Erdős Problem #14

Round-5 Objective

Path pursued: (C) obstruction/correction, by settling the upper-construction side rigorously.

Round 4 isolated the main unresolved issue as the behaviour near the exponent $1/2$: it produced structural obstructions on any putative counterexample to the lower-bound direction, but it did not rigorously prove the matching upper construction $|[1,N]\setminus B|\ll_\varepsilon N^{1/2+\varepsilon}$.

In this round I prove exactly that upper-construction statement. Thus the remaining open part of Erdős Problem #14 is not the existence of near-optimal constructions from above, but whether one can beat the square-root threshold from below, i.e. whether $o(N^{1/2})$ is possible or whether a lower bound $\gg_\varepsilon N^{1/2-\varepsilon}$ always holds.

Round-4 Foundation Used

I use the following Round 4 conclusions.

• The Round 4 global density obstruction showed that if $E_N:=|[1,N]\setminus B|$ were as small as $o(N^{1/2})$, then the counting function of $A$ would have to lie in a narrow window, roughly between $N^{1/2}$ and $N^{3/4}$.
• More importantly for the present round, Round 4 left open the exact construction side: produce a single infinite set $A\subseteq \mathbb N$ with \[ E_N\ll_\varepsilon N^{1/2+\varepsilon}. \]

No Round 4 lemma is needed inside the new proof; Round 4 is used to identify the exact remaining gap and to interpret the significance of the new theorem.

New Insight / Tool

The new tool is a balanced digit-splitting construction in base $s^2$.

For an integer $s\ge 2$, let \[ q:=s^2. \] I construct two sets $F_s,G_s\subseteq \mathbb N_0$ such that:

1. every nonnegative integer has a unique representation as $f+g$ with $f\in F_s$ and $g\in G_s$;
2. the self-sum sets $F_s+F_s$ and $G_s+G_s$ have counting function \[ \ll_s N^{\beta_s},\qquad \beta_s:=\frac{\log(2s-1)}{\log(s^2)}=\frac{\log(2s-1)}{2\log s}; \]
3. since $\beta_s\downarrow 1/2$ as $s\to\infty$, this yields, for every $\varepsilon>0$, an infinite set $A\subseteq\mathbb N$ with \[ |[1,N]\setminus B|\ll_\varepsilon N^{1/2+\varepsilon}. \]

The case $s=2$ recovers the classical exponent $\log 3/\log 4$; the new point is that allowing arbitrary $s$ pushes the exponent down to $1/2+\varepsilon$.

Attack Plan

The exact Round 4 gap is the absence of a rigorous infinite construction matching the heuristic square-root upper scale.

The plan is:

1. define $F_s$ and $G_s$ by digit restrictions in base $q=s^2$;
2. prove that every $m\in\mathbb N_0$ has a unique decomposition $m=f+g$ with $f\in F_s$, $g\in G_s$;
3. prove that $F_s+F_s$ and $G_s+G_s$ are sparse, with exponent $\beta_s$;
4. define \[ A_s:=(F_s+1)\cup(G_s+1)\subseteq\mathbb N \] and show that every $n\ge 2$ has one canonical representation coming from $F_s+G_s$, while every failure of uniqueness is forced into the sparse exceptional sets $(F_s+F_s)+2$ or $(G_s+G_s)+2$.

This gives a full, gap-free proof of the upper-construction statement.

Work

For a set $X\subseteq\mathbb N_0$ and $N\ge 0$, write \[ X(N):=|X\cap [0,N]|. \] For $A\subseteq\mathbb N$, as before, \[ r_A(n):=|\{\{a,b\}:a,b\in A,\ a\le b,\ a+b=n\}|, \qquad B(A):=\{n\in\mathbb N:r_A(n)=1\}. \]

Lemma 5.1 (digit splitting)

Fix an integer $s\ge 2$ and put $q=s^2$. Define \[ F_s:=\left\{\sum_{j\ge 0} u_j q^j: 0\le u_j\le s-1\ \text{for all $j$, and $u_j=0$ for all but finitely many $j$}\right\}, \] and \[ G_s:=sF_s=\left\{\sum_{j\ge 0} sv_j q^j: 0\le v_j\le s-1\ \text{for all $j$, and $v_j=0$ for all but finitely many $j$}\right\}. \] Then every integer $m\ge 0$ has a unique representation \[ m=f+g,\qquad f\in F_s,\ g\in G_s. \] Moreover, \[ F_s\cap G_s=\{0\}. \]

Proof.

