443] 
ON THE SOLUTION OF A QUARTIC EQUATION. 
129 
so that identically X 2 + 1 2 + Z s = 0, the expression aX + /3F + yZ will be a square if 
only a? + ¡3 2 + <y 2 = 0. (To see this observe that in virtue of the equation X 2 + F 2 + Z' 2 = 0, 
we have X + iY, X — iY each of them a square, and thence 
aX + /3Y+yZ, = \ (a + i/3) (X-iY) + \(a- i/3) (Z - iY) - yi VZ^fF 2 , 
is a square if the condition in question be satisfied.) 
Hence in particular writing 
Vo> 2 — &) 3 Va/ + 6/3(0! J,..., V&>! — eo 2 VaI + 6/3co 3 J, 
for a, /3, 7, we have 
(eo 2 — co 3 ) V a I + 60a) X J VIH + a> x JU + ... + (co x — a> 2 ) Va/ + 60a> 3 J V 7_ff + (o 3 JU 
a perfect square, and since the product of the four different values is a multiple of 
(all+ 6/3Hy (this is most readily seen by observing that for aU + 6/3H = 0, the 
irrational expression omitting a factor is (co 3 — co 3 ) (al + 60co x J) +...+(&>! — &> 2 ) (al + 6/3&> 3 J), 
which vanishes identically) it follows that the expression in question is the square of 
a linear factor of a U + 60H. 
It thus appears that the radicals (other than those arising from the solution of 
JJ =0) contained in the solution of the equation aU+6/3H = 0 are the three roots 
0 al + 6/3&>i J, Va/ + 6/3(öo J, 0al + 6ßm 3 J. 
Cambridge, September 2, 1868. 
C. VH. 
17