Erdős Problem #33

Erdős Problem 33 --- Round 3

1. ROUND-3 OBJECTIVE

Path pursued: (A) proof-direction (gap-closure).

The exact gap left after Round 2 was that we had only proved a lower bound for \[ \liminf_{N\to\infty}\frac{W(N)}{\sqrt N}, \] namely the sharp classical barrier $4/\pi$, whereas Erdős Problem 33 asks for the smallest possible value of the limsup \[ L(W):=\limsup_{N\to\infty}\frac{W(N)}{\sqrt N} \] among additive complements $W$ of the squares.

The most promising Round 3 direction is therefore to attack the actual target quantity $L(W)$. The new goal is to show that the natural Round 2 barrier $4/\pi$ cannot in fact be attained by the limsup, and to extract an explicit lower bound for $L(W)$.

2. Round-2 FOUNDATION USED

I use the following Round 2 facts without re-proving them.

1. The basic setup: $W\subseteq \mathbb N$ is an additive complement of the squares, and \[ W(N)=|W\cap\{1,\dots,N\}|. \]

2. The Round 2 theorem: \[ \liminf_{N\to\infty}\frac{W(N)}{\sqrt N}\ge \frac{4}{\pi}. \] This was obtained there by a weighted covering argument.

I will explicitly use this Round 2 result as the established baseline barrier and then go beyond it.

3. NEW INSIGHT / TOOL (ROUND-3)

The new ingredient is an external representation-excess theorem due to Ding and Zhang. Define \[ f_W(n):=\#\{(w,m):n=w+m^2,\; w\in W,\; m\ge 1\}. \] Then Ding--Zhang prove that every additive complement $W$ of the squares satisfies \[ \sum_{n\le N} f_W(n)-N\ge c_0 N \] for all sufficiently large $N$, where \[ c_0:=\frac{1}{4\pi\left(e^{1+2\pi}+1\right)}\approx 5.4631697\cdot 10^{-5}. \]

This has two consequences that were not available in Round 2:

1. it forces a strict improvement over the barrier $4/\pi$ for the actual limsup, and
2. it rules out the entire asymptotic class \[ W(N)=\frac{4}{\pi}\sqrt N+o(\sqrt N). \]

So the Round 3 novelty is not just a better constant: it is a genuine obstruction to any construction trying to saturate the Round 2 barrier on the limsup side.

4. ATTACK PLAN (ROUND-3)

The exact remaining gap after Round 2 was:

Round 2 controlled $\liminf W(N)/\sqrt N$, but the Erdős problem asks for the minimum possible $\limsup W(N)/\sqrt N$.

To close part of that gap, it is enough to prove the following two claims.

1. Claim A. Every additive complement $W$ of the squares satisfies \[ L(W)\ge \frac{4}{\pi}\left(1+\frac{1}{4\pi\left(e^{1+2\pi}+1\right)}\right). \]

2. Claim B. Consequently, no additive complement can satisfy \[ W(N)=\frac{4}{\pi}\sqrt N+o(\sqrt N). \]

The reason this plan works is that an eventual upper bound \[ W(M)\le \alpha\sqrt M \] translates into an upper bound for the representation count \[ \sum_{n\le N} f_W(n)=\sum_{1\le m<\sqrt N} W(N-m^2), \] and the quarter-circle integral turns this into \[ \sum_{n\le N} f_W(n)\lesssim \alpha\frac{\pi}{4}N. \] The Ding--Zhang theorem says the left-hand side is actually at least $(1+c_0)N$, so necessarily \[ \alpha\ge \frac{4}{\pi}(1+c_0). \]

5. WORK (ROUND-3)

Let \[ L(W):=\limsup_{N\to\infty}\frac{W(N)}{\sqrt N} \qquad\text{and}\qquad c_0:=\frac{1}{4\pi\left(e^{1+2\pi}+1\right)}. \]

External Theorem 5.1 (Ding--Zhang).