Write the base-$q$ expansion of $m$ as \[ m=\sum_{j\ge 0} d_j q^j,\qquad 0\le d_j\le q-1=s^2-1. \] For each $j$, divide $d_j$ by $s$: \[ d_j=u_j+sv_j,\qquad 0\le u_j\le s-1,\quad 0\le v_j\le s-1. \] This choice is unique. Therefore \[ f:=\sum_{j\ge 0} u_j q^j\in F_s, \qquad g:=\sum_{j\ge 0} sv_j q^j\in G_s, \] and $m=f+g$.

Uniqueness is digitwise: if $m=f+g=f'+g'$ with $f,f'\in F_s$ and $g,g'\in G_s$, then comparing base-$q$ digits yields the same quotient-and-remainder decomposition $d_j=u_j+sv_j$, so $f=f'$ and $g=g'$.

Finally, if $n\in F_s\cap G_s$, then each base-$q$ digit of $n$ lies both in $\{0,1,\dots,s-1\}$ and in $\{0,s,2s,\dots,(s-1)s\}$. Their intersection is $\{0\}$, so every digit is $0$ and hence $n=0$. $\square$

Lemma 5.2 (sparse self-sums)

With $F_s,G_s$ as above, define \[ \beta_s:=\frac{\log(2s-1)}{\log(s^2)}=\frac{\log(2s-1)}{2\log s}. \] Then \[ (F_s+F_s)(N)\ll_s N^{\beta_s} \qquad\text{and}\qquad (G_s+G_s)(N)\ll_s N^{\beta_s}. \]

Proof.

Take $f,f'\in F_s$, say \[ f=\sum_{j\ge 0} u_j q^j, \qquad f'=\sum_{j\ge 0} u'_j q^j, \qquad 0\le u_j,u'_j\le s-1. \] Then \[ f+f'=\sum_{j\ge 0}(u_j+u'_j)q^j, \qquad 0\le u_j+u'_j\le 2s-2<q. \] So there are no carries. Hence $F_s+F_s$ consists exactly of those integers whose base-$q$ digits all lie in $\{0,1,\dots,2s-2\}$.

If $q^m\le N<q^{m+1}$, then every element of $(F_s+F_s)\cap[0,N]$ is determined by at most $m+1$ base-$q$ digits, each having $2s-1$ possibilities. Therefore \[ (F_s+F_s)(N)\le (2s-1)^{m+1} \ll_s N^{\log_q(2s-1)} = N^{\beta_s}. \]

Since $G_s=sF_s$, we have \[ G_s+G_s=s(F_s+F_s), \] so \[ (G_s+G_s)(N)\le (F_s+F_s)(N/s)\ll_s N^{\beta_s}. \] This proves the lemma. $\square$

Proposition 5.3 (explicit infinite construction)

For each integer $s\ge 2$, set \[ A_s:=(F_s+1)\cup(G_s+1)\subseteq\mathbb N. \] Then $A_s$ is infinite and \begin{equation} \label{wsref-0-65-71-3a-41-73-2d-75-70-70-65-72-2d-62-6f-75-6e-64} |[1,N]\setminus B(A_s)|\ll_s N^{\beta_s}, \qquad \beta_s=\frac{\log(2s-1)}{2\log s}. \end{equation}

Proof.

First, $A_s$ is infinite because $F_s$ is infinite.

Let $n\ge 2$. By Lemma 5.1, $n-2$ has a unique representation \[ n-2=f+g,\qquad f\in F_s,\ g\in G_s. \] Therefore \[ n=(f+1)+(g+1), \] so every $n\ge 2$ is represented as a sum of two elements of $A_s$.

Now suppose that $n\notin B(A_s)$. Then either $n=1$, or $n\ge 2$ and there is some representation \[ n=(x+1)+(y+1),\qquad x,y\in F_s\cup G_s, \] which is different from the canonical unordered pair $\{f+1,g+1\}$ coming from $n-2=f+g$.

