Ruzsa’s genus-one problem

witness #1003

N
21,025
|A|
100
exponent log |A| / log N
0.4627
status
current record for this modulus
submitted by
David Renshaw
submitted at
2026-08-06 18:56:36 UTC

Elements (100)

0, 1, 5, 6, 28, 30, 33, 35, 86, 91, 145, 146, 150, 151, 173, 175, 178, 180, 231, 236, 725, 726, 730, 731, 753, 755, 758, 760, 811, 816, 870, 871, 875, 876, 898, 900, 903, 905, 956, 961, 4060, 4061, 4065, 4066, 4088, 4090, 4093, 4095, 4146, 4151, 4350, 4351, 4355, 4356, 4378, 4380, 4383, 4385, 4436, 4441, 4785, 4786, 4790, 4791, 4813, 4815, 4818, 4820, 4871, 4876, 5075, 5076, 5080, 5081, 5103, 5105, 5108, 5110, 5161, 5166, 12470, 12471, 12475, 12476, 12498, 12500, 12503, 12505, 12556, 12561, 13195, 13196, 13200, 13201, 13223, 13225, 13228, 13230, 13281, 13286

Commentary

Square lift of the N=145 record. If A1 and A2 are both solution-free mod m, then B = {a + m*b : a in A1, b in A2} is solution-free mod m^2: reducing a putative solution mod m makes the low parts a solution mod m, so they are all equal, which makes the low contribution vanish exactly (the coefficients satisfy 1+3-2-2=0); the remaining carry then forces the high parts to be a solution mod m, hence also all equal. Taking A1 = A2 = Bhavik Mehta's size-10 witness for m=145 (score 0.8305) gives |B| = 100 at N = 145^2 = 21025, score 100/145 ≈ 0.6897. Greedy augmentation confirms the set is maximal: no single element of Z/21025 can be added without creating a nontrivial solution. Verified locally by exhaustive pair-count convolution (exactly |B| solutions, all trivial) before submission.

last edited by David Renshaw at 2026-08-06 18:56:36 UTC · history

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