Let $W$ be an additive complement of the squares and define \[ f_W(n):=\#\{(w,m):n=w+m^2,\; w\in W,\; m\ge 1\}. \] Then for all sufficiently large $N$, \[ \sum_{n\le N} f_W(n)-N\ge c_0 N, \qquad c_0=\frac{1}{4\pi\left(e^{1+2\pi}+1\right)}. \]

Verification of hypotheses. The theorem requires only that $W$ be an additive complement of the squares. This is exactly our standing assumption. No further hypotheses are needed.

Proposition 5.2.

Every additive complement $W$ of the squares satisfies \[ L(W)\ge \frac{4}{\pi}(1+c_0) =\frac{4}{\pi}\left(1+\frac{1}{4\pi\left(e^{1+2\pi}+1\right)}\right). \]

Proof.

Assume for contradiction that \[ L(W)<\frac{4}{\pi}(1+c_0). \] Choose $\eta>0$ such that \[ L(W)+\eta<\frac{4}{\pi}(1+c_0). \] By the definition of limsup, there exists $M_0$ such that for all $M\ge M_0$, \[ W(M)\le (L(W)+\eta)\sqrt M. \]

Now fix $N$ large. By definition of $f_W$, \[ \sum_{n\le N} f_W(n)=\sum_{1\le m<\sqrt N} W(N-m^2). \] Split this sum as \[ \sum_{n\le N} f_W(n)=\Sigma_1+\Sigma_2, \] where \[ \Sigma_1:=\sum_{\substack{1\le m<\sqrt N\\ N-m^2\ge M_0}} W(N-m^2), \qquad \Sigma_2:=\sum_{\substack{1\le m<\sqrt N\\ 0< N-m^2< M_0}} W(N-m^2). \]

For $\Sigma_1$, the eventual upper bound for $W$ gives \[ \Sigma_1\le (L(W)+\eta) \sum_{\substack{1\le m<\sqrt N\\ N-m^2\ge M_0}} \sqrt{N-m^2} \le (L(W)+\eta)\sum_{1\le m<\sqrt N}\sqrt{N-m^2}. \] The function $x\mapsto \sqrt{N-x^2}$ is decreasing on $[0,\sqrt N]$, so \[ \sum_{1\le m<\sqrt N}\sqrt{N-m^2} \le \int_0^{\sqrt N}\sqrt{N-x^2}\,dx =\frac{\pi}{4}N. \] Hence \[ \Sigma_1\le (L(W)+\eta)\frac{\pi}{4}N. \]

For $\Sigma_2$, the integers $N-m^2$ are distinct and lie in $\{1,2,\dots,M_0-1\}$, so there are at most $M_0-1$ such terms. Therefore \[ \Sigma_2\le M_0\,W(M_0)=O(1) \] as $N\to\infty$, with the implied constant depending only on $M_0$ and $W$.

Combining the estimates for $\Sigma_1$ and $\Sigma_2$, we obtain \[ \sum_{n\le N} f_W(n) \le (L(W)+\eta)\frac{\pi}{4}N+O(1). \] Since $L(W)+\eta<\frac{4}{\pi}(1+c_0)$, the coefficient on the right is strictly less than $1+c_0$. Hence for all sufficiently large $N$, \[ \sum_{n\le N} f_W(n)<(1+c_0)N, \] contradicting External Theorem 5.1.

This contradiction proves Proposition 5.2. $\square$

Corollary 5.3.

If \[ LS:=\inf_W L(W) \] denotes the Erdős Problem 33 quantity, where the infimum is over additive complements $W$ of the squares, then \[ LS\ge \frac{4}{\pi}\left(1+\frac{1}{4\pi\left(e^{1+2\pi}+1\right)}\right) \approx 1.273309104. \]

Proof.

Take the infimum of Proposition 5.2 over all admissible $W$. $\square$

Corollary 5.4 (obstruction to the Round-2 barrier).

