170 
Proceedings of the Royal Society of Edinburgh. [Sess. 
in other words, to find the cofactor of 
(a 1 + a 2 + • • • + <hntj( a \ - a 2 + a s - ... - a 2m ) in C(a x . a 2 , ... , a 2m ) . 
Thus, in the case where 2 m = 8, from the equation 
ax 7 + bx Q + cx 5 + dx 4 + ex 3 + fx 2 + ^^- 1 -^ = 0 
the seventh and sixth powers of x are removed with the help of the other 
equation 
x Q + a: 4 + x 2 + 1 = 0 , 
the result being 
(c - a)x 5 + (d- b)x 4 + (e - a)x 3 + (/ - b)x 2 + (g- a)x + (h - b) = 0 , 
whence by cyclical substitution and elimination we obtain 
c - a 
d-b 
e - a 
f-b 
g-a 
h-b 
d — b 
e - c 
f~b 
g-c 
h-b 
a - c 
e - c 
f-d 
g-c 
h-d 
a - c 
b-d 
f-d 
g-e 
h-d 
a - e 
b-d 
c-e 
g-e 
*-f 
a - e 
b~f 
c-e 
d-f 
h-f 
a-g 
b-f 
c-g 
d-f 
e-g 
which, again, can be expressed persymmetrically, namely, 
P(c - e,d-f,e-g,f-h,g-a,h-b,a-c,b-d,c-e,d -f , e - g ) . 
In the next place, the factor of C(a x , a 2 , <x 3 , a 4 , a 5 , <x 6 ) corresponding 
to the factor x 2 -\-x + l of x Q — 1 is obtained. We simply, as before, remove 
in order from the equation 
a Y x b + a 2 x 4 + a z x z + a^x 2 + agx + a 6 = 0 , 
with the help of the equation 
x 2 + x + 1 = 0 , 
the fifth, fourth, third, second powers of x, with the result 
(a 5 - a 4 + a 2 - a x )x + (a 6 - a 4 + a 3 - a x ) = 0 , 
whence the eliminant 
a 5 ~ + a 2 ~ a i °6 “ a 4 + a 3 ~ 
a 6 ~ °5 + a 3 ~ a 2 a i “ a 5 + a 4: ~ a 2 > 
which, as before, can be expressed persymmetrically. 
Proceeding in this way, the prime factors of the circulants up to and 
including that of the tenth order are calculated, and are tabulated for 
reference. By reason of the persymmetry and the notation for it, the 
table occupies only one page; and it is pointed out how it might have 
been made still shorter by using the fact that when once the first element 
of any of the persymmetric determinants is obtained, the others follow 
