OF THE HIGHER DEGREES, WITH APPLICATIONS, 105 
Let us reasnn now on the assumption that the cycle (8) is not made 
up of pairs of reciprocal roots. It contains in that case no reciprocal 
; al 
roots. By the same reasoning as above we get Yj = — s. As re- 
gards the terms in (33) after the first, one of the terms C1, C2, etc., 
say C, , must be such that the 2 roots of unity of which it is the 
sum are reciprocals of those of which C; is the sum. In passing from 
Ci to C,, we change 7; intor,. In fact, C; being 71, C, is 7, | 
This being kept.in view, we get, by the same reasoning as above, 
Fs =n—s. But, if any of the expressions C;, C2, ete, except 
C, be selected, say (’, , none of the roots in (8) are reciprocals of any 
of those of which (, is the sum. Therefore a = —s. Therefore, 
from (34) 
22 eS Le 
m m = 
4 ee (0 ey t 
-1 z—1 
2 m 
Le \(a + 4 + tls +t bene, =e 
In like manner every one of the expressions in (34) can be shown to 
have the value n. 
§42. Two numerical illustrations of the law established in the 
preceding section may be given. The reducing Gaussian equation of 
the third degree to the equation «!9 — 1 = 0 is a — a? — 6a —7T=0; 
which gives 
3 3 
m=4(-— 1 + 4) + A2 ), 
24,;=19(7+ 3 ¥ 3), 
242 = 19 (7 — 3 ¥ 3), 
ds 19) 
The next example is taken from Lagrange’s Theory of Algebraical 
Equations, Note XIV., §30. The Gaussian of the fifth degree to the 
equation «1 — 1 = Ois a 4+ at — 4a3 — 322 4 32 + 1 = 0; 
which gives 
SS reer 
my=2(—1+4+ 454 4¢4+ 434+ 44); 
4 4,;= 11 (— 89 — 25 Y5 + 5p — 459), 
4 4g = 11 (— 89 + 25 4 5 — 45p — 54), 
11 (— 89 — 25 4 5 — 5p + 459), 
11 (— 89 + 25 45+ 45p 4+ 5q), 
99 Sf) AS 
oP 
BS eS 
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