If any multitude whatsoever of numbers is in continued proportion, (starting) from a unit, then the third from the unit will be square, and (all) those (numbers after that) which leave an interval of one (number), and the fourth (will be) cube, and all those (numbers after that) which leave an interval of two (numbers), and the seventh (will be) both cube and square, and (all) those (numbers after that) which leave an interval of five (numbers). * Let any multitude whatsoever of numbers, $A$, $B$, $C$, $D$, $E$, $F$, be in continued proportion, (starting) from a unit. * I say that the third from the unit, $B$, is square, and all those (numbers after that) which leave an interval of one (number). * So I also say that the fourth (number) from the unit, $C$, is cube, and all those (numbers after that) which leave an interval of two (numbers). * So, similarly, we can show that all those (numbers after that) which leave an interval of five (numbers) are (both) cube and square.
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Proofs: 1