660 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xxn 



R. D. Carmichael 246 showed that one solution of x*+ay 4J rbz*= D leads 

 to a second. 



E. N. Barisien 247 noted that N=(a 2 +b 2 )(c 2 +d 2 )(a 2 c 2 +b 2 d 2 ) equals 



|a&(c 2 Td 2 ) } 2 + {cd(a 2 b 2 ) } 2 +(a 2 c 2 +6 2 d 2 ) 2 . 



Let N' be derived from N by interchanging c and d. Then NN' is a sum 

 of nine squares in four ways, in two of which two of the nine squares are 

 biquadrates. 



See papers 178, 188, 206-7, 219-20, 287-8, 292; also Gerardin, p. 38; 

 Lucas 880 of Ch. XXIII. 



MISCELLANEOUS SINGLE EQUATIONS OF DEGREE FOUR. 



C. Wolf 248 treated X 2 y 2 +x 2 +y 2 = D. First, make zy+a; 2 =n, i.e., 

 2/ 2 +l = ^= (ty) 2 , whence y = (t 2 l)/2t. Since X 2 y 2 +x 2 =x 2 v 2 , it remains to 

 make zV+2/ 2 = D, say (z vx) 2 ; we thus obtain x. 



L. Euler 249 made P = (p 2 q 2 ) (q 2 r 2 ) a biquadrate by setting p = o+6+2c, 

 g = a -f- 6, r = ab, whence P = 1 6a6c (a + 6 + c) . Consider therefore 



xyz(x+y+z)=s\ 

 Take s 4 = (x+y+z) 2 p 2 . Thus 



I>2 = (z+?/)p 2 , D(x+y+z)=xy(x+y), Ds*=xyp(x+y}, D = xyp z . 

 Set x = nq 2 } y = nr 2 , nqrp = k(q 2 -\-r 2 ) and eliminate p. Thus 



n{ nqr-\-k(q 2 +r 2 )} _ 

 ~ = ' 



For n = 2k, F = 2(g-r) 2 /(g 2 +r 2 -4gr). As Euler omitted the factor 2, it is 

 not sufficient to make the denominator a square. Next, let n = k. Then 

 F=(q 2 -\-r 2 qr)l(q r) 2 . Equate the numerator to the square of q+rf/g. 

 Thus q : r = g z f 2 : g 2 +2fg. Or we may begin by taking p=2xyl(x+y), 

 whence s 2 = 2xy (x + y) 2 / (xy} 2 ; take x = 2q 2 , q = r*to make 2o;?/ = D . 

 Euler 250 treated (p 2 +l) 2 +(g 2 +l) 2 = D by setting 



p 2 +l=o: 2 y 2 , q 2 -{-l = 2xy, p = xz. 



Thus 2zx = z 2 +y 2 +l. Take y = 2z. Then g 2 = 10 2 +l, which is satisfied 

 by (2, g) = (2/3, 7/3), (2/9, 11/9), (6, 19). 



Euler 251 treated LJ=D, where L = A+^+Cs 2 , l = a+bz+cz 2 . Take 

 Ll = p 2 l 2 . Then L = pH, which can be solved if one solution is known. 



J. L. Lagrange 252 treated the more general problem to solve 



F(x, y}=f(x)+s(x)y+cy 2 = 0, 

 where / is of the fourth degree and s of the second. If F(p, q) = 0, set 



246 Diophantine Analysis, 1915, 44. 



2 " Nouv. Ann. Math., (4), 16, 1916, 390-1. 



M8 Elementa Matheseos Universac, Halae, 1, 1742, 380. 



249 Opera postuma, 1, 1862, 239 (about 1769). Extract in Bull. Soc. Philomathique, (10), 3, 



1911, 240-3. Cf. Euler, 187 Gerardin, 266 Kommerell. 270 

 260 Opera postuma, 1, 1862, 215-6 (about 1774). 

 281 Ibid., 218-9 (about 1777). 

 252 Nouv. Me"m. Acad. Sc. Berlin, annexe 1777, 1779; Oeuvres, IV, 397. 



