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2, 14, 10is not a solution? \$\endgroup\$2, 14, 10,1,7,5etc. have one thing in common: the x and y coordinates of the intersection are equal, and both chords are split equally. These solutions are in some sense trivial. I understood "a diagonal cannot be a solutionm+n<2r" to mean "a diameter cannot be a solution" but it seems the intent may indeed have been to exclude solutions where the intersection falls on a diagonal as well as ones where a chord is a diameter. I will wait for OP response but that's the only way I can explain the test cases. \$\endgroup\$(m,n,r)=(1,7,5)(with CS=BS) now looks valid. For X=100, I believe this would lead to 74 solutions instead of 37. \$\endgroup\$