Page:EB1911 - Volume 09.djvu/750

Rh (L3 &minus; M3)2 are symmetrical functions of the roots, and consequently rational functions of the coefficients; hence

{L3 + M3 + &radic;(L3 &minus; M3)2}

is a rational function of the coefficients, which when these are replaced by their values as functions of the roots becomes, according to the sign given to the quadric radical, = L3 or M3; taking it = L3, the cube root of the expression has the three values L, L, 2L; and LM divided by the same cube root has therefore the values M, 2M, M; whence finally the expression

[p + ∛{ (L3 + M3 + &radic;(L3 &minus; M3)2) } + LM ÷ ∛{{{EB1911 tfrac|1|2}} L3 + M3 + &radic;(L3 &minus; M3)2) }]

has the three values

(p + L + M), (p + L + 2M),  (p + 2L + M);

that is, these are = a, b, c respectively. If the value M3 had been taken instead of L3, then the expression would have had the same three values a, b, c. Comparing the solution given for the cubic x3 + qx &minus; r = 0, it will readily be seen that the two solutions are identical, and that the function r 2 &minus; q3 under the radical sign must (by aid of the relation p = 0 which subsists in this case) reduce itself to (L3 &minus; M3)2; it is only by each radical being equal to a rational function of the roots that the final expression can become equal to the roots a, b, c respectively.

20. The formulae for the cubic were obtained by J. L. Lagrange (1770–1771) from a different point of view. Upon examining and comparing the principal known methods for the solution of algebraical equations, he found that they all ultimately depended upon finding a “resolvent” equation of which the root is a + b + 2c + 3d + ..., being an imaginary root of unity, of the same order as the equation; e.g. for the cubic the root is a + b + 2c, an imaginary cube root of unity. Evidently the method gives for L3 a quadric equation, which is the “resolvent” equation in this particular case.

For a quartic the formulae present themselves in a somewhat different form, by reason that 4 is not a prime number. Attempting to apply it to a quintic, we seek for the equation of which the root is (a + b + 2c + 3d + 4e), an imaginary fifth root of unity, or rather the fifth power thereof (a + b + 2c + 3d + 4e)5; this is a 24-valued function, but if we consider the four values corresponding to the roots of unity, 2, 3, 4, viz. the values

any symmetrical function of these, for instance their sum, is a 6–valued function of the roots, and may therefore be determined by means of a sextic equation, the coefficients whereof are rational functions of the coefficients of the original quintic equation; the conclusion being that the solution of an equation of the fifth order is made to depend upon that of an equation of the sixth order. This is, of course, useless for the solution of the quintic equation, which, as already mentioned, does not admit of solution by radicals; but the equation of the sixth order, Lagrange’s resolvent sextic, is very important, and is intimately connected with all the later investigations in the theory.

21. It is to be remarked, in regard to the question of solvability by radicals, that not only the coefficients are taken to be arbitrary, but it is assumed that they are represented each by a single letter, or say rather that they are not so expressed in terms of other arbitrary quantities as to make a solution possible. If the coefficients are not all arbitrary, for instance, if some of them are zero, a sextic equation might be of the form x6 + bx4 + cx2 + d = 0, and so be solvable as a cubic; or if the coefficients of the sextic are given functions of the six arbitrary quantities a, b, c, d, e, f, such that the sextic is really of the form (x2 + ax + b)(x4 + cx3 + dx2 + ex + f) = 0, then it breaks up into the equations x2 + ax + b = 0, x4 + cx3 + dx2 + ex + f = 0, and is consequently solvable by radicals; so also if the form is (x &minus; a) (x &minus; b) (x &minus; c) (x &minus; d) (x &minus; e) (x &minus; f) = 0, then the equation is solvable by radicals,—in this extreme case rationally. Such cases of solvability are self-evident; but they are enough to show that the general theorem of the non-solvability by radicals of an equation of the fifth or any higher order does not in any wise exclude for such orders the existence of particular equations solvable by radicals, and there are, in fact, extensive classes of equations which are thus solvable; the binomial equations xn &minus; 1 = 0 present an instance.

22. It has already been shown how the several roots of the equation xn &minus; 1 = 0 can be expressed in the form cos 2s/n + i sin 2s/n, but the question is now that of the algebraical solution (or solution by radicals) of this equation. There is always a root = 1; if be any other root, then obviously, 2, ... n&minus;1 are all of them roots; xn &minus; 1 contains the factorx&minus; 1, and it thus appears that, 2, ... n−1 are the n−1 roots of the equation

xn&minus;1 + xn&minus;2 + ...x+ 1 = 0;

we have, of course, n&minus;1 + n&minus;2 + ... + + 1 = 0.

