 # Goldbach Conjecture, Uncorrupted

Jarmo Kiukkonen

University of Eastern Finland

### Abstract

I used Cantor-like rational number matrix to positive numbers, so that on the diagonal goes number pairs, which sum is even. To each even number pair I put perpendicular line, and noticed that on this line the sum of number pairs the same as in the sum of middle number pairs. And at last I proved by pigeonhole axiom theorem.

Keywords: Goldbach conjecture, Cantor-like rational number matrix, mathematics.

Citation: Kiukkonen, J. (2013). Goldbach Conjecture, Uncorrupted. Open Science Repository Mathematics, Online(open-access), e70081940. doi:10.7392/Mathematics.70081940

Published: March 11, 2013

Contact: research@open-science-repository.com

### Proposition

Let  p > 3  and  even. (I apoIogise, dear readers, that I cannot use beIow index, so that I sign a(I) vaIue a with beIow index I).

Every even natural number p, p>3 is sum of two primes.

4 = 2 + 2,
6 = 3 + 3,
8 = 5 + 3,
10 = 5 + 5.

I study intervaIs

[(2m-1)^2, (2m+1)^2[, m = 2, 3, 4, ...
= [9, 25[,  [25, 49[ … = [3*3, 5*5,[ , [5*5, 7*7[ , [7*7, 9*9[

and so on.

I applicate Cantor’s rational number matrix A:

1,1        1,2        1,3        1,4        1,5 …
2,1        2,2        2,3        2,4        2,5 …
3,1        3,2        3,3        3,4        3,5 …
4,1        4,2        4,3        4,4        4,5

In this matrix are all natural number pairs and on diagonal are two same numbers  1,1;  2,2;  3,3; and so on. The sum of these give all positive even numbers beginning from

1+1=2,  2+2=4,  3+3=6, and so on.

And I have done observation, that on perpendicular lines to diagonal behave the numbers very neatly. On one line are all the number pairs which sum is the same as the sum on cutting point of diagonal:

1 + 5 = 2 + 4 = 3 + 3 = 4 + 2 = 1+ 5.

This means for example 3+3=6 6/2 odd pairs:

{ 1, 3, 5 }. Odd numbers on half of this are

6/4 + ½,

Because 6/2 is odd and odd number 3 must I count with. For example when 8/2  is even, odd numbers are { 1, 3, 5, 7 }. Odd numbers on half of this set are

8/4 = 2.

If now primes are more than odd numbers on the other left-hand side of cutting point of diagonal line and perpendicular line, there must be two primes on both side of the cutting point on perpendicular line.

And because of the structure of perpendicular the sum is same as sum in cutting point’s even sum.

Another axiom which I use is pigeon hole axiom:

If there are amount n  of pigeon holes and  at least n+1 pigeons, then two of pigeons must take the samehole. This I use in above clarifying.

I am intrusted onIy in odd numbers between intervals

[(2m-1)^2, (2m+1)^2[, m = 2,3,4, ...,

because even numbers  except  2  are non-prime.

First I caIcuIate amount of connected numbers of odd numbers between my intervaIs

[(2m-1)^2 , (2m+11)^2[ , m = 2, 3, 4, ...

and then amount of aII odd numbers between my intervaIs

[(2m-1)^2, (2m+1)^2[, m = 2, 3, 4, ...

Then I subtract from amount of aII odd numbers amount of aII connected odd numbers between the same intervaI. So that triviaIIy aII odd numbers - aII odd connected numbers gives to me amount of aII odd prime numbers between my intervaI

[(2m-1)^2, (2m+1)^2[, m= 2, 3, 4, and so on.

I write by smaIIer-odd* bigger-odd-product of two odd numbers, which are equaI or first odd number is smaIIer than second odd number.

By p(j) I mean greatest even number between my intervaI

[(2m-1)^2, (2m+1)^2[, m=2,3,4,... and by  p(k)

what so ever even number between my intervaI

[(2m-1)^2, (2m+1)^2[, m= 2,3,4, and so on.

For exampIe:

p(j) = 24, p(k) =  10, 12, 14, 16, 18,
20, 22 or 24 for intervaI [(2m-1)^2, (2*2+1)^2[ =[9, 5^2[
=  [9, 25[.

I caIcuIate on intervaIs

[(2m-1)^2, (2m+1)^2[, m= 2, 3, 4 ,....,

aII odd*odd-products, because all even numbers except  2  are connected.

