Can every nonzero number in a finite arithmetic system be arranged around a circle under a demanding rule? Multiply each pair of neighbours, add a fixed number, and the result must be primitive. That means its powers visit every nonzero number in the system.

Sun conjectured that this was always possible for fields with more than seven elements. This candidate gives a precise correction: there is exactly one exception, in arithmetic modulo eleven when the fixed number is two.

The obstruction is small enough to see. Two vertices, labelled two and nine, each have only two possible neighbours: three and eight. Any circle visiting every vertex would have to use all four of those links. But those links already close a four-vertex circle. The other six vertices cannot join it.

Why should every other case work? Earlier research settled all sufficiently large fields and the zero-parameter case. An improved bound leaves five hundred and ten field sizes. A search then generates circular orderings for every required nonzero parameter, and a separate checker verifies the field arithmetic, every vertex and every neighbouring product. The complete witness streams are regenerated during replay rather than stored in the archive.

The revision also gives a limited robustness result: where the bound has enough room, deleting some vertices and some edges still leaves a Hamiltonian circle. It is a sufficient condition, not a sharp limit.

This remains an unrefereed computer-assisted candidate. Its exact classification is a specialist contribution built on substantial earlier work; the checks do not establish formal verification or external peer review.

This Evidence Press briefing is dated the twelfth of October, twenty twenty-six. The paper and replay package are linked on the release page. This synthetic voice explanation is not additional evidence.