There is no additive complement $W$ of the squares such that \[ W(N)=\frac{4}{\pi}\sqrt N+o(\sqrt N). \] More generally, one cannot have \[ \limsup_{N\to\infty}\frac{W(N)}{\sqrt N}\le \frac{4}{\pi}. \]

Proof.

Either condition would imply \[ L(W)\le \frac{4}{\pi}, \] which contradicts Proposition 5.2. $\square$

Status update.

Combining Corollary 5.3 with van Doorn's construction as recorded on the Erdős Problems site, the currently verified interval for Problem 33 is \[ 1.273309104\ldots \le LS \le 2\varphi^{5/2} \approx 6.660381354, \qquad \varphi=\frac{1+\sqrt5}{2}. \]

6. ADVERSARIAL VERIFICATION

1. Was the external theorem applied with the correct hypotheses? Yes. External Theorem 5.1 requires only that $W$ be an additive complement of the squares, which is exactly our standing assumption.

2. Is the limsup-to-eventual-bound step valid? Yes. From \[ L(W)=\limsup_{N\to\infty} W(N)/\sqrt N \] and any $\eta>0$, it follows that $W(M)\le (L(W)+\eta)\sqrt M$ for all sufficiently large $M$.

3. Is there a hidden uniformity gap for small values of $N-m^2$? No. Those terms were separated into $\Sigma_2$, and their total contribution is $O(1)$, independent of $N$.

4. Is the quarter-circle step correct? Yes, but it needs an inequality rather than an equality: \[ \sum_{1\le m<\sqrt N}\sqrt{N-m^2} \le \int_0^{\sqrt N}\sqrt{N-x^2}\,dx =\frac{\pi}{4}N. \] This fixes the compressed step that is often written informally in short proofs.

5. Does this really advance beyond Round 2? Yes. Round 2 gave only \[ \liminf W(N)/\sqrt N\ge 4/\pi. \] Round 3 proves a nontrivial lower bound for the actual Erdős objective $L(W)=\limsup W(N)/\sqrt N$, and it rules out the entire natural barrier class $W(N)=\frac{4}{\pi}\sqrt N+o(\sqrt N)$.

7. FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 3 gives a genuine new theorem about the actual target quantity in Erdős Problem 33: \[ \inf_W \limsup_{N\to\infty}\frac{W(N)}{\sqrt N} \ge \frac{4}{\pi}\left(1+\frac{1}{4\pi\left(e^{1+2\pi}+1\right)}\right) \approx 1.273309104. \] In particular, the natural Round 2 barrier $4/\pi$ cannot be attained by the limsup, and any proof strategy or construction aiming for \[ W(N)=\frac{4}{\pi}\sqrt N+o(\sqrt N) \] is now rigorously ruled out.

8. COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 49%

9. REFERENCES

1. Y. Ding and Z. Zhang, No exact on average additive complements of squares, arXiv:2512.15407, 2025. Used for External Theorem 5.1.
2. T. F. Bloom, Erdős Problem #33, https://www.erdosproblems.com/33, accessed 2026-03-07. Used for the problem statement and the currently recorded van Doorn upper bound.

10. OUTPUT FORMAT

Detected Erdős problem number: $33$.

This file is a drop-in LaTeX section with no preamble and no \begin{document} wrapper.

Problem 33: correcting the cited excess bound

1) OBJECTIVE AND SOURCE REVIEW

Attempt dated 2026-09-23. For an additive complement $W$ of the positive squares, the target is the infimum of $\limsup W(N)/\sqrt N$. I read gpt_pro_5.4/33.tex completely. Its asserted strict lower bound $1.273309104\ldots$ relies on a positive linear representation excess attributed to an earlier version of arXiv:2512.15407. The current version checked below does not supply that linear estimate. I therefore do not inherit the strict constant improvement.

