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

198

THIEOIEY OF EQUATIONS

Ex. 1. Show that the discriminant of III is a perfect square,
D = 92(m2 + m + 1)2.
Ex. 2. For the equation III determine the function 0 in the relation
Get = 0(a).
Ex. 3. Any cyclic equation of the fourth degree can be reduced to the
form ye - 2 b(2 s + r2)y2 - 4 br(l + bs2)y + lb(r2 -2 s)2- b(1 + bs2)2 = 0,
where b, r, s, are rational numbers and b is not a perfect fourth power.
See Ex. 11, ~ 159. Prove that this equation can be solved without the
extraction of cube roots.
CYCLOTOMIC EQUATIONS; GEOMETRIC CONSTRUCTIONS
177. Introduction. In ~ 63 and ~ 64 it was shown that the
roots of xh -1 = 0 may be represented thus,
2T.. 2k7rr
ak = cos + i sin
IL 't
where kc takes successively the values 0, 1, *.., n-1, and that
the solution of " - 1 = 0 is geometrically equivalent to the
division of the circumference of a circle into n equal parts.
The solution of xn - 1 = 0, given in ~ 63, is trigonometric. We
proceed to show that it is always possible to give an algebraic
solution. We shall point out how this solution can be effected
and shall consider the cases in which the division of the circle
into equal parts can be effected with the aid of the ruler and
compasses.
178. Cyclotomic Equations. If we remove the root 1 from
xn - 1 0 by dividing by x- 1, we obtain
x"-1 + x"-2 +-... + X + 1i =. I
If n is a prime number, equation I is called a cyclotomic equation.
In the domain 2(1) the cyclotomic equation is irreducible, ~ 130,
and cyclic, ~ 170.
If n is a composite number, we know from ~ 66 that the solution of -- 1 = 0 can be reduced to the solution of binomial