If one of $x,y$ lies in $F_s$ and the other lies in $G_s$, then $x+y=n-2$ is a mixed decomposition of $n-2$. By the uniqueness in Lemma 5.1, this forces \[ \{x,y\}=\{f,g\}, \] so it gives the same unordered representation of $n$ and does not create a second one.

Hence any genuine failure of uniqueness must come from a same-type representation: either \[ x,y\in F_s \qquad\text{or}\qquad x,y\in G_s. \] Consequently, \[ [1,N]\setminus B(A_s) \subseteq \{1\}\cup \left((F_s+F_s)+2\right)\cup\left((G_s+G_s)+2\right). \] Therefore, by Lemma 5.2, \[ |[1,N]\setminus B(A_s)| \le 1+(F_s+F_s)(N-2)+(G_s+G_s)(N-2) \ll_s N^{\beta_s}. \] This proves (eq:As-upper-bound). $\square$

Corollary 5.4 (the $N^{1/2+\varepsilon}$ construction)

For every $\varepsilon>0$ there exists an infinite set $A\subseteq\mathbb N$ such that \[ |[1,N]\setminus B(A)|\ll_{\varepsilon} N^{1/2+\varepsilon}. \]

Proof.

For $s\ge 2$, \[ \beta_s =\frac{\log(2s-1)}{2\log s} =\frac12+\frac{\log(2-1/s)}{2\log s} <\frac12+\frac{\log 2}{2\log s}. \] Choose $s$ so large that $\beta_s<\tfrac12+\varepsilon$; for instance any \[ s>2^{1/(2\varepsilon)} \] suffices. Then Proposition 5.3 gives an infinite set $A_s\subseteq\mathbb N$ with \[ |[1,N]\setminus B(A_s)|\ll_s N^{\beta_s}\ll_{\varepsilon} N^{1/2+\varepsilon}. \] $\square$

Consequence for the original problem

This round proves the full upper-construction side that had remained missing in the previous rounds: \[ \forall\varepsilon>0\ \exists\, A\subseteq\mathbb N\ \text{infinite such that}\ |[1,N]\setminus B(A)|\ll_{\varepsilon}N^{1/2+\varepsilon}. \] Therefore the unresolved part of Erdős Problem #14 is now concentrated entirely on the lower side: whether every infinite $A\subseteq\mathbb N$ must satisfy \[ |[1,N]\setminus B(A)|\gg_{\varepsilon}N^{1/2-\varepsilon} \] for all large $N$, or whether one can in fact achieve $o(N^{1/2})$.

Adversarial Verification

I check the new construction against the main possible failure modes.

• Carry issue in the self-sums. For $F_s+F_s$, the largest digit sum is $2s-2$, and since $2s-2<s^2=q$, there are no carries. This is the key point making the exponent $\beta_s$ exact. For $G_s+G_s$, I avoid a separate carry analysis by using the identity $G_s=sF_s$.

• Overlap of the two parts. We proved $F_s\cap G_s=\{0\}$. After shifting by $1$, the two parts intersect only at $1$. This does not create extra unordered representations: if a mixed representation of $n-2$ exists, Lemma 5.1 makes it unique, so it contributes exactly one unordered pair.

• Possibility of missing sums. There are no missing sums for $n\ge 2$, because every $n-2\in\mathbb N_0$ has a mixed representation $f+g$. The integer $1$ is the only inevitable missing sum in the shifted positive setting.

• Boundary case $n=2$. Here $n-2=0=0+0$, so the canonical representation is $2=1+1$. This is still exactly one unordered representation.

• Dependence on $s$. All constants in Proposition 5.3 depend only on the fixed parameter $s$. After choosing $s=s(\varepsilon)$, this becomes the claimed $\ll_\varepsilon$ bound.

No hidden assumption beyond the adopted unordered-representation convention is used.

Final

UNRESOLVED (BUT STRICTLY ADVANCED).

