1896-97.] Dr Muir on Resolution of Circulants. 
373 
required may be considerably condensed. In fact, what has been 
done is really equivalent to multiplying 
x 2 + x 4- 1 
by 
cqx 3 + (a 2 - afx 2 + ( a 3 - a 2 )x -f (a 4 - a 3 + <q) , 
and subtracting the product 
a 4 x 5 + <x 2 x 4 + a 3 x s + a 4 x 2 + (<x 4 — a 2 + cq)x + (<x 4 - a 3 + cq) 
from 
cqx 5 + <x 2 x 4 + <x 3 x 3 + a 4 x 2 + a b x + a 6 . 
Moreover, when put in this way, the process closely resembles that 
followed in the previous cases. 
The factor corresponding to + 1 is similarly found to be 
a b d- « 4 — <% 2 — eq cIq — a 4 — a b + <q 
Uq 4" d b CCq (X 2 Cl-y d b ciq 4~ a g j 
or, in persymmetric form, 
— d 3 — d 4 4~ (Xq + oq — d 4 — d b + cq + a 2 
- a 4 - a 5 + a 4 + a 2 - a h - a 6 + a 2 + a 3 , 
so that as a final result we have 
C(cq , a 2 , a s , a 4 , d b , a 6 ) = (a 1 + a 2 + a 3 + a 4 + a b + a 6 ) 
. (cq - a 2 + a 3 - a 4 - <x 5 + a 6 ) 
u 3 - a 4 + a 6 - cq a 4 — a b + 6q — a 2 
a 4 — a 5 + «i - u 2 « 3 — + a 2 -- a 3 
— cl 3 ~ d 4 4- cLq + cq — cq — ci b + cq + ci 2 
— d 4 — cq + cq + cq — <x b — (Xg 4° cq + u 3 o 
(7) The case where n = 7 presents no new feature, 7 being like 5 
a prime. The result thus is 
C(cq $2 , . . . , Cl-]) = (<q 1' $ 2 1" • • • • 4" <q) * P ($2 <q , $3 , . « « , $7 ^4 , (Xj $2 > • • • > ^5 — ^6 )• 
(8) The lengthened process given in § 6 suffices to show not only 
that for every rational factor of x n - 1 there must be a rational 
factor of C(a 4 , d 2 , . . . . , d n ), but also that the degree of correspond- 
ing factors must be the same. 
Thus, to take the next case, the mere fact that x 8 - 1 is resolv- 
able into 
(x- l)(x4- l)(x 2 4- l)(x 4 4- 1) 
