518 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xix 



and showing that when r is replaced by its value from the preceding 

 equation, (s-l) 2 O+l) becomes (2 2 -p-s-l) 2 . 



C. 0. Boije af Gennas 94 gave the solution 

 r, s(rs+2), (s + l)(rs+r+2), 4(rs + l)(rs+r+l)(rs 2 +r5+2s + l). 



For r = l, s = 2, we get 1, 8, 15, 528. 

 J. Knirr 95 took as the four numbers 



n, a?n+2a, tfn+2b, p 2 n+2p. 

 The product of the second and third, increased by unity, is 



{abn+(a+b}} 2 +{l+4ab-(a+b} 2 } 



and is a square if the final part is zero, whence b = al. The product of 

 the second and fourth, increased by unity, is then the square of l-\-pq if 



p (q 2 a 2 n 2 2ari) = 2a?n -\-4a-2q. 



The coefficient of p is unity if q = an-\-l. G. H. F. Nesselmann 95 " took 

 fe = a-f-l, p = a-\-2. 



C. C. Cross 96 gave the set due to Boije 94 with r, s replaced by m, n 1. 

 He and others failed to find five such numbers. He 97 later took the fifth 

 number to equal the first one m, the only new condition being m 2 +l = D, 

 for example, m= (k 2 I}/ (2k}. 



M. A. Gruber 98 noted a special case of Euler's 92 five numbers. 



A. Gerardin" obtained special solutions by recurring series. 



Fermat 100 treated the problem to find four numbers such that the product 

 of any two increased by the sum of those two gives a square. He made 

 use of three squares such that the product of any two increased by the sum 

 of the same two gives a square. Stating that there is an infinitude of such 

 sets of three squares, he cited 4, 3504384/d, 2019241/d, where d = 203401. 

 However, he actually used the squares 25/9, 64/9, 196/9, of Diophantus V, 5, 

 which have the additional property that the product of any two increased 

 by the third gives a square. Taking these three squares as three of our 

 numbers and x as the fourth, we are to satisfy 



3J:~_1_2_5 f-| I .64._ r-i 205 I 196 [-[ 



9 X~T 9 1 I, 9 X~r 9 J, 9 X-\~ 9 J. 



This " triple equation' 1 ' with squares as constant terms is readily solved. 

 T. L. Heath 101 found x to be the ratio of two numbers each of 21 digits. 



L. Euler 102 gave a more general treatment of the latter problem. Let 

 A, B, C, D denote the numbers increased by unity. Then AB 1, , 

 CD I are to be squares. Take AB = p z +l, 

 A+B+2(ap+a} 



r- -rr- -, - 7^ -- 



(ap a) 2 ((3p 



94 Nouv. Ann. Math., (2), 19, 1880, 278-9; E. Lucas, Thdorie des nombres, 1891, 129. 

 86 Die Auflosung der Gleichung z 2 -cz 2 = l, 18. Jahresbericht Oberrealschule, 1889, 31. 

 960 Zeitschr. Math. Phys., Hist.-lit. Abt., 37, 1892, 167. 

 99 Amer. Math. Monthly, 5, 1898, 301-2. 



97 Ibid., 6, 1899, 85-87. 



98 Ibid., 122-3. 



99 L'interm&Haire des math., 23, 1916, 14-15. 



D0 Oeuvres, III, 242-3. A special case of our main problem since xy+x+y = (x + l)(y + 1) 1. 

 11 Diophantus of Alexandria, ed. 2, 1910, p. 163. 

 102 Posth. paper, Comm. Arith., II, 579-582; Opera postuma, 1, 1862, 131-4. 



