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Section 5. Mathematics
Section 5. Mathematics
Druzhinin Victor Vladimirovich Lazarev A. A. National Research Nuclear University «MEPHI» Sarov Physical-Technical Institute, Department of Mathematics E-mail: vvdr@newmail.ru
Quantitatively Exact proof of the Euler-Goldbach hypothesis
Abstract: It is shown that any even number can always be composed by two primes. Algorithms that give approximate number of pair of Euler-Goldbach and comparison with real data are presented. Keywords: primes, a pair of Euler-Goldbach, Euler v - function.
The hypothesis of the Euler-Goldbach were formulated in 1742: every even integer can be represented by a sum of two primes (PR) 2N = pa + pb. A couple of these two PR we call a pair of Euler-Goldbach (PEG). It is a binary version of hypothesis. Trinary version — any odd number is the sum of three PR was proved by academician Vinogradov in 1937 for PR, greater than 10 6 846168 [1; 2]. Proof ofbinary version does obvious the trinary hypothesis. In the works of the authors [3; 4], the binary hypothesis has been proven using a simplified version ofprobability theory. It gave correct qualitative and semi-quantitative result with a controlled error. Although our theory of probabilities confirms the hypothesis of the Euler — Goldbach, accurate quantitative evidence was not given.
In this work the hypothesis of the Euler-Goldbach were proved without the theory of probability by comparing the arithmetic progressions (AP), the introduction of the generalized Euler ty -function for relatively PR and by finding majority. The logic of the previous works [3; 4] is briefly the following. l.Take an even number 2N, and we find the greatest PR pm <V 2N - 3. 2 . The generated basis set (BS) consecutive PR am = { = 3; p2 = 5;...;pm } . 3.Find the number of columns n = (N -1) / 2 (the largest integer not exceeding (N -1) / 2 ) a three line of the matrix D (2N) and the first buffer zone (1BZ) Ln = [1, n ]. The numbers of 1BZ are a middle (second) row of the matrix D(2N) . 4. Recorded top line (2N) xt = (2t + l),t e LN . This is a set of odd numbers from "3" to (2n+l).5. Record bottom line D(2N)yt = (2N - 2t -l), te LN .This isa set of decreasing odd numbers from « 2N - 3 » to (2N - 2n -1) .6. Compose the m AP on the top row of the matrix (2N) \xt = at + pt (k +1)} ,where ai =(pj -1) / 2. There is an integer k > 0 and further. 7. Compose the m AP on the bottom row of the matrix (2N) {yi = $ + pk} . The initial number ¡it is from the decision of the comparison ((at +1)- +ai) Pt (modpt ),where Pt are included in a complete system of deductions [1,pt]. The number ^ ((0 < ^ < pt)) is the remainder when dividing 2N into pt, i. e. 2N = (modpi) .8. Delete from 1BZ Ln all the values of these 2m AP. 9. Remained not deleted numbers t give PEG {2t + 1;2N - 2t -1}.
As example, take 2N = 56, m = 3, p3 = 7,n = 13,
^ 3 5 7 9 11 13 15 17 19 21 23 25 27^
D (56 ) =
1 2 3 4 5 6 7 8 9 10 11 12 13 53 51 49 47 45 43 41 39 37 35 33 31 29
In the matrix D (2N) contains all possible sums of two odd numbers, giving 2N . Our task is to make the columns with two simple numbers. To make this, first in the top row delete all composite numbers that are multiples of pj e am (2N) . At the same time it will delete entire column. After the "cleaning" of the top row, composite numbers will be deleted in the bottom row, with the deletion of the entire column again. The remaining matrix contains only PEG. Dynamics of changes of the matrix are shown below.
D (56 ) =
; -D(56) =
1 6 9 53 43 37
.(2)
(3)
(1)
1 2 3 5 6 8 9 11 53 51 49 454339 37 33
V y v y
Thus, the number 2N = 56 has three PEG. The same result we get on the second row of the matrix, i. e. with 1BZ. Delete the numbers of the top AP {4 + 3k ;7 + 5k;10 + 7k}. At the same time delete the numbers of the bottom AP {2 + 3k;5 + 5k;3 + 7k}. In the middle row it will remain three numbers {1; 6; 9} ,which give the same good columns set of PEG. Therefore, we will use the second version: 1BZ and the ri ght set ofAP (RAP) fa, + pi (k +1);
I P, + Pk
The number t, not deleted by RA.P from 1BZ gives PEG {2t + 1;2N - 2t -1}.
To proceed further, we introduce a generalized y (m) — the Euler function. The original the Euler function ) gives the number of integers that are not deleted by a single AP {y + Pik},0 < Ji < Pi at the interval of the second buffer zone (2BZ)
m m
Cm = [l,Tm], where Tm = n• This number V(m) = n( -1) .
i=i i=i
For example, we have two PR { = 3, p2 = 5} ,and two single AP {x = 2 + 3k; y = 1 + 5k}. Here 2BZ C2 =[1,72 ] = [1,15]. Deleted numbers are Not removed the number
. There are eight of them in accordance with the Euler's function (p(2) = (3 -1)(5 -1) = 8. Next, for each PR pt it will enter the index 8t = 1 or 2. If pt has two mismatched (double) AP, Sj = 2, if pt has one AP, St = 1. For example, when AP {2 + 3k;1 + 3k;2 + 5k} } ^ = 2,82 = 1 . In this case the generalized
m
Euler function y (m) = H (pt - Sj ) gives the number of all not de-
i=1
leted these AP in 2BZ. The example. In C2 this three AP removed
' 3 5 7 111317 19 23 >
' 3 13 19 ^
Quantitatively Exact proof of the Euler-Goldbach hypothesis
. There are four not removed numbers {3;6;9;15} in accordance with y (2) = (3- 2)(5 -1) = 4.
