primes p1 = n/2 − ǫ and p2 = n/2 + µ with ǫ, µ to
be positive numbers. Because n = p1 + p2 = n/2 + µ + n/2 − ǫ ⇒ µ = ǫ.
ii) Previously by (i), we assume that there is an integer, j > 0 so j = µ = ǫ, and then implies that n = p1 +p2
with p1 = n/2 − j and p1 = n/2 + j. We can now transform the relationship n = p1 + p2 in accordance
with previously to n/2 − (p2 − n/2) = n − p2 = p1 and also n/2 + (p2 − n/2) = p2 . From the mathematical
logic is clear that if j = 2w with n/2 odd and j = 2w + 1 if n/2 is even, the p1 , p2 results always odd
numbers. Of course if may be primes, as you know it will always be odd numbers. Therefore there is always
a positive j integer such that n/2 − j and n/2 + j is odd numbers and possibly primes numbers and apply
that n = (n/2 − j) + (n/2 + j).
We define as π(x) is the function that determines the number √
of primes that are less than
√ the equal by x.
Specifically applicable for the sieve of Eratosthenes that k = π[ 2x] integer part of 2x. We prove then
that there may be at least a couple of primes such that there the sum is equal to n.
2
II. The First Conjecture of Goldbach’s.
Theorem 3.
√
Given an even, n > 2, n = 2h, h, w ∈ N, k = π( n), k ≥ 2 and pk is prime corresponding to an index k. Then there are positive integers j so in order to apply in each case conditions:
a. If n/2 is odd. then j = 2w and
b. If n/2 is even, then j = 2w + 1
so that there is at least a pair, and for which each of the pair n/2 − j and n/2 + j to be prime and
not divisible by any of the primes 2, 3, 5, 7, . . . , pk .
Proof
We start with the n > 2, n = 2h, h >= 0, h, w ∈ N from Theorem 1., and Theorem 2. occurs that
there then n/2 − j and n/2 + j is odd and possibly also prime numbers with j = 2w. In this case the maxi1
mum number of values w ∈ N will occur from inequality n/2+2w ≤ n−3 ⇒ w ≤ ·(n/2−3)∧2w ≤ n/2−3.
2
It is obvious as total solutions that represent all cases for an n/2 = odd is S = {0, 2, 4, 6, , n/2 − 3} while
when n/2 = even is S = {1, 3, 5, 7, n/2 − 3}. The set that we get andp
the inequality
√ mod 2, i.e. Accordance
with the sieve of Eratosthenes is n/2 ± j 6= 0(mod2) that so k = π[ 2n/2] = π( n). Must will first need
from the set S, remove or reject all cases of individual systems modulo.
< ms >
s1
s2
n/2 ± j = 0 (modp2 )
n/2 ± j 6= 0(mod2)
n/2 ± j = 0 (modp3 )
n/2 ± j 6= 0(mod2)
.........
.........
.........
n/2 ± j = 0 (modpk )
sf
n/2 ± j 6= 0(mod2)
The comprehensive solution of all systems, is from the Union of the individual solutions of each system
s1 , s2 , s3 , n, sf , 1 ≤ f ≤ k − 1 and is the unique solution of systems, with the respective sets Sf , i.e
f
S
R=
Si .
i=1
3
The clear solution such as we seek will be the difference of sets S and R, i.e Ω = S − R. So the solution Ω
that we are interested coincides with the solution of the system unequally mod i.e.
n/2 ± j = 0 (modp1 )
n/2 ± j 6= 0(modp2 )
.........
< mS >
.........
.........
n/2 ± j 6= 0(modpv )
.........
.........
