486 HlSTOEY OF THE THEORY OF NUMBERS. [CHAP. XVII 



(II) Find two rectangles of equal perimeters such that the area of the 

 second is 4 times the area of the first. The sum of two sides of the first 

 rectangle is taken to be 4 3 1 = 63, and one side 4 1=3, so that the other 

 is 63 3 = 60. For the second rectangle one side is 4 2 1 = 15 and the other 

 63 - 15 = 48. The areas are 180, 720. 



Pappus of Alexandria 9 (end , of third century) discussed the simpler 

 (determinate) problem: Given a parallelogram, find a second whose sides 

 have a given ratio to the sides of the first, while the areas have a given ratio. 



Maximus Planude 10 (about 1260-1310) discussed the problem to find 

 two rectangles of equal perimeters such that their areas have a given 

 ratio 6:1. The solution is given in words; expressed algebraically, it 

 states that the sides of one are 6 1 and 6 3 6, those of the other 6 2 1 

 and 6 3 6 2 . 



G. Valla x applied the last rule for 6 = 3, 4. 



M. Cantor 12 noted that the general [but see Zeuthen 15 ] solution of 

 Planude's problem is as follows: sides of first a and 6 (6+1) a, sides of second 

 (6+1) a and 6 2 a. 



P. Tannery 13 discussed the generalization of Heron's two problems: 



(1) a(x-\-y)=u-\-v, xy = buv, 



and stated that the general solution, obtained by setting a = pq, a 2 b l = rs, 

 &b = mn, is 



u = aj3q, v = a(x+y)u, x = abu -\-orq~ur, y=abu+^ms. 

 For a b, Heron gave x = 2a 3 , y = 2a? l, u = a, v = 2a(2a? 1); for a = l, 



(2) x = b*-l, ?/ = 6 2 (6-l), u = b-l, y = 6(6 2 -l). 



* Ad. Steen 14 discussed the rational solutions of Planude's problem. 



H. G. Zeuthen 15 noted that, to obtain Heron's solution of (1) for a = 6, 

 it suffices to assume that u = a, whence by the first equation v is a multiple 

 za of a. Then x-\-y = l+z, xy = a?z. Eliminating z, we get 



(x a y ](ya z ) =a 3 (a 3 1), 



which holds if x a 3 = a 3 1, y a 3 = a 3 . Next, for a=l, try v = bx, y = tfu. 

 Then the first equation (1) gives (6 !)#= (6 2 l)u, which is satisfied by (2). 

 If we replace the common factor 6 1 in (2) by a, we get Cantor's solution, 

 which is however not the general one. If we use v/x = m in place of the 

 earlier v/x = b, we get y mbu by (1 2 ) and find that the general solution of 



9 Sammlung, Buch III, Pappus ausgabe (cd., Hultsch), Berlin, 1875, 1877, 1878, 126. Cf. 

 M. Cantor, Geschichte Math., ed. 3, 1, 1907, 454. 



10 Computation (Rechenbuch, Livre de Calcul). Greek text by C. I. Gerhardt, Halle, 1865, 



pp. 46, 47. German transl. (inadequate) by H. Waeschke, Halle, 1878. M. Cantor, 

 Geschichte Math., cd. 3, 1, 1907, 513. 



11 De Expetendis et fugiendis rebus opus, Aldus, 1501, Liber IV ( = Arithmeticae, III), Cap. 13, 



12 Die Romischen Agrimensoren, 1875, 62-3, 194-5. 



13 L'Arith. des Grecs dans H6ron d'Alexandrie, Me"m. soc. sc. phys. et nat. Bordeaux, (2), 4, 



1882, 192. 



14 Tidsskrift for Math., (5), II, 139-147. 



15 Bibliotheoa Math., (3), 8, 1907-8, 118-120, 127-9. 



