CHAP, xxi] Two EQUAL SUMS OF Two CUBES. 555 



in the explicit form 



= pff'--p rz , 



p'=/' 2 +3</ 2 , a, a'=ff 



He stated that we may set/' = 1, 0' = without loss of generality and hence 

 express the general solution of (10) in the form 



(11) x = k 2 -l, y=-k z +?n, z = km-l, u=-kl+l, 



where & = a 2 +36 2 , l = a-3b, m = a+3b. We may take a = m/3, |8= -Z/3 as 

 new parameters in place of a, 6, and get 



where t = kl% = c?+& 1 -ap. The case a = p = l gives 3 3 +4 3 +5 3 = 6 3 . 



* V. Bouniakowsky 59 treated (4) . 



C. Richaud 60 noted that in (x+l) 3 -o; 3 = ?/ 3 +z 3 , y+z is of the form 

 i 2 +3w 2 , whence 2x = l l, 2y = s+v, 2z = s v, where 



o3_1 



.0 O * 



*-* 8- 



From one solution of the last equation we get the second solution 



,_ ,_ 



t . v 



., , ., . 



s 1 s 1 



Hence from one solution a,b = dl,c ) d of (4), by replacing x, y, z by 

 d 1, c, a, and hence f, s, y by 2d 1, c+o, c o, respectively, we get 

 another solution : 



_c(a+c)-a+2d-l ^ 



C ~ ' 



o+c-1 ' o+c-1 



since A = i(s-y'), C = Ks+y ; ), D = K^+1) Thus the solution 3, 5, 4; 6 

 leads to 1, 8, 6; 9 and -8, 50, 29; 53. 

 H. Grassmann 61 reduced (10) to 



by setting x = a+c, y = a c,z = b d, u = b+d, and stated that a/b must be 

 a square, whence a = moc, b = ra/3 2 , 



Giving artibrary integral values to a, |8, ???, and expressing the left member 

 as a product pg, we get d, c from fidac = p, q. 



C. Hermite 62 derived Binet's solution (11) of (10) from a general property 

 of cubic surfaces. Let co be an imaginary cube root of unity. The lines 



69 Memoirs Imper. Acad. Sc., St. Petersburg, 6, 1865, 142 (In Russian). 



60 Atti Accad. Pont. Nuovi Lincei, 19, 1865-6, 183-6. 



61 Archiv Math. Phys., 49, 1869, 49; Werke, 2, pt. I, 1904, 242-3. Error indicated by *A. 



Hurwitz, Jahresber. d. Deutschen Math.-Vereinigung, 27, 1918, 55-56. 

 Nouv. Ann. Math., (2), 11, 1872, 5-8; Oeuvres, III, 115-7. 



