550 Proceedings of Royal Society of Edinburgh. [sess. 
this being clearly contained in the first h terms of the expression 
and in each of the succeeding groups of li terms. Again, since 
n is a multiple of h, 1 - x n has 1 — x h for a factor and therefore 
also 1 4- x + • • • + x h ~ l . The existence of this common factor 
necessitates the vanishing of the eliminant. 
( 6 ) But we may go a step further if we recall our attention to 
the fact that any circulant 
CK , J| ... , a n ) 
is resolvable into as many factors as there are in 1 - x n , and that 
the factors of the one have an intimate relation with those of the 
other. (See Proc. Roy . Soc. Edinburgh , xxi. pp. 369-382.) 
Thus not only is the circulant C(a 1 , a 2 , ... , a 6 ) , — that is, the 
eliminant of 
a 1 + a 2 x 4 - a s x 2 + « 4 a ? 3 + agp + a e x 5 = 0 (a) 
and 1 - x Q = 0 (/?) 
—resolvable into the factors 
a-^ + a 2 + a 3 4 - $4 + $5 4 - ttg , 
- Ci 2 + a B - a 4 + a 5 - a 6 , 
(u 3 — cij) ( a ^ cLq 4 ~ cl 2 cl^j (c & 4 — > 
(-a s -a /L + a 6 + a 1 )(-a 5 -a 6 + a 2 + a 3 )-(-a±-a 5 + a 1 + a 2 ) 2 , 
but the first of these factors is the eliminant of (a) and 1 -x = 0 , 
the second is the eliminant of (a) and 1 + a? = 0 , the third is the 
eliminant of (a) and 1 4- x + a ? 2 = 0 , and the fourth the eliminant 
of (a) and 1 —x + x 2 — 0 . Not only, therefore, do we know that, 
when p and n have a common factor other than 1 , the circulant 
vanishes, but we also know which factor of the circulant is the 
cause of the evanescence. For example, let us look again at the 
circulant of § 2. It is the eliminant of 
- 1 - a? 4- a ? 2 4- a ? 3 4- a ? 4 4- a ? 5 = 0 ) 
and 1 - a ? 6 = 0 f 
which have the common factor 1 +a*. Now the factor of 
<C(a x , a 2 , ... , a 6 ) which corresponds to 1 +x is 
a Y - a 2 4- a 3 - a 4 4- a b - a Q 
