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\centerline{\Largebf THEOREMS RESPECTING ALGEBRAIC}

\vskip12pt

\centerline{\Largebf ELIMINATION}

\vskip12pt

\centerline{\Largebf By}

\vskip24pt

\centerline{\Largebf William Rowan Hamilton}

\vskip24pt

\centerline{\vbox{\halign{\largerm #\hfil\cr
(Philosophical Magazine (3rd series):\cr
   \qquad vol.~viii (1836), pp.\ 538--543,\cr
   \qquad vol.~ix (1836), pp.\ 28-32.)\cr}}}

\vskip36pt

\vfill

\centerline{\largerm Edited by David R. Wilkins}

\vskip 12pt

\centerline{\largerm 2000}

\vskip36pt\eject

\null\vskip36pt

\noindent
{\largeit Theorem respecting Algebraic Elimination, connected
with the Question of the Possibility of resolving in finite Terms
the general Equation of the Fifth Degree.}
{\largerm Extracted by Permission, from a Communication
recently made to the Royal Irish Academy.}
{\largeit By Professor Sir\/}
{\largesc William Rowan Hamilton,}
{\largeit Astronomer Royal of Ireland\/}\footnote*{Communicated
by the Author.}.

\bigbreak

\vskip 12pt

\centerline{[{\it The London, Edinburgh and Dublin Philosophical
Magazine and Journal of Science,}}
\centerline{3rd series, vol.~viii (1836), pp.\ 538--543.]}

\bigskip

{\it Theorem~I.}
If $x$ be eliminated between two equations, of the following
forms, namely, 1st, an equation of the fifth degree, of the form
$$0 = x^5 + {\rm D} x + {\rm E},
   \eqno (1.)$$
in which the roots are supposed to be all unequal, and the
coefficients ${\rm D}$ and ${\rm E}$ to be, both of them,
different from $0$, and, 2nd, an equation of the form
$$y = {\rm Q} x + f(x),
   \eqno (2.)$$
in which $f(x)$ denotes any rational function of $x$, whether
integral or fractional,
$$f(x) = { {\rm M}' x^{\mu'} + {\rm M}'' x^{\mu''} + \hbox{\&c.}
           \over {\rm K}' x^{\kappa'} + {\rm K}'' x^{\kappa''}
             + \hbox{\&c.}};
   \eqno (3.)$$
and if, in the result of this elimination, which will always be
an equation of the fifth degree in $y$, of the form
$$0 = y^5 + {\rm A}' y^4 + {\rm B}' y^3 + {\rm C}' y^2
            + {\rm D}' y + {\rm E}',
   \eqno (4.)$$
we suppose that the coefficients are such as to satisfy,
{\it independently of\/}~${\rm Q}$, the second as well as the
first of the two conditions
$${\rm A}' = 0,\quad {\rm C}' = 0,
   \eqno (5.)$$
in virtue of the values of the constants
$${\rm M}', {\rm M}'',\ldots, \mu', \mu'',\ldots,
  {\rm K}', {\rm K}'',\ldots, \kappa', \kappa'',\ldots
   \eqno (6.)$$
in the rational function~$f(x)$; I say that then those
constants~(6.) must be such as to admit of our reducing that
rational function to the form
$$f(x) = q x + ( x^5 + {\rm D} x + {\rm E}) \mathbin{.} \phi(x),
   \eqno (7.)$$
$q$ being some new constant, and $\phi(x)$ being some new
rational function of $x$, which does not contain the polynome
$x^5 + {\rm D} x + {\rm E}$ as a divisor.

