CHAP. XXI] TWO EQUAL SUMS OF TWO CUBES. 553
there exists the simpler solution x = 6, y = 5. To treat (4), he set
(5) A=p+q, B = p-q, C = r~s, D-r+s. Thus
(6) p(p2+3g2)==s(s2+3r2). Taking
p = ax+3by} q = bx—ay, s = 3cy — dx, r = dy-i-cx} we have
Hence our equation becomes (3(axJr3by)==y(3cy~dx)J whence
x — — 3nbj3 +3nc'Y) y = na/3 + ndj. Writing X, n = 3aciL3bc::Fad+3bd, we get
The abbreviatons ft 7, X, n were not used by Euler ; but their introduction49 enables us to point out the identity which underlies his solution. In
it is the final factor which vanishes, and this in view of the identity
which in turn follows from
(a+6 V^3) (d+c V^3) = ad
Euler noted (p. 206) that we may solve similarly Z?r = Xp, where ir = mp*+nqz, p = ?7ir2+?w2, while I, X are any linear functions of p} q, r, s, by setting
p = nfx+gy, q=*mfy-gx, r = nhx+ky, s=*mhy — kx. Then
^ — (f + win/2) (nx2 + my*) , p = (^2 + mnh2) (nx2 + my2} . Hence x/y is ratonal.
Euler80 treated (4) by setting, without loss of generality,
A = (m — ri)p+q2, B = (m+ri)p—q2, C = p2 — (m+ri)q, D = p2+(m—ri)q.
Then (A+B)(A2-AB+B2)=:(D-C)(D2+DC+C2) becomes, after division by 2ra(p3-g3), m2+3n2 = 3pg. Thus m=3k, where pg = n2+3/b2. But he had proved in the same paper that every divisor of n2+3&2, in which n and k are relatively prime, is of like form. Thus
while n is ac^FS&d or its negative.
<9 L. E. Dickson, Amer. Math. Monthly, 18, 1911, 110-111.
60 Novi Comm. Acad. Petrop., 8, annees 1760-1, 1763, 105; Comm. Arith., I, 287; Opera Omnia, (1), II, 556.