OBJECTIVE AND SOURCE REVIEW
Erdős #203; independent attempt, 2026-09-23. The original target requires $m>1$ with $(m,6)=1$ (a condition omitted from the supplied formulation) for which every $2^k3^\ell m+1$, $k,\ell\ge0$, is composite. I read the complete
203.tex, including its finite-ray theorem and prime strengthening. Those results quantify over a new $m$ for each finite collection of rays. This attempt isolates precisely what a finite modular certificate for the full two-dimensional problem would have to prove.
WORK: FINITE PERIODIC CERTIFICATE
Take a finite set $Q$ of primes greater than three and choose residues $c_q\in\mathbb Z/q\mathbb Z$. Put \[
U=\operatorname{lcm}_{q\in Q}\operatorname{ord}_q(2),\qquad
V=\operatorname{lcm}_{q\in Q}\operatorname{ord}_q(3).
\] Suppose the following finite check succeeds: \[
\forall (i,j)\in\{0,\ldots,U-1\}\times\{0,\ldots,V-1\},\quad
\exists q\in Q:\quad 2^i3^j c_q\equiv-1\pmod q.
\tag{*}
\] Then the Chinese remainder theorem gives positive $m\equiv1\pmod6$ with $m\equiv c_q\pmod q$ for all $q$. Choose such an $m>\max Q$. Reducing $(k,\ell)$ modulo $(U,V)$ and using $*$ gives $q\mid 2^k3^\ell m+1$. This integer exceeds $q$, so it is composite. The pair $(0,0)$ is also covered (or simply handled by parity). Conversely, if a proposed fixed finite family of primes $Q>3$ divides at least one of the values for
every exponent pair for some fixed $m$, its residues $c_q=m\bmod q$ satisfy $*$. Hence $*$ is an exact finite test for this specified covering-prime strategy, including the proper-divisor check via sufficiently large CRT representatives.
ADVERSARIAL VERIFICATION AND REMAINING GAP
The multiplicative orders exist because $q>3$. Choosing one residue per prime avoids incompatible CRT demands. Separate periods $U,V$ cover the full lattice rather than merely finitely many rays. No choices satisfying $*$ have been found or claimed here. The existence of successful finite-ray constructions does not imply one finite $Q$ meeting $*$. Nor is every hypothetical composite-valued $m$ guaranteed to have a finite covering-prime certificate. Compactness in a profinite completion would produce a profinite integer, not necessarily an ordinary positive integer. Thus neither direction silently exchanges $\forall D\,\exists m$ with $\exists m\,\forall D$.
SOURCES CHECKED
Complete local source:
gpt_pro_5.4/203.tex. Filaseta's cited primary record was checked:
https://arxiv.org/abs/2305.09219. The certificate theorem uses only elementary multiplicative orders and CRT, and does not reassert unverified details of its prime-string corollary.
FINAL
UNRESOLVED. A candidate finite covering can now be checked on an explicit rectangle, but no successful certificate or unrestricted impossibility proof is supplied.
COMPLETION ESTIMATE
Conservative subjective progress toward one integer handling every exponent pair.
COMPLETION: 10%