\bigbreak

{\it Demonstration}.---Let $x_1 \, x_2 \, x_3 \, x_4 \, x_5$
denote the five roots of the equation~(1.), which are supposed to
be all unequal among themselves, and different from $0$; and let
us put for abridgement
$$\left. \eqalign{
f(x_1) - {x_1 \over x_5} f(x_5) &= h_1, \cr
f(x_2) - {x_2 \over x_5} f(x_5) &= h_2, \cr
f(x_3) - {x_3 \over x_5} f(x_5) &= h_3, \cr
f(x_4) - {x_4 \over x_5} f(x_5) &= h_4, \cr
{f(x_5) \over x_5} = q,\quad
{\rm Q} + q &= {\rm Q}'. \cr}
   \right\}
   \eqno (8.)$$
We shall then have
$$\left. \eqalign{
   & f(x_1) = h_1 + q x_1,\quad
     f(x_2) = h_2 + q x_2,\cr
   & f(x_3) = h_3 + q x_3,\quad
     f(x_4) = h_4 + q x_4,\quad
     f(x_5) =       q x_5,\cr}
   \right\}
   \eqno (9.)$$
and the result~(4.) of the elimination of $x$ between the
equations (1.) and (2.), may be expressed as follows:
$$0 =    (y - {\rm Q}' x_1 - h_1)
         (y - {\rm Q}' x_2 - h_2)
         (y - {\rm Q}' x_3 - h_3)
         (y - {\rm Q}' x_4 - h_4)
         (y - {\rm Q}' x_5).
   \eqno (10.)$$
Comparing (10.) with (4.), and observing that the form of the
equation~(1.) gives the relations
$$0 = x_1 + x_2 + x_3 + x_4 + x_5,
   \eqno (11.)$$
$$0 = x_1 x_2 + x_2 x_3 + x_3 x_4 + x_4 x_5 + x_5 x_1
       + x_1 x_3 + x_2 x_4 + x_3 x_5 + x_4 x_1 + x_5 x_2,
   \eqno (12.)$$
$$\eqalignno{
0  &= x_1 x_2 x_3 + x_2 x_3 x_4 + x_3 x_4 x_5
          + x_4 x_5 x_1 + x_5 x_1 x_2 \cr
   &\mathrel{\phantom{=}}
       + x_1 x_3 x_4 + x_2 x_4 x_5 + x_3 x_5 x_1
          + x_4 x_1 x_2 + x_5 x_2 x_3,
   &(13.)\cr}$$
we easily find these expressions for ${\rm A}'$ and ${\rm C}'$,
namely,
$${\rm A}' = - (h_1 + h_2 + h_3 + h_4),
   \eqno (14.)$$
and
$$\eqalignno{
{\rm C}'
   &= - {\rm Q}'^2
         ( h_1 x_1^2 + h_2 x_2^2 + h_3 x_3^2 + h_4 x_4^2 ) \cr
   &\mathrel{\phantom{=}}
      + {\rm Q}'  \left\{ \matrix{
               h_1 h_2 (x_1 + x_2)
             + h_1 h_3 (x_1 + x_3)
             + h_1 h_4 (x_1 + x_4) \cr
             + h_2 h_3 (x_2 + x_3)
             + h_2 h_4 (x_2 + x_4)
             + h_3 h_4 (x_3 + x_4) \cr}
            \right\} \cr
   &\mathrel{\phantom{=}}
       - ( h_1 h_2 h_3 + h_1 h_2 h_4
          + h_1 h_3 h_4 + h_2 h_3 h_4).
   &(15.)\cr}$$