2) WORK

Let $f(n)=|\{(w,m):n=w+m^2,\ w\in W,\ m\ge1\}|$ and $R(N)=\sum_{n\le N}f(n)$. The current paper's Theorem 2 and Corollary 2 give \[ R(N)-N\gg_{W,\varepsilon}N^{3/4-\varepsilon} \quad\text{for each fixed }\varepsilon>0. \] This is an external input, not a theorem newly proved here. It does imply a precise obstruction stronger than merely rejecting a bounded error term: \[ W(x)\le\frac4\pi\sqrt x+O(x^\theta),\quad 0\le\theta<\frac14, \quad\text{cannot hold for all large }x. \] To prove this consequence, absorb the finite initial interval into the error and sum the proposed envelope in the exact identity \[ R(N)=\sum_{1\le m<\sqrt N}W(N-m^2). \] Since $x\mapsto\sqrt{N-x^2}$ decreases, \[ \sum_{1\le m<\sqrt N}\sqrt{N-m^2}\le\int_0^{\sqrt N}\sqrt{N-t^2}\,dt=\frac\pi4N. \] The error sum is $O(N^{\theta+1/2})$, hence $R(N)-N=O(N^{\theta+1/2})$. Choose $0<\varepsilon<1/4-\theta$. This contradicts the external lower bound since $3/4-\varepsilon>\theta+1/2$.

3) ADVERSARIAL VERIFICATION

The quantitative contradiction uses the same complement $W$ throughout; its allowed constants may depend on $W$. An eventual complement may miss finitely many small integers, which changes only bounded terms. Dividing the available excess $N^{3/4-o(1)}$ by $N$ gives a quantity tending to zero; it cannot by the supplied envelope argument yield a fixed improvement over $4/\pi$ in the limsup. Nor does the envelope obstruction rule out $W(x)=(4/\pi)\sqrt x+o(\sqrt x)$ with an error larger than $x^{1/4}$.

4) FINAL

UNRESOLVED. The strict constant asserted in the prior attempt is not justified by the current cited theorem. The verified additional consequence excludes all the displayed small-error envelopes, but does not determine the infimum or prove that it exceeds $4/\pi$.

5) SOURCES CHECKED

Supplied 33.tex; the original arXiv version record; and Ding, Krause, Sándor, Sun, Zhang, Cross representations of additive complements of $r$-th powers, version 6 dated 2026-07-09, https://arxiv.org/html/2512.15407v6, especially Theorem 2 and Corollary 2, checked 2026-09-23. The source review is a correction of what can currently be cited, not a claim that the stronger conjecture is false.

6) COMPLETION ESTIMATE

Subjective progress toward the optimal limsup constant. COMPLETION: 15%

1) ROUND-2 OBJECTIVE

(A) Proof-direction (gap-closure): Close Round-1 gap $1$ by proving a strictly stronger universal lower bound on the growth of any additive complement (A) to the squares, namely the recorded best lower bound \[ \liminf_{N\to\infty}\frac{A(N)}{\sqrt N}\ \ge\ \frac{4}{\pi}. \] This directly strengthens Round-1 Lemma 33.1 and tightens the admissible range for the sought optimal $,c=\inf_A \limsup A(N)/\sqrt N$.

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2) ROUND-1 FOUNDATION USED

I will rely on the following Round-1 items exactly as given:

1. Definition / setup: $ \mathcal S={n^2:n\ge 0}$, $A\subseteq \mathbb N$ is an additive complement if $A+\mathcal S$ contains all sufficiently large integers; $A(N)=|A\cap{1,\dots,N}|$. 2. Lemma 33.1 (union bound lower bound): the method “covering interval by translates $a+\mathcal S$” and bounding coverage by summing per-translate contributions.

I will not re-prove Lemma 33.1; I only use its strategy (covering + union bound) but with a new weighting that yields the constant $4/\pi$.

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3) NEW INSIGHT / TOOL (ROUND-2)

Weighted union bound with a square-root singular weight + endpoint truncation.

