An introduction to the modern theory of equations, by Florian Cajori.

194

THEORY OF EDQUATIONS

We can now prove that f(x) can be broken up into three
factors of the fourth degree each, thus,
f(x) =F( F( (, y) ) F (x, y) (, ) II
where F(x, y)= 0 is a quartic cyclic equation, in which the
coefficients of x are numbers in the domain W(y. For, let
F1(x) ==(x -) (x - ) (x -- a) (x - ag), III
then each coefficient of x in III admits the circular substitution c; hence it admits also the substitutions of what becomes
the Galois group of f(x) = 0 after the adjunction of y. This
group must consist only of powers of c, cl, c2. Therefore, these
coefficients of x are functions of y, ~ 162, and we have F1(x)
= F(x, y). Moreover, F(x, y) = 0 is a cyclic equation in l2,,
since the cyclic functions of its roots lie in this domain.
If in n = e. f, e or f are composite numbers, then we repeat
the process upon the new cyclic equations until all the factor
equations are of prime degree.
Thereby the solution of cyclic equations of any degree n is
made to rest on the solution of cyclic equations whose degrees
are prime numbers.
Ex. 1. As an illustration, take x4 + X3 + x2 + + 1 = 0, where a = o,
ai=w2, a(2=-4, 0C3= 8= -. Hence s= (a(1ama:2(3) =( WJ243), c=(cWW4),
C1 = (Cw2w). Take y = aa22 + a2a2 = w4 + o, then Yl = ala32 + a3al2
= 3 + CW2, y + yl =-, yyi = -1, (t - y) (t - y1) = t2 + t- I = 0,
2 t =- i - v, f(x) (t2+ ( - 1v 5)t + 1)(t+ ()t + v ) =
F(x, y) F(x, y1). Each quadratic factor, equated to zero, is a cyclic
equation.
Ex. 2. Given that f(x) _= 6 + x- 5 X4-4x3 + 6x2 +3 - 1=0 is
a cyclic equation in which a c = 2 c os a, al = 2 c os na, *, = 2 cos nua,
a5 = 2 cos n5a, where n = 2 and a = 2 - In illustration of the theorem,
13
we have s = (aoa2a3aC4aa), c = (aa2oC4), cl = (oaao35). Take y = aoa22
+- a2a42 + a4a2, yi = a1a32 + aC3(X2 + as5a12. With some effort we find
y+Y1=-5, YY1=3. Hence (t-y)(t-yl)=t2+5t+3=0, 2t=-5+V/13.
We get f(x) (t3 - dt2 - t + d - 1) (t3 + (d + 1)t2 - t- d - 2)= 0, where
2 d =- 1 ~ V13.