# 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

 $u_{0}v_{0}w_{0}=-\frac{q}{8}.\\$
 $\displaystyle(y_{1}-y_{2})(y_{1}-y_{3})(y_{1}-y_{4})(y_{2}-y_{3})(y_{2}-y_{4})% (y_{3}-y_{4})=$ $\displaystyle-2^{6}(v_{0}^{2}-w_{0}^{2})(w_{0}^{2}-u_{0}^{2})(u_{0}^{2}-v_{0}^% {2})$ $\displaystyle=$ $\displaystyle-64(z_{2}-z_{3})(z_{3}-z_{1})(z_{1}-z_{2}),$

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