I discovered a new (to me) way of approaching the Goldbach problem; this is a brief overview.

Scratch page [ HTML | PDF ]

Trying to prove there are always primes p, q such that p+q = N for any even N > 2, where p <= q.

N.B. Largely ignoring 2 as a prime as it seems irrelevant here. There are a couple of 2-related edge cases (N=4=2+2) which I'm setting aside as trivial for now. Also disregarding meta-knowledge that this solution is too simple.

Update 2015.09.04

For generating R, \sqrt(N) insufficient except for small N. N/2 works as an upper bound, but conceivably something as low as N/3.

Other thoughts: setting N//2 to a primorial may be useful for generating primes. For example, letting N//2 = #(3) = 2 \cdot 3 \cdot 5 = 30 and running through the below with R = {3,5,7} gives you S = {7, 13, 17, 19, 23, 29} yielding Q = {53, 47, 43, 41, 37, 31}, all the primes in [N//2, N] save 59.

I assert that you can obtain all values of p for any N by the following:

"Let " S = "set of all primes" <= N//2.
"Let "R = "set of all primes" <= sqrt(N).

For each prime r∈R, create a set constrained within the bounds [3, N//2 − 1], starting with the value N mod r and incrementing by r. Even and non-prime values may be omitted.

e.g. If N=42, r=3, then 42 -= 0" (mod 3)", and generate {3, 6, 9, 12, 15, 18}, or simply {3}.

Repeat this for every value of r in the domain.

In the above example, the only other value is r=5, yielding 42 -= 2" (mod 5)", and consequently {7, 12, 17} or just {7, 17}.

Once complete, remove the generated terms from S, your list of all primes in 2 < s < N//2. We should have:

S = {3, 5, 7, 11, 13, 17, 19}" \ "{3, 7, 17} = {5, 11, 13, 19}.

And we have obtained all values for p:

5+37=42
11+31=42
13+29=42
19+23=42

I further assert that one can use this technique to prove Goldbach's Strong Theorem if it is possible to demonstrate that a) this algorithm is universally sound within its domain, and that b) it will always return at least one value.

###### Below this line are messy self-notes. Ignore it all for now.

(Except maybe the conclusion.)

For a), it seems to work and I'm taking it on faith for now pending further scrutiny.

For b), I've had several thoughts.

I suspect there may be a basic line of reasoning that is sufficient. As I understand it, the only time an r value can be removed is if r | N, as happened with 3 in the example. Attempts to remove things straightforwardly appear self-defeating, as in N = 2*3*5*7 = 210, which does remove {3,5,7} but lets in loads more terms.

Here are some example reference calculations, the terms (up to 50) that would be removed in each case with the given r and modulus.

N mod "__"

##### 3:

0=>3 1=>7, 13, 19, 31, 37, 43 2=>5, 11, 17, 23, 29, 41, 47

##### 5:

0=>5 1=>11, 31, 41 2=>7, 17, 37, 47 3=>3, 13, 23, 43 4=>19, 29

##### 7:

0=>7 1=>29, 43 2=>23, 37 3=>3, 17, 31 4=>11 5=>5, 19, 47 6=>13, 41

Each value has the same number of primes, a unique and full set! Let's see…

3+2 gives 7, 5+2 | 3 gives 4, 7+3|5 gives 3
3+2, 5+2, 7+3
5, 7, 10
10, 24, 38, 52, 66, 80, 94, 108, 122
3,5,7,11,17,23,29,31,37,41,47

3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47
3, 7, 13, 19, 31, 37, 43
7, 13, 19, 37, 43
13, 19, 43

6's, 10's, 14's

This all raises the question… if you take the primes from √N and offset each one however you like, can you hit all the primes up to N/2??

N √N Offense π(N∕2)−1 Defense Gaps
10 3 3 2 3, 5 1
12 3 3 2 3, 5 1
26 5 3, 5 5 3, 5, 7, 11, 13 1, 1, 2, 1
50 7 3, 5, 7 8 3, 5, 7, 11, 13, 1, 1, 2, 1, 17, 19, 23 2, 1, 2

3, 5, 7, 11, 13, 17, 19, 23
3, 7, 13, 19 = 2, 3, 3
5 eats 3, 13
7 eats 7 (or 19)

Surely it's too random. Contradiction proof available?

3k+A
5k+B
7k+C

They can never start on their own spot.

What if we go big?

Let N=1000.

10 attackers, 3 through 31.
94 p-side, 73 q-side.

Or start small:

N = 10, 12
R={3}
S={3, 5}

N=14, 16, 18, 20
S={3, 5, 7}

N=22, 24
S={3, 5, 7, 11}

N=26
R={3, 5}
S={3, 5, 7, 11, 13}

S terms are generated as the holes made by the R sieve.
How can you 'invert' the R offsets systematically to ensure ideal coverage of those previous holes? (And then prove that's not sufficient to get them all.)

N=170
R={3, 5, 7, 11, 13}
S={3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83}

N=173 gives ... mod 5 -= 3

S={5, 7, 11, 17, 19, 29, 31, 37, 41, 47, 59, 61, 67, 71, 79}
then 3
S={7, 19, 31, 37, 61, 67}
then 7
S={7, 31, 37, 67}
and 11 and 13 do nothing.

N=122
R={3,5,7,11}
S={3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61}
3@61 --> {7, 13, 19, 31, 37, 43, 61}
S={3,5,11,17,23,29,41,47,53,59}
5@59 --> {29, 59}
S={3,5,11,17,23,41,47,53}
7@57 --> {┤
11@53 --> {53}
S={3,5,11,17,23,41,47}

In order not to return some successful value, one would have to craft an N such that the union of any applicable sets chosen from the relatively limited domain in R cover every single prime up to N//2, which I am convinced is an impossibility but haven't yet proven.

If nothing else, there is a strong case to be made where the number of primes in S grows strictly faster than the maximum worst-case size of the removal sets (i.e.  |S| ~= (2/3) * (4/5) * (6/7) * ... * |S|  with increasing size of R, but seems to lose to growth of  pi (N // 2)  ).