Diophantos of Alexandria a Study in the History of Greek Algebra

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(1751), pp. 49 sqq. Commentationes arithmeticae, I. Pp. 62-72).
BOOK VI The first equation then gives 54 15 - or 1 5; 2 36 = a square.
This equation we cannot solve because \ 5 is not tfie sum of two squares*. Therefore we must change the assumed triangle. Now (with reference to the triangle 3, 4, 5) I5w 2 = the continued product of a square less than the area, the hypotenuse, and one perpendicular ; while 36 = the continued product of the area, the perpen- dicular, and the difference between t
...he hypotenuse and the perpendicular. Therefore we have to find a right-angled triangle (h, p, b, say) and a square (w 2 ) less than 6 such that m^hp ^pb . P(]i p} is a square. If we form the triangle from two numbers X lt X 3 and suppose that p=2. X^X^ and if we then divide throughout by (X l X^f which is equal to // -/, we must find a square, 5 s [= m-l(X^ - XJf\ such that z*/ip ^pb . P is a square. Ttie problem can be solved if X^, X^ are "similar plane numbers*? Form the auxiliary triangle from similar plane numbers accordingly, say 4, i.

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