If, then, the coefficient~${\rm C}'$, as well as ${\rm A}'$, is
to vanish independently of ${\rm Q}$, and consequently of ${\rm
Q}'$, we must have the four following equations:
$$0 = h_1 + h_2 + h_3 + h_4;
   \eqno (16.)$$
$$0 = h_1 x_1^2 + h_2 x_2^2 + h_3 x_3^2 + h_4 x_4^2;
   \eqno (17.)$$
$$\eqalignno{
0  &=    h_1 h_2 (x_1 + x_2)
       + h_1 h_3 (x_1 + x_3)
       + h_1 h_4 (x_1 + x_4) \cr
   &\mathrel{\phantom{=}}
       + h_2 h_3 (x_2 + x_3)
       + h_2 h_4 (x_2 + x_4)
       + h_3 h_4 (x_3 + x_4);
   &(18.)\cr}$$
$$0 = h_1 h_2 h_3 + h_1 h_2 h_4 + h_1 h_3 h_4 + h_2 h_3 h_4;
   \eqno (19.)$$
which give, by elimination of $h_4$,
$$\eqalignno{
0  &=    h_1 (x_1^2 - x_4^2)
       + h_2 (x_2^2 - x_4^2)
       + h_3 (x_3^2 - x_4^2), &(20.)\cr
0  &=    h_1^2 x_1 + h_2^2 x_2 + h_3^2 x_3
       + (h_1 + h_2 + h_3)^2 x_4, &(21.)\cr
0  &= (h_2 + h_3) (h_3 + h_1) (h_1 + h_2). &(22.)\cr}$$
Of the three factors of the last of these equations, it is
manifestly indifferent {\it which\/} we employ; since the
conclusions which can be drawn from the consideration of any one
of these three factors can also be drawn from the consideration
of either of the other two, by merely interchanging two of the
three roots $x_1 \, x_2 \, x_3$, without altering the other of
those three roots, or the two remaining roots $x_4 \, x_5$ of the
equation~(1.).  We shall therefore take the first of the three
factors of (22.), namely, the equation
$$0 = h_2 + h_3;
   \eqno (23.)$$
which reduces the two equations (20.) and (21.) to the two
following, obtained by elimination of $h_3$,
$$\eqalignno{
0  &= h_1 (x_1^2 - x_4^2) + h_2 (x_2^2 - x_3^2), &(24.)\cr
0  &= h_1^2 (x_1 + x_4)   + h_2^2 (x_2 + x_3).   &(25.)\cr}$$
These two last equations give, by elimination of $h_2$,
$$0 = h_1^2 (x_1 + x_4)
         \{    (x_1 + x_4) (x_1 - x_4)^2
             + (x_2 + x_3) (x_2 - x_3)^2 \};
   \eqno (26.)$$
in which we cannot suppose the factor $x_1 + x_4$ to vanish,
because the relations
$$0 = x_1^5 + {\rm D} x_1 + {\rm E},\quad
  0 = x_4^5 + {\rm D} x_4 + {\rm E},
   \eqno (27.)$$
give
$$\left. \eqalign{
{\rm D}
   &= - ( x_1^4 + x_1^3 x_4 + x_1^2 x_4^2
             + x_1 x_4^3 + x_4^4),\cr
{\rm E}
   &= (x_1 + x_4) (x_1^2 + x_4^2) x_1 x_4,\cr}
   \right\}
   \eqno (28.)$$
and we have supposed that ${\rm E}$ does not vanish; and since,
for a similar reason, we cannot suppose that $x_2 + x_3$
vanishes, we see that we must conclude
$$h_1 = 0,\quad
  h_2 = 0,\quad
  h_3 = 0,\quad
  h_4 = 0,
   \eqno (29.)$$
unless we can suppose that the third factor of (26.) vanishes,
that is, unless
$$(x_1 + x_4) (x_1 - x_4)^2 + (x_2 + x_3) (x_2 - x_3)^2 = 0.
   \eqno (30.)$$
Let us then examine into the meaning of this last condition, and
the circumstances under which it can be satisfied.

