A vertex of the strengthened problem: a visible lattice point with both coordinates exceeding 1 and at least one coordinate composite.

Sanity check for Valid: the vertex $(4, 3)$ is valid — both coordinates exceed $1$, they are coprime, and $4$ is composite.

Two lattice points are adjacent iff they differ by exactly 1 in exactly one coordinate.

Let $G$ be the graph with vertex set those pairs $(x,y)\in \mathbb{N}^2$ with $\mathrm{gcd}(x,y)=1$, in which we join two vertices if the differ in only one coordinate, and there by $\pm 1$.

Is there a path going to infinity on $G$, say $P$, such that for all $(x,y)\in P$ both $\min(x,y)>1$ and at least one of $x$ or $y$ is composite?

The weaker version (only $\min(x,y) > 1$) was solved by C. Stewart via the prime-pair path $(p_k, p_{k+1}) \to (p_{k+1}, p_{k+2})$, as recounted in [Er80]; the compositeness condition forbids those anchors and the question is open.

Core of the composite-anchor reduction: vertical-leg vertices $(a, s)$ for $b \le s \le c$ are valid vertices of the strengthened problem, given the anchor $a$ is composite and coprime to the whole leg.

Core of the composite-anchor reduction: horizontal-leg vertices $(s, c)$ for $a \le s \le b$ are valid, given the anchor $c$ is composite and coprime to the whole leg.

Roughness criterion (sufficiency for the anchor conditions): if $a < s$ for all $s$ in the leg and the leg stays below $a + P^-(a)$, then $a$ is coprime to the whole leg. Stated via divisibility: no prime factor of $a$ divides any $s$ with $a < s < a + p$ for all prime factors $p$ of $a$.

Isolation lemma, right neighbour (core of the no-periodic-certificate theorem): if every prime in $P$ divides $x$ and none divides $y$, then no prime of $P$ divides either coordinate of $(x+1, y)$.

Isolation lemma, left neighbour.