134 
where a is an imaginary fifth root of unity ; and any transforma- 
tion, by which the right side of this expression can be made to 
contain only symmetrical functions of x Q x l x 2 x 3 a? 4 , will be, by 
proper reading of roots, a correct solution of F=iO, as x 0 is any one 
of the five. Let us try to effect this. We can write 
x 0 g Hx “|“ g ^ (^0 ~V a V^\~\~ <*2^2 ”1” a 3%3 "4" a 4^4 ) 5 ^ 
g ^(^0+ a 2^3"h a 3^4“l“ a 4^l) 5 ^ 
g ^ (^od” a 1*^*3 “h a 2^4 ~J“ a 3^1 d* a 4^2 ) 5 ^ 
g ^ (%0 d” a 1^4 d“ a 2^1 d“ a 3^2 H" a 4^3) 5 ^ ) 
=^+K p O i+ §( Pi ) i+ B( p ^ + K Pt ) i 
= i &+ 1(s + h 1 )4J(s + h 2 )^ 
+1 
+j 
+; 
d- 
J(s + H 3 ) i +J(s+H 4 ) i 4 
where S is a symmetric function of the roots of F=0, and Hi H 2 
H 3 H 4 are four values of a 24-valued function of these roots. 
H 2 is formed, from H x by the substitution ffif f? effected on the 
subindices of the af s, leaving undisturbed the a a 2 , &c. ; H 3 from 
H 1 by the substitution ffiif, and H 4 from H 4 by -stiff. Hi is 
unaltered by the cyclical permutation of x 0 x x x 2 x 3 a? 4 . 
Another set of four values, H 5 H 6 H 7 H 8 , can be formed by 
operating on Hj H 2 H 3 H 4 by the substitution flif-f, or, which 
is the same thing, by operating on H x only, by the four substitutions 
01342, 03421, 04213, 02134, omitting the suffixed 01234. And 
if we write in the above expression for x 0 , H 5 H 6 H 7 H 8 in the 
places of H, H 2 H 3 H 4 , we shall have still a correct expression of 
the value of x 0 . Is not this a sufficient and a new demonstration 
of the absurdity of supposing a solution of the quintic ? 
Four more sets each of four values of H 4 can be made by operating 
on H 1 H 2 H 3 H 4 , by the four substitutions, 01423, 01243, 01432, 
01324 ; and the sum of the 24 values of H 1? thus obtained in sets 
