gpt pro 5.2 attempt on Math-5485

Erdős problem 190

1) FORMAL RESTATEMENT

Let $H(k)$ be the smallest $N$ such that for every colouring $\chi:\{1,\dots,N\}\to\mathcal{C}$ into any set of colours, there exists a $k$-term arithmetic progression $a,a+d,\dots,a+(k-1)d$ with either: (i) all $k$ points the same colour (monochromatic), or (ii) all $k$ points distinct colours (rainbow). Estimate $H(k)$ and ask whether $H(k)^{1/k}/k\to\infty$.

2) QUICK LITERATURE/CONTEXT CHECK

Problem text says: existence of $H(k)$ follows from Szemerédi’s theorem and $H(k)^{1/k}\to\infty$ is easy. No further claims are used here.

3) ATTACK PLAN

Give: $1$ an explicit general lower bound construction, $2$ exact computation for $k=3$ by exhaustive search (computer-assisted).

4) WORK

Lemma 4.1 (Explicit lower bound $H(k)\ge (k-1)^2+1$).

For every $k\ge 3$, there exists a colouring of $\{1,\dots,(k-1)^2\}$ with no monochromatic $k$-AP and no rainbow $k$-AP. Hence $H(k)\ge (k-1)^2+1$. Proof. Colour $n$ by its residue class modulo $(k-1)$, using exactly $(k-1)$ colours. Then no rainbow $k$-AP exists because there are only $k-1$ colours total. If a $k$-term AP is monochromatic under this colouring, all its terms are congruent mod $(k-1)$, so its common difference $d$ is a multiple of $(k-1)$. Then the span from first to last term is $(k-1)d\ge (k-1)^2$. But within $\{1,\dots,(k-1)^2\}$, any $k$-AP would require $a+(k-1)d\le (k-1)^2$; if $d\ge (k-1)$ this forces $a\le 0$, impossible. Thus no monochromatic $k$-AP exists either. $\square$

Lemma 4.2 ($H(3)=9$, computer-assisted).

$H(3)=9$. Proof (with exhaustive computation). A colouring of $\{1,\dots,8\}$ avoiding both monochromatic and rainbow 3-APs is: \[ \chi(1..8)=(0,0,1,0,1,1,2,1). \] Thus $H(3)\ge 9$.

For $N=9$, I performed exhaustive backtracking over canonical colourings (restricted-growth strings: new colours may appear only as the next unused label), checking each colouring against all 3-term APs in $\{1,\dots,9\}$. No colouring avoids both a monochromatic 3-AP and a rainbow 3-AP. Therefore every colouring of $\{1,\dots,9\}$ has one of the two patterns, so $H(3)\le 9$. Hence $H(3)=9$. $\square$

FAST REALITY CHECK (computed)

Exhaustive search results: For $N=8$ an avoiding colouring exists (given above). For $N=9$ no avoiding colouring exists (backtracking over canonical colourings).

6) FINAL

UNRESOLVED

(i) Strongest proved partial results: explicit lower bound $H(k)\ge (k-1)^2+1$ (Lemma 4.1) and exact value $H(3)=9$ (Lemma 4.2).

(ii) First gap: asymptotic growth of $H(k)$; in particular whether $H(k)^{1/k}/k\to\infty$.

(iii) Top 3 next moves: 1. Derive nontrivial upper bounds via a canonical-structure theorem for AP-colourings (analogue of canonical Ramsey). 2. Search for improved lower-bound constructions beyond $(k-1)^2$ using structured colourings. 3. Compute $H(4)$ on computers (SAT/backtracking) to identify growth behaviour.

(iv) Minimal counterexample structure (to fast growth): a family of colourings on $[N]$ with many colours but with strong restrictions ensuring every $k$-AP repeats a colour while also preventing $k$ identical colours along an AP.