n/2 ± j 6= 0(modpk )
for which each of the pair n/2−j and n/2+j is primes and not divisible by any of the primes 2, 3, 5, 7, . . . , py ,
. . . , pk , with 1 ≤ v ≤ k. By this logic therefore determine that the possibility that n/2 − j and n/2 + j
be prime numbers, it is now safe to say there may be at least a pair (p, p2 ) of primes, such that their sum
equals n, because the set Ω is never empty set because S is never empty set and R 6= S with R ⊂ S. If
N (Ω} = {xi : xi ∈ Ω} with {i = 1, 2, 3, . . . , λ} where λ the cardinality of the set Ω, then the N (Ω) define
the elements of the set Ω. The pairs of primes arising from the set Ω, given by the relations:
1 pi
′
1 pi
= n/2 − xi
with 1 pi + 1 p′i = n
= n/2 + xi
Further to previous we symbolized to P (Ω) = {(1 p1 , 1 p′1 ) , (1 p2 , 1 p′2 ) , . . . (1 pλ , 1 p′λ )} the set of pairs that
correspond to the set Ω, which is derived from the system with unequally mod, < mS > and constitutes
the first part of the set of solutions of couples (1 pi , 1 p′i ) with {i = 1, 2, . . . , λ} so that the sum to equals
n and cardinality λ. Finally we define as T = {2, 3, 5, 7, . . . , pk } to set the of primes so that the couples
n/2 − j and n/2 + j who are primes not divisible by any of the primes 2, 3, 5, . . . , pk , as defined multitude of
k. More specifically define the set of primes P (T ) = {(2 pν , 2 p′ν ) :2 pν ∈ T and, 2 pν / (2 pν · m + vν ) m ∈ Z:
vν ∈ {n/2 ± j = 0 (modpν ) ∧ n/2 ± j 6= 0(mod2)}} with 1 < ν ≤ k, k ≥ 2, where vν the residues arising from the system {n/2 ± j = 0 (modpν ) ∧ n/2 ± j 6= 0(mod2)} with cardinality σ. Finally the complete
set of pairs of the primes to sum n will be the set L2 = P (Ω) ∪ P (T ) as defined above, with cardinality σ + λ.
III.The Second Conjecture of Goldbach’s.
Theorem 4. For n > 2, n = 2h + 1, h ∈ N and n = p1 + p2 + p3 , with p1 , p2 , p3 prime numbers. Namely
every odd number is obtained as the sum of 3 prime numbers.
Proof. According to the relations n = p1 + p2 + p3 ⇒ p1 + p2 = n − p3 , if call n = 1 pi + 2 pi + 3 pi and if
we have many different values 3 pi and other primes, then we can call 2 Si = 1 pi + 2 pi = n − 3 pi the sums
of 2 primes 2 Si . According to Theorem 3, we know the cardinality of cases are 2 Si , with σi + λi , in each
case n − 3 pi of different values 3 pi according to each preset selection. The method of finding all the cases
is precisely from selected values that are 3 pi which is the first of the primes less of n, to the greatest prime
of the two is closer to the value n/3. The fact leads the order of the Triad p1 , p2 , p3 in relation to n/3.
i) In this case, if the value of the lower limit of the selected values 3 pi will be the value p3 of that is
obviously the second greater price series during the first of prime numbers after i.e Round (n/3). In the
mathematica language will define this number N extPrime[N extPrime[Round[n/3]]].
4
ii) If p1 < p2 < n/3 < p3 ∨ p1 ≤ p2 < n/3 < p3 ∨ p1 < p2 < n/3 ≤ p3 . In this case the value of the
lower limit of the selected values 3 pi will be the value p3 of which is obviously a next-largest number first
value after Round(n / 3). In language of mathematica will define this figure as N extP rime[Round[n/3]].
iii) If p1 = p2 = n/3 = p3 . The value of the lower limit of the selected values 3 pi will be the value of
p3 which is obviously the n/3 = Round(n/3).
And for the three cases in relation to each device, we symbolized as the prime minimum δmin Finally,
the biggest 3 pi the first of the primes below, will be denoted by mathematica as NextPrime [n, −µ] ≤ n − 6,
µ ≥ 1 and will be the largest prime number before from n less than or equal to n − 6, which symbolizes
δmax . If you now call the multitude of the primes, π ([δmin , δmax ]) = δ we will know how many sums of
2 will have, i.e {i = 1, 2, 3, ., δ}. Therefore the set of summation of total solutions of 2 primes will be
L3 = {2 S1 , 2 S2 , . . . , 2 Sδ }. If we now call κi = σi + λi with {i = 1, 2, 3, . . . δ} by Theorem 3, the final number
of the individual triads κt ie. the final number κt = κ1 + κ2 + . . . + κδ will be the total number of pairs of
primes, for sums of 2, 2 Si = 1 pi + 2 pi = n − 3 pi , with multitude of δ and hence the total number of triads
equal κt ,using δ multitude sums with two primes, making δ -multiple use of the method Theorem 3. The
permissible cases 2 Si , i = 1, 2, . . . , δ therefore the will is the total L3 = {2 S1 , 2 S2 , . . . , 2 Sδ } . Namely every
odd number is obtained as the sum of 3 of primes numbers after respectively switched to multiple totals 2
primes, δ multitude, the existence of which fully ensures the Theorem 3.