This round completely settles the upper-construction side left open in the previous rounds: \[ \forall\varepsilon>0\ \exists\, A\subseteq\mathbb N\ \text{infinite such that}\ |[1,N]\setminus B(A)|\ll_{\varepsilon}N^{1/2+\varepsilon}. \] The proof is explicit and gap-free.

So the surviving open core of Erdős Problem #14 is now precisely this:

Can one prove a matching lower bound of order $N^{1/2-o(1)}$ for every infinite $A$, or can one construct a single infinite $A$ with $|[1,N]\setminus B(A)|=o(N^{1/2})$?

In other words, the exponent $1/2$ is now rigorously established as the correct upper threshold, and only the lower-threshold question remains open.

Completion Estimate

COMPLETION: 84%

References

• Internal previous-round files: 14_v2.tex, 14_v2_solutions.tex, 14_solutions.tex.
• T. F. Bloom, Erdős Problem #14, erdosproblems.com/14, accessed 2026-03-07.
• A. Sárközy and V. T. Sős, On Additive Representation Functions, in The Mathematics of Paul Erdős I, Springer, 1997.

Problem 14 --- Round 3

1) ROUND-3 OBJECTIVE

I continue on path (A): proof track. A full resolution of Q1/Q2 still appears out of reach, so the Round-3 goal is to obtain a strictly stronger obstruction than the Round-2 near-Sidon bound. The specific target is to rule out an entire class of potential counterexamples: those for which the counting function \[ A(x):=|A\cap\{1,\dots,x\}| \] stays below a constant multiple of $\sqrt{x}$ with constant strictly smaller than a critical threshold. This is the most promising direction because Round 2 already ruled out sets that are too dense; the missing complementary step is to rule out sets that are too sparse.

2) Round-2 FOUNDATION USED

I use the following Round-2 results explicitly.

1. Round-2 Theorem 14.5 and Corollary 14.6: if \[ K(N):=|\{1,\dots,N\}\setminus B|, \] then \[ A(x)\le 2\sqrt{x}+K(2x). \] In particular, any hypothetical set with $K(N)=o(N^{1/2})$ satisfies \[ A(x)\le (2+o(1))\sqrt{x}. \]
2. Round-2 Corollary 14.8: if $A(x)\ge (2+\delta)\sqrt{x}$ on some scale, then $K(2x)\gg_\delta \sqrt{x}$. This ruled out the high-density side.

Round 3 builds on this by proving the complementary low-density obstruction.

3) NEW INSIGHT / TOOL (ROUND-3)

The new ingredient is a different counting functional: \[ P_A(N):=\left|\{\{a,b\}:a,b\in A,\ a\le b,\ a+b\le N\}\right|, \] the number of unordered pairs from $A$ with sum at most $N$.

Every integer $n\in B\cap[1,N]$ contributes one distinct pair to $P_A(N)$, so small $K(N)$ forces $P_A(N)$ to be close to $N$. The Round-3 innovation is to prove a sharp upper bound for $P_A(N)$ under a global envelope \[ A(x)\le C\sqrt{x}+D. \] The main term turns out to be \[ P_A(N)\le \frac{\pi}{8}C^2N+O_{C,D}(\sqrt N). \] This yields a new critical constant \[ \sqrt{\frac{8}{\pi}}\approx 1.59577, \] strictly stronger than the Round-1 lower-density threshold $\sqrt2\approx 1.41421$ coming from the crude total-pair bound.

4) ATTACK PLAN (ROUND-3)

After Round 2, the exact remaining gap was the near-extremal regime in which \[ A(x)=O(\sqrt{x}) \quad\mbox{and}\quad K(N)=o(\sqrt N). \] Round 2 ruled out densities above $2\sqrt{x}+o(\sqrt{x})$, but did not exclude sparser quadratic-growth models.

The Round-3 plan is:

1. express $P_A(N)$ exactly in terms of the increasing sequence of elements of $A$;
2. prove that the envelope $A(x)\le C\sqrt{x}+D$ forces \[ P_A(N)\le \frac{\pi}{8}C^2N+O_{C,D}(\sqrt N); \]
3. combine this with the trivial lower bound \[ N-K(N)=|B\cap[1,N]|\le P_A(N) \] to show that if $C<\sqrt{8/\pi}$, then necessarily $K(N)$ is linear in $N$;
4. combine that new sparse-side obstruction with the Round-2 dense-side obstruction to isolate a narrower asymptotic density window for any hypothetical counterexample to Q2.

