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\begin{document}

\title{Essays on Quintic Equations}
\author{Prof. Francesco Brioschi\\
{\small Translated by Kevin Iga}}
\date{1858\\
{\small from Annali di Matematica, ser. I, t.1, 1858}}
\maketitle

\section{On the Multiplier Equation for Transformations of Elliptic Functions,\\
Prof. Francesco Brioschi, pp. 175-177}

It is known how substituting $n$ [dispari] number is satisfied by the equation:

\[\dk{y}{\lambda}{}=\dk{z}{x}{k}\]

by means of the substitution $y=\frac{U}{V}$, in which $U, V$ are two
polynomials in $x$ of order $n$ and $n+1$; and how putting:

\[K=\int_0^1\dk{x}{k}{}, K'=\int_0^1\dk{x}{k'}{}, k^2+k'^2=1\]

and

\[\omega=\frac{mK+m'iK'}{n},\]

where $m, m'$ are two integers, and $i=\sqrt{-1}$, one has the relations:

\begin{equation}
\left\{\begin{array}{l}
\sqrt[4]{\lambda}=k^{n/4}\sin\coam.4\omega\sin\coam.8\omega....\sin\coam.2(n-1)\omega\\
\sqrt{z}=\sqrt{(-1)^\frac{n-1}{2}}
\frac{\sin\am.4\omega\sin\am.8\omega....\sin\am.2(n-1)\omega}
{\sin\coam.4\omega\sin\coam.8\omega....\sin\coam.2(n-1)\omega}
\end{array} \right.
\label{eq:o1}
\end{equation}

If $n$ is prime, one has $n+1$ transformations of the [ennesimo] ordinary
differentials between them, the ones correspoding to $n+1$ values of $\omega$:

\[\frac{K}{n},\frac{iK'}{n},\frac{K+iK'}{n},\frac{K+2iK'}{n},...,\frac{K+(n-1)iK'}{n};\]

and which substitutes into \ref{eq:o1} values corresponding to the
modulus $\lambda$ and $n+1$ to the multiplier $z$.  Then the
equations, the roots of which are $n+1$ values of $\lambda$ or of $z$,
<saranno> of degree $n+1$.  The first equations are denominate equations
of the modulus or moduli; to the second we give the name multiplier equations.

Indicating with $z_1, z_2, ..., z_{n+1}$, the $n+1$ values of the multipliers;
and with $\Lambda_1, \Lambda_2, \Lambda_3, ...$ the values of the
complete function:

\[\Lambda = \int_0^1\dk{x}{\lambda}{}\]

corresponding to the $n+1$ values of the modulus are the:

\[z_1=(-1)^\frac{n-1}{2}\frac{n\Lambda_1}{K},z_2=\frac{\Lambda_2}{K},
..., z_{n+1}=\Lambda{n+1}{K};\]

or observing to be: ({\em Fundamenta Nova}, p. 184)

\[\sqrt{\frac{2K}{\pi}}=1+2q+2q^4+2q^9+\cdots = \sum_{m=\infty}^{m=-\infty}q^{m^2}\]

[si avra] by a theorem by Jacobi\footnote{Giornale di Crelle T. 3, p193}
demonstrating from Sohnke\footnote{Idem T. 16, p 103}

\[\sqrt{z_1}=\sqrt{(-1)^\frac{n-1}{2}}\frac{\sum_{m=\infty}^{m-\infty}q^{nm^2}}
{\sum_mq^{m^2}}\]

and the values of $\sqrt{z_2}$, $\sqrt{z_3}$, ... are obtained putting into:

\[\frac{\sum_mp^{m^2}}{\sum_mq^{m^2}}\]

in place of $p$ ordinarily $q^\frac{1}{n}, \alpha q^\frac{1}{n},
\alpha^2 q^\frac{1}{n}, ... ,\alpha^{n-1} q^\frac{1}{n}$; $\alpha$ being
a root of the equation $a^n-1=0$.  

\section{On the Determination of coefficients of the Tschirnhaus Transformation\\
Applications to the Reduction of Jerrard of the general Quintic Equation, pp. 238-249}

\section{On the resolution of the Quintic Equation, pp.256-259\\
Hermite--from the French, Comptes Rendus--n.11. Mars. 1858}

\subsection{}

In a note ``On Multiplier Equations, etc.'' [inserita] in 3rd [fascicolo]
of which Annali we have demonstrated that the square roots of the roots of
the multiplier equation corresponding to a transformation of order $n$ prime
can be expressed linearly by the $\frac{n+1}{2}$ quantities $A_0, A_1,... ,
A_\frac{n-1}{2}$.  Considering [per ora] in particular the equations of the
sixth degree:

\begin{equation}
z^6+a_1z^5+a_2z^4+\cdots+A_5z+A_6=0
\label{eq:3.1}
\end{equation}

corresponding to the quintic transformation, and indicated by $z_1, z_2,
..., z_6$ the roots of the [medesima] are:

\begin{equation}
\left\{\begin{array}{lll}
\sqrt{z_1}=A_0\sqrt{5}, & \sqrt{z_2}=A_0+A_1+A_2, & \sqrt{z_3}=A_0+\alpha A_1
+\alpha^4 A_2\\
\sqrt{z_4}=A_0+\alpha^2A_1+\alpha^3A_2, & \sqrt{z_5}=A_0+\alpha^3A_1+
\alpha^2A_2, & \sqrt{z_6}=A_0+\alpha^4 A_1+\alpha A_2
\end{array}\right.
\label{eq:3.2}
\end{equation}

in which $\alpha$ is an imaginary root of the equation $\alpha^5-1=0$; and
$A_0, A_1, A_2$ have the values [trovati] in the note cited.