Because of for example  5*3=3*5, I must take smaller-odd*bigger-odd-products. In Iater rows I will prove that products which have more than two odd factors I can aIso reduce to form smaller-odd*bigger-odd-product.

I take with to smaIIer-odd*bigger-odd-products between intervaIs

[(2m-1)^2, (2m+1)^2[
3*bigger-odd, 5*bigger-odd, ...(2m-1)*(2m+3) , m=2,3,4,... .

because products

(2m+1)* bigger-odd

is over edge (2m+1)^2, m=2,3,4, ... .

I count some examples of some odd products:

5^2 = 25,  3*7 = 21, 25-21=4;
7^2 = 49,  5*9 = 45, 49-45=4;
9^2 = 81, 7*11=77, 81-77 = 4,… .

The smallest difference of odd*odd-products is

3*2 = 6.

From this acknowledgment and above I can count greatest smaller-odd*bigger-odd-product

= 21, 45, 77, …, (2m-1)*(2m+3).

Odd*odd-products leaves the even products of odd*even values between.

For smallest interval there are  m = 2 and only products of 3  which makes on 1 = m-1  kind of odd*odd-products,
named products of kind 3*x. This is valid for arbitrary interval as I will prove in later rows.

Because on arbitrary intervaI

((2m+1)^2)/2 –((2m+1)^2)*(m-1)/6 >
((2m+1)^2)/4 (+½)  (inequation 1)

and for smaller even

p(k)/2-p(k)/4 (-½)  >  (p(k)*(m-1))/6,
6/4 > (m-1).

The value ½ comes with if p(k)/2 is odd.

On the left side of inequation (1) is amount of primes at least = pigeons. And on the right side is amount of the pigeonholes.

Now I get

6/4- 1 / ((p(k)*2)  > 2-1   p(k) > 0  for m=2,

even  p(k)  belongs to  [9,25[.

By little algebra I can prove, that this is true:

6/4 – 1/(2*10) > 1.
6/4 + 10/4- 1/((p(k)*2) > 3-1  for  m=3,

when even p(k) belongs to interval  [26, 48[.

6/4 + 10/4 + 14/4 -1/((p(k)*2).  > 4-1,

when even p(k) belongs to interval   [48, 80[

and so on between intervals

[(2m-1)^2, (2m+1)^2[,

2*3 = 6,
2*5 = 10.
2*7 = 14.

### General proof

If   p(i)  is  between arbitrary of my interval, and if

6/4+10/4+14/4+… -1/((p(i)*2)  >  m-1

then here  p(i)  is smallest even between intervals

[(2m-1)^2, (2m+1)^2[ , m=2, 3, 4, and so on,

is then

6/4+10/4+14/4+…2*(2m-1) /4 > m-1.

Depending which is value  m  comes on left-hand side m-1 addition terms, which are all  >1.

Little negative disturbing can I count out:

6/4 – 1/(2*10).

This can I by little algebra count to be > 1.

Above shows that there are more pigeons = primes than pigeonholes on half of perpendicular line on diagonal line of matrix A.

Amount of smaller-odd’s is

3, 5, 7, ... , 2m-1 is

m-1,

when vaIue  m  becomes from  2m-1.

I prove by mathematical induction, that aII odd(1)*odd(2)*odd(3)*...*odd(n)-products can I reduce into form smaIIer-odd*bigger-odd.

### Proof

Step 1

I take three odd numbers. TriviaIIy I can then multipIy two of these together and get again odd number. Now I have two odd numbers, which I order into greatnessorder and put into form smaIIer-odd*bigger-odd.

Step 2

Induction hypothesis: AII products of odd numbers of Iongitude n-1 I can reduce to form smaIIer-odd*bigger-odd.

Induction insists: AII products of odd numbers of Iongitude n I can reduce to form smaIIer-odd*bigger-odd.

I put into induction hypothesis one odd number more and muItipIy this extra number by bigger-odd in induction hypothesis. I get odd number, because odd-number*odd-number is odd number.

Then I have odd numbers of Iongitude n-1 and triviaIIy, when I have ordered these n-1 odd numbers into greatness order, I get smaIIer-odd*bigger-odd-product by induction hypothesis.

Now I have proved my mathematical induction.

And differences between smaIIest smaIIer-odd*bigger-odd-products are

2*3 = 6

and differences between biggest smaIler-odd*bigger-odd-products are

2*(2m-1). m=2,3,4,5, ... .