5) WORK (ROUND-3)

Retain the notation \[ K(N):=|\{1,\dots,N\}\setminus B|, \qquad A(x):=|A\cap\{1,\dots,x\}|. \] Define the low-sum pair count \[ P_A(N):=\left|\{\{a,b\}:a,b\in A,\ a\le b,\ a+b\le N\}\right|. \] Then every $n\in B\cap[1,N]$ determines a unique pair $\{a,b\}$ with $a+b=n$, and different integers $n$ give different pairs because their sums differ. Therefore \[ N-K(N)=|B\cap[1,N]|\le P_A(N). \]

Lemma 14.9 (exact decomposition of $P_A(N)$). Write \[ A=\{a_1<a_2<a_3<\cdots\}. \] Let $m:=A(N/2)$. Then \[ P_A(N)=\sum_{i=1}^{m}\left(A(N-a_i)-i+1\right). \]

Proof. Every pair counted by $P_A(N)$ has a unique smaller element $a_i\le N/2$. Fix such an index $i\le m$. The larger element $b$ must satisfy \[ b\in A, \qquad b\ge a_i, \qquad b\le N-a_i. \] The number of admissible $b$ is therefore \[ A(N-a_i)-A(a_i-1). \] Since $a_i$ is the $i$-th element of $A$, we have $A(a_i-1)=i-1$. Hence the number of pairs with smaller element $a_i$ is exactly $A(N-a_i)-i+1$. Summing over $i=1,\dots,m$ proves the formula. $\square$

Theorem 14.10 (sparse-envelope bound for low-sum pairs). Assume that for some fixed constants $C>0$ and $D\ge 0$ one has \[ A(x)\le C\sqrt{x}+D \qquad (x\ge 1). \] Then \[ P_A(N)\le \frac{\pi}{8}C^2N+O_{C,D}(\sqrt N). \] Consequently, \[ N-K(N)\le \frac{\pi}{8}C^2N+O_{C,D}(\sqrt N). \]

Proof. Let $d:=\lceil D\rceil$ and write $m:=A(N/2)$. If $m\le d$, then trivially \[ P_A(N)\le \binom{m+1}{2}=O_{C,D}(1), \] so the stated estimate is immediate. Thus assume $m>d$.

By Lemma 14.9, \[ P_A(N)=\sum_{i=1}^{m}\left(A(N-a_i)-i+1\right). \] Split this at $d$: \[ P_A(N)=\sum_{i=1}^{d}\left(A(N-a_i)-i+1\right) +\sum_{i=d+1}^{m}\left(A(N-a_i)-i+1\right). \] The first sum has only $d=O_D(1)$ terms, and each term is at most $A(N)\le C\sqrt N+D$, so it contributes \[ O_{C,D}(\sqrt N). \]

Now fix $i>d$ and set $j:=i-d\ge 1$. Since $i=A(a_i)\le C\sqrt{a_i}+D\le C\sqrt{a_i}+d$, we get \[ j=i-d\le C\sqrt{a_i}, \qquad\mbox{hence}\qquad a_i\ge \frac{j^2}{C^2}. \] Also $i\le m=A(N/2)$ implies $a_i\le N/2$, so $j\le C\sqrt{N/2}$. Therefore \[ A(N-a_i)-i+1 \le C\sqrt{N-a_i}+D-i+1 \le C\sqrt{N-j^2/C^2}-j+1, \] because $D-d\le 0$.

