TITLE: Covering the primes by 3-term APs ?
QUESTION [11 upvotes]: Hello, the Green-Tao theorem says infinitely many k-term Arithmetic Progressions exist for any integer k.  
My question is: can we actually partition the primes into 3-term APs only (or is there a simple reason why it cannot be expected) ? And if it were possible, then what would it mean for the set of primes ?
For example fix a large integer (say M=10,000) then take the primes (except 2): 3, 5, 7, 11, 13... and remove the longest possible AP with common difference less than M as you go along. It provides the partition:
3    5    7 -- 11    17    23    29 -- 13    37    61 -- 19    31    43 -- 41    47    53    59 -- 67    73    79 -- 71    89    107 -- 83    131    179    227 -- 97    103    109 -- 101    137    173 -- ... Numerical data for the first 10,000 primes with that M shows that the average length of APs so defined is 3.2 (which is thus quite close to 3, so at least numerically there exists partitions where 3-terms APs cover a large fraction of the primes, hence the question).

REPLY [8 votes]: Using the greedy algorithm, this would follow if for any fixed prime q, there exist infinitely prime "pairs" of the form p and 2p-q. This follows from standard (difficult) conjectures if q is odd (for example, the case q = -1 corresponds to "Sophie Germaine Primes"). On the other hand, it would be an implication of such a partition that for each odd prime q, there either:
(i) exists at least one prime pair (p,2p-q),
(ii) 2q is the sum of two primes,
(iii) exists at least one prime pair (p,q-2p).
Almost all proofs showing that there exist primes of a certain form also prove that there are infinitely many such primes. Thus, I suspect, one could not prove this result without also proving the difficult conjectures alluded to above.