13 
not to undervalue his research, but because at p. 87 he says 
that the fact that the Y’s are perfect fifth powers ‘ probably 
cannot be proved otherwise’ than from, I suppose, a priori 
argument as opposed to calculation — his conjecture may 
nevertheless be a justifiable one) is not borne out by fact 
than (a), yet the latter affords some ground for remark. 
Rx must be unsymmetric (as indeed Hargreave seems to 
imply), for if we consider Ri as symmetric and solve (a) for 
R 2 we shall find R 2 = a two valued function, which cannot 
be, for R 2 is of two dimensions only in the roots, and cannot 
therefore be two valued for a quintic. Again, solving (a) 
with respect to R x we see that the number of values of Ri is 
a multiple of 5, and therefore that R x is not cyclical. And 
we know it is linear. If therefore, since S 5 is symmetrical, 
we are to regard the bracketed quantity on the sinister of 
(a), viz. RH-10R?R 2 + 5R| as the product of the remaining 
values of the R x outside the bracket we must regard Rx as 
of the form A-j- Bz r , z being some one root, and the bracketed 
quantity Rf as equal to 
(A+B<)(A+B 0 ,) . . .=7r(A+B?)-(A+B? r ), 
2 S z t , &c., being the remaining roots. Hence (a) would be- 
come 7 t(A+B^)=S 5 . But how is such a form consistent 
with the fact of R 2 being unsymmetric ? It must not be 
said that R 2 may be symmetric, for if that were so, then 
Rx being a five-valued function, the sinister of (b) would be 
a five-valued function, and therefore R 10 would be five- 
valued. But that is not so ; for 
^,/R l0 = A— £/ tt{x x — #o) 2 
Therefore Rxo=tt(#i — x 2 ) = a two-valued function, But 
apart from this argument, which may however (as it stands 
or with more elaboration) be conclusive, let us turn to (b). 
What ground is there for saying that R 10 , or what is the 
same thing, tt(xi — x 2 ) or, the same, 
{ X 1 X i) ( X l X :i) ( x i x :>) 
is of the form P 2 Q as indicated by (b) ? (b) is equivalent to 
