446 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xv 



Thus, if a = 2, 6 = 3, we substitute 2x 2 -\-2x for x to satisfy the first identically ; 

 then the other two become 6x~-\-6x-\-l = D, 10:E 2 -|-10+1 = D, one solution 

 being x= 1; but this makes the unknown 2x 2 -\-2x zero [von Schaewen 81 ]. 

 Although the method fails for a = 5, 6 = 16, x = 3 is a solution. For 

 a = 1, 6 = 2, there is no solution, whence four squares (the first^being taken 

 as unity) cannot be in A. P. 



M. Petrus 12 found three squares A 2 , B 2 , C 2 such that the difference of 

 any two is a square and the difference of the sides of any two is a square. 

 He first gave a process to find four numbers p, s, t, q such that p 2 -\-s 2 , 

 t 2 +q 2 and pstq are squares, while p/s>t/q, solutions being 112, 15, 35, 12 

 and 364, 27, 84, 13. From the former he derived the answer to the first 

 problem: 



A = 26633678, B = 29316722, C = 40606322. 



In general, we have the answer 13 



|= (pt-sq) 2 +(pq-st) 2 , 



since 



C+A = (pt+sq) 2 , C-A = 

 B-A=4(pq-st) 2 , C-B 



Renaldini 14 (1615-1698) treated Petrus' 12 initial problem (in Part II) 

 and (in Sec. 1 of Part III) duplicate and triplicate equalities. 



J. Prestet 15 treated the problem of Diophantus III, 7. Let the sum of 

 the first and third be x~, that of the first and second y 2 , that of all three z 2 . 

 Then the numbers are x 2 -\-y 2 z 2 , z 2 x 2 , z 2 y 2 . The sum of the last two is 

 not easily made a square. Since 2 = 1/25+49/25, set x = z/5. Then the 

 sum of the last two is 49z 2 /25-2/ 2 =(a-7z/5) 2 if 2 = 5(a 2 +?/ 2 )/14. But the 

 numbers obtained this way are larger than those of Diophantus and Vieta. 



For Diophantus III, 9, he used (p. 326) z, z+d, z-\-2d, with 2z+d = y 2 , 

 2z+3d = z 2 , which give z and d. To avoid fractions, multiply the numbers 

 by 4. Hence the numbers are 3y 2 x 2 , y 2 -\-x 2 , y 2 +3x 2 . It remains to 

 make the sum 2y 2 -\-2x 2 of the first and third a square. Express 2 as a sum 

 of two squares, the smaller between 1/2 and 1. By Diophantus II, 10, 

 the root of the smaller is (c 2 2c l)/(c 2 +l). By trial, 9 is the first integer 

 c giving a fraction (31/41) >3/4. Thus 2(31 2 +49 2 ) =S2 2 . Hence z 2 = 2401, 

 ?/ 2 = 961. He gave also a less special solution. He treated (p. 329) 

 analogously Diophantus III, 10. 



J. Ozanam 16 found two numbers, such that each when increased by a 

 square (say unity) gives a square, and such that their sum and their dif- 

 ference increased by another square (say t 2 = x 2 -\-2x-\-l) shall give squares. 

 The required numbers are taken to be 168 2 and I20t 2 . Then the final 



12 Arithmeticae Rationalis Mcngoli Petri, Bononiae, 1674, 1st Pref. Cf. Euler. 28 



13 Reconstructed from the author's inadequate notes on Petrus. 



14 Carol! Renaldinii Mathematum Analyticae Artis Pars Tertia, 16S4; reviewed in Acta 



Eruditorum, 1685, p. 178. 



16 Elemens des Math, ou Principes Generaux . . . , Paris, 1675, 325. 

 Letter, Oct., 13, 1676, to de Billy, Bull. Bibl. Storia Sc. Mat. e Fis., 12, 1879, 517. 



