By Jacques Sesiano
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Here we write y where Diophantus, who has only a single symbol for the unknown, writes the equivalent of our x. He thus uses the same unknown for the main problem (see below) and this preliminary problem. But no confusion between these unknowns is possible as the two calculations are clearly separate. 40 Diophantus does not justify his choice. However, if we set the square to be (my + 1)2 , 2m we obtain y = 26−m 2 ; then m = 5 is indeed the most appropriate integer value, because y must be large for the fraction added to 26 to be small.
I always 25 multiply the hypotenuse by itself, there results 625. I add to this quantity 4 areas, which makes 600 feet; both together, there results 1225 feet. I take the side of this, there results 35. 26 I subtract from this 4 areas, there results 25 feet; I take the side of this, there results 5; it will be the diﬀerence. I always add this to the two added (straight lines), that is, to 35, there results 40 feet. I always take half of this, there results 20 feet. It will be the base of the triangle.
The reason that Fermat had to reestablish this condition was that the text transmitted through all surviving manuscripts is corrupted at this place due to the negligence or carelessness of a copyist; however, the remaining fragments show that Diophantus did indeed know the elements of the condition as reconstructed by Fermat37 . 37 It may seem surprising that a corruption of the text would be common to all the manuscripts. However, ancient texts that were less commonly used, as were mathematical texts of higher level, often survived at the end of antiquity through a single specimen that served as the source of all subsequent copies.
An Introduction to the History of Algebra by Jacques Sesiano