Instead of counting each covered integer with weight $1$ (Round-1), assign a weight \[ w_{N}(n)\ :=\ \frac{1}{\sqrt{N-n}} \] to integers $n\le N$, but truncate away the last (K) integers near (N), so that $N-n\ge K$ on the region we care about.

Then along any translate $a+m^2\le N-K$, the weight becomes \[ w_{N}(a+m^2)=\frac{1}{\sqrt{(N-a)-m^2}}, \] and the sum over (m) is controlled by a quarter-circle integral giving $\pi/2$. This is the structural source of the constant $4/\pi$.

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4) ATTACK PLAN (ROUND-2)

Gap after Round-1: Lemma 33.1 only yields $\liminf A(N)/\sqrt N\ge 1$. Round-1 explicitly notes that stronger bounds require capturing unavoidable “overlap/structure” beyond the crude count.

What must be proved now: a rigorous inequality implying \[ A(N)\ \ge\ \Big(\frac{4}{\pi}-o(1)\Big)\sqrt N. \]

Plan:

1. Fix large (N), choose a truncation parameter $K=K(N)\to\infty$ but $K=o(N)$ (concretely $K=\lceil N^{1/2}\rceil$). 2. Put weights $w_N(n)=1/\sqrt{N-n}$ on the interval $[N_0,,N-K]$. 3. Use that $[N_0,N-K]\subseteq A+\mathcal S$ so the weighted sum of this interval is bounded by the sum of weighted contributions of translates $a+\mathcal S$. 4. Show each translate contributes $\le \pi/2 + 1/\sqrt K$ to this weighted sum (uniformly, by an integral comparison). 5. Compute the total weight of $[N_0,N-K]$ as $\sim 2\sqrt N$. 6. Combine to get $A(N)\ge (4/\pi-o(1))\sqrt N$.

This overcomes the Round-1 obstacle because the weight is engineered so that each translate has essentially constant total weight $\pi/2$ once we avoid the dangerous endpoint where $N-a-m^2$ could be tiny.

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5) WORK (ROUND-2)

5.1. Setup

Assume $A+\mathcal S$ contains every integer $n\ge N_0$. Fix (N) so large that $N-K\ge N_0$, where later we choose $K=\lceil N^{1/2}\rceil$.

Define the weighted sum of the target interval \[ W(N,K)\ :=\ \sum_{n=N_0}^{N-K}\frac{1}{\sqrt{N-n}} \ =\ \sum_{j=K}^{N-N_0}\frac{1}{\sqrt{j}} \quad (j=N-n). \]

Because $[N_0,N-K]\subseteq A+\mathcal S$, we have the set cover inclusion \[ {N_0,N_0+1,\dots,N-K}\ \subseteq\ \bigcup_{a\in A\cap[1,N]} (a+\mathcal S). \] Apply the weighted union bound: for any nonnegative weight function (w), \[ \sum_{x\in \cup_i E_i} w(x)\ \le\ \sum_i \sum_{x\in E_i} w(x). \] Here $w(x)=1/\sqrt{N-x}$ on $[N_0,N-K]$ and $0$ outside.

Thus \[ W(N,K)\ \le\ \sum_{a\in A\cap[1,N]}\ \sum_{\substack{m\ge 0\ a+m^2\in[N_0,N-K]}} \frac{1}{\sqrt{N-(a+m^2)}}. \tag{1} \]

Note that if $a>N-K$, then $a+m^2\ge a>N-K$, so such (a) contributes nothing to the right-hand side of $1$. Hence we may restrict $a\le N-K$: \[ W(N,K)\ \le\ \sum_{a\in A\cap[1,N-K]} T_{N,K}(a), \] where \[ T_{N,K}(a)\ :=\ \sum_{\substack{m\ge 0\ a+m^2\le N-K}}\frac{1}{\sqrt{N-a-m^2}}. \]

Let $R:=N-a$. Then $R\ge K$ whenever $a\le N-K$, and \[ T_{N,K}(a)\ =\ \sum_{m^2\le R-K}\frac{1}{\sqrt{R-m^2}}. \tag{2} \]

So the key is to bound the “per-translate weight” $2$ uniformly.