If we put, for abridgement,
$$x_2 + x_3 = - \alpha,\quad x_2 x_3 = \beta,
   \eqno (31.)$$
the condition~(30.) will become
$$0 = x_4^3 - x_4^2 x_1 - x_4 x_1^2 + x_1^3
       - \alpha^3 + 4 \alpha \beta;
   \eqno (32.)$$
and we shall have, in virtue of the relations (11.) (12.) (13.),
two other equations between $x_4$, $x_1$, $\alpha$, $\beta$,
namely,
$$0 = x_4^2 + x_4 (x_1 - \alpha) + x_1^2 - x_1 \alpha
       + \alpha^2 - \beta,
   \eqno (33.)$$
and
$$0 = x_1^3 - x_1^2 \alpha + x_1 (\alpha^2 - \beta)
       - \alpha^3 + 2 \alpha \beta;
   \eqno (34.)$$
between which three equations, (32.) (33.) (34.), we shall now
proceed to eliminate $x_4$ and $x_1$.  For this purpose we may
begin by multiplying (33.) by $x_1$, and adding the product to
(32.); a process which gives, by (34.),
$$0 = x_4^3 - x_4 x_1 \alpha + x_1^3 + 2 \alpha \beta,
   \eqno (35.)$$
a relation more simple than (32.).  In the next place we may
observe that, in general, the result of elimination of any
variable~$x$ between any two equations of the forms
$$\left. \eqalign{
  0 &= p' + q' x + r' x^2 + s' x^3,\cr
  0 &= p'' + q'' x + r'' x^2,\cr}
   \right\}
   \eqno (36.)$$
is
$$\eqalignno{
0  &= p'^2 r''^3 - p' q' q'' r''^2 - 2 p' r' p'' r''^2
       + p' r' q''^2 r'' + 3 p' s' p'' q'' r''
       - p' s' q''^3 + q'^2 p'' r''^2 \cr
   &\mathrel{\phantom{=}}
       - q' r' p'' q'' r'' - 2 q' s' p''^2 r''
       + q' s' p'' q''^2 + r'^2 p''^2 r''
       - r' s' p''^2 q'' + s'^2 p''^3.
   &(37.)\cr}$$
Applying this general formula to the elimination of $x_4$ between
the equations (35.) and (33.), and making, for that purpose,
$$\left. \eqalign{
   &
p' = x_1^3 + 2 \alpha \beta,\quad
q' = - x_1 \alpha,\quad
r' = 0,\quad
s' = 1,\cr
   &
p'' = x_1^2  - x_1 \alpha + \alpha^2 - \beta,\quad
q'' = x_1 - \alpha,\quad
r'' = 1,\cr}
   \right\}
   \eqno (38.)$$
we find, after some easy reductions,
$$\eqalignno{
0  &= 4 x_1^6 - 4 x_1^5 \alpha + x_1^4 (8 \alpha^2 - 6 \beta)
       + x_1^3 (-8 \alpha^3 + 14 \alpha \beta)
       + x_1^2 (6 \alpha^4 - 12 \alpha^2 \beta + 3 \beta^2) \cr
   &\mathrel{\phantom{=}}
       + x_1 (-2 \alpha^5 + 7 \alpha^3 \beta - 7 \alpha \beta^2)
       + \alpha^6 - 7 \alpha^4 \beta + 13 \alpha^2 \beta^2
       - \beta^3;
   &(39.)\cr}$$
which is easily reduced by (34.) to the form
$$0 = x_1^2 (2 \alpha^4 - 2 \alpha^2 \beta + \beta^2)
       + x_1 (2 \alpha^5 - 7 \alpha^3 \beta + \alpha \beta^2)
       + \alpha^6 - 3 \alpha^4 \beta + 5 \alpha^2 \beta^2
       - \beta^3.
   \eqno (40.)$$
Again, applying the same general formula~(37.) to the elimination
of $x_1$ between the equations (34.) and (40.), by making now
$$\left. \eqalign{
   &  p' = - \alpha^3 + 2 \alpha \beta,\quad
      q' = \alpha^2 - \beta,\quad
      r' = - \alpha,\quad
      s' = 1,\cr
   &  p'' = \alpha^6 - 3 \alpha^4 \beta + 5 \alpha^2 \beta^2
                - \beta^3,\quad
      q'' = 2 \alpha^5 - 7 \alpha^3 \beta + \alpha \beta^2,\cr
   &  r'' = 2 \alpha^4 - 2 \alpha^2 \beta + \beta^2,\cr}
   \right\}
   \eqno (41.)$$
we find after reductions,
$$ 0 =     25 \alpha^{18}
       -  250 \alpha^{16} \beta
       +  975 \alpha^{14} \beta^2
       - 1850 \alpha^{12} \beta^3
       + 1725 \alpha^{10} \beta^4
       -  700 \alpha^8    \beta^5
       +  100 \alpha^6    \beta^6,
   \eqno (42.)$$
that is,
$$0 = 25 \alpha^6 (\alpha^2 - 2 \beta)^2
         (\alpha^4 - 3 \alpha^2 \beta + \beta^2)^2.
   \eqno (43.)$$
But this condition cannot be satisfied, consistently with the
supposition which we have already made that neither ${\rm D}$ nor
${\rm E}$ vanishes; because, by expressions similar to (28.), we
have
$${\rm D} = - (\alpha^4 - 3 \alpha^2 \beta + \beta^2),\quad
  {\rm E} = - \alpha \beta (\alpha^2 - 2 \beta).
   \eqno (44.)$$

We must therefore reject the supposition (30.), and adopt the
only other alternative, namely, (29.); and hence we have, by
(9.),
$$f(x_1) = q x_1,\quad
  f(x_2) = q x_2,\quad
  f(x_3) = q x_3,\quad
  f(x_4) = q x_4,\quad
  f(x_5) = q x_5.
   \eqno (45.)$$

In this manner we find, that, under the circumstances supposed in
the enunciation of the theorem, the function
$$f(x) - qx$$
vanishes, for every value of $x$ which makes the polynome
$x^5 + {\rm D} x + {\rm E}$ vanish; and since these values have
been supposed unequal, we must have, therefore,
$$f(x) - qx = (x^5 + {\rm D} x + {\rm E}) \mathbin{.} \phi(x),
   \eqno (46.)$$
the function $(\phi x)$ being rational, like $f(x)$, and not
containing $x^5 + {\rm D} x + {\rm E}$ as a divisor; which was
the thing to be proved.

