OF THE HIGHER DEGREES, WITH APPLICATIONS. 89 
A ; A A 
g (w1) = ¢ (a; ). Therefore, by §12, changing into #1, ¢g (w; )= 
e 2 
r r 
yg (w ). Therefore ¢ (v1) = ¢ (w, ). And thus ultimately g (#1) = 
Ne MZ 
¢ ( ), or g (#1) = ¢g (#1 +), 2 being any whole number positive or 
2 
rN 
negative. But, includes all the terms in (8). Therefore each of 
these terms is a term in (7). Suppose if possible that there is a term 
h 
in (7), say @; , which does not occur in (8). Then, just as we deduced 
mz ™m 
¢ (¥;) = ¢(o1 ) from the equation g (#1) = ¢ (v1 ), we can, 
B h pm + hu 
because still farther g (w1) = ¢ (1 ), deduce ¢ (#1) = ¢ (1 i} 
h 
Because w; lies outside the cycle (8), A is not a multiple of m. And 
™ 
ie Bris : 1D 
it is not less than m, because w; is the first term in (9) after 1, 
which, when substituted for w; in g (a1), leaves g (#1) unaltered. 
Therefore h = gm + v, where g and v are whole numbers, and v is 
less than m but not zero. Put 
Bp” 
z= — (h+q), andu=m+1.:. mz+hu=v.°. g(o1)=¢ (1 ); 
™ 
which, because v is less than m but not zero, and a is the first term 
in (9) after #; which, when substituted for in ¢g (v1), leaves g (1) 
unaltered, is impossible. Hence, no term in (7) lies outside the cycle . 
(8), while it has also been shown that all the terms in (8) are terms 
in (7). Therefore the terms in (7) are identical with those constituting 
the cycle (8). We have now to determine the form of /’ (x). The 
expressions, (), Co, etc., taken together, are the sum of the terms in 
(9). Therefore (.+0C.+....+¢Cm= —1. (11) 
Because (9) contains all the primitive n** roots of unity, we may put 
B 
a= a4 —}p + (p+ pi) 1 + (p + pe) o1 F+ ete.  et-14 ete, ; (12) 
where p, 71, etc., are clear of w;. But / (x) remains unaltered when 
UL 
d : B 
w1 1s changed into w; . Therefore 
‘ p™” 
F(x) =2t—{p+(ptp)or + ete.} wt—1 +4 etc. (13) 
Therefore, equating the coefficients of #4—1 in (12) and (13), 
- p™ 
(p— pi) + ..-. + (Pm ¢1—PI 1 +ete. = 0. 
Here, by §13, the coefficients of the different powers of wy have all 
the same value. And one of them, p — pi, is zero. Therefore 
