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Explore Stack InternalI find a workaround which can find all that exact self-intersections:
sol = Solve[(100 (t1 - t2) == 2 k1 \[Or] 100 (t1 + t2) == (2 k1 + 1)) && (99 (t1 - t2) == 2 k2 \[Or] 99 (t1 + t2) == (2 k2 + 1)), {t1, t2}]; Flatten[({t1, t2} /. # /. Solve[(0 <= t1 <= 2 && 0 <=< t2 <=< 2) /. #, {k1, k2}, Integers]) & /@ sol, 1] I find a workaround which can find all that exact self-intersections:
sol = Solve[(100 (t1 - t2) == 2 k1 \[Or] 100 (t1 + t2) == (2 k1 + 1)) && (99 (t1 - t2) == 2 k2 \[Or] 99 (t1 + t2) == (2 k2 + 1)), {t1, t2}]; Flatten[({t1, t2} /. # /. Solve[(0 <= t1 <= 2 && 0 <= t2 <= 2) /. #, {k1, k2}, Integers]) & /@ sol,1] I find a workaround which can find all that exact self-intersections:
sol = Solve[(100 (t1 - t2) == 2 k1 \[Or] 100 (t1 + t2) == (2 k1 + 1)) && (99 (t1 - t2) == 2 k2 \[Or] 99 (t1 + t2) == (2 k2 + 1)), {t1, t2}]; Flatten[({t1, t2} /. # /. Solve[(0 <= t1 < t2 < 2) /. #, {k1, k2}, Integers]) & /@ sol, 1] I find a workaround which can find all that exact self-intersections:
sol = Solve[(100 (t1 - t2) == 2 k1 \[Or] 100 (t1 + t2) == (2 k1 + 1)) && (99 (t1 - t2) == 2 k2 \[Or] 99 (t1 + t2) == (2 k2 + 1)), {t1, t2}]; Flatten[({t1, t2} /. # /. Solve[(0 <= t1 <= 2 && 0 <= t2 <= 2) /. #, {k1, k2}, Integers]) & /@ sol,1]