516 



HISTORY OF THE THEORY OP NUMBERS. 



[CHAP, xix 



Take y and 4?/+3 as two of the numbers, which each increased by unity 

 have a ratio which is a square 1/4. From 



= D = 



we get ?/ = 3/10. For the numbers 3/10, 42/10, x, the conditions are 



2 ,2__ i-i 

 5*15 ' - 1 I. 



__._ 

 1 ~ 



By the usual method (Ch. XV), a; = 7/10. Cf. Nesselmann, 95a pp. 142-4. 



N. Saunderson 87 found three numbers a, b, c such that the product of any 

 two increased by t times their sum is a square. Since (a +)(&+) =n 2 -\-t 2 , 

 express ri*-\-t 2 as a product of two factors, say n+r, n s, each >t. Then 

 and 



~2 I /2 



=r-s- or 



rs rs 



When t = l, take r s = l, whence a = r 2 , & = s 2 , c = or 2a+2fr-|-2. 



The same numbers are such that the product of any two increased by t 

 times the third is a square (Diophantus, III, 14). 



P. Cossali 82 noted that if the product of any two of z 2 , z 2 , 

 2{z 2 +2 2 +(2 z) 2 } be increased by (z x)~ times the sum of the two or by 

 (z x) 2 times the third, we get a square. On adding 2(z x) z to each of 

 these three numbers, we get three numbers such that the product of any 

 two diminished by (z #) 2 times either their sum or the third gives a square. 



Diophantus 88 , V, 3 [4], required three numbers such that any one of 

 them or the product of any two of them increased [diminished] by a given 

 number a is a square. He quoted from his Porisms that if x+a = m?, 

 y-\-a = n?, xy+a=[3, then m and n are consecutive numbers. 89 Thus if 

 a = 5 we take x = (2+3) 2 5, y (2+4) 2 5 as two of the required numbers, 

 and 2(x+y) 1 = 42 2 + 282+29 as the third. We are to make 



42 2 +282+34=D, 



say (22 - 6) 2 . Hence z = 1/26. 



For V, 4, Diophantus took a = 6, # = 2 2 +6 and 2/ = (2+l) 2 -f 6 as two of 

 the numbers, and 2(x-\-y) l as the third. The latter less 6 is 



if 2 = 17/28. 



Diophantus' method shows that xy+a, xz-\-a, yz+a, x+a, y+a are all 

 squares if x = m z a, y = (171+1)- a, z = (2m-\-l) 2 4a. To make also 



2+a= D, say (2m r) 2 , we have w = 



l)/{4(l+r)}. 



Fermat (Oeuvres, III, 250) gave a solution for the case a = l. In 



_-_ 

 5 1 84 



. 



36) 



7225 | 85 

 5 184X-T3G, 



s 



the constant terms increased by unity give squares; further, xy+1, 

 2/2+1 are squares. The "triple equation " x +1 = D, y+l = D, 2 + 1 = 

 readily solved since the constant terms are squares (Ch. XV). 



7 The Elements of Algebra, 2, 1740, 399-405; French transl., Sphinx-Oedipe, 1908-9, 3-9. 



19 But this is incorrect; mn = 1 is a sufficient but not necessary condition for xy-\-a= D. 

 In fact, by eliminating x, y, we get ?/i 2 H 2 a(m-+it? l)+a 2 = D. While this is satisfied 

 if wi 2 +n*l=2mn, whence m=nl, it can be satisfied as usual by setting m= 



