Is there a polynomial $f:\mathbb{Z}\to \mathbb{Z}$ of degree at least $2$ and a set $A\subset \mathbb{Z}$ such that for any $n\in \mathbb{Z}$ there is exactly one $a\in A$ and $b\in \{f(k):k\in \mathbb{Z}\}$ such that $n=a+b$?

A proven wall: The missing input is an obstruction lemma for degree >= 3 that replaces the quadratic difference-set argument. Concretely: prove that for every polynomial f in Z[X] of degree >= 3 (smallest open case f = X^3), no set A subset Z can be a perfect additive complement of the value set {f(n)} — i.e. Z cannot be tiled exactly-once by A (+) f(Z). The degree-2 case is SOLVED because f(Z)-f(Z) contains every multiple of a fixed q (q = 2b if b != 0, q = 4a if b = 0), so any (necessarily infinite) A has A-A hitting a multiple of q, yielding a forced double representation n = a+f(k) = b+f(-k). For deg >= 3 the difference set f(Z)-f(Z) is sparse and is NOT a union of full arithmetic progressions, so it contains the multiples of no fixed modulus; the pigeonhole on A-A collapses. The precise bottleneck object is therefore: a structural impossibility certificate for exact additive tilings of Z by a sparse polynomial value set of degree >= 3 — e.g. a counting/density contradiction (the value set {f(n) : |n| <= N} has size ~N while a perfect complement would need ~ X/N elements of A in [0,X], and exact-once representation must hold for ALL n, a condition the sparse non-AP difference structure should preclude). Equivalently, prove the cubic (X^3) variant erdos_477.variants.X_pow_three. There is no candidate construction believed to exist (owner and Erdős/Graham expect the answer NO), so closure = a proof of universal non-tileability for deg >= 3, not a search for an example.

number theory · open · formalized (Lean) · 1 attempt

machinery: additive-complement-tiling,perfect-difference-tiling,polynomial-value-set,difference-set-arithmetic-progressions,Sidon/B_h,pigeonhole-density-counting,covering-system,exact-cover-uniqueness