418 
[855 
855. 
SOLUTION OF (a, b, c, d) = (a\ b\ c\ d 2 ). 
[From the Messenger of Mathematics, vol. xv. (1886), pp. 59—61.] 
It is required to find four quantities (no one of them zero) which are in some 
order or other equal to their squares, say 
(a, b, c, d) = (a 2 , b 2 , c 2 , d 2 ). 
Supposing that the required quantities (a, b, c, d) are the roots of the biquadratic 
equation 
x 4 4- px 3 + qx 2 + rx + s=0, 
(s not = 0), then the function 
(¿r 4 + qx 2 4- s) 2 — (px* 4- rx) 2 must be = a? 4- px s 4- qx 4 4- rx 2 + s; 
and we have thus the conditions 
2 q—p 2 = p, 2 s + q 2 — 2pr = q, 2 qs — r 2 = r, s 2 = s, 
the last of which (since s is not = 0) gives s = 1, and the others then become 
2q—p 2 +p, 2 (pr— 1) = q 2 — q, Zq=r 2 + r; 
viz. regarding p, q, r as the coordinates of a point in space, this is determined as 
the intersection of three quadric surfaces, and the number of solutions is thus = 8. 
We, in fact, have 2q =p 2 +p = r 2 4- r; that is, p 2 4- p — r 2 4- r, or (p — r) (p 4- r 4-1) = 0 ; 
hence r = p or r = — 1 — p. 
First, if r=p\ here 2q=p 2 +p, 2 {p 2 - 1) = q 2 - q: the last equation multiplied by 
4 gives 
8(p 2 — 1) = (p 2 + p)(p 2 +p — 2), =p(p 2 - 1) (p 4- 2), 
that is, p 2 — 1 = 0 or p 2 4- 2p — 8 = 0.