Version: 10/7/04 by Gregory Mazur (greg@tell-all.com)
The Goldbach Conjecture: Every even number may be written as the sum of two odd primes.
Assume even integer C is a counter-example to the Goldbach conjecture.
A fixed sequence of primes, p_1 thru p_n, are each less than C.
Given our assumption, there is a matched set of odd composites, r_1 thru r_n,
such that C = p_i + r_i for n matched sets.
Let p_1 = 3; C = p_1 + r_1; Also, C = p_n + r_n
Let g1 be the first prime gap.
Thus g1 = (5 – 3) = 2
C = (p_1 + g1) + (p_n + r_n – g1 – p_1)
C = p_2 + (p_n + r_n – p_2)
If (r_n – p_2) > 0 then (p_n + r_n – p_2) = r_2
If (r_n – p_2) < 0 then (p_n + r_n – p_2) = p_n-1 + s_1 (s is even).
Each C will have a set of s from s_1 thru s_j (j < n). Also a large gap, g_i , may result in some p_n-i to be
the next calculated smaller prime but not the next sequentially smaller prime as we step down from p_n to p_1.
To favor the counter-example case, assume that we can encounter subsequent s at the same rate as r_2 thru r_(n/2).
Also assume that we can encounter each sequentially smaller prime, p_n-i .
Since we can construct every prime starting with p_1, neither of these two false assumptions are material to the outcome.
On the contrary, the more primes, p_n-i that fail our test, the greater our confidence in the existence of a counter-example.
r_2 = p_n-1 + s_1
g2 = (7 – 5) = 2
C = (p_2 + g2) + (r_2 – g_2)
C = p_3 + (p_n-1 + s_1 – g2)
If (s_1 – g2) > 0 then (p_n-1 + s_1 – g2) = r_3
If (s_1 – g2) < 0 then (p_n-1 + s_1 – g2) = (p_n-2 + s_2) = r_3
p_1 + g1 + g2 + … g_((n/2)-1) = p_(n/2)
C = p_(n/2) + r_(n/2)
Thus r_(n/2) = (p_((n/2)-1) + s_((n/2)-1)
Next step past the median prime:
C = (p_(n/2) + g_(n/2)) + [(p_((n/2)-1) + s_((n/2)-1)] - g_(n/2)
C = p_((n/2) + 1) + [(p_((n/2)-1) + s_((n/2)-1)] - g_(n/2)
If [s_((n/2)-1) - g_(n/2)] > 0 then (p_((n/2)-1) + s_((n/2)-1) - g_(n/2)] = r_((n/2)+1)
If [s_((n/2)-1) - g_(n/2)] < 0 then (p_((n/2)-2) + s_((n/2)-2) = r_((n/2)+1)
C = p_((n/2)+1) + r_((n/2)+1)
But we have previously computed p_((n/2)-2)
It is the sum of p_1 + ((n/2)-3) sequential prime gaps.
Also, calculated differently above,
p_((n/2)-2) = r_((n/2)+1) - s_((n/2)-2) and inserting into C we have:
C = [r_((n/2)+1) - s_((n/2)-2)] + r_((n/2)-2)
p_((n/2)+1) = r_((n/2)-2) - s_((n/2)-2) because C = p_((n/2)+1) + r_((n/2)+1)
As p increases from p_1 thru p_n, the remainder r, decreases from r_1 thru r_n.
r_((n/2)+1) < r_((n/2)-2)
r_((n/2)+1) < r_((n/2)-2) - s_((n/2)-2)
r_((n/2)+1) < p_((n/2)+1)
r_((n/2)+1) > p_((n/2)-2) because p_((n/2)-2) = r_((n/2)+1) - s_((n/2)-2) as noted above.
Therefore we find a logical inconsistency:
The frequency and distribution of primes declines after p_n/2 for larger composites, C.
Therefore, we must expect the relation
r_((n/2)+1) > p_((n/2)+1) as opposed to the calculated result above.
Since the logic of a counter-example fails, we must conclude
no counter-example exists in proof of the Goldbach Conjecture.
Q.E.D.