\bigbreak

{\it Corollary}.
It is evident that, under the circumstances above supposed, the
coefficients ${\rm B}'$,~${\rm D}'$,~${\rm E}'$ of (4.) will be
expressed as follows:
$${\rm B}' = 0,\quad
  {\rm D}' = {\rm Q}'^4 {\rm D},\quad
  {\rm E}' = {\rm Q}'^5 {\rm E};
   \eqno (47.)$$
that is, the equation of the fifth degree in $y$ will be of the
form
$$0 = y^5 + {\rm Q}'^4 {\rm D} y + {\rm Q}'^5 {\rm E}.
   \eqno (48.)$$
At the same time the relation between $y$ and $x$ will reduce
itself, by (2.) and (7.), to the form
$$y = {\rm Q}' x
       + (x^5 + {\rm D} x + {\rm E}) \mathbin{.} \phi(x),
   \eqno (49.)$$
${\rm Q}'$ still denoting ${\rm Q} + q$.  If then, we were to
establish this additional supposition
$${\rm D}' = {\textstyle {1 \over 5}} {\rm B}'^2,
   \eqno (50.)$$
in order to complete the reduction of (4.) to De Moivre's
solvible form, we should have
$${\rm Q}'^4 = 0,
   \eqno (51.)$$
that is,
$${\rm Q}' = 0;
   \eqno (52.)$$
the equation of the fifth degree in $y$ would become
$$y^5 = 0,
   \eqno (53.)$$
and the relation between $y$ and $x$ would become
$$y = (x^5 + {\rm D} x + {\rm E}) \mathbin{.} \phi(x);
   \eqno (54.)$$
and thus, although the equation in $y$ would indeed by easily
solvible, yet it would entirely fail to give the least assistance
towards resolving the proposed equation of the fifth degree in
$x$.

\nobreak\bigskip

\quad Observatory, Dublin, May 13, 1836.

\vfill\eject

\null\vskip36pt

\noindent
{\largeit Second Theorem of Algebraic Elimination, connected
with the Question of the Possibility of resolving, in finite Terms,
Equations of the Fifth Degree.
By Professor Sir\/}
{\largesc William Rowan Hamilton,}
{\largeit Astronomer Royal of Ireland.}

\bigbreak

\vskip 12pt

\centerline{[{\it The London, Edinburgh and Dublin Philosophical
Magazine and Journal of Science,}}
\centerline{3rd series, vol.~ix (1836), pp.\ 28--32.]}

\bigskip


\bigbreak

{\it Theorem~II}.
If $x$ be eliminated between a proposed equation of the fifth
degree,
$$x^5 + {\rm A} x^4 + {\rm B} x^3 + {\rm C} x^2
       + {\rm D} x + {\rm E} = 0,
   \eqno (55.)$$
and an assumed equation, of the form
$$y = {\rm Q} x + f(x),
   \eqno (2.)$$
in which $f(x)$ denotes any rational function of $x$,
$$f(x) = { {\rm M}' x^{\mu'} + {\rm M}'' x^{\mu''} + \ldots
           \over {\rm K}' x^{\kappa'} + {\rm K}'' x^{\kappa''}
               + \ldots};
   \eqno (3.)$$
and if the constants of this function be such as to reduce the
result of the elimination to the form
$$y^5 + {\rm B}' y^3 + {\rm D}' y + {\rm E}' = 0,
   \eqno (56.)$$
{\it independently of\/}~${\rm Q}$: then not only must we have
$${\rm A} = 0,\quad {\rm C} = 0,
   \eqno (57.)$$
so that the proposed equation of the fifth degree must be of the
form
$$x^5 + {\rm B} x^3 + {\rm D} x + {\rm E} = 0,
   \eqno (58.)$$
but also the function $f(x)$ must be of the form
$$f(x) = qx + (x^5 + {\rm B} x^3 + {\rm D} x + {\rm E})
           \mathbin{.} \phi(x),
   \eqno (59.)$$
$q$ being some constant multiplier, and $\phi(x)$ some rational
function of $x$, which does not contain the polynome
$x^5 + {\rm B} x^3 + {\rm D} x + {\rm E}$ as one of the factors
of its denominator; unless we have either, first,
$${\rm E} = 0;
   \eqno (60.)$$
or else, secondly,
$$5 {\rm D} = {\rm B}^2,
   \eqno (61.)$$
or, as the third and only remaining case of exception,
$$5^5 {\rm E}^4
       + 2^2 {\rm B} {\rm E}^2
         ( 2^2 5^3 {\rm D}^2 - 3^2 5^2 {\rm B}^2 {\rm D}
             + 3^3 {\rm B}^4 )
       + 2^4 {\rm D}^3
         (2^4 {\rm D}^2 - 2^3 {\rm B}^2 {\rm D} + {\rm B}^4) = 0.
   \eqno (62.)$$

