The result of Helfgott: every odd integer $ n geq $ 7 is the sum of $ 3 $ (odd) bonuses. I saw somewhere in the newspaper the mention of odd prime numbers. It means that every single integer $ n geq $ 10 is the sum of 4 odd prime numbers (you just added a $ 3 $ to each odd number).

Let's first assume something bold, to simplify the discussion, that for each integer $ x geq $ 13, the even number $ 2x $ can be expressed as the sum of $ 4 $ distinct prime numbers: $ 2x = p + q + r + s $. To handle all cases, I'm sure it's possible, because it would take either an enumeration of the small set of cases, or a combinatorial argument. Reason is $ x geq $ 13 is because: $ 3 + 5 + 7 + 11 = 26 = 2 cdot $ 13 is the smallest even number which is the sum of $ 4 $ distinct odd primes.

If you are not familiar with the basics of simplicial homology but you know what the free $ Bbb {Z} $-module on a set of symbols is, then google "introduction to simplicial homology". You basically need limit cards: $ partial_n: C_n to C_ {n-1} $ who take each element of the module $ C_n $ at its "limit" in $ C_ {n-1} $, its border being represented as a formal element in the free module, whatever the border map, it must be a $ Bbb {Z} $homomorphism -module and satisfy also: $ partial_ {n-1} circ partial_n = 0 $.

So, according to Helfgott's result (and the net distinction hypothesis), each integer $ x geq $ 13 is such that the even number $ 2x = p + q + r + s $, or $ {p, q, r, s } $ are distinct odd prime numbers. As a result, there is $ {4 choose 2} = $ 6 ways of forming even smaller numbers (by adding two of the prime numbers involved in the sum of $ x $) $ 2x_i, i in {1, dots, 6 } $ such as $ 2x_i + 2x_j = 2x $ or equivalent $ x_i + x_j = x $. Now, there is another problem with duplicates as $ x_i = x_j $ is a possibility. But for now, suppose that $ x_i neq x_j $, for reasons of discussion. So, to recap, everyone $ x_i = u + v $ or $ u, v in {p, q, r, s } $.

So, if we associate each $ x_i $ with the official symbol $ bar {x_i} $ which might not be an integer (for the moment) then we will call $ bar {x_i} $ the *faces* of the *simplex* $ bar {x} $. But $ bar {x} $ could also be a simplex face of another integer! It is therefore possible to obtain a long-chain complex eventually.

So we have at least $ Delta = { bar {x}, bar {x_i}: i in {1, points, 6 } } $ like a complex. This does not necessarily correspond to an abstract simplicial complex in which $ forall sigma in Delta, $ if $ tau subset sigma $then $ tau in Delta $. Since for one, we have not defined the $ bar {x_i} $ to be sets – these are just formal symbols.

Another approach would be?

remember that $ sum limits_ {i = 1} ^ 6 alpha_i x_i = x $ or $ sum limits_ {i = 1} ^ 6 alpha_i = 2, alpha_i in {0, 1 $. It's a bit like a "convex combination". Anway, as you can see, I took elements of simplicial homology and tried to adapt it. Do you have any ideas on how to proceed?