√
For example sum 2 primes, let’s say when m = n/2 = 159 is odd and n = 17.83 therefore i.e get to
mod, the first 2, 3, 5, 7, 11, 13, 17. The corresponding program with mathematica and according to the theory would be:
Program 1.
n/2 := 159; m := n/2
Cases[Reduce[ Mod[m +j, 17] 6= 0&& Mod[m -j, 17] 6= 0&&
Mod[m +j, 13] 6= 0&& Mod[m -j, 13] 6= 0&&
Mod[m +j, 11] 6= 0&& Mod[m -j, 11] 6= 0&&
Mod[m +j, 7] 6= 0&& Mod[m -j, 7] 6= 0&&
Mod[m +j, 5] 6= 0&& Mod[m -j, 5] 6= 0&&
Mod[m +j, 3] 6= 0&& Mod[m -j, 3] 6= 0&&
Mod[m +j, 2] 6= 0&& Mod[m -j, 2] 6= 0&&
0 ≤ j ≤ m-3, {j}, Integers], Except[j]]
Results: j=8, j=20, j=22, j=32, j=52, j=70, j=80, j=92, j=98, j=112, j=118, j=122} are the values for
j. The N (Ω} = {xi , xi ∈ Ω} = {8, 20, 22, 32, 52, 70, 80, 92, 98, 112, 118, 122} with λ = 12. Therefore the
set of pairs will be P (Ω) = {(1 p1 , 1 p′1 ) , (1 p2 , 1 p′2 ) , . . . , (1 pλ , 1 p′λ )} = {(37, 281), (41, 277), (47, 271), (61, 257),
(67, 251), (79, 239), (89, 229), (107, 211), (127, 191), (137, 181), (139, 179), (151, 167)} with thereafter the T =
{2, 3, 5, 7, . . . , 17} with k = 7 and shuts off after looking at how many primes values we receive from all T.
The corresponding program will be mathematica according to relevant theory.
Program 2.
√
′
n := 2 · 159;m:=n;c:=IntegerPart[ m];Cases[Table[Reduce[Mod[m-p ,k]==0 &&
′
′
Mod[m+p ,2] 6= 0 && Mod[m-p ,2] 6= 0 && p= < p’ && p= < c && 0 <= p’ <= (m-3) &&
′
p==m-p′ , {p,p },Primes],{k,1,c}],Except[False]].
Final result: {(5, 313)(7, 311)(11, 307)} with multitude of σ = 3. So all of the pairs of primes will be
λ + σ = 12 + 3 = 15, according to the theorem 3 & 4 mentioned above. Something similar happens for
example if n = 2∗ 158 i.e when arising that n/2 = m (even) therefore we have N (Ω} = {xi : xi ∈ Ω} =
{9, 21, 69, 75, 99, 105, 111, 135} and λ = 8 using the same programs 1,2 in language mathematica. Therefore the set of pairs is {(23, 293), (47, 269), (53, 263), (59, 257), (83, 233), (89, 227), (137, 179), (149, 167)}, then
5
T = {2, 3, 5, 7, . . . , 17} with k = 7 (is the multitude of primes) for which accept to be with multitude of σ = 2,
which is {(5, 311), (3, 313)}. So all of the pairs of primes will be λ+σ = 8+2 = 10. b) For example the sum of
3 primes, consider the case when n = 111. According to Theorem 4, and iii case, that the δmin = n/3 = 37.
With the language mathematica as N extPrime[n, −µ] ≤ n − 6, µ ≥ 1 with n = 111 and µ = 3, and therefore
the largest prime number before n = 111, less than or equal to n − 6, will be δmax = 103. So the respectively
π ([δmin , δmax ]) = 16, that 16 is the number of each sum, for which the program will seek to find all the
triples after verifying the relation 2 Si = 1 pi + 2 pi = n − 3 pi , i = 1, 2, . . . , 16. The allowed cases according to
the assumptions 2 Si , i = 1, 2, . . . , 16 with δ = 16 are, L3 = {2 S1 , 2 S2 , . . . , 2 Sδ } = {111 − 103, 111 − 101, 111 −
97, 111−89, 111−83, 111−79, 111−73, 111−71, 111−67 111−61, 111−59, 111−53, 111−47, 111−43, 111−
41, 111−37} = {8, 10, 14, 22, 28, 32, 40, 44, 50, 52, 58, 64, 68, 70, 74}. By Theorem 4, the final multitude of the
individual triads, the multitude κt which would occur with multiple use of Theorem 3., without iterations
we will obtain the relation, kt = k1 +k2 +. . .+kδ = 1+2+2+3+3+2+2+3+3+4+3+4+2+1+1+1 = 36,
thus the relevant full program in mathematica will be:
Program 3.
