A proof of Goldbach's hypothesis that all even numbers greater
than four are the sum of two primes.
By Kent G Slinker
Abstract
In this paper I introduce a model which allows one to prove Goldbachs
hypothesis. The model is produced by studying Goldbach partitions as displayed
by an inverted mirror image of all the primes up to some even number equal to
the last prime plus three. The bottom half of the model is then moved to the right
in steps of two which exhibit the Goldbach partitions for the next even number.
As long as the model contains all the primes up to the resulting even number
minus three, then Goldbach’s hypothesis can be proven if it can be shown that
each move must produce a Goldbach partition until one reaches the next prime
plus one. I show that this must be the case.
Notation
k
m
p
pa
pn
pn +1
pn −a
ra
rt
Any even number
A specific even number, always three more than some prime.
A prime number (2, the only even prime, is not considered in this
paper).
Any prime in the list which makes up a Goldbach block.
The last prime (also the greatest) in a Goldbach block.
The next prime after pn .
The mirror image counter-part of pa in a Goldbach block.
The gap between pa and pn −a in a Goldbach block.
An even number added to m to obtain k.
Introduction
Christian Goldbach (1690-1764) proposed, in a now famous letter to Euler in
1742, that every even number greater than four can be expressed as the sum of
two primes. This was also part of Hilbert’s 8th conjecture at the 2nd International
Congress in Paris in 1900. Hardy and Littlewood stated in a paper which
appeared in Acta Mathematica in 1922, “There is no reasonable doubt that the
theory is correct, and that the number of representations is large when m is large;
but all attempts to obtain a proof have been completely unsuccessful.”
Several computational results have shown that Goldbach's hypothesis is true for
a large number of even numbers. Javier Echevarria verified the hypothesis up to
232 , and Matti Sinisalo to 4 x1011 (RG). Sheldon calculated that the probability of
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some even number m > 4 x1014 not having a Goldbach partition to be
approximately 10−150,000,000,000 (NS).
Impressive as this may be, it is not a proof, and the possibility remains that there
does exist some even number which is not the sum of two primes. The seeming
random distribution of primes have made the conjecture quite difficult to prove,
along with the lack of techniques related to the additive properties of primes. It is
certainly possible that, even though very rare, there are an infinite number of
exceptions to Goldbach's conjecture.
The question is open as to why no proof of Goldbach's hypothesis has appeared.
I suggest that the fault does not rest in the existing methods of analysis, but
rather in the lack of a model to guide that analysis. If I am correct, and the
following model presented in this paper does provide a proof of Goldbach's
conjecture, then this will be no small victory for model theory and should reemphasize the need for the construction of models as an aid to existing
techniques of analysis.
The Model
To prove Goldbach's conjecture I will employ a model based on the following
simple equation:
Where p is some odd prime, k an even number, and 3 ≤ p ≤ (k - 3):
(k − p ) + p = k
This equation supplies one of the necessary primes; if it can be demonstrated
that (k − p ) can always be equal to another prime for every even number k, then
the hypothesis is proven.
Example:
Take for instance the number 26. It has three Goldbach partitions:
23 + 3 = 26
19 + 7 = 26
13 + 13 = 26
For each prime in the partition: {3, 7, 13, 19}, the differences:
{(26 - 3), (26 - 7), (26 - 13), (26 - 19} are also prime.
We can now construct a model that expresses these partitions.
We first write down each prime as a block up to (26 - 3), since 23 is prime it
forms the top prime row. The top row contains the integers up to 26 for reference.
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Figure 1
If we now mirror and invert the prime rows of this model, we can see the
Goldbach partitions as listed above.
Figure 2
Let us now move the right side of the mirror image 2 moves to the right. This will
yield the Goldbach partition for (26 + 2):
Figure 3
The resulting configuration will show all the Goldbach partitions for 28, which has
only two: 5 + 23 , and 11 + 17 . Figure 3 shows these partitions along with their
commutative equivalents: 17 + 11, and 23 + 5 .
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We could move the Goldbach block two more moves to the right to obtain all the
Goldbach partitions for 30, however, as soon as we move the block 26 + 6 we
lose one Goldbach partition for 32, namely 29 + 3 . This is because our model
does not contain all the primes up to 32 − 3 .
If we restrict our moves to the right such that from the initial position at m every
move to the right contains all the primes up to the resulting even number minus
three, we can prove Goldbach’s hypothesis by showing that every such move
must produce a Goldbach partition.
Goldbach Blocks
Since this entire analysis rests on what I have termed a “Goldbach block”, let us
now define them.
