405] 
AN EIGHTH MEMOIR ON QUANTICS. 
155 
where 
O' _ n ft / ny 
P — , y = —- , &c.; 
1 — m/3 1 — my ’ 
hence m may be so determined that /3' + y may be = 0; viz. this will be the case 
if /3 + 7 = 2mfty, or m = . In order that 8'+ e' may be =0, we must of course 
3 -j- £ 
have m=s-gg- , and hence the condition that simultaneously /3' + y' = 0 and 8' + e' = 0 is 
-|- ry 8 + 6 . 
2^ = 9g e ‘ 5 fh a t 1S > + y) Se — /Sy (8 + e) = 0. Or putting x — a for ¿c and /3 — a, 
y — a, &c. for /3, 7, &c., we have the equation 
{x - a) (x - ft) (x — y) (x — 8) (x — e) = 0, 
which is by the transformation x — a into ——f-— changed into 
m (x — a) + n ° 
(x — a) (x — /3') (x — 7') (x — 8') {x — e) = 0 
(where a' = a), and the condition in order that in the new equation it may be possible 
to have simultaneously /3' + y — 2a' = 0, 8' + e' — 2a' = 0, is 
(/3 + 7 — 2a) (8 — a) (e — a) — (8 + e — 2a) (/3 — a) (7 — a) = 0, 
or, as this may be written, 
1, 
1, 
1, 
2a 
a 2 
= 0. 
/3 + 7, f3y 
8 + e, 8e 
Hence writing x + a for x, the last-mentioned equation is the condition in order that 
the equation 
(x — a)(x — ft) (x — 7) (x — 8) (x — e) = 0 
may be transformable into 
x{x — ft') (x — y) (x — 8') (x — e) = 0, 
where ft' + 7' = 0, 8' + e = 0, that is, into the form x (x 2 — ft' 2 ) (x 2 — 8' 2 ) = 0. Or replacing 
y, if we have 
(a, b, c, d, e, /$>, y) 5 = a (x - ay) (x- fty)(x - yy) (x - 8y) (x- ey), 
then the equation in question is the condition in order that this may be transformable 
into the form (a', 0, c', 0, e', 0^x, y) 5 \ that is, in order that the 18-thic invariant I 
may vanish. Hence observing that there are 15 determinants of the form in question, 
and that any root, for instance a, enters as a 2 in 3 of them and in the simple power 
a in the remaining 12, we see that the product 
a 18 II 1, 2a , a 2 
1, /3 + 7, fty 
1, 8+e, 8e 
20—2