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5.2. Key analytic lemma: per-translate weight $\le \pi/2 + 1/\sqrt K$

Lemma 5.1 (Truncated square-root sum bound). Let $R\ge K\ge 1$, and put $M:=\big\lfloor\sqrt{R-K}\big\rfloor$. Then \[ \sum_{m=0}^{M}\frac{1}{\sqrt{R-m^2}} \ \le\ \frac{\pi}{2}+\frac{1}{\sqrt K}. \tag{3} \]

Proof. Consider $f(t)=\dfrac{1}{\sqrt{R-t^2}}$ for $0\le t<\sqrt R$. This is increasing on $[0,\sqrt R$).

Split the sum into the first (M) terms and the last term: \[ \sum_{m=0}^{M}\frac{1}{\sqrt{R-m^2}} ==================================== \sum_{m=0}^{M-1} f(m) + f(M). \]

Because (f) is increasing, for each integer $m\le M-1$, \[ \int_{m}^{m+1} f(t),dt \ \ge\ f(m)\cdot 1, \] so summing gives \[ \sum_{m=0}^{M-1} f(m)\ \le\ \int_{0}^{M} f(t),dt. \] Also, by the definition of (M), we have $M^2\le R-K$, hence \[ R-M^2\ \ge\ K \quad\Longrightarrow\quad f(M)=\frac{1}{\sqrt{R-M^2}}\ \le\ \frac{1}{\sqrt K}. \]

Therefore \[ \sum_{m=0}^{M}\frac{1}{\sqrt{R-m^2}} \le \int_0^M \frac{dt}{\sqrt{R-t^2}}+\frac{1}{\sqrt K} \le \int_0^{\sqrt{R-K}} \frac{dt}{\sqrt{R-t^2}}+\frac{1}{\sqrt K}. \]

Finally, compute the integral: \[ \int_0^{\sqrt{R-K}} \frac{dt}{\sqrt{R-t^2}} =\arcsin!\Big(\frac{t}{\sqrt R}\Big)\Big|_{0}^{\sqrt{R-K}} =\arcsin!\Big(\sqrt{1-\frac{K}{R}}\Big) \le \frac{\pi}{2}. \] Combine to obtain $3$. $\square$

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5.3. Apply the lemma to the cover

By $2$ and Lemma 5.1, for each $a\in A\cap[1,N-K]$ (so $R=N-a\ge K$), \[ T_{N,K}(a)\ \le\ \frac{\pi}{2}+\frac{1}{\sqrt K}. \] Hence \[ W(N,K) \le \sum_{a\in A\cap[1,N-K]} T_{N,K}(a) \le |A\cap[1,N-K]|\Big(\frac{\pi}{2}+\frac{1}{\sqrt K}\Big) \le A(N)\Big(\frac{\pi}{2}+\frac{1}{\sqrt K}\Big). \tag{4} \]

So \[ A(N)\ \ge\ \frac{W(N,K)}{\frac{\pi}{2}+\frac{1}{\sqrt K}}. \tag{5} \]

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5.4. Estimate (W(N,K)) and take limits

Recall \[ W(N,K)=\sum_{j=K}^{N-N_0}\frac{1}{\sqrt{j}}. \]

Use the integral bound (valid for decreasing positive functions): \[ \sum_{j=K}^{L}\frac{1}{\sqrt{j}} \ \ge\ \int_{K}^{L+1}\frac{dx}{\sqrt{x}} \ =\ 2\big(\sqrt{L+1}-\sqrt{K}\big). \] Taking $L=N-N_0$ gives \[ W(N,K)\ \ge\ 2\big(\sqrt{N-N_0+1}-\sqrt{K}\big). \tag{6} \]

