[383 
383] 
PROBLEMS AND SOLUTIONS. 
611 
(part 
of the 
points 
point 
roots, 
hat is, 
ontact 
form 
real 
et of 
a, c) 
», the 
в foi 
es of 
circles 
of a 
then considering x : у as the unknown quantity, it is required to find the roots of 
the equation IU 2 — 3H 2 =0 in terms of the roots (a, /3, y, 8) of the equation U = 0 ; 
or, what is the same thing, it is required to find the linear factors of the function 
IU 2 ~3H 2 . The function in question is the product of four quadratic factors, rational 
functions of (a, /3, y, 8); and these being known, the four pairs of linear factors can 
be determined each of them by the solution of a quadratic equation. In fact, writing 
= a {(/3 - a) (x - уу) (x - By) + (y- a) (x- By) (x - fry) + (8 - a) (x - /Згу) (x - угу)}, 
Цр=а {(у - /3) {x - By) (x-ay) + (8-f3) (ж - ay) (x - y y) + (a -/3) (x - y y) (x - Вгу)}, 
<P) y =a {(B - y) (x - ay) (x - /3гу) + (a-y)(x- /Згу) (x - Вгу) + (/3 - y) (x - Вгу) {x - агу)}, 
0 5 = a {(a - B) (x- /Згу) (x - y y) + (0-B)(x- y y) (x - ay) + (y - B) (x- ay) (x - /Згу)}, 
we have identically 25(5(IU 2 — SH 2 ) = © а ©у©э®г ; so that the quadratic factors of IU 2 —3H 2 
are © a , ©0, © y , @ s . To show that this is so, it is to be remarked that the product 
© a ©0© y ©s is a symmetrical function of the roots a, ¡3, y, 8, and consequently a rational 
and integral function of the coefficients (a, b, c, d, e) of U ; moreover © a , ©p, © Y , ©5 being 
each of them a covariant (an irrational one) of U, the product in question must be 
a covariant. But a covariant is completely determined when the leading coefficient is 
given ; hence it will be sufficient to show that the leading coefficients, or coefficients 
of x 8 , in the functions © a © /3 © v ©5 and 256 (IÜ 2 — 3H 2 ) are equal to each other. Writing 
for a moment %a = p, %a/3 = q, %a/3y = r, a/3y8=s, the coefficient of x 2 in а -1 0 а is 
/3 + у + 8 — За, which =p — 4a ; we have thence the product (p — 4a)(p — 4/3)(p — 4y) (p — 48), 
which is =р л — 4p s . p + 16p 2 . q — 64p . r + 2565, = 256s — 64pr + 16p 2 q — 3p 4 . 
Hence, restoring the omitted factor a 4 , and observing that we have p, q, r, s equal 
to — 4b, 6c, — 4d, e, each divided by a, the coefficient of ж 8 in © 0 ©^© Y © S is 
256 (a 3 e — 4a 2 bd + Qab 2 c — 3b 4 ), or 256 {(ae — 4bd + 3c 2 ) a 2 — 3 (ac — b' 2 ) 2 }, 
and is consequently equal to the coefficient of ж 8 in 256 (IU 2 — 3H 2 ); which proves the 
theorem. 
It may be remarked that the leading coefficient of IU 2 — 3H 2 is =a~ 2 (a, b, c, d, e][b, — a) 4 ; 
and that for a quantic U — (a, b,. .) (x, гу) п of the order n we have a corresponding 
covariant of the order n(n — 2), the leading coefficient of which is =а -1 (а, b,... .$&, — a) n . 
For 11= 2, this is the invariant (discriminant) ac — b 2 ; for n = 3 it is the cubicovariant 
{a 2 d — 3abc+ 2b z ,. . .\x, y) 3 ; for n = 4 it is, as we have seen, the covariant IU 2 — 3H 2 . 
For n = 5, the leading coefficient a 4 /— 5a 3 be + 10a 2 b 2 d — 10ab 3 c + 46® is = a 2 (a 2 f— 5abe + 2acd 
+ 8b 2 d — 6be 2 ) — 2 (ac — b 2 ) (a 2 d — 3abc + 2b 3 ), which shows that the covariant in question 
(of the order 15) is = U 2 (No. 17) — 2 (No. 15) (No. 18), where the Nos. refer to the 
Tables of my Second Memoir on Quantics, Phil. Tra7is., vol. cxlvi. (1856), pp. 101—126, 
[141; in the notation there explained, the expression for the covariant is A 2 E —2CF\ 
[The roots of © a = 0 are readily found to be 
a (/3 + у + 8) - (у8 + 8/3 + /Зу) j [| [(a — /3) 2 (y - 8) 2 + (a - y) 2 (8 - /3) 2 + (a - 8) 2 (/3 - y) 2 ]}* 
3a — (/3 + у + 8) 
these then, with three similar pairs, express the eight roots as required.] 
77—2