gpt pro 5.2 attempt on Math-5382, version 2

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 $,c=\inf_A \limsup A(N)/\sqrt N$.

---

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 $4/\pi$.

---

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 $4/\pi$.

---

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.

---

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.

---

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 $3$. $\square$

---

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

---

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 $c_*=\inf_{A}\limsup A(N)/\sqrt N$, \[ c_*\ \ge\ \frac{4}{\pi}. \]

---

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.

---

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 $c\ge 4/\pi$.

---

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 40%

---

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.