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\title{On\\ The Quintic Equation}
\author{Charles Hermite\\
{\small Translated by Kevin Iga}}
\date{1865}
\maketitle

The theory of elliptic functions leads to two methods of resolving the
quintic equation.  The first is based on the possibility of reducing
the equation to:

\[x^5-x-\frac{2}{\sqrt[4]{5^5}}\frac{1+k^2}{k'\sqrt{k}}=0\]

of the modular equation relative to the quintic transformation.  In
the second, which is due to Mr. Kronecker\footnote{{\em Comptes rendus
de l'Acad\'emie des Sciences}, t. XLVI, 1858, and {\em Journal de
Crelle}, t. 59, 1861.}, one starts with certain cyclic functions of
the roots whose squares take on only six values, and which can be
represented by the expression:

\[\frac{\cos \am 2\omega}{\cos \am 4\omega}-\frac{\cos\am
4\omega}{\cos\am 2\omega}\]

where $\omega$ is, successively, $\frac{K}{5}$, $\frac{iK'}{5}$,
$\frac{K\pm iK'}{5}$, $\frac{2K\pm iK'}{5}$.  From these functions we
then rationally deduce the roots themselves, which are obtained thus
from the explicit form of their own quantities.  These two methods
have been, for me, the subject of a long study, the results of which I
now have the honor of presenting to the Academy.  I will describe the
reduction of the general quintic equation to the trinomial form
$x^5-x-a=0$, and, at that time, I will describe the research of the
conditions of reality of the roots, on which Mr. Sylvester published
recently one of his more beautiful memoires\footnote{{\em On the real
and imaginary roots of algebraical equations: a trilogy} ({\em
Philosophical Transactions}. Part III, 1864).}.  I will then try to
extend Mr.  Kronecker's method and compare it to the former, using as
a basis of the remarkable and inventive work described by Mr.
Brioschi.  \footnote{{\em Sul methodo di Kronecker per la risoluzione
della equazione di quinto grado ( Atti dell' Instituto Lombardo, t. I,
1858, p. 275).}}.  This method permits, in effect, the reduction of
the general quintic equation to the one of the form:

\[x^5-10x^3+45x-4\frac{(1-2k^2)(1+32k^2k'^2)}{kk'}=0\]

which reduces to the equation:

\[z^6+\frac{5.2^5}{k^2k'^2}z^3-2^8\frac{1-16k^2k'^2}{k^4k'^4}z+\frac{5.2^8}{k^4k'^4}=0\]

Which has roots of the form $\frac{1}{2}\left(\frac{\cos\am
2\omega}{\cos\am 4\omega} - \frac{\cos\am 4\omega}{\cos\am
2\omega}\right)^2$.

My primary goal will be to completely carry out the work of the
substitution calculus which gives these results their importance; and
just as the invariant and covariant algebraic elements of quintic
forms serve as the base of this calculus, hence of the reduction to the
trinomial form, I will recall the approach in this regard the notions
of which I *** usual technique.

\section{}

Let

\[f(x,y)=\alpha x^5+5\beta x^4y+10\gamma x^3y^2 + 10\gamma'x^2y^3 +
5\beta'xy^4 +
\alpha'y^5=(\alpha,\beta,\gamma,\gamma',\beta',\alpha')(x,y)^5\]

be the proposed form and

\begin{eqnarray*}
x&=&mX+m'Y,\\
y&=&nX+n'Y
\end{eqnarray*}

a substitution $S$ with determinant {\em one}, which reduces the
covariant quadratic

\[(\alpha\beta'-4\beta\gamma'+3\gamma^2)x^2+(\alpha\alpha'-3\beta\beta'
+ 2\gamma\gamma')xy + (\alpha'\beta-4\beta'\gamma+3\gamma'^2)y^2\]

with monomial $\sqrt{A}.XY$, where I set

\[A=(\alpha\alpha'-3\beta\beta'+2\gamma\gamma')^2 - 4(\alpha\beta' -
4\beta\gamma' + 3\gamma^2)(\alpha'\beta - 4\beta'\gamma + 3\gamma'^2)\].

This substitution is thus not completely determined; but one knows that one
*** the general expression in the *** search for all those which produce
$\sqrt{A}.XY$.  These are comprised of the proper substition:

\begin{eqnarray*}
X&=&\omega X',\\
Y&=&\frac{1}{\omega}Y',
\end{eqnarray*}

and the improper substition:

\begin{eqnarray*}
X&=&\omega Y',\\
Y&=&\frac{1}{\omega}X';
\end{eqnarray*}

but these can be rejected if we require the determinant of the
substition $S$ to remain at $+1$.  With this imposed, let:

\[f(mX+m'Y,nX+n'Y)=F(X,Y)=(a,b,c,c',b',a')(X,Y)^5,\]

so that that one has:

\[F(\omega X,\frac{1}{\omega}Y)=(a\omega^5,b\omega^3,c\omega,c'\omega^{-1},
b'\omega^{-3},a'\omega^{-5})(X,Y)^5.\]

