Kitap 10
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And the product of AB, BC is square, inasmuch as it was proved that, if two similar plane numbers by multiplying one another make some number the product is square. [IX. 1]
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Therefore two square numbers, the product of AB, BC, and the square on CD, have been found which, when added together, make the square on BD.
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And it is manifest that two square numbers, the square on BD and the square on CD, have again been found such that their difference, the product of AB, BC, is a square, whenever AB, BC are similar plane numbers.
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But when they are not similar plane numbers, two square numbers, the square on BD and the square on DC, have been found such that their difference, the product of AB, BC, is not square. Q. E. D.
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LEMMA 2. To find two square numbers such that their sum is not square.
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For let the product of AB, BC, as we said, be square, and CA even, and let CA be bisected by D.
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It is then manifest that the square product of AB, BC together with the square on CD is equal to the square on BD. [See Lemma 1]
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Let the unit DE be subtracted; therefore the product of AB, BC together with the square on CE is less than the square on BD.
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I say then that the square product of AB, BC together with the square on CE will not be square.
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For, if it is square, it is either equal to the square on BE, or less than the square on BE, but cannot any more be greater, lest the unit be divided.