The probability to find not delete number in 2BZ is am =W (m) / Tm . We do not know the distribution probability density of not deleted numbers in 2BZ. But to evaluate the result it will make a uniform distribution. Then the probabilistic number of PEG S (2N) = (n-y(m)/Tm).
This number is of course different from
Table 1. - The exact S(2N) and the calculated S (2N) on the
numbers
the true value PEG S (2N ) . In the example matrix in D(56) have Tm = 105;i//(3) = (3 - 2 )(5 - 2)(7-l) = I8;rnm = 0,171;S(2N ) = 2,223; S (2N ) = 3 . This approach allowed us to qualitatively explain the growth of S(2N) by growth of 2N, sharp almost twice increase of S(2N) for 2N with a multiple of «3» compared to other 2N, and the existence of PEG for any arbitrarily large 2N. There is good quantitative proportion of numbers of PEG with large 2N, see table № 1.
theory of probability the amount of PEG for different even 2N .
2N 94 600 1000 2000 3000 4000 5000 6000 10000 17000 30000
S (2N ) 5 23 31 37 104 68 75 178 127 215 602
S (2N) 3,25 21,42 20,61 35,46 98,07 61,31 74,11 168,28 127,71 207,17 607,31
The question arises whether or not these 2m ^AP to delete all numbers in 1BZ, i. e. all numbers in the middle row of the matrix D (2N ) , and not deleted numbers will not be. Therefore, there will not be PEG.
Let's move to another, the quantitative proof of the hypothesis ofEuler-Goldbach. There are two intervals of 2BZ Cm and 1BZ Lm . Always for N > 7 Lm c Cm , since n(m) < Tm i. e. 1BZ is inside 2BZ. We know the exact number of not deleted numbers in 2BZ V (m) . Could it be that the smallest not deleted sm number was greater than n(m) (at the smallest possible S(2N) and were not included in 1BZ, i. e. are there not PEG at all? To answer this question, we conducted a numerical experiment. Took a special set ofpairs of arithmetic progressions (SAP), which maximally shifts the not deleted numbers in 2BZ and therefore it shifts sm to right.
Set of SAP looks as follows. The first four AP {1 + 3k ;2 + 3k ;1 + 5k;3 + 5k} C2 = [1,15] give the maximum shift of not deleted numbers {9;12;15} to right, i. e. s2 = 9 . Also fix the fol-
lowing nearest to sm not deleted number dm. Here we have d2 = 12 . Immediately find n(2) from 2N = 28, n = 6, i. e. n < s2 and there are not PEG at this specific set of SAP. Take the next pairs of SAP by the principle to delete the first two not deleted the number of sm and dm. It AP {9 + 7k;12 + 7k}, C3 = [1,105], s3 = 15,d3 = 24 . By taking 2N = 52, n(3) = 12 < 15. Again, there are not PEG. Take the next pairs of {15 + 11k; 24 + 11k} todelete the first smallest not deleted number. Here C4 = [1,1155], s4 = 27, d4 = 39 . The smallest even number 2N containing p4 = 11 2N = 124 , has n(4) = 30 . There is already «(4) > s4 = 27, i. e., the not deleted number by SAP locate into 1BZ and PEG are appeared. Another pair of SAP has the form {sm + Pm+ik'>dm + pm+ik} and always the direct calculation gives n (m) > sm. This suggests that even in the worst ofpossible sets ofpairs of RAP in 1BZ there is a PEG. We conducted test until pm = 17393,sm = 32784, dm = 32787.2N = 302516452. « = 75629112.
Statistics shows that sm = (pm) ,ym = (1,4 ^ 1). See table № 2.
Table 2. - Dependence of ym from pm by SAP
m 2 3 4 10 25 168
Pm 5 7 11 31 101 1009 2003 3001 4001 5003 10007 13001 17011
Y m 1,365 1,391 1,374 1,421 1,419 1,364 1,288 1,234 1,199 1,172 1,105 1,085 1,066
The dependence of n(m) from pm with large m has asymptotic behavior n{m)~ p2m / 4, where it is clear that n(m) > sm, wherein the difference (n(m) - sm ) with increasing m increases. This indicates that PEG arise with any even numbers.
Now explain why the real set of RA.P giving PEG has the lowest not deleted number that is less than sm. At first, in the top row of the matrix the first half of RAP xi =((p -1)/2) + pt (k +1) do not delete the PR, and deletes all composite numbers. On the bottom row the second half of the RAP is taken by the formula and not to delete
the first not deleted numbers. Second, often in RAP at one pt there is only one AP, which increases the number of not deleted numbers. Third, a starting set of the first four RARAP with p = {3;5} and RAP cannot give the first not deleted number «9», as experimentally with 2N = 28 there are two PEG {5;23},{11,17}.
The authors thank the collaborator of Mathematical Institute of RAS V. A. Steklov and Institute of mathematics and mechanics N. N. Krasovsky, Ural branch of RA.S for seminars and useful suggestions on this task.
References:
1. Sizykh S. V. lectures on the theory of numbers, FIZMATGIZ, - M., 2007.
2. Sushkevich A. K. number Theory, Kharkov, KSU, 1956.
3. Druzhinin V. V. Scientific and technical Gazette of the Volga region -NTIT, No. 3, 14-17, 2014.
4. Druzhinin V. V., Lazarev A. A. Austrian Journal Techical and Natural Science, No. 9-10, 29-19, 2014.