\bigbreak

{\it Demonstration}.---If we denote by
$x_1 \, x_2 \, x_3 \, x_4 \, x_5$ the five roots of the proposed
equation of the fifth degree, and put, as is permitted,
$$\left. \eqalign{
   & f(x_1) = h_1 + q x_1,\quad
     f(x_2) = h_2 + q x_2,\quad
     f(x_3) = h_3 + q x_3,\cr
   & f(x_4) = h_4 + q x_4,\quad
     f(x_5) =       q x_5,\cr}
   \right\}
   \eqno (9.)$$
and
$${\rm Q} + q = {\rm Q}',
   \eqno (8.)$$
the result of the elimination of $x$ between the two equations
(55.) and (2.) may be denoted thus,
$$       (y - {\rm Q}' x_1 - h_1)
         (y - {\rm Q}' x_2 - h_2)
         (y - {\rm Q}' x_3 - h_3)
         (y - {\rm Q}' x_4 - h_4)
         (y - {\rm Q}' x_5)
   =  0;
   \eqno (10.)$$
and if this result is to be of the form (56.), independently of
${\rm Q}$, and therefore also of ${\rm Q}'$, we must have the six
following relations:
$$x_1 + x_2 + x_3 + x_4 + x_5 = 0,
   \eqno (11.)$$
$$\eqalignno{
   &\mathbin{\phantom{+}}
            x_1 x_2 x_3 + x_2 x_3 x_4 + x_3 x_4 x_5
          + x_4 x_5 x_1 + x_5 x_1 x_2 \cr
   &      + x_1 x_3 x_4 + x_2 x_4 x_5 + x_3 x_5 x_1
          + x_4 x_1 x_2 + x_5 x_2 x_3 = 0,
   &(13.)\cr}$$
$$h_1 + h_2 + h_3 + h_4 = 0,
   \eqno (16.)$$
$$h_1 ( x_2 x_3 + x_2 x_4 + x_2 x_5
             + x_3 x_4 + x_3 x_5 + x_4 x_5 )
       + h_2(\hbox{\&c.})
       + h_3(\hbox{\&c.})
       + h_4(\hbox{\&c.})
   =  0,
   \eqno (63.)$$
$$h_1 h_2 (x_3 + x_4 + x_5)
       + h_1 h_3 (\hbox{\&c.})
       + h_1 h_4 (\hbox{\&c.})
       + h_2 h_3 (\hbox{\&c.})
       + h_2 h_4 (\hbox{\&c.})
       + h_3 h_4 (\hbox{\&c.})
   =  0,
   \eqno (64.)$$
$$h_1 h_2 h_3 + h_1 h_2 h_4 + h_1 h_3 h_4 + h_2 h_3 h_4 = 0;
   \eqno (19.)$$
of which the two first give
$${\rm A} = 0,\quad {\rm C} = 0,
   \eqno (57.)$$
and the last three may, by attending to the first and third, and
by eliminating $h_4$, be written thus:
$$       h_1 (x_1^2 - x_4^2)
       + h_2 (x_2^2 - x_4^2)
       + h_3 (x_3^2 - x_4^2)
   =  0,
   \eqno (20.)$$
$$h_1^2 x_1 + h_2^2 x_2 + h_3^2 x_3 + (h_1 + h_2 + h_3)^2 x_4
   =  0,
   \eqno (21.)$$
$$(h_2 + h_3) (h_3 + h_1) (h_1 + h_2) = 0.
   \eqno (22.)$$
Selecting, as we are at liberty to do, the first of the three
factors of (22.), namely,
$$h_2 + h_3 = 0,
   \eqno (23.)$$
and eliminating $h_3$ by this, we reduce the two conditions (20.)
and (21.) to the two following:
$$\eqalignno{
h_1 (x_1^2 - x_4^2) + h_2 (x_2^2 - x_3^2) &= 0, &(24.)\cr
h_1^2 (x_1 + x_4)   + h_2^2 (x_2 + x_3)   &= 0, &(25.)\cr}$$
which give, by elimination of $h_2$,
$$h_1^2 (x_1 + x_4) \{
         (x_1 + x_4) (x_1 - x_4)^2 + (x_2 + x_3) (x_2 - x_3)^2 \}
   =  0.
   \eqno (26.)$$
And from these equations (of which some occurred in the
investigation of the former theorem, but are now for greater
clearness repeated,) we see that we must have
$$h_1 = 0,\quad
  h_2 = 0,\quad
  h_3 = 0,\quad
  h_4 = 0,
   \eqno (29.)$$
and therefore, by (9.),
$$f(x_1) = q x_1,\quad
  f(x_2) = q x_2,\quad
  f(x_3) = q x_3,\quad
  f(x_4) = q x_4,\quad
  f(x_5) = q x_5,
   \eqno (45.)$$
unless we have either
$$x_1 + x_4 = 0,
   \eqno (65.)$$
or else
$$(x_1 + x_4) (x_1 - x_4)^2 + (x_2 + x_3) (x_2 - x_3)^2 = 0,
   \eqno (30.)$$
or at least some one of those other relations into which (65.)
and (30.) may be changed, by changing the arrangement of the
roots of the proposed equation of the fifth degree.