Thus \[ P_A(N) \le O_{C,D}(\sqrt N) +\sum_{j=1}^{\lfloor C\sqrt{N/2}\rfloor} \left(C\sqrt{N-j^2/C^2}-j+1\right). \] Set \[ f_N(x):=C\sqrt{N-x^2/C^2}-x+1 \qquad\mbox{for }0\le x\le C\sqrt{N/2}. \] A direct differentiation gives \[ f_N'(x)=-\frac{x/C}{\sqrt{N-x^2/C^2}}-1<0, \] so $f_N$ is decreasing on its domain. Hence \[ \sum_{j=1}^{\lfloor C\sqrt{N/2}\rfloor} f_N(j) \le f_N(0)+\int_0^{C\sqrt{N/2}} f_N(x)\,dx. \] Now $f_N(0)=C\sqrt N+1=O_C(\sqrt N)$, while \[ \int_0^{C\sqrt{N/2}} f_N(x)\,dx = \int_0^{C\sqrt{N/2}}\left(C\sqrt{N-x^2/C^2}-x\right)dx +O_C(\sqrt N). \] Substitute $x=C\sqrt N\,t$. Then \[ \int_0^{C\sqrt{N/2}}\left(C\sqrt{N-x^2/C^2}-x\right)dx = C^2N\int_0^{1/\sqrt2}(\sqrt{1-t^2}-t)\,dt. \] Finally, \[ \int_0^{1/\sqrt2}(\sqrt{1-t^2}-t)\,dt = \left(\frac{\pi}{8}+\frac14\right)-\frac14 =\frac{\pi}{8}. \] Therefore \[ P_A(N)\le \frac{\pi}{8}C^2N+O_{C,D}(\sqrt N). \] Combining with the basic inequality $N-K(N)\le P_A(N)$ gives \[ N-K(N)\le \frac{\pi}{8}C^2N+O_{C,D}(\sqrt N). \] This proves the theorem. $\square$

Corollary 14.11 (low-density envelopes force a linear bad set). Suppose there exist constants $C<\sqrt{8/\pi}$ and $D\ge 0$ such that \[ A(x)\le C\sqrt{x}+D \qquad (x\ge 1). \] Then \[ K(N)\ge \left(1-\frac{\pi}{8}C^2\right)N-O_{C,D}(\sqrt N). \] In particular, \[ K(N)\gg_{C,D} N. \]

Proof. By Theorem 14.10 and the basic inequality $N-K(N)\le P_A(N)$, \[ N-K(N)\le \frac{\pi}{8}C^2N+O_{C,D}(\sqrt N). \] Rearranging gives the stated inequality. Since $C<\sqrt{8/\pi}$, the coefficient $1-\pi C^2/8$ is positive. $\square$

Corollary 14.12 (new asymptotic obstruction for Q2). If \[ K(N)=o(\sqrt N), \] then necessarily \[ \limsup_{x\to\infty}\frac{A(x)}{\sqrt x} \ge \sqrt{\frac{8}{\pi}}. \] Equivalently, for every $\delta>0$ there exist arbitrarily large $x$ such that \[ A(x)\ge \left(\sqrt{\frac{8}{\pi}}-\delta\right)\sqrt x. \]

Proof. Assume for contradiction that \[ L:=\limsup_{x\to\infty}\frac{A(x)}{\sqrt x}<\sqrt{\frac{8}{\pi}}. \] Choose $C$ with \[ L<C<\sqrt{\frac{8}{\pi}}. \] Then there exists $X_0$ such that $A(x)\le C\sqrt x$ for all $x\ge X_0$. Define \[ D:=\max_{1\le x<X_0}\left(A(x)-C\sqrt x\right). \] This is finite, and now \[ A(x)\le C\sqrt x + D \qquad (x\ge 1). \] Hence Corollary 14.11 applies and yields \[ K(N)\gg N, \] contradicting the assumption $K(N)=o(\sqrt N)$. Therefore \[ \limsup_{x\to\infty}\frac{A(x)}{\sqrt x}\ge \sqrt{\frac{8}{\pi}}. \] The equivalent formulation is immediate. $\square$

Corollary 14.13 (Round-2 $+$ Round-3 density window for any putative counterexample). If \[ K(N)=o(\sqrt N), \] then \[ \sqrt{\frac{8}{\pi}} \le \limsup_{x\to\infty}\frac{A(x)}{\sqrt x} \le 2. \]

Proof. The lower bound is Corollary 14.12. For the upper bound, apply Round-2 Corollary 14.6: \[ A(x)\le 2\sqrt{x}+K(2x)=(2+o(1))\sqrt x, \] so the limsup is at most $2$. $\square$

Interpretation. Round 2 ruled out sets that are too dense; Round 3 rules out sets that are too sparse. Thus any genuine counterexample to Q2 must live in the narrower asymptotic window \[ 1.59577\ldots=\sqrt{\frac{8}{\pi}} \le \limsup_{x\to\infty}\frac{A(x)}{\sqrt x} \le 2. \] This is a strict strengthening over the pre-Round-3 lower threshold $\sqrt2\approx1.41421$.

