Proposition 10.28
To find medial straight lines commensurable in square only which contain a medial rectangle.
To find medial straight lines commensurable in square only which contain a medial rectangle.
Let the rational straight lines A, B, C commensurable in square only be set out; let D be taken a mean proportional between A, B, [VI. 13] and let it be contrived that, as B is to C, so is D to E. [VI. 12]
Since A, B are rational straight lines commensurable in square only, therefore the rectangle A, B, that is, the square on D [VI. 17], is medial. [X. 21]
Therefore D is medial. [X. 21]
And since B, C are commensurable in square only, and, as B is to C, so is D to E, therefore D, E are also commensurable in square only. [X. 11]
But D is medial; therefore E is also medial. [X. 23, addition]
Therefore D, E are medial straight lines commensurable in square only.
I say next that they also contain a medial rectangle.
For since, as B is to C, so is D to E, therefore, alternately, as B is to D, so is C to E. [V. 16]
But, as B is to D, so is D to A; therefore also, as D is to A, so is C to E; therefore the rectangle A, C is equal to the rectangle D, E. [VI. 16]
But the rectangle A, C is medial; [X. 21] therefore the rectangle D, E is also medial.
Therefore medial straight lines commensurable in square only have been found which contain a medial rectangle. Q. E. D.
LEMMA I.
To find two square numbers such that their sum is also square.
Let two numbers AB, BC be set out, and let them be either both even or both odd.
Then since, whether an even number is subtracted from an even number, or an odd number from an odd number, the remainder is even, [IX. 24, 26] therefore the remainder AC is even.
Let AC be bisected at D.
Let AB, BC also be either similar plane numbers, or square numbers, which are themselves also similar plane numbers.
Now the product of AB, BC together with the square on CD is equal to the square on BD. [II. 6]
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]
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.
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.
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.