n:=111;Reduce[p1 + p2 + p3 == n && p1 > 2 && p2 > 2 && p3 > 2 && p1 <= p2 &&
p2 <= p3 , {p1 , p2 , p3 }Primes];
Count[Reduce[p1 + p2 + p3 == n && p1 > 2 && p2 > 2 && p3 > 2 && p1 <= p2 &&
p2 <= p3 , {p1 , p2 , p3 },Primes],Except[False]]
Total 36 Triads will therefore be in ascending order:
(p1 = 3, p2 = 5, p3 = 103) , (p1 = 3, p2 = 7, p3 = 101) , (p1 = 3, p2 = 11, p3 = 97) , (p1 = 3, p2 = 19, p3 = 89) ,
(p1 = 3, p2 = 29, p3 = 79) , (p1 = 3, p2 = 37, p3 = 71) , (p1 = 3, p2 = 41, p3 = 67) , (p1 = 3, p2 = 47, p3 = 61) ,
(p1 = 5, p2 = 5, p3 = 101) , (p1 = 5, p2 = 17, p3 = 89) , (p1 = 5, p2 = 23, p3 = 83) , (p1 = 5, p2 = 47, p3 = 59) ,
(p1 = 5, p2 = 53, p3 = 53) , (p1 = 7, p2 = 7, p3 = 97) , (p1 = 7, p2 = 31, p3 = 73) , (p1 = 7, p2 = 37, p3 = 67) ,
(p1 = 7, p2 = 43, p3 = 61) , (p1 = 11, p2 = 11, p3 = 89) , (p1 = 11, p2 = 17, p3 = 83) , (p1 = 11, p2 = 29, p3 = 7) ,
(p1 = 11, p2 = 41, p3 = 59) , (p1 = 11, p2 = 47, p3 = 53) , (p1 = 13, p2 = 19, p3 = 79) , (p1 = 13, p2 = 31, p3 = 67) ,
(p1 = 13, p2 = 37, p3 = 61) , (p1 = 17, p2 = 23, p3 = 71) , (p1 = 17, p2 = 41, p3 = 53) , (p1 = 17, p2 = 47, p3 = 47) ,
(p1 = 19, p2 = 19, p3 = 73) , (p1 = 19, p2 = 31, p3 = 61) , (p1 = 23, p2 = 29, p3 = 59) , (p1 = 23, p2 = 41, p3 = 47) ,
(p1 = 29, p2 = 29, p3 = 53) , (p1 = 29, p2 = 41, p3 = 41) , (p1 = 31, p2 = 37, p3 = 43) , (p1 = 37, p2 = 37, p3 = 37) .
Conclusions.
The purpose of this project was to solve the Goldbach conjecture using the same statement: For every integer
n ≥ 2, there is an integer j such that n/2 + j and n/2 − j are prime numbers. The two approaches were made
using this statement, which resulted in the development and resolution of the two assumptions. Hopefully
these speculations reveal new approaches to solve the problems and make solving this problem of old, a
much easier task. Since the Goldbach put the two speculations, have passed 259 years. Illustrating this is
”a unfathomable enigma”, many mathematicians and amateurs from many countries have expended great
energy. But until now, no one could find a suitable method to perform this research,since the opportunity
to prove anyone, this problem depends only on the available theories. Obviously, the available theories
are not very effective, because they could not fully take to prove this very complex problem of existence
and distribution and therefore all these can not be used to prove the conjecture of Goldbach [3, 4, 5, 6, 7].
Therefore, we must actively open a new path, find a new way and method to we analyze and methodically
finding and selective union of those sets that comprise this seemingly simple but so complex problem. Thus,
with the new technique is clearly shown that success is easier and more understand and practically more
possible.
6
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