Extending $(0,u)$, $u\equiv3$, by $t\equiv2 \pmod 4$ with one helper works iff $3t=2u$

Fix $\varepsilon:\mathbb{N}\to\{0,1\}$ and the $2$-adic total order on $\mathbb{Z}$ given by $x\prec_\varepsilon y$ iff $x\neq y$ and the $k$-th two's-complement digit of $x$ equals $\varepsilon_k$, where $k=\nu_2(y-x)$. A finite word $Q$ of integers is $\varepsilon$-safe if the sequence "the letters of $Q$ in order, then all remaining integers in $\prec_\varepsilon$-order" contains no monotone $4$-term arithmetic progression. Let $u\equiv 3\pmod 4$ and $t\equiv 2\pmod 4$ be integers. The following are equivalent:

Thus, outside the single exceptional relation $3t=2u$, absorbing a new target $t\equiv 2\pmod 4$ into the safe word $(0,u)$ with $u\equiv 3\pmod 4$ requires at least two auxiliary letters, wherever they are placed. This is a sharp case of the one-step extension problem for Ho-style constructions of $4$-AP-free permutations of $\mathbb{Z}$ (Erdős Problem 195); e.g. for $u=-1$ every $t\equiv 2\pmod 4$ needs at least two helpers.

Formalization note: "begins with $0,u$" is [0, u] <+: Q; the letter bound is ∀ x ∈ Q, x = 0 ∨ x = u ∨ x = t ∨ x = h; Safe, epsLt come from Definitions.Def_Erdos195_defs.