GPT 6 Astra Ultra attempt on Math-827
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Erdős Problem #477: a polynomial value set that tiles the integers
Known positive construction with an explicit analytic input, 24 September 2026.Statement and choice of polynomial
We seek a polynomial $f:\mathbb Z\to\mathbb Z$, of degree at least two, and $A\subseteq\mathbb Z$ such that every integer has exactly one expression $a+b$ with $a\in A$ and $b\in f(\mathbb Z)$. We use \[f(t)=t^{13},\qquad B=\{t^{13}:t\in\mathbb Z\},\qquad D=B-B.\] The thirteenth-power construction appears in the linked pipeline-math exposition. The current page also reports positive results for even powers. No such even-degree strengthening is needed here. The older local attempt treating squares negatively does not rule out this construction.The external input is the diagonal case of Heath-Brown's Theorem 2, stated precisely as follows: for a fixed degree $d\ge3$ and positive integer $N\ll X$, the equation \[x_1^d\mathbin{\pm}x_2^d\mathbin{\pm}x_3^d=N,\qquad \max_i|x_i|\le X\] has $O_d(X^{10/d})$ integer solutions apart from the special solutions in which one signed individual power equals $N$. This theorem, not its determinant-method proof, is used below.
Bad shifts have zero density
Fix $c\in\mathbb Z\setminus B$; in particular $c\ne0$. Call $t$ bad if $t^{13}-c=u^{13}-v^{13}$ for some integers $u,v$. Then $u\ne v$. There is an absolute constant $\eta>0$ such that \[\left|u^{13}-v^{13}\right|\ge \eta\max(|u|,|v|)^{12}\qquad(u\ne v\text{ integers}).\tag{1}\] Indeed the polynomial quotient $(u^{13}-v^{13})/(u-v)$ is positive on real pairs other than $(0,0)$, including its continuous diagonal value $13u^{12}$. Its positive minimum on the compact set $\max(|u|,|v|)=1$, followed by homogeneity and $|u-v|\ge1$, proves $1$.Thus, if $|t|\le T$, every corresponding pair satisfies \[\max(|u|,|v|)\ll_c T^{13/12}.\] The equation becomes $u^{13}+(-v)^{13}+(-t)^{13}=-c$. If $-c<0$, negate all three coordinates. It is now a sum of three thirteenth powers equal to the positive integer $|c|$, with all coordinates bounded by $X=C_c T^{13/12}$. No special solution is possible: if one power equals $|c|$, then, since the exponent is odd, $c$ itself is an integer thirteenth power, contrary to its choice.
For all sufficiently large $T$, the condition $|c|\ll X$ holds. Heath-Brown's theorem therefore bounds the number of triples by \[O_c(X^{10/13})=O_c(T^{5/6}).\tag{2}\] The number of bad $t$ is at most the number of triples, so $2$ is $o(T)$. Constants may depend on the fixed $c$; no unjustified uniformity in an increasing collection of shifts is used.
A greedy tiling lemma and the construction
The preceding estimate implies that for every finite $C\subseteq\mathbb Z\setminus B$ there is $b\in B$ with \[(C-b)\cap D=\varnothing.\tag{3}\] For each $c\in C$, the excluded choices $b=t^{13}$ with $|t|\le T$ number $O_c(T^{5/6})$. The union over fixed finite $C$ is still $o(T)$, whereas $2\lfloor T\rfloor+1$ distinct thirteenth powers are available. The equivalence of the exclusion in $3$ with $2$ uses $D=-D$.Enumerate the integers as $z_1,z_2,\ldots$, and start with $A_0=\varnothing$. Maintain that the translates $a+B$, $a\in A_j$, are pairwise disjoint and cover $z_1,\ldots,z_j$. If the next integer $z$ is already covered, leave the finite set unchanged. Otherwise $C=z-A_j$ is a finite subset of $\mathbb Z\setminus B$. Choose $b$ as in $3$ and adjoin $a_*=z-b$.
The new translate contains $z$. For every old $a$, its difference $a_*-a=(z-a)-b$ is outside $D$. An intersection $(a_*+B)\cap(a+B)$ would put this difference in $B-B$, a contradiction. In particular $a_*$ is new, since $0\in D$. This proves the induction. The union $A=\bigcup_j A_j$ covers every integer exactly once together with $B$, as required.