The alternative (65.), combined with (57.), gives evidently
$${\rm E} = 0;
   \eqno (60.)$$
but the meaning of the alternative~(30.) is a little less easy to
examine, now that we do not suppose the coefficient~${\rm B}$ to
vanish, as we did in the investigation of the former theorem.
However, the following process is tolerably simple.  We may
conceive that $x_1 \, x_2 \, x_3 \, x_4$ are roots of a certain
biquadratic equation,
$$x^4 + a x^3 + b x^2 + c x + d = 0,
   \eqno (66.)$$
and may express, by means of its coefficients $a \, b \, c \, d$,
the symmetric functions of $x_1 \, x_2 \, x_3 \, x_4$ which enter
into the development of the product formed by multiplying
together the condition~(30.), and these two other similar
conclusions,
$$\eqalignno{
(x_1 + x_3) (x_1 - x_3)^2 + (x_2 + x_4) (x_2 - x_4)^2
   &= 0, &(67.)\cr
(x_1 + x_2) (x_1 - x_2)^2 + (x_3 + x_4) (x_3 - x_4)^2
   &= 0. &(68.)\cr}$$
If we put, for abridgement,
$$x_1^3 + x_2^3 + x_3^3 + x_4^3 = {\rm f},
   \eqno (69.)$$
$$- x_1 x_2 (x_1 + x_2) - x_3 x_4 (x_3 + x_4) = {\rm g},
   \eqno (70.)$$
$$     - x_1 x_3 (x_1 + x_3)
       - x_2 x_4 (x_2 + x_4)
       - x_1 x_4 (x_1 + x_4)
       - x_2 x_3 (x_2 + x_3)
   =  {\rm h},
   \eqno (71.)$$
$$    \{ x_1 x_3 (x_1 + x_3)
       + x_2 x_4 (x_2 + x_4) \}
      \{ x_1 x_4 (x_1 + x_4)
       + x_2 x_3 (x_2 + x_3) \}
   =  {\rm i},
   \eqno (72.)$$
the condition~(68.) will become
$${\rm f} + {\rm g} = 0,
   \eqno (73.)$$
and the product of the two other conditions (67.) and (68.) will
become
$${\rm f}^2 + {\rm h} {\rm f} + {\rm i} = 0,
   \eqno (74.)$$
so that the product of all the three conditions becomes
$${\rm f}^3 + ({\rm g} + {\rm h}) {\rm f}^2
       + ( {\rm g} {\rm h} + {\rm i} ) {\rm f}
       + {\rm g} {\rm i}
   =  0;
   \eqno (75.)$$
and the symmetric functions ${\rm f}$, ${\rm g} + {\rm h}$,
${\rm g} {\rm h} + {\rm i}$, ${\rm g} {\rm i}$, may be expressed
as follows:
$${\rm f} = - a^3 + 3ab - 3c,
   \eqno (76.)$$
$${\rm g} + {\rm h} = ab - 3c,
   \eqno (77.)$$
$${\rm g} {\rm h} + {\rm i} = a^3 c - 4 a^2 d - 2 abc + 3 c^2,
   \eqno (78.)$$
$${\rm g} {\rm i} = a^5 d - 4 a^3 b d + 4 a^2 cd + abc^2 - c^3.
   \eqno (79.)$$
Again, the proposed equation of the fifth degree,
$$x^5 + {\rm B} x^3 + {\rm D} x + {\rm E} = 0,
   \eqno (58.)$$
must be exactly divisible by the quadratic equation~(66.),
because all the roots of the latter are also roots of the former;
and therefore we must have
$${\rm B} = b - a^2,\quad
  {\rm D} = d - a^2 b,\quad
  {\rm E} = - ad,
   \eqno (80.)$$
and
$$c = ab.
   \eqno (81.)$$

