If any multitude whatsoever of numbers is in continued proportion, (starting) from a unit, and the (number) after the unit is not square, then no other (number) will be square either, apart from the third from the unit, and all those (numbers after that) which leave an interval of one (number). And if the (number) after the unit is not cube, then no other (number) will be cube either, apart from the fourth from the unit, and all those (numbers after that) which leave an interval of two (numbers). * Let any multitude whatsoever of numbers, $A$, $B$, $C$, $D$, $E$, $F$, be in continued proportion, (starting) from a unit. * And let the (number) after the unit, $A$, not be square. * I say that no other (number) will be square either, apart from the third from the unit [and (all) those (numbers after that) which leave an interval of one (number)]. * And so let $A$ not be cube. * I say that no other (number) will be cube either, apart from the fourth from the unit, and (all) those (numbers after that) which leave an interval of two (numbers).
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Proofs: 1