Equivalently, Round 3 rules out the entire class of constructions satisfying a global envelope \[ A(x)\le c\sqrt x+O(1) \qquad\mbox{with}\qquad c<\sqrt{\frac{8}{\pi}}. \] Any eventual counterexample to Q2 would therefore have to be both

1. dense enough to beat the new $\sqrt{8/\pi}$ threshold infinitely often, and
2. sparse enough to stay below the Round-2 ceiling $2\sqrt x+o(\sqrt x)$.

This sharply narrows the surviving regime.

6) ADVERSARIAL VERIFICATION

1. Distinct-sum issue. The inequality $N-K(N)\le P_A(N)$ uses only that different integers $n$ correspond to different pair-sums. There is no hidden injectivity assumption beyond equality of sums.
2. Diagonal pairs. Lemma 14.9 counts $a_i=a_i$ correctly because $a_i\le N/2$ implies $a_i\le N-a_i$, so the diagonal is included exactly once when admissible.
3. Envelope inversion. From $i=A(a_i)\le C\sqrt{a_i}+D\le C\sqrt{a_i}+d$ we indeed obtain $a_i\ge ((i-d)/C)^2$ for $i>d$. No monotonicity gap is hidden here.
4. Range of the new index $j=i-d$. Since $i\le A(N/2)$ implies $a_i\le N/2$, the bound $j\le C\sqrt{a_i}\le C\sqrt{N/2}$ is valid. This prevents the function $f_N(j)$ from entering the region where the square root would be undefined or where the bound would become vacuous.
5. Monotone sum-to-integral step. The derivative \[ f_N'(x)=-\frac{x/C}{\sqrt{N-x^2/C^2}}-1<0 \] shows $f_N$ is decreasing on $[0,C\sqrt{N/2}]$, so the upper Riemann-sum estimate used in Theorem 14.10 is legitimate.
6. Finite initial segment correction. The constant $D$ absorbs all irregularities coming from finitely many small values of $x$. Since $D$ is fixed in Theorem 14.10 and Corollary 14.11, the resulting error term remains $O_{C,D}(\sqrt N)$ and does not affect the main linear coefficient.
7. Compatibility with Round 2. There is no conflict with Round-2 Corollary 14.6. In fact, the two rounds are complementary: Round 2 gives the upper-density ceiling $2$, and Round 3 gives the new lower-density floor $\sqrt{8/\pi}$.
8. What remains unproved. The new argument still does not decide whether the surviving density window \[ \sqrt{\frac{8}{\pi}}\le \limsup A(x)/\sqrt x\le 2 \] actually contains an admissible infinite construction with $K(N)=o(\sqrt N)$, nor does it prove the conjectural lower bound $K(N)\gg_\varepsilon N^{1/2-\varepsilon}$. The near-extremal regime remains open.

No internal contradiction was found in the new work.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED)

The strict Round-3 advance is the new sparse-side obstruction: \[ A(x)\le C\sqrt x+O(1) \quad\mbox{with}\quad C<\sqrt{\frac{8}{\pi}} \Longrightarrow K(N)\gg N. \] Hence any putative counterexample to Q2 must satisfy \[ \limsup_{x\to\infty}\frac{A(x)}{\sqrt x} \ge \sqrt{\frac{8}{\pi}}. \] Combining this with the Round-2 dense-side obstruction yields the narrowed density window \[ \sqrt{\frac{8}{\pi}} \le \limsup_{x\to\infty}\frac{A(x)}{\sqrt x} \le 2. \]

So Round 3 rules out an entire class of counterexample strategies---all globally subcritical $c\sqrt x$ envelopes with $c<\sqrt{8/\pi}$---and isolates the remaining difficulty more sharply than Round 2 did.

