Euler’s derivation of the quartic formula

Let us consider the quartic equation

\displaystyle y^{4}+py^{2}+qy+r=0,

(1)

where $p,\,q,\,r$ are arbitrary known complex numbers. We substitute in the equation

\displaystyle y:=u+v+w.

(2)

We get firstly
$y^{2}=(u^{2}+v^{2}+w^{2})+2(vw+wu+uv),$
$y^{4}=(u^{2}+v^{2}+w^{2})^{2}+4(u^{2}+v^{2}+w^{2})(vw+wu+uv)+4(v^{2}w^{2}+w^{2% }u^{2}+u^{2}v^{2})+8uvw(u+v+w).$

Thus (1) attains the form

4(v^{2}w^{2}+w^{2}u^{2}+u^{2}v^{2})+(u^{2}+v^{2}+w^{2})^{2}+p(u^{2}+v^{2}+w^{2% })+r\qquad\;

+(vw+wu+uv)[4(u^{2}+v^{2}+w^{2})+2p]+(u+v+w)[8uvw+q]=0.

When $u,\,v,\,w$ are determined so that

\displaystyle u^{2}+v^{2}+w^{2}=-\frac{p}{2},

(3)

\displaystyle uvw=-\frac{q}{8},

(4)

the expressions in the brackets vanish and our equation shrinks to the form

\displaystyle v^{2}w^{2}+w^{2}u^{2}+u^{2}v^{2}=\frac{p^{2}-4r}{16}.

(5)

Squaring (4) gives

\displaystyle u^{2}v^{2}w^{2}=\frac{q^{2}}{64}.

(6)

The left hand sides of (3), (5) and (6) are the elementary symmetric polynomials of $u^{2}$ , $v^{2}$ , $w^{2}$ , whence these three squares are the roots $z_{1}$ , $z_{2}$ , $z_{3}$ of the so-called cubic resolvent equation

\displaystyle z^{3}+\frac{p}{2}z^{2}+\frac{p^{2}-4r}{16}z-\frac{q^{2}}{64}=0.

(7)

Therefore we may write

u=\pm\sqrt{z_{1}},\quad v=\pm\sqrt{z_{2}},\quad w=\pm\sqrt{z_{3}}.

All 8 sign combinations of those square roots satisfy the equations (3), (5), (6). In order to satisfy also (4) the signs must be chosen suitably. If $u_{0},\,v_{0},\,w_{0}$ is some suitable combination of the values of the square roots, then all possible combinations are

u_{0},\,v_{0},\,w_{0};\quad u_{0},\,-v_{0},\,-w_{0};\quad-u_{0},\,v_{0},\,-w_{% 0};\quad-u_{0},\,-v_{0},\,w_{0}.

Accordingly, we have the

Theorem (Euler 1739). The roots of the equation (1) are

\displaystyle\begin{cases}y_{1}=\;\;u_{0}+v_{0}+w_{0},\\ y_{2}=\;\;u_{0}-v_{0}-w_{0},\\ y_{3}=-u_{0}+v_{0}-w_{0},\\ y_{4}=-u_{0}-v_{0}+w_{0},\end{cases}

(8)

where $u_{0},\,v_{0},\,w_{0}$ are square roots of the roots of the cubic resolvent (7). The signs of the square roots must be chosen such that

u_{0}v_{0}w_{0}=-\frac{q}{8}.\\

The equations (8) imply an important formula

	$\displaystyle(y_{1}-y_{2})(y_{1}-y_{3})(y_{1}-y_{4})(y_{2}-y_{3})(y_{2}-y_{4})% (y_{3}-y_{4})=$	$\displaystyle-2^{6}(v_{0}^{2}-w_{0}^{2})(w_{0}^{2}-u_{0}^{2})(u_{0}^{2}-v_{0}^% {2})$
	$\displaystyle=$	$\displaystyle-64(z_{2}-z_{3})(z_{3}-z_{1})(z_{1}-z_{2}),$

which yields the

Corollary. A quartic equation has a multiple root always and only when its cubic resolvent has such one.

References

1 Ernst Lindelöf: Johdatus korkeampaan analyysiin. Fourth edition. Werner Söderström Osakeyhtiö, Porvoo ja Helsinki (1956).
2 K. Väisälä: Lukuteorian ja korkeamman algebran alkeet. Tiedekirjasto No. 17. Kustannusosakeyhtiö Otava, Helsinki (1950).

Title	Euler’s derivation of the quartic formula
Canonical name	EulersDerivationOfTheQuarticFormula
Date of creation	2013-03-22 17:51:58
Last modified on	2013-03-22 17:51:58
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	10
Author	pahio (2872)
Entry type	Theorem
Classification	msc 12D10
Synonym	quartic formula by Euler
Related topic	TchirnhausTransformations
Related topic	CasusIrreducibilis
Related topic	ZeroRuleOfProduct
Related topic	ErnstLindelof
Related topic	KalleVaisala
Related topic	BiquadraticEquation2
Related topic	SymmetricQuarticEquation