GPT 6 Astra Ultra attempt on Math-5648
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Erdős Problem #438: sets whose two-term sums avoid squares
Exact modular construction and asymptotic deduction, 24 September 2026.1) Statement
Let $M(N)$ be the largest size of $A\subseteq\{1,\ldots,N\}$ such that $A+A$ contains no perfect square. Equal summands are included in $A+A$. We prove the lower construction explicitly and deduce \[M(N)=\left(\frac{11}{32}+o(1)\right)N \] from the stated Khalfalah--Lodha--Szemerédi upper theorem. This is an asymptotic determination, not an exact formula for each $N$.2) Verify the eleven residue classes
Take the residue set modulo 32 \[R=\{1,5,9,13,14,17,21,25,26,29,30\}. \] Its eight odd members are precisely the classes congruent to 1 modulo 4, and its three even members are $14,26,30$, all congruent to 2 modulo 4. Two odd members sum to 2 modulo 4, and an odd and an even member sum to 3 modulo 4. Neither can be a square.For two even members, including repetitions, the possible sums modulo 32 are \[14+14\equiv28,\quad14+26\equiv8,\quad14+30\equiv12, \] \[26+26\equiv20,\quad26+30\equiv24,\quad30+30\equiv28. \] The square residues modulo 32 are $0,1,4,9,16,17,25$, disjoint from this list. Thus $R+R$ contains no square residue.
It follows that $A_N=\{a\in[1,N]:a\bmod32\in R\}$ has no square in $A_N+A_N$. Every complete block of 32 contributes eleven elements, so \[M(N)\ge|A_N|=\frac{11}{32}N+O(1).\tag{1}\] The verification includes diagonal sums $2a$; no convention about distinct summands has been hidden.
3) An elementary coarse upper bound
There is a useful direct comparison before importing the sharp theorem. Let $Q=\lfloor\sqrt N\rfloor^2$. In $[1,Q-1]$, the involution $a\mapsto Q-a$ pairs integers whose sum is the square $Q$. At most one member of each two-element pair belongs to $A$, and any fixed point $Q/2$ is itself forbidden because equal summands are allowed. Therefore \[|A\cap[1,Q-1]|\le\lfloor(Q-1)/2\rfloor. \] There are at most $N-Q+1=O(\sqrt N)$ additional elements in $[Q,N]$, proving $M(N)\le N/2+O(\sqrt N)$. This bound alone does not give the coefficient $11/32$.4) Declared sharp input and limiting conclusion
The Khalfalah--Lodha--Szemerédi theorem states that for every fixed $\delta>0$, all sufficiently large $N$, and every $A\subseteq[1,N]$ with $|A|\ge(11/32+\delta)N$, two elements of $A$ sum to a perfect square. The published result for distinct elements is already sufficient for the present, stronger avoidance condition. The proof of this theorem is external; modular optimality by itself would not justify its application to arbitrary integer sets.For each fixed $\delta>0$, an extremal avoiding set must therefore have size less than $(11/32+\delta)N$ once $N$ is sufficiently large. Letting $\delta$ decrease to zero gives $\limsup M(N)/N\le11/32$. Construction $1$ supplies the reverse limit inferior, proving the stated asymptotic.
5) Outcome and sources
SOLVED at the asymptotic scale, relative to the stated sharp upper theorem. The residue construction, count, treatment of repeated summands, elementary half-density bound and limit deduction are proved. The difficult upper theorem is clearly separated from these elementary arguments.Sources: https://www.erdosproblems.com/438, checked 24 September 2026; A. Khalfalah, S. Lodha and E. Szemerédi, Tight bound for the density of sequence of integers the sum of no two of which is a perfect square, https://doi.org/10.1016/S0012-365X(01)00435-6; the author's statement of the quantitative theorem at https://static.uni-graz.at/fileadmin/veranstaltungen/additive2016/Talks/szemeredi_e_slides.pdf. The eleven-class example is attributed to Massias. No novelty or formal verification is claimed.
The estimate refers to the complete asymptotic deduction with its external dependency.