If [mediante] which expressions of the roots are formed the coefficients
in the \ref{eq:3.1}, we easily obtain the relations:

\begin{equation}
\left\{\begin{array}{lll}
a_1=-10A,&a_2=35A^2,&a_3=-60A^3+10B\\
a_4=55A^4-30AB,&a_5=-26A^5+30A^2B-C,&a_6=5(A^3-B)^2
\end{array}\right.\label{eqn:3.3}
\end{equation}

being [posto] for brevity:

\begin{equation}
\left\{\begin{array}{l}
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_1^3+A_2^3)\\
C=320A_0^6A_1^2A_2^2-160A_0^4A_1^3A_2^2+20A_0^2A_1^4A_2^4+6A_5^5A_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{array}\right.
\label{eqn:3.4}
\end{equation}

But it is known that for $n=5$ the multiplier equations are:

\[z^6-10z^5+35z^4-60z^3+55z^2-2(13-2^8k^2k'^2)z+5=0\]

of which the first, third, and fifth of \ref{eqn:3.3} give:

\begin{equation}
A=1, B=0, C=-2^8k^2k'^2
\label{eqn:3.5}
\end{equation}

and the others are satisfied by these values.  [Dalle] preceding relations
[deduconsi] that which we have given in the note cited.

\subsection{}

We put [ora]:

\begin{equation}
\left\{\begin{array}{ll}
x_1=(z_2-z_1)(z_3-z_6)(z_4-z_5),&x_2=(z_3-z_1)(z_4-z_2)(z_5-z_6)\\
x_3=(z_4-z_1)(z_5-z_3)(z_6-z_2),&x_4=(z_5-z_1)(z_6-z_4)(z_2-z_3)\\
x_5=(z_6-z_1)(z_2-z_5)(z_3-z_4);
\end{array}\right.
\end{equation}

and we express [mediante] the \ref{eq:3.2} the second member of this equation
as a function of $A_0, A_1, A_2$.  We have easily:

\begin{equation}
\left\{\begin{array}{ll}
x_1=\sqrt{5}(B_0+B_1+B_2+B_3+B_4)m &x_2=\sqrt{5}(B_0+\alpha B_1+\alpha^2 B_2+
\alpha^3 B_3+\alpha^4 B_4\\
x_3=\sqrt{5}(B_0+\alpha^2 B_1+\alpha^4 B_2+\alpha B_3+\alpha^3 B_4&
x_4=\sqrt{5}(B_0+\alpha^3 B_1+\alpha B_2+\alpha^4 B_3+\alpha^2 B_4\\
x_5=\sqrt{5}(B_0+\alpha^4 B_1+\alpha^3 B_2+\alpha^2 B_3+\alpha B_4&
\end{array}\right.
\label{eqn:3.7}
\end{equation}

in which $B_0=4B$ and:

\begin{eqnarray}
B_1&=&8A_0A_1^3A_2^2-16A_0^3A_1^2A_2-2A_1A_2^5-A_1^6+4A_0^2A_2^4\\
B_4&=&8A_0A_1^2A_2^3-16A_0^3A_1A_2^2-2A_1^5A_2-A_2^6+4A_0^2A_1^4\\
B_2&=&16A_0^3A_2^3-16A_0^4A_1^2+A_1^4A_2^2-4A_0A_1A_2^4\\
B_3&=&16A_0^3A_1^3-16A_0^4A_2^2+A_1^2A_2^4-4A_0A_1^4A_2.
\end{eqnarray}

Thus:

\[x^5+p_1x^4+\cdots+p_4x+p_5=0\]

the [avente] equation of the roots $x_1, x_2, ..., x_5$; put briefly,

\[P_2=B_1B_4+B_2B_3, P_3=B_1^2B_3+B_2^2B_1+B_3^2B_4+B_4^2B_2\]
\[P_4=B_1^3B_2+B_2^3B_4+B_3^3B_1+B_4^3B_3+3B_1B_2B_3B_4\]

[dalle] \ref{eqn:3.7} we have:

\[p_1=-5\sqrt{5}B_0, p_2=-5^2P_2, p_3=-5^2\sqrt{6}P_3, p_4=-5^3(P_4-P_2^2);\]

but [pei] upper values of $B_1, B_2 ...$ reducing \label{eqn:3.4} is obtained:

\[P_2=2(AC-3B^2), P_3=4B(AC_5B^2), P_4=15B^4-A^2C^2+2AB^2C\]

where [pei] values \label{eqn:3.5} we have:

\[p_1=0, p_2=2^9.5^2k^2k'^2, p_3=0, p_4=2^16.5^4k^4k'^4/\]

Of that last observing that indicating with 

\end{document}