I will define the substitution entirely in determining $\omega$ by the
condition $c\omega=c'\omega^{-1}$; that done, I will take $F( \omega
X, \frac{1}{\omega}Y)$ by a $canonical$ transformation of the general
quintic form, and, in brief, sets it to:

\[
\begin{array}{cccccc}
a\omega^5=\lambda,&b\omega^3=\mu,&c\omega=\sqrt{k},&c'\omega^{-1}=\sqrt{k},
&b'\omega^{-3}=\mu',&a'\omega^{-5}=\lambda',
\end{array}
\]

I will designate it by ${\cal F}$, of the form:

\[{\cal F}(X,Y)=(\lambda,\mu,\sqrt{k},\sqrt{k},\mu',\lambda')(X,Y)^5\]

It is easy to see that one obtains the same canonical form by all
the transformations deduced from the proposed $f(x,y)$ by a
typical substitution $\Sigma$ of determinant {\em one}; since it suffices to
apply the inverse substitution $\Sigma^{-1}$ which will return to the
form proposed, and then do the same for $S$, to retrieve ${\cal F}(X,Y)$, the
determinant of the composed substitution $\Sigma^{-1}S$ would furthermore be
equal to unity.  It follows from this that the coefficients of the canonical
form are invariants of the proposed form, and it is important to study
these coefficients with care.

\section{}

I said, in the first place, that they can be expressed as a function
of three quanitities: $\lambda\lambda'=g$, $\mu\mu'=h$, and $k$.  The
quadratic covariant of ${\cal F}(X,Y)$ can effectively be written as
$\sqrt{A}.XY$, one has the two relations:

\[\lambda\mu'-4\mu\sqrt{k}+3k=0\]
\[\lambda'\mu-4\mu'\sqrt{k}+3k=0\]

where one takes the two quadratic equations

\[36\sqrt{k^5}\lambda^2-[h(g-16k)^2-9k^2(g+16k)]\lambda+36g\sqrt{k^5}=0,\]

\[12\sqrt{k^3}\mu^2-(9k^2+16hk-gh)\mu+12h\sqrt{k^3}=0.\]

In these they resolve and put, in short,

\[\Delta = (9k^2+16hk-gh)^2-24^1hk^3,\]

one has a first set of solutions, knowing

\[72\sqrt{k^5}\lambda=h(g-16k)^2-9k^2(g+16k)+(g-16k)\sqrt{\Delta},\]

\[24\sqrt{k^3}\mu=9k^2+16hk-gh-sqrt{\Delta},\]

and the second will be deduced by changing $\sqrt{\Delta}$ to be imaginary in
the two formulas.  But, after the product of the roots of the equations
in $\lambda$ and $\mu$, this second system may reduce thus is represented by
$\frac{g}{\lambda}$, $\frac{h}{\mu}$, this is to say by $\lambda'$ and $\mu'$,
of sort that one has

\section{XII}

The transformation of elliptic functions leads to two kinds of
algebraic equations of the greatest importance for Algebra and Number
theory; some are between the moduli $\lambda$ and $k$, the others
between the multiplier $\frac{1}{M}$ and the modulus.  It is these
last ones which have been, from the point of view of the resolution of the
quintic equation, the subject of beautiful work of M. Kronecker and
Mr. Brioschi, and as the results obtained by these eminent geometers
provide me with a point of departure, I will recount them briefly.

According to Jacobi's theorem, demonstrated in the 3rd of the {\em Annali di
Mathematica}, 1. Im 1858, p. 175, by M. Brioschi\footnote{}, we know
that the roots of the sextic equation between $\frac{1}{M}=x$ and $k$, are

\[ \sqrt{x} = \frac{\sin\am 4\omega\sin\am 8\omega}{\sin\coam 4\omega\sin\coam 8\omega}\],

$\omega$ being sucessively $\frac{K}{5}$ and $\frac{\nu K+iK'}{5}$ for
$\nu=0,1,2,3,4$, they are expressed in the following manner:

We designate by $\sqrt{x_\nu}$ that which corresponds to $\omega=\frac{\nu K
+iK'}{5}$, which naturally leads us to adopt the notation $\sqrt{x_\infty}$
for the first which is given by setting $\omega=\frac{K}{5}$; one has:

\begin{eqnarray}
\left\{\begin{array}{lll}
\sqrt{x_\infty}&=&A_0\sqrt{5},\\
\sqrt{x_0}&=&A_0+A_1+A_2,\\
\sqrt{x_1}&=&A_0+\rho A_1+\rho^4 A_2,
\end{array}\right.&&
\left\{\begin{array}{lll}
\sqrt{x_2}&=&A_0+\rho^2 A_1+\rho^3 A_2,\\
\sqrt{x_3}&=&A_0+\rho^3 A_1+\rho^2 A_2,\\
\sqrt{x_4}&=&A_0+\rho^4 A_1+\rho A_2,
\end{array}\right.
\end{eqnarray}

