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[parent] Euler's derivation of the quartic formula (Theorem)

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
$y^2 = (u^2+v^2+w^2)+2(vw+wu+uv),$
$y^4 = (u^2+v^2+w^2)^2+4(u^2+v^2+w^2)(vw+wu+uv)+4(v^2w^2+w^2u^2+u^2v^2)+8uvw(u+v+w).$

Thus (1) attains the form $$4(v^2w^2+w^2u^2+u^2v^2)+(u^2+v^2+w^2)^2+p(u^2+v^2+w^2)+r \qquad\;$$ $$+(vw+wu+uv)[4(u^2+v^2+w^2)+2p]+(u+v+w)[8uvw+q] = 0.$$ 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
$\displaystyle v^2w^2+w^2u^2+u^2v^2 = \frac{p^2-4r}{16}.$ (5)

Squaring (4) gives
$\displaystyle u^2v^2w^2 = \frac{q^2}{64}.$ (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

\begin{align*}\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}\end{align*} (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 $$u_0v_0w_0 = -\frac{q}{8}.\\$$

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.

Bibliography

1
ERNST LINDELÖF: Johdatus korkeampaan analyysiin. Fourth edition. Werner Söderström Osakeyhtiö, Porvoo ja Helsinki (1956).
2
K. V¨AISÄLÄ: Lukuteorian ja korkeamman algebran alkeet. Tiedekirjasto No. 17.    Kustannusosakeyhtiö Otava, Helsinki (1950).




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See Also: Tschirnhaus transformations, casus irreducibilis, zero rule of product, Ernst Lindelöf, Kalle Väisälä, biquadratic equation, symmetric quartic equation

Other names:  quartic formula by Euler

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Cross-references: multiple root, formula, imply, Euler, theorem, combination, order, square roots, combinations, cubic resolvent, roots, squares, elementary symmetric polynomials, left hand sides, vanish, expressions, equation, complex numbers, quartic equation

This is version 7 of Euler's derivation of the quartic formula, born on 2008-02-27, modified 2008-03-03.
Object id is 10344, canonical name is EulersDerivationOfTheQuarticFormula.
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AMS MSC12D10 (Field theory and polynomials :: Real and complex fields :: Polynomials: location of zeros )

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