Ruzsa’s genus-one problem

Constructed by affine gap compression from the 14-element witness at (N=301) (witness #17). Multiplication by (102\pmod{301}) transforms the original set into [{6,12,20,26,92,96,106,110,225,239,270,272,284,286}.] The interval between (110) and (225) contains no elements. Compressing this empty gap by (14)—shifting the upper block down by (14)—reduces the modulus from (301) to (287) while retaining all 14 elements. Translating the resulting set by (-211\pmod{287}) gives the displayed witness. Exhaustive verification of all (14^4=38{,}416) ordered quadruples finds exactly 14 solutions to [ a+3b\equiv2c+2d\pmod{287}, ] all of which are diagonal.