GPT 6 Astra Ultra attempt on Math-762

Erdős Problem 323: a checked attempt

GPT_6_Astra_Ultra

25 September 2026

Statement and scope

Let $1\le m\le k$ and let $f_{k,m}(x)$ count integers $0\le n\le x$ representable as a sum of $m$ nonnegative $k$th powers. Is $f_{k,k}(x)\gg_\epsilon x^{1-\epsilon}$ for every $\epsilon>0$? For $m<k$, is $f_{k,m}(x)\gg x^{m/k}$? Source: https://www.erdosproblems.com/323.

The general questions are unresolved here. We prove the one-summand case, a recursive separated-sums lower bound for every $m$, and a precise sufficient second-moment condition for the diagonal question. This replaces the earlier GPT-5.2 bound whose written constant calculation lost a factor of two and which did not use the number of available summands.

Elementary upper bound and the exact first case

Put $L=\lfloor x^{1/k}\rfloor$. Every representation can be put in nondecreasing order and all its entries lie in $[0,L]$. The number of such tuples is $\binom{L+m}{m}$, so \[ f_{k,m}(x)\le\binom{L+m}{m}=O_m(x^{m/k})\quad(x\ge1). \tag{1} \] For $m=1$ there are exactly $L+1$ values, proving the proposed order in that case. The case $k=1$ is also immediate.

A recursive construction of disjoint blocks

For $k\ge2$, define \[ \beta_m=1-\left(1-\frac1k\right)^m. \] We prove, by induction on $m$, that for a constant $c_{k,m}>0$ and all sufficiently large $x$, \[ f_{k,m}(x)\ge c_{k,m}x^{\beta_m}. \tag{2} \] The base $m=1$ follows from the exact count. Suppose the assertion holds for $m-1$. Let $N=\lfloor(x/2)^{1/k}\rfloor$ and \[ y=\frac12(N/2)^{k-1}. \] For each integer $a$ with $\lceil N/2\rceil\le a\le N$, add $a^k$ to every value at most $y$ representable by $m-1$ powers. All resulting values lie in $[a^k,a^k+y]$ and are at most $x$ for sufficiently large $x$, since $N^k\le x/2$ and $y=O_k(x^{(k-1)/k})$.

These intervals are pairwise disjoint. Indeed, for successive eligible $a$, \[ (a+1)^k-a^k\ge k a^{k-1}\ge k(N/2)^{k-1}>y. \] There are at least $N/2$ eligible integers. As $N\asymp_k x^{1/k}$ and $y\asymp_k x^{(k-1)/k}$, induction gives \[ f_{k,m}(x)\ge\frac N2 f_{k,m-1}(y) \gg_{k,m}x^{1/k+(k-1)\beta_{m-1}/k}=x^{\beta_m}. \] This proves $2$, with rounding absorbed only into fixed positive constants after the explicit disjointness inequalities. For example, the three-cube case gives exponent $19/27$, and each additional summand improves the bound. These are elementary estimates, weaker than known results in several cases, not claims of new records.

The collision estimate that would bridge the gap

Let $r_{k,m}(n)$ count ordered nonnegative representations and set \[ T(x)=\sum_{n\le x}r_{k,m}(n),\qquad E(x)=\sum_{n\le x}r_{k,m}(n)^2. \] A box of side $\lfloor(x/m)^{1/k}\rfloor$ supplies $T(x)\gg_{k,m}x^{m/k}$ tuples; the containing box of side $\lfloor x^{1/k}\rfloor$ gives the matching upper order. Cauchy--Schwarz over represented values gives \[ f_{k,m}(x)\ge\frac{T(x)^2}{E(x)}. \tag{3} \] Consequently $E(x)\ll_{k,m,\epsilon}x^{m/k+\epsilon}$ would imply $f_{k,m}(x)\gg x^{m/k-\epsilon}$. For $m=k$ this is exactly the sufficient Hypothesis $K^*$ second-moment scale recorded with Problem 322, and would answer the first question here. For $m<k$, removing every epsilon loss would require further control; $3$ with an epsilon loss does not give the requested constant-order bound.

Remaining gap

The separated intervals use smaller and smaller scales for the later summands, leading to $\beta_m<m/k$ when $m\ge2$. This construction cannot reach either target exponent merely by optimizing its fixed constants. Estimate $3$ identifies the additional average collision control that would suffice; no such general sharp bound has been proved in this attempt. Finite counts are not substituted for that missing asymptotic argument.

Completion rate estimate: 25%.