It is proper to distinguish the cases n prime and n composite; and in the latter case there is a distinction according as the prime factors of n are simple or multiple. By way of illustration, suppose successively n = 15 and n = 9; in the former case, if be an imaginary root of x3 &minus; 1 = 0 (or root of x2 +x+ 1 = 0), and an imaginary root of x5 &minus; 1 = 0 (or root of x4 + x3 + x2 +x+ 1 = 0), then may be taken = ; the successive powers thereof,, 22, 3, 4, 2, , 2, 23, 4,, 2, 2, 3, 24, are the roots of x14 + x13 + ... +x+ 1 = 0; the solution thus depends on the solution of the equations x3 &minus; 1 = 0 and x5 &minus; 1 = 0. In the latter case, if be an imaginary root of x3 &minus; 1 = 0 (or root of x2 +x+ 1 = 0), then the equation x9 &minus; 1 = 0 gives x3 = 1,, or 2; x3 = 1 givesx= 1, , or 2; and the solution thus depends on the solution of the equations x3 &minus; 1 = 0, x3 &minus; = 0, x3 &minus; 2 = 0. The first equation has the roots 1,, 2; if be a root of either of the others, say if 3 =, then assuming = , the successive powers are , 2,, , 2, 2, 2, 22, which are the roots of the equation x8 + x7 + ... +x+ 1 = 0.

It thus appears that the only case which need be considered is that of n a prime number, and writing (as is more usual) r in place of , we have r, r2, r3,...rn&minus;1 as the (n &minus; 1) roots of the reduced equation

xn&minus;1 + xn&minus;2 + ... +x+ 1 = 0;

then not only rn &minus; 1 = 0, but also rn&minus;1 + rn&minus;2 + ... + r + 1 = 0.

23. The process of solution due to Karl Friedrich Gauss (1801) depends essentially on the arrangement of the roots in a certain order, viz. not as above, with the indices of r in arithmetical progression, but with their indices in geometrical progression; the prime number n has a certain number of prime roots g, which are such that gn&minus;1 is the lowest power of g, which is &equiv; 1 to the modulus n; or, what is the same thing, that the series of powers 1, g, g2, ... gn&minus;2, each divided by n, leave (in a different order) the remainders 1, 2, 3, ...n&minus; 1; hence giving to r in succession the indices 1, g, g2,. . . gn&minus;2, we have, in a different order, the whole series of roots r, r2, r3,...rn&minus;1.

In the most simple case, n= 5, the equation to be solved is x4 + x3 + x2 +x+ 1 = 0; here 2 is a prime root of 5, and the order of the roots is r, r2, r4, r3. The Gaussian process consists in forming an equation for determining the periods P1, P2, = r + r4 and r2 + r3 respectively;—these being such that the symmetrical functions P1 + P2, P1P2 are rationally determinable: in fact P1 + P2 = &minus;1, P1P2 = (r + r4) (r2 + r3), = r3 + r4 + r6 + r7, = r3 + r4 + r + r2, = &minus;1. P1, P2 are thus the roots of u2 + u &minus; 1 = 0; and taking them to be known, they are themselves broken up into subperiods, in the present case single terms, r and r4 for P1, r2 and r3 for P2; the symmetrical functions of these are then rationally determined in terms of P1 and P2; thus r + r4 = P1, r·r4 = 1, or r, r4 are the roots of u2 &minus; P1u + 1 = 0. The mode of division is more clearly seen for a larger value of n; thus, for n= 7 a prime root is = 3, and the arrangement of the roots is r, r3, r2, r6, r4, r5. We may form either 3 periods each of 2 terms, P1, P2, P3 = r + r6, r3 + r4, r2 + r5 respectively; or else 2 periods each of 3 terms, P1, P2 = r + r2 + r4, r3 + r6 + r5 respectively; in each case the symmetrical functions of the periods are rationally determinable: thus in the case of the two periods P1 + P2 = &minus;1, P1P2 = 3 + r + r2 + r3 + r4 + r5 + r6, = 2; and the periods being known the symmetrical functions of the several terms of each period are rationally determined in terms of the periods, thus r + r2 + r4 = P1, r·r2 + r·r4 + r2·r4 = P2, r·r2·r4 = 1.

The theory was further developed by Lagrange (1808), who, applying his general process to the equation in question, xn&minus;1 + xn&minus;2 + ... +x+ 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 + b + 2c + ...)n&minus;1 was in this case a given function of with integer coefficients.

Reverting to the before-mentioned particular equation x4 + x3 + x2 +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 root of the equation 4 &minus; 1 = 0 (whence is = 1, &minus;1, i, or &minus;i, at pleasure), and consider the expression

(a + b + 2c + 3d)4,

the developed value of this is