EQUATION.
517
and, the periods being known, the symmetrical functions of the several terms of each
period are rationally determined in terms of the periods, thus
■y» —- jp^ ^ V • V • T* 4" V 2 . = JPo, T • V 2 T^ = 1.
The theory was further developed by Lagrange (1808), who, applying his general
process to the equation in question, x n ~ l + x n ~ 2 +...+«+1=0, the roots a, b, c,... being
the several powers of r, the indices in geometrical progression as above, showed that
the function (a + o>5 +&) 2 c+was in this case a given function of w with integer
coefficients. Reverting to the before-mentioned particular equation « 4 + « 3 + « 2 + x +1 = 0,
it is very interesting to compare the process of solution with that for the solution
of the general quartic the roots whereof are a, b, c, d.
Take a, a root of the equation eo 4 —1 = 0 (whence &> is =1,-1, i, or —i, at
pleasure), and consider the expression
(a + cob + co 2 c + co 3 dy.
The developed value of this is
= a 4 + b 4 + c 4 + d 4 + 6 (a 2 c 2 + b 2 d 2 ) + 12 (a 2 bd + b 2 ca + c 2 db + d 2 ac)
+ to {4 (a?b + b 3 c + c 3 d + d 3 a) + 12 (a 2 cd + b 2 da + c 2 ab + d 2 bc)}
+ (o 2 {6 {a 2 b 2 + b 2 c 2 + c 2 d 2 + d 2 a 2 ) + 4 (a 3 c + b 3 d + c 3 a + d 3 b) + 24abed)
+ to 3 {4 (a 3 d + b 3 a + c 3 b + d 3 c) +12 (a 2 bc + b 2 cd + c 2 da + d 2 ab)} ;
that is, this is a 6-valued function of a, b, c, d, the root of a sextic (which is, in
fact, solvable by radicals; but this is not here material).
If, however, a, b, c, d denote the roots r, r 2 , r 4 , r 3 of the special equation, then
the expression becomes
r 4 + r 3 + r + r 2 + 6 (1 + 1) + 12 (r 2 + r 4 + r 3 + r)
+ to {4(1 + 1 +1 + 1) + 12 (r 4 + r 3 + r + r 2 )}
+ ft) 2 {6 (r + r 2 + r 4 + r 3 ) + 4 (r 2 + r 4 + r 3 + r )}
+ a) 3 {4 (r + r 2 + r 4 + r 3 ) + 12 (r 3 + r + r 2 + r 4 )];
viz. this is
= — 1 + 4 to + 14 eo 2 — 16&) 3 ,
a completely determined value. That is, we have
(r + <wr 2 + &) 2 r 4 + g) 3 r 3 ) 4 = — 1 + 4ft) + 14ft) 2 — 16ft) 3 ,
which result contains the solution of the equation. If &> = 1, we have (r+ r 2 + r 4 + r 3 ) 4 = 1,
which is right; if w = — 1, then (r + r 4 — r 2 — ?- 3 ) 4 = 25 ; if &) = i, then we have
[r — r 4 + f (r 2 — «3)} 4 = _ 15 + 20i; and if &) = — i, then [r — r 4 — i (r 2 — r^)} 4 = — 15 — 20f; the
solution may be completed without difficulty.
The result is perfectly general, thus:—n being a prime number, r a root of the
equation x n ~ 4 + x n ~ 2 + ...+« + 1=0, w a root of a> n ~ l —1 = 0, and g a prime root of
g 1l ~ 1 = 1 (mod. n), then
(r + u>r g + ... + u) n ~ 2 r gn ') n_1