For integers 4 ≤ k < l, some difference occurs three times in {0, ±1, ±k, ±l} iff l ∈ {k+1, k+2, 2k−1, 2k, 2k+1, 3k}

Let $4 \le k < l$ be integers and $S = \{0, \pm 1, \pm k, \pm l\}$. Some $d > 0$ has $\#\{s \in S : s + d \in S\} \ge 3$ if and only if $l \in \{k + 1, k + 2, 2k - 1, 2k, 2k + 1, 3k\}$.

Proof. The six cases have explicit witnesses (for $l = 2k$: $d = k$, $s \in \{0, -k, k\}$). Conversely, $s \mapsto -s - d$ is an involution of $\{s \in S : s + d \in S\}$, so three elements contain two, $x$ and $y$, with $y \notin \{x, -x - d\}$; the pairs $(x, x + d)$ and $(y, y + d)$ then force a coincidence among the differences $1, 2, k \pm 1, k, 2k, l - k, l \pm 1, l, l + k, 2l$, and a case analysis closed by linear arithmetic leaves exactly the six values of $l$.

Novelty. Elementary.