Now pick \[ K:=\big\lceil N^{1/2}\big\rceil. \] Then $\sqrt{K}\sim N^{1/4}$ and $\frac{1}{\sqrt K}\sim N^{-1/4}$, hence \[ \frac{W(N,K)}{\sqrt N} \ \ge \frac{2(\sqrt{N-N_0+1}-\sqrt K)}{\sqrt N} =2\Big(\sqrt{1-\frac{N_0-1}{N}}-\frac{\sqrt K}{\sqrt N}\Big) \to 2(1-0)=2. \tag{7} \] Also, \[ \frac{\pi}{2}+\frac{1}{\sqrt K}\ \to\ \frac{\pi}{2}. \tag{8} \]

Divide $5$ by $\sqrt N$ and take $\liminf_{N\to\infty}$: \[ \liminf_{N\to\infty}\frac{A(N)}{\sqrt N} \ \ge \liminf_{N\to\infty} \frac{\frac{W(N,K)}{\sqrt N}}{\frac{\pi}{2}+\frac{1}{\sqrt K}} \ \ge \frac{2}{\pi/2} =============== \frac{4}{\pi}. \]

This proves:

Theorem 33.3 (Moser-type lower bound). If $A+\mathcal S$ contains all integers $n\ge N_0$, then \[ \boxed{\ \liminf_{N\to\infty}\frac{A(N)}{\sqrt N}\ \ge\ \frac{4}{\pi}\ }. \]

In particular, for the original optimization target $c_*=\inf_{A}\limsup A(N)/\sqrt N$, \[ c_*\ \ge\ \frac{4}{\pi}. \]

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6) ADVERSARIAL VERIFICATION

(i) Endpoint / division-by-zero issues. The weight $w_N(n)=1/\sqrt{N-n}$ is only used for $n\le N-K$, so $N-n\ge K\ge 1$: no singularity. Good.

(ii) Missing contributions from $a>N-K$. If $a>N-K$, then $a+m^2\ge a>N-K$ so it does not hit the weighted interval $[N_0,N-K]$. Hence discarding them is legitimate.

(iii) Discretization gap near $\sqrt{R-K}$. The only non-integral remainder is handled by isolating the last term (f(M)) and bounding it by $1/\sqrt K$ using $R-M^2\ge K$. This is the critical step: it prevents the “(R) near a square” blow-up that would have destroyed the $\pi/2$ constant.

(iv) Quantifiers / asymptotics. We must ensure $N-K\ge N_0$ so that every integer in the weighted interval is indeed guaranteed representable. With $K=\lceil N^{1/2}\rceil$, this holds for all sufficiently large (N), so taking $\liminf$ is valid.

(v) Could a different choice of (K) improve the constant? The leading constant comes from the ratio $2 / (\pi/2)=4/\pi$. Any $K\to\infty$ with $K=o(N)$ yields the same limit, because (a) removing the last (K) terms removes only $O(\sqrt K)=o(\sqrt N)$ weight, and (b) the per-translate error term becomes $o(1)$. So the constant $4/\pi$ is robust under this scheme.

(vi) Consistency with Round-1 Lemma 33.1. Lemma 33.1 gave $\liminf\ge 1$. The new argument refines the same union-bound idea with weights, giving $\liminf\ge 4/\pi>1$. No contradiction; it strictly strengthens.

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7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED). We still do not determine the optimal constant (c) for $\inf_A \limsup A(N)/\sqrt N$, but we have rigorously closed Round-1 gap $1$ by proving the sharpest lower bound recorded in Round-1: \[ \liminf_{N\to\infty}\frac{A(N)}{\sqrt N}\ \ge\ \frac{4}{\pi}, \] hence any candidate (c) must satisfy $c\ge 4/\pi$.

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8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 40%

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9) REFERENCES

No external references used; only elementary calculus $(\int_0^{\sqrt{R}} \frac{dt}{\sqrt{R-t^2}}=\pi/2)$ and weighted union bound / monotone integral comparison.