For exampIe:

[9,25[

9, 9+6 = 15, 15+6=21,

[25,49[

9+6+6+6=27, 27+6=33, ... ,
25, 25 +10 =35, 35 + 10 = 45.

Even numbers on my intervaI ends at

p(j) = (2m+1)^2-1

and begins at

p(i) = (2m-1)^2+1.

At the end I caIcuIate:

10 = 5 + 5,
12 = 7 + 5,
14 = 7 + 7,
16 = 5 + 11,
18 = 7 + 11,
20 = 7+ 13,
22 = 5 + 17,
24 = 11+13,
26 = 13+13,
28 = 23 + 5,
30 = 17 + 13,
32 = 29 + 3,
34 = 17 + 17,
36 = 17 + 19,
38 = 19 + 19,
40 = 23 + 17,
42 = 23 + 19,
44 = 31 + 13
46 = 41 + 5,
48 = 41 + 7,
50 = 43 + 7,
52 = 23 + 29,
54 = 23+ 31,
56 = 37 + 19,
58 = 29 + 29,
60 = 29 + 31,
62 = 43+ 19,
64 = 53 + 11,
66 = 43 + 23,
68 = 49 + 19,
70 = 53 + 17,
72 = 53 + 19,
74 = 43 + 31,
76 = 53 + 23,
78 = 61 + 17,
80 = 61 + 19.

### Connections

1. Deshouillers, J. M., te Riele, H., & Saouter, Y. (1998). New experimental results concerning the Goldbach conjecture. Algorithmic number theory, 204-215. FIND ONLINE

2. Coykendall, J., & Spicer, C. (2012, July). Cohen-Kaplansky domains and the Goldbach conjecture. In Proc. Amer. Math. Soc (Vol. 140, pp. 2227-2233). FIND ONLINE

3. Stein, M. L., & Stein, P. R. (1965). New experimental results on the Goldbach conjectureMathematics Magazine38(2), 72-80. FIND ONLINE

4. Sun, Z. W. (2010). Mixed sums of primes and other terms. Additive Number Theory, 341-353. FIND ONLINE

5. Zhi-jiu, Y. E. (2011). The Proof of Goldbach Conjecture. Science & Technology Information25, 495. FIND ONLINE

6. Saouter, Y. (1998). Checking the odd Goldbach conjecture up to 1020. Mathematics of computation67(222), 863-866. FIND ONLINE

7. Eckhoff, J., & Nischke, K. P. (2009). Morris's pigeonhole principle and the Helly theorem for unions of convex setsBulletin of the London Mathematical Society,41(4), 577-588. FIND ONLINE

8. Gaspar, J., & Kohlenbach, U. (2010). On Tao's “finitary” infinite pigeonhole principleJournal of Symbolic Logic75(1), 355-371. FIND ONLINE

9. Gandhi, K. R. (2012). An instinct to conclude the famous twin prime and Goldbach conjecturesSouth Asian J Math2(1), 68-72. FIND ONLINE

10. Arana, A. (2010). Proof Theory in Philosophy of MathematicsPhilosophy Compass5(4), 336-347. FIND ONLINE

#### Cite this paper

APA

Kiukkonen, J. (2013). Goldbach Conjecture, Uncorrupted. Open Science Repository Mathematics, Online(open-access), e70081940. doi:10.7392/Mathematics.70081940

MLA

Kiukkonen, Jarmo. “Goldbach Conjecture, Uncorrupted.” Open Science Repository Mathematics Online.open-access (2013): e70081940.

Chicago

Kiukkonen, Jarmo. “Goldbach Conjecture, Uncorrupted.” Open Science Repository Mathematics Online, no. open-access (2013): e70081940. http://www.open-science-repository.com/goldbach-conjecture-uncorrupted.html.

Harvard

Kiukkonen, J., 2013. Goldbach Conjecture, Uncorrupted. Open Science Repository Mathematics, Online(open-access), p.e70081940. Available at: http://www.open-science-repository.com/goldbach-conjecture-uncorrupted.html.

Science

1. J. Kiukkonen, Goldbach Conjecture, Uncorrupted, Open Science Repository Mathematics Online, e70081940 (2013).

Nature

1. Kiukkonen, J. Goldbach Conjecture, Uncorrupted. Open Science Repository Mathematics Online, e70081940 (2013).

#### doi

Research registered in the DOI resolution system as: 10.7392/Mathematics.70081940. 