GPT 6 Astra Ultra attempt on Math-5578

Erdős Problem 325: a checked attempt

GPT_6_Astra_Ultra

25 September 2026

Statement and scope

For $k\ge3$, let $f_{k,3}(x)$ count integers at most $x$ that are sums of three nonnegative $k$th powers. Is its order $x^{3/k}$, or at least $\gg_{k,\epsilon}x^{3/k-\epsilon}$ for every $\epsilon>0$? The question remains unresolved here. We prove the elementary partial bound \[ f_{k,3}(x)\gg_{k,\epsilon}x^{3/k-2/k^2-\epsilon}. \tag{1} \] Source: https://www.erdosproblems.com/325. This improves the exponent in the earlier GPT-5.2 attempt by using many distinct two-power sums inside separated blocks. It is weaker than known results in important cases; no literature improvement or novelty is claimed.

Two-power sums with subpolynomial multiplicity

Let $r_{k,2}(n)$ count ordered pairs of nonnegative integers with $u^k+v^k=n$. For odd $k$ and $n>0$, $s=u+v$ divides $n$. At a fixed $s$, strict convexity and symmetry of $u^k+(s-u)^k$ give at most two ordered pairs. Hence $r_{k,2}(n)\le2\tau(n)$.

For even $k$, map a pair to $(u^{k/2},v^{k/2})$, an injective map into integer representations of $n$ as two squares. Unique factorization in the Euclidean ring $\mathbb Z[i]$ gives at most $4\tau(n)$ such representations: a split prime of exponent $e$ offers $e+1$ conjugate allocations, inert primes offer none unless their exponent is even, the ramified prime $2$ offers one, and there are four units. This standard Gaussian-integer factorization is the explicit algebraic input.

For every $\delta>0$, $\tau(n)\ll_\delta n^\delta$. To verify the usual divisor bound, write $\tau(n)=\prod_p(e_p+1)$. For sufficiently large $p$, $e+1\le2^e\le p^{\delta e}$; for the finitely many smaller primes the ratios $(e+1)p^{-\delta e}$ have finite maxima. Multiplication proves the bound. Also $r_{k,2}(0)=1$.

All pairs $0\le u,v\le\lfloor(y/2)^{1/k}\rfloor$ have sum at most $y$, and there are $\gg_k y^{2/k}$ such pairs. Dividing this number by the maximum multiplicity $O_\delta(y^\delta)$ shows that the number of distinct two-power sums at most $y$ satisfies \[ f_{k,2}(y)\gg_{k,\delta}y^{2/k-\delta}. \tag{2} \] This is a proved weaker form of the sharp two-power count; the latter is not being silently assumed.

Separated translates produce three-power sums

Let $N=\lfloor(x/2)^{1/k}\rfloor$ and $y=(N/2)^{k-1}/2$. For every integer $a\in[\lceil N/2\rceil,N]$, translate all values counted in $2$ by $a^k$. They lie in $[a^k,a^k+y]$. These intervals are disjoint because \[ (a+1)^k-a^k\ge k a^{k-1}\ge k(N/2)^{k-1}>y. \] For sufficiently large $x$ their right endpoints are at most $N^k+y\le x$. There are at least $N/2$ eligible choices of $a$, so \[ f_{k,3}(x)\ge\frac N2 f_{k,2}(y) \gg_{k,\delta}x^{1/k+(k-1)(2/k-\delta)/k}. \] Taking $\delta=\epsilon k/(k-1)$ proves $1$. For cubes this gives exponent $7/9-\epsilon$, rather than the old draft's $1/2$ exponent. All constants and thresholds may depend on $k,\epsilon$ as indicated.

The exact obstruction in this method

The interval width is $O(x^{(k-1)/k})$, forced by the spacing between successive large $a^k$. Even replacing $2$ by the sharp lower bound $f_{k,2}(y)\gg y^{2/k}$ only yields exponent $3/k-2/k^2$, still strictly below $3/k$. To reach the proposed scale one must control overlaps between a much larger family of translates, rather than keeping them disjoint.

Counting all triples gives the upper bound $f_{k,3}(x)\le\binom{\lfloor x^{1/k}\rfloor+3}{3}=O_k(x^{3/k})$. This shows the target is the natural counting scale, but the existence of that many input triples supplies no lower bound without multiplicity control. The missing overlap estimate is not supplied by finite enumerations or by the known two-summand theorem alone.

Completion rate estimate: 30%.