12 
serves (see my paper on Equations of the 5th degree in 
Diary, art. 32) that stich a sfim would be the root of an 
equation of 625 dimensions [not of 25 only]. Let (sc) be an 
algebraic root (assumed for a moment to exist) of a trino- 
mial quintic. We know that a (sc), a 2 (x), &c. ( a being an 
unreal fifth root of 1), will be the roots of the correlated 
quintics. But this is no more than the very form of the 
quintics shows (p. 86). The multiplication of (sc) by a 
makes no alteration in the internal distribution of the radi- 
cals in (sc), and it seems to me that all that can be said is 
that if (sc) be a root of sc 5 + A 4 sc-f- A 5 , then a(sc) is a root or 
the root of sc 5 + aA 4 sc-f- A 5 . Using A to denote the discrimi- 
nant of the quintic (p. 68), we may of course put A, and 
hence get h — +^A S 2 = — Is = + y/~l y/ 
— \/>— 1 v 4/ A. Hence, for instance, — £ 2 = 2 ^ A, and if v 
denote the discriminant of the quartic in £ we see that 
when A=0 then v— 0 (compare pp. 63, 89, 90). But as 
Hargreave probably (or rather certainly, see p. 70) was 
aware, there is nothing in this. What Hargreave relies 
on is — and now I fall in with his notation — that when 
%, — y — (51) we then have the system of p. 85 and the result 
mentioned at the top of p. 86. In estimating Hargreave’s 
argument we ought now not to forget the argument of Mr. 
Cayley as given at p. 100 of Mr. Salmon’s Higher Algebra 
(1859), art. 130. But how far can such an argument 
be carried ? Turning to p. 91 and taking y 7r =B 1 -f- we 
have by involution 
; y==:Ri+10R^R 2 +5R 1 R2 + (5R4 +10R 2 R2+R 2) ^ 
and therefore (see p. 91), 
R 1 (R;+10R 1 2 R 2 +5R^)=S 5 ..... (a), 
(5R1+10R^R 2 -1-R^ 2 R 2 =R 10 (b). 
Ro being of two dimensions only, cannot have so few as 
six values. How is the sinister of (a) symmetric ? Now, 
though I think that perhaps (b) furnishes the stronger 
ground for conjecture that Hargreave’s conjecture (I so call it 
