gpt pro 5.4 attempt on Math-643, version 2

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.