# Euler’s derivation of the quartic formula

Let us consider the quartic equation

${y}^{4}+p{y}^{2}+qy+r=0,$ | (1) |

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

$y:=u+v+w.$ | (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

$$4({v}^{2}{w}^{2}+{w}^{2}{u}^{2}+{u}^{2}{v}^{2})+{({u}^{2}+{v}^{2}+{w}^{2})}^{2}+p({u}^{2}+{v}^{2}+{w}^{2})+r\mathit{\hspace{1em}\hspace{1em}\hspace{0.25em}}$$ |

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

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

${u}^{2}+{v}^{2}+{w}^{2}=-{\displaystyle \frac{p}{2}},$ | (3) |

$uvw=-{\displaystyle \frac{q}{8}},$ | (4) |

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

${v}^{2}{w}^{2}+{w}^{2}{u}^{2}+{u}^{2}{v}^{2}={\displaystyle \frac{{p}^{2}-4r}{16}}.$ | (5) |

Squaring (4) gives

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

${z}^{3}+{\displaystyle \frac{p}{2}}{z}^{2}+{\displaystyle \frac{{p}^{2}-4r}{16}}z-{\displaystyle \frac{{q}^{2}}{64}}=0.$ | (7) |

Therefore we may write

$$u=\pm \sqrt{{z}_{1}},v=\pm \sqrt{{z}_{2}},w=\pm \sqrt{{z}_{3}}.$$ |

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

$${u}_{0},{v}_{0},{w}_{0};{u}_{0},-{v}_{0},-{w}_{0};-{u}_{0},{v}_{0},-{w}_{0};-{u}_{0},-{v}_{0},{w}_{0}.$$ |

Accordingly, we have the

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

$\{\begin{array}{cc}{y}_{1}={u}_{0}+{v}_{0}+{w}_{0},\hfill & \\ {y}_{2}={u}_{0}-{v}_{0}-{w}_{0},\hfill & \\ {y}_{3}=-{u}_{0}+{v}_{0}-{w}_{0},\hfill & \\ {y}_{4}=-{u}_{0}-{v}_{0}+{w}_{0},\hfill & \end{array}$ | (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^{}

$({y}_{1}-{y}_{2})({y}_{1}-{y}_{3})({y}_{1}-{y}_{4})({y}_{2}-{y}_{3})({y}_{2}-{y}_{4})({y}_{3}-{y}_{4})=$ | $-{2}^{6}({v}_{0}^{2}-{w}_{0}^{2})({w}_{0}^{2}-{u}_{0}^{2})({u}_{0}^{2}-{v}_{0}^{2})$ | ||

$=$ | $-64({z}_{2}-{z}_{3})({z}_{3}-{z}_{1})({z}_{1}-{z}_{2}),$ |

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 |