Each Goldbach block will start at some initial state represented by some even
number m, and consist of two mirror images of the list of all primes
{p0 , p1, p2 … pn } , with the first prime being p0 = 3 and the last prime, pn always
equal to m − 3 (see Figures 1 and 2), hence:
[Def. 1.0] The first prime p0 .
p0 = 3
[Def. 1.1] The last prime in the Goldbach block pn .
pn = m − p0
[Def. 1.2] The value of m at the Goldbach block’s initial position.
m = p0 + pn
For example, in Figure 4 below, the prime list goes from { p0 , p1, p2 … p7 } , so that
pn = p7 which has a value of 23. The value of p3 is 11, and the value of p7 −3 = p4
is 13. The value of m is p0 + pn , or 3 + 23 = 26 .
We will now designate the "gap" between each prime, pa , and its mirror image
counter part, pn-a , as ra at the Goldbach block's initial configuration at m.
Figure 4 shows the different ra as applied to Figure 2:
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Figure 4
From the fact that m = pn + p0 , r0 is always equal to zero. The value of each
individual ra can be seen to be:
[Eq. 1.0]
The value of each gap in the Goldbach block.
ra = m − ( pn −a + pa ) , or solving for m:
[Eq. 1.1]
m = ra + pn −a + pa
Example:
In the above figure m = 26 , p2 = 7 , and p7-2 = 17 , hence r2 = 26 − (17 + 7 ) = 2 .
By definition, Goldbach blocks at the initial configuration at m have at least one
Goldbach partition, since m = p0 + pn .
Let us examine a Goldbach block with the right hand mirror image shifted to the
right 4 spaces to obtain m + 4 .
Figure 5 shows this configuration:
Figure 5
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Since r0 = 0 by definition, we designate rt as the number of moves to the right
from the initial position at m = pn + p0 , hence rt at m is zero.
We will define the new sum, represented by m + rt as k, hence:
[Eq. 2.0]
The value of a Goldbach block shifted to the right.
k = m + rt
Where m + rt ≤ pn +1 + 1
Solving for rt in terms of k and m:
[Eq. 2.1]
k − m = rt
Where rt ≤ ( pn +1 + 1) − m
Definition of a Goldbach Partition
Let us now define a Goldbach partition for these Goldbach blocks.
Since a partition occurs when a prime in the bottom mirror image matches a
prime in the top image, then if some prime ( pa ) in the prime list plus the original
gap ( ra ) plus the move to the right ( rt ) equals another prime higher up on the list,
then the resulting Goldbach block exhibits a Goldbach partition.
Let pm be another prime in the prime list { p0 , p1, p2 … p7 } such that pm > pa then:
[Theorem. 1.0]
Definition of a Goldbach Partition
If pa + ra + rt = pm , then k has a Goldbach partition equal to pm + (k − pm ) .
An easy Corollary to [Theorem 1.0] is:
[Corollary. 1.1]
If k − ( pa + ra + rt ) = pq , then k has a Goldbach partition equal to pq + (k − pq )
Proof:
From [Theorem 1.0], pa + ra + rt = pm which allows us to re-write [Corollary 1.1] as
k − pm = pq , which leads immediately to k = pq + pm
Example:
Specifically, in the above example for 30 (Figure 5), one Goldbach partition is
11 + 19 .
Hence from Theorem 1.0:
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pa + ra + rt = pm
Instantiating pa + ra → p1 + r1
p1 + r1 + rt = pm
Replacing each variable with the values given in Figure 5:
5 + 2 + 4 = 11
Hence, from [Theorem 1.0], the Goldbach partition for 30 is given by:
11 + (30 − 11) = 30 , or:
11 + 19 = 30
Proof of Goldbach’s Hypothesis
We can now prove that each k must have a Goldbach partition.
Proof:
From [Eq. 2.0]
k = m + rt
If rt = 0 , then k = m , and hence has a Goldbach partition from [Def. 1.2]
Suppose rt ≠ 0 , then from [Eq. 2.1]
k − m = rt
But from [Eq. 1.1]
m = ra + pn −a + pa
So we can write:
k − ( ra + pn −a + pa ) = rt
Rearranging terms:
k − ( pa + ra + rt ) = pn −a
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But from [Corollary 1.1], if k − ( pa + ra + rt ) = pq , then k has a Goldbach partition
equal to pq + (k − pq ) , hence k has a Goldbach partition at pn −a + (k − pn −a ) .
Kent Slinker
Pima Community College
kslinker@pima.edu
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References
[NS]
NEIL SHELDON. A Statistician's Approach to Goldbach's
Conjecture. Teaching Statistics. Vol. 25, No. 1. Spring 2003.
[RG] RICHARD K. GUY. Unsolved Problems in Number Theory.
Springer-Verlag. New York, 1994.
[WY] WANG YUAN (editor). Goldbach Conjecture.
World Scientific Publishing Co Pte Ltd. Singapore, 1984.
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