Return to Vignettes of Ancient Mathematics

The text used is the edition of Tannery (1893), but I have also consulted the translation of ver Eecke (1959) and the paraphrase of Heath (1910).

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For notes on translation, go to the introduction to Book I.

Book II

8. To divide the prescribed square into two squares.

In fact, let it be prescribed to divide 16 into two squares. And let the 1st be assigned ΔΥ 1, therefore the other will be units 16 minus ΔΥ. It will be required, therefore, for units 16 minus ΔΥ to be equal to a ⬜.

I form the ⬜ from however many arithmos’s minus Μ’s that are the side of 16 units. Let it be arithmos 2 minus units 4. Therefore, the ⬜ will be ΔΥ 4 units 16 minus arithmos 16. These are equal to units 16 minus ΔΥ. Let a common be added, the deficiency and similars from similars.

Therefore, ΔΥ 5 are equal to arithmos 16, and arithmos becomes 16 of fifths.

One will be 26425´, another 14425´, and the two added make 40025´, or units 15, and each is a square.

Differently

In fact let it again be required to divide square 16 into two squares. Let again the side of the 1st be assigned arithmos 1, that of the other however many arithmos’s minus units of as many as the side of the divided. In fact, let it be arithmos 2 minus units 4.

Therefore, the ⬜’s will be, one ΔΥ 1 and the other ΔΥ 4 units 16 minus arithmos 16. Remaining, I want the two added equal to units 16.

Therefore, ΔΥ 5 units 16 minus arithmos 16 are equal to units 16. And arithmos becomes 16.

The sd. of the 1st will be 16; therefore, it will be 25625´. An the sd. of the 2nd 12; therefore, it will be 14425. And the demonstration is obvious.

9. With given arithmos which is composed from two squares to divide it again into two different squares.

Let it be required with 13, which is composed from squares 4 and 9, to divide it again into two other squares.

Let there be taken of the mentioned squares the sds, units 2, units 3, and let there be assigned the sds of the prescribed squares, one arithmos 1 units 2, the other arithmos of as many it may be minus units of as many as is the side of the remaining. Let it be arithmos 2 minus units 3. And the squares become, one ΔΥ 1 arithmos 4 units 4, the other ΔΥ 4 units 9 minus arithmos 12.

Remaining is to make the two added units 13. But the two added make ΔΥ 5 units 13 minus arithmos 8. These are equal to units 13. and arithmos becomes 85. For the actualities: I assigned the sd of the 1st, arithmos 1 units 2. It will be 18Μ5. The sd of the 2nd I assigned arithmos 2 minus units 3. It will be the side of one of them. And the ⬜’s themselves will be, one 32425´, which is of one. And the two added make 32525´, which brings together the prescribed units 13.

10. To find two square arithmoi in the given excess

In fact, let it be prescribed that the excess of them is units 60.

Let it be assigned one whose side is arithmos 1 and the other whose side is arithmos 1 and however many units’s you want, merely so that the ⬜ from the units’s does not exceed the excess that’s given, nor, truly, that it be equal. For, when one form is left equal to one form, the problem will be established.

Let it be arithmos 1 units 3. Therefore the squares will be ΔΥ 1 and ΔΥ 1 arithmos 6 units 9. And the excess of them is arithmos 6 units 9. These are equal to units 60. And arithmos becomes units 8 𐅵. The side of the 1st will be units 8 𐅵, and that of the 2nd units 11 𐅵. The ⬜’s themselves will be, one 72 4fraction, the other 132 4fraction, and things of the proposition are obvious.

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