$\rho$ being a fifth root of unity.  These remarkable expressions do
not only occur, furthermore, in the equation between $x$ and $k$: they
occur again with respect to the roots of the analogous
equations\footnote{M. Brioschi, {\em Sul methodo di Kronecker per la
risoluzione delle equazioni di quinto grado, Ati dell' Instituto
Lombardo}, 1. I, 1858, p. 275 \S II, and P. Joubert, {\em Comptes
rendus}, I. XLVII, 1858, p.341.} that the theory of elliptic functions
gives between the inverse of the multiplier $x$ and the quantities
$x\frac{\sqrt{k}}{\sqrt{\lambda}}, x\frac{k}{\lambda},
x\left(\frac{k}{\lambda}+\frac{k'}{\lambda'}\right)^2, ...$.  The bold
idea, which should prove to be fruitful, of studying in general all
sextic equations which give the roots expressed in this manner, which
are $A_0, A_1, A_2$, is due entirely to M. Kronecker.  A primary
result, obtained but not published by the knowledgable geometer, has
been given by M. Brioschi on page 256 of the {\em Annali di
Matematica}, 1. I, 1858, and consists in the formation itself of that
sextic equation.  If we set

\begin{eqnarray}
A&=&A_0^2+A_1A_2,\\
B&=&8A_0^4A_1A_2-2A_0^2A_1^2A_2^2+A_1^3A_2^3-A_0(A_1^5+A_2^5),\\
C&=&320A_0^6A_1^2A_2^2-160A_0^4A_1^3A_2^3+20A_0^2A_1^4A_2^4+6A_1^5A_2^5\\
&&-4A_0(32A_0^4-20A_0^2A_1A_2+5A_1^2A_2^2)(A_1^5+A_2^5)+A_1^{10}+A_2^{10}
\end{eqnarray}

it has this remarkable form\footnote{M. Kronecker, {\em Ueber die
Gleichungen f\"unften Grades (Journal de Crelle}, t. 59, 1861).}:

\[(x-A)^5(x-5A)+10B(x-A)^3-C(x-A)+5B^2-AC=0\].

A second very important result is in the lowering of this equation to the
fifth degree\footnote{}.  The expressions given by the formula

\[y\sqrt{5}=(x_\infty-x_\nu)(x_{\nu+1}-x_{\nu-1})(x_{\nu+2}-x_{\nu-2})\],

the index $\nu$ being always an integer modulo 5, are in effect the
roots of the following equation, where $5^5\Pi$ represents the
discriminant\footnote{One easly discovers, by $A=0$,

$\Pi=(C^3-12^3B^5)^2$

and, for $B=0$,

$\Pi=(C^3+2^6C^2A^5)^2$}, is:

\[y^5+20By^4+10(9B^2-AC)y^3+100B(9B^2-AC)y^2+25(9B^2-AC)^2y-\sqrt{\Pi}=0\].

Finally, this small equation is simplified, and, upon setting $\sqrt{y}=4x$,
M. Brioschi ultimately arrives at this beautiful result\footnote{No. 5 of
the {\em Annali di Matematica}, 1858, 1. I, p. 326}:

\[x^5+\frac{5B}{8}x^3+\frac{5(9B^2-AC)}{4^4}x-\frac{1}{4^5}\sqrt[4]{\Pi}=0\].

It serves, in effect, as the origin of these remarkable quintic
equations whose roots can be obtained in their explicit form with the
help of elliptic transcendentals and which, in this regard, offers
two principal cases.  The first obtained puts

\[A=1, B=0, C=-2^8k^2k'^2\];

then the sextic equation in $x$ is none other than the relation between the
multiplier and the greater modulus considered, and, as it is said, we have

\[\sqrt{x}=\frac{\sin\am\omega\sin\am 8\omega}{\sin\coam 4\omega\sin\coam 8\omega}\].

As for the equation, it takes the trinomial form

\[x^5+5k^2k'^2x-2k^2k'^2(1-2k^2)=0\],

to which I had for aim, in the first part of this research, of reducing
all equations of the fifth degree.

The second case obtained puts

\[A=0, B=-\frac{16}{k^2k'^2}, C=2^8\frac{1-16k^2k'^2}{k^4k'^4}\];

it leads to the equation

\[x^5-\frac{10}{k^2k'^2}x^3+\frac{45}{k^4k'^4}x+\frac{4}{k^6k'^6}(1-2k^2)(1+32k^2k'^2)=0\],

which more simply, on substituing $\frac{x}{kk'}$ for $x$,

\[x^5-10x^3+45x+4\frac{(1-2k^2)(1+32k^2k'^2)}{kk'}=0,\]


\end{document}