This relation $c = ab$ reduces the expression
(76.)~$\ldots$~(79.) to the following,
$$\left. \eqalign{
  {\rm f} &= - a^3,\cr
  {\rm g} + {\rm h} &= - 2ab,\cr
  {\rm g} {\rm h} + {\rm i} &=  a^4 b + a^2 b^2 - 4 a^2 d,\cr
  {\rm g} {\rm i} &= a^5 d;\cr}
   \right\}
   \eqno (82.)$$
and thereby reduces the condition (75.), that is, the product of
the three conditions (66.) (67.) (68.), to the form
$$- a^9 - 3 a^7 b - a^5 b^2 + 5 a^5 d = 0,
   \eqno (83.)$$
which gives either
$$a = 0,
   \eqno (84.)$$
or else
$$a^4 + 3 a^2 b + b^2 = 5d,
   \eqno (85.)$$
and therefore, by (80.), either
$${\rm E} = 0,
   \eqno (60.)$$
or else
$$5 {\rm D} = {\rm B}^2.
   \eqno (61.)$$

Thus,when we set aside these two particular cases, we see by
(45.), that under the circumstances supposed in the enunciation
of the theorem, the function $f(x) - qx$ vanishes, for every
value of $x$ which makes the polynome
$x^5 + {\rm B} x^3 + {\rm D} x + {\rm E}$
vanish; and that therefore if we set aside the third and only
remaining case of exception, namely, the case in which the
proposed equation of the fifth degree has two equal roots, and in
which consequently the condition~(62.) is satisfied, the
function~$f(x)$ must be of the form (59.); which was the thing to
be proved.

\bigbreak

{\it Corollary}.---Setting aside the three excepted cases
(60.) (61.) (62.), the coefficients of the equation~(50.) of the
fifth degree in $y$ will be expressed as follows,
$${\rm B}' = {\rm Q}'^2 {\rm B},\quad
  {\rm D}' = {\rm Q}'^4 {\rm D},\quad
  {\rm E}' = {\rm Q}'^5 {\rm E};
   \eqno (86.)$$
and if we attempt to reduce it to De Moivre's solvible form, by
making
$${\rm D}' = {\textstyle {1 \over 5}} {\rm B}'^2,
   \eqno (50.)$$
we find
$${\rm Q}'^4 = 0,
   \eqno (51.)$$
that is,
$${\rm Q}' = 0,
   \eqno (52.)$$
so that the relation between $y$ and $x$ reduces itself to the
form
$$y = ( x^5 + {\rm B} x^3 + {\rm D} x + {\rm E})
            \mathbin{.} \phi(x),
   \eqno (87.)$$
which can give no assistance towards resolving the proposed
equation~(58.) of the fifth degree in $x$.

\nobreak\bigskip

\quad Observatory, Dublin, June 11, 1836.

\bye
