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938. 
ON TWO CUBIC EQUATIONS. 
[From the Messenger of Mathematics, vol. xxn. (1893), pp. 69—71.] 
Starting from the equations 
2 + a = b 2 , 
2 + b = c 2 , 
2 + c = a?, 
then eliminating h, c, we find 
that is, 
(a 4 — 4a 2 + 2) 2 — (a + 2) = 0, 
a 8 — 8a 6 + 20a. 4 — 16a 2 — a + 2 = 0 ; 
we satisfy the equations by a = b = c, and thence by 
a 2 — a — 2 = (a — 2) (a + 1) = 0 : 
there remains a sextic equation breaking up into two cubic equations; the octic 
equation may in fact be written 
(a — 2) (a 4- 1) (a 3 + a 2 — 2a — 1) (a 3 — 3a + 1) = 0, 
and we have thus the two cubic equations 
x 3 + x 1 — 2# — 1 = 0, of — 2>x + 1 = 0, 
for each of which the roots (a, b, c) taken in a proper order are such that 2 + a = b 2 , 
2 + b = c 2 , 2 + c = a 2 . 
It may be remarked that starting from y 3 + y 2 — 2y — l = Q, y 2 = x+ 2, the first 
equation gives (y 3 — 2y) 2 — (y 2 — l) 2 = 0, that is, y 6 — 5y* + 6i/ 2 — 1 = 0, whence 
that is, 
(x + 2) 3 — 5 {x + 2) 2 + 6 {x + 2) — 1 = 0, 
¿c 3 + x 2 — 2x — 1 =0.