135 
of four, will be that of all the values that can have by permu- 
tation of the x% leaving untouched the a a 2 , &c. We thus obtain 
six distinct expressions for a 0 , of which no two contain a common 
value of the 24-valued All this is true if x 0 —(f)t 0 , x x =(pt l9 
x 2 =(j)t 2 , &c., Hi being a function of the 0£’s, and S also. 
It is plain that we can form the equation 
y uj r ay™+by‘ 12 + &c. =0=U, 
whose roots are these 24 values of H b and that a, b , <fcc., in this 
equation will be rational functions of the coefficients of F=0. If 
we knew howto solve this equation, U=0, we should have the 
correct solution of the quintic F=:0, by writing, instead of 
HiHaHgEft, the algebraic root of U=0, under each of the four 
quintic radical signs. The ambiguity arising from the different 
roots of unity read with all the radicals would of course remain, as 
in the lower equations. 
I see but two conceivable chances of success, either that U=0 
should break up readily into six biquadratics, or that it should be 
somehow reducible to the form 
y 24 +Ky 18 +Ly 12 +My 6 +N=:0=U 1 , 
or to some other biquadratic in z, z being a cubic in y l . 
Such reduction would be possible if we had a sufficient number 
of constants at our disposal in x 0 x l x. 2 x 3 x 4 . Let us then suppose 
that, instead of the general quintic F=0, we have a quintic F^O 
to solve, whose roots are (f>x 0 , <px h 0# 2 , <px 3 , <ftx 4 , x 0 x x &c. being the 
roots of F=0, and 
(pxz=hx i -^-kx s -\-lx 2 -\-mx-\-?i, 
h k l m n being undetermined quantities, surd or rational. We 
can write 
^ 0 =gS^+g(s+H,^+^S+H 2 ^+^S+H 3 ^+^S+H 4 ^ 
where S is a symmetrical, and H 2 H 3 H 4 are values of a 
24-valued, function of the 0afs. We can form, as before, the equa- 
tion U=0, whose roots are the 24 values of this H 4 and in all whose 
coefficients the constants h Jc l m n are at our disposal. If by 
solutions of equations among these quantities, of degrees below 