The unresolved core is now even more specific: one must analyze the narrow quadratic-growth regime corresponding roughly to densities between $1.59577\sqrt x$ and $2\sqrt x$.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 58%

9) REFERENCES

No external references were used in the new argument. The proof is self-contained and uses only the Round-2 results supplied in the prompt.

Problem 14: one mixed-radix construction at exponent $1/2+o(1)$

1) OBJECTIVE AND SOURCE REVIEW

Attempt dated 2026-09-23. For $A\subseteq\mathbb N$, let $r_A(n)$ count unordered representations $n=a+b$, allowing $a=b$, and put $E_A(N)=|\{1\le n\le N:r_A(n)\ne1\}|$. The original lower-side questions concern how small $E_A$ can be, in particular the square-root threshold. I read 14.tex and 14_v2.tex in gpt_pro_5.4 completely. The former gives an envelope obstruction; the latter constructs a different fixed-base set for each exponent $1/2+\varepsilon$. I extend the latter to a single set working for all positive $\varepsilon$.

2) WORK

For $j\ge0$ put $s_j=j+2$, $q_j=s_j^2$, $Q_0=1$, $Q_{j+1}=q_jQ_j$, and define finite-support digit sets \[ F=\left\{\sum_j u_jQ_j:0\le u_j<s_j\right\},\qquad G=\left\{\sum_j s_jv_jQ_j:0\le v_j<s_j\right\}. \] Every nonnegative integer has a unique mixed-radix expansion with $j$th digit less than $q_j$. Splitting that digit uniquely as $u_j+s_jv_j$ proves that every $m\ge0$ has exactly one ordered representation $m=f+g$, $f\in F,g\in G$. Also $F\cap G=\{0\}$. Take $A=(F+1)\cup(G+1)$.

Every $n\ge2$ has its canonical mixed representation. Any different unordered representation must use two members of the same set; consequently \[ E_A(N)\le1+|(F+F)\cap[0,N]|+|(G+G)\cap[0,N]|. \] Choose $k$ with $Q_k\le N<Q_{k+1}$. Nonnegativity means that summands contributing a sum at most $N$ have no nonzero digits above $k$. For each $j\le k$ there are at most $2s_j-1$ possible sums of the corresponding digits. For $G+G$ carries may occur, but the number of resulting integers is at most the number of digit-sum vectors. Hence each of the last two counts is at most \[ \prod_{j=0}^k(2s_j-1)\le2^{k+1}\prod_{j=0}^ks_j =2^{k+1}\sqrt{Q_{k+1}} \le2^{k+1}(k+2)\sqrt N. \] Since $Q_k=((k+1)!)^2$, elementary factorial estimates give $k=O(\log N/\log\log N)$. Therefore this single infinite set satisfies \[ E_A(N)\le\sqrt N\exp\!\left(O\left(\frac{\log N}{\log\log N}\right)\right)=N^{1/2+o(1)}. \] This strengthens the quantifier order of the supplied fixed-base construction, without claiming a new published result.

3) ADVERSARIAL VERIFICATION

The low-$m$ case of Theorem 14.10 in 14.tex asserts $P_A(N)\le\binom{m+1}{2}$ for $m=A(N/2)$; this misses pairs whose larger member exceeds $N/2$. The correct bound is $P_A(N)\le mA(N)=O_{C,D}(\sqrt N)$ when $m\le\lceil D\rceil$, which repairs that case without changing its main term. Here we do not use that theorem. Mixed representations remain unique despite the common element 1. Counting digit vectors is an upper bound even when different vectors coalesce through carries.

4) FINAL

UNRESOLVED. The displayed upper bound does not give $E_A(N)=o(\sqrt N)$, nor prove a universal $N^{1/2-o(1)}$ lower bound. Those original lower-side issues remain untouched.

5) SOURCES CHECKED

Both supplied versions, read in full on 2026-09-23. The mixed-radix proof is self-contained. The official problem-page request returned HTTP 403, so no live statement or literature-priority claim is inferred from it.

6) COMPLETION ESTIMATE

Subjective coverage of the original problem; known upper-side constructions are not counted as a solution. COMPLETION: 25%