442 15. MODERN ALGEBRAwhere a^3 = 1, á ø 1. He argued that since the original equation was symmetric in
a, b, and c, the resolvent would have to admit this y as a root, no matter how the
letters o, b, and c were permuted. It therefore followed that the resolvent would in
general have six different roots.
For the quartic equation with roots a, b, c, and d, he showed that the resolvent
cubic equation would have a root
ab + cd
t=
~i~-Since this expression could assume only three different values when the roots were
permuted—namely, half of ab + cd, ac + bd, or ad + be—it would have to satisfy an
equation of degree three.
Proceeding to equations of fifth degree, Lagrange noted the only methods pro-
posed up to that time, by Tschirnhaus and Euler-Bezout, and showed that the
resolvent to be expected in all cases would be of degree 24. Pointing out that even
Tschirnhaus, Euler, and Bezout themselves had not seriously attacked equations of
degree five or higher, nor had anyone else tried to extend their methods, he said, "It
is therefore greatly to be desired that one could estimate á priori the success that
is to be expected in applying these methods to degrees higher than the fourth." He
then set out to provide proof that, in general, one could not expect the resolvent
equation to reduce to lower degree than the original equation in such cases, at least
using the methods mentioned.
To prove his point, Lagrange analyzed the method of Tschirnhaus from a more
general point of view. For cubic and quartic equations, in which only two coefficients
needed to be eliminated (the linear and quadratic terms in the cubic, the linear and
cubic terms in the quartic) the substitution y = x^2 + ax + b would always work,
since the elimination procedure resulted in linear and quadratic expressions in á
and b in the coefficients that needed to be eliminated. Still, as Lagrange remarked,
that meant two pairs of possible values (a, b) and hence really two cubic resolvents
to be solved. The resolvent was therefore once again an equation of degree 6,
which happened to factor into the product of two cubics. He noted what must
be an ominous sign for those hoping to solve all algebraic equations by algebraic
methods: The construction of the coefficients in the resolvent for an equation of
degree ç appeared to require solving ç - 1 equations in ç — 1 unknowns, of degrees
1, 2,..., ç — 1, so that eliminating the variable ÷ in these equations therefore led
to an expression for ÷ that was of degree (ç — 1)! in y, and hence to a resolvent
equation of degree n! in y.
Lagrange summed up his analysis as follows:
To apply, for example, the method of Tschirnhaus to the equation
of degree 5, one must solve four equations in four unknowns, the
first being of degree 1, the second of degree 2, and so on. Thus
the final equation resulting from the elimination of three of these
unknowns will in general be of degree 24. But apart from the
immense amount of labor needed to derive this equation, it is clear
that after finding it, one will be hardly better off than before, unless
one can reduce it to an equation of degree less than 5; and if such
a reduction is possible, it can only be by dint of further labor, even
more extensive than before.