Quintic Equations

To understand how to solve the general quintic equation, one must first reconsider what it means to “solve” an equation at all. Consider the familiar analytic formulation of an ordinary radical:

\displaystyle \sqrt[n]{a} =\text{exp} \big ( \frac{1}{n} \int_1^a \frac{dt}{t} \big ).

This formula says that an ordinary radical is produced in two stages. First, integrate the elementary differential dt/t. Second, apply the exponential function. The Hermite–Kronecker–Brioschi solution to the quintic has the same broad pattern, except that the logarithm is replaced by an elliptic integral and the exponential is replaced by a modular function.

For cubic and quartic equations, the passage to a simpler normal form is almost the whole solution. A cubic is shifted so that the quadratic term disappears, and Cardano then reduces the remaining problem to a quadratic auxiliary equation followed by radical extraction. A quartic is shifted so that the cubic term disappears, and Ferrari reduces the problem to a resolvent cubic and then to quadratics. The quintic behaves differently. It can also be simplified very drastically, and the simplification can be carried out by transformations whose auxiliary parameters are found through equations of degree at most three. But the final one-parameter equation \displaystyle w^5+w+a=0 is not, in general, solvable by ordinary radicals. The classical reduction of the quintic should therefore not be read as a solution by radicals. Its purpose is to isolate, as sharply as possible, the one genuinely new algebraic function which remains after every elementary simplification has been made.

We begin with a monic quintic

\displaystyle F(x)=x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0=0.

Let its roots be x_1,\ldots,x_5. The basic operation is a Tschirnhaus transformation of the roots. One does not simply substitute a new expression into the equation and hope that the degree falls. Instead, one chooses a polynomial, or sometimes a rational function, T(x), and defines new quantities y_i=T(x_i), i=1,\ldots,5. The five transformed quantities are roots of another monic quintic,

\displaystyle G(y)=\prod_{i=1}^5\bigl(y-T(x_i)\bigr).

Equivalently, one eliminates x from F(x)=0,\quad y-T(x)=0, and obtains

\displaystyle G(y)={\text{Res}}_x(F(x),y-T(x)).

The resultant is useful conceptually, but it is rarely the best way to calculate. The coefficients of G are symmetric functions of the five numbers T(x_i), so they can be found from Newton sums. This is what makes the reduction manageable.

Suppose more generally that five numbers u_1,u_2, \ldots,u_5 are roots of

\displaystyle  G(u)=u^5-e_1u^4+e_2u^3-e_3u^2+e_4u-e_5.

Write p_m=u_1^m+\cdots+u_5^m. Therefore p_1=0 makes the coefficient of u^4 vanish; if also p_2=0, then the coefficient of u^3 vanishes; and if p_3=0 as well, then the coefficient of u^2 vanishes. This is the organizing principle of the whole construction. p_1=p_2=0 produces a principal quintic,

\displaystyle u^5+ Au^2+Bu+C=0,

whereas p_1=p_2=p_3=0 produces a Bring–Jerrard quintic,

\displaystyle u^5+Pu+Q=0.

The reduction is therefore a controlled cancellation of the first few power sums of a transformed set of roots.

Tschirnhaus transformations

The first step is to use shifts to reduce to a depressed quintic. Put x=X-\frac{a_4}{5}. This translates every root by a_4/5, so that the new roots have sum zero. Indeed, since x_1+\cdots+x_5=-a_4, the transformed roots X_i=x_i+\frac{a_4}{5} satisfy X_1+\cdots+X_5=0. The equation becomes

\displaystyle X^5+AX^3+BX^2+CX+D=0.

We now use a quadratic transformation of the roots.

\displaystyle  y_i=X_i^2+\alpha X_i+\beta.

The reason for choosing a quadratic is transparent. We want to remove two more coefficients, namely the future y^4 and y^3 coefficients. These are controlled by two conditions, \sum_i y_i=0,~ \sum_i y_i^2=0, and the quadratic transformation has two parameters, \alpha and \beta. Let T_m=\sum_{i=1}^5y_i^m. If T_1=T_2=0, then the polynomial with roots y_1,\ldots,y_5 has the form y^5+cy^2+dy+e=0. The first condition determines \beta. We have

T_1=\sum_i(X_i^2+\alpha X_i+\beta)=S_2+\alpha S_1+5\beta.

Using S_1=0 and S_2=-2A, this becomes T_1=-2A+5\beta. Hence \beta=\frac{2A}{5}. This is simply a centering operation. The constant term in the quadratic transformation shifts all five transformed roots by the same amount, and it is chosen so that their sum vanishes. The second condition determines \alpha. Squaring the transformation gives

y_i^2=X_i^4+2\alpha X_i^3+(\alpha^2+2\beta)X_i^2 +2\alpha\beta X_i+\beta^2.

Therefore T_2=S_4+2\alpha S_3+(\alpha^2+2\beta)S_2 +2\alpha\beta S_1+5\beta^2. The S_1 term vanishes. Substituting the known values of S_2,S_3,S_4 and then substituting \beta=2A/5 gives

T_2=-2\left(A\alpha^2+3B\alpha+2C-\frac{3A^2}{5}\right).

Thus the required equation for \alpha is A\alpha^2+3B\alpha+2C-\frac{3A^2}{5}=0. If A\ne0, then

\displaystyle \alpha= \frac{-3B\pm\sqrt{9B^2-4A\left(2C-\frac{3A^2}{5}\right)}}{2A}.

If A=0 but B\ne0, the condition is linear. \alpha=-\frac{2C}{3B}. If A=B=0, then the depressed quintic was already X^5+CX+D=0, which is itself Bring–Jerrard form. In that case there is nothing further to remove. After choosing one root \alpha of the auxiliary quadratic and setting \beta=2A/5, the transformed roots satisfy a principal quintic. Its coefficients are obtained from the later Newton sums.

At this stage nothing has been solved. We have changed the five roots to a new five-tuple whose first two power sums vanish. But we have done so using only a quadratic auxiliary equation, and the general quintic has become the principal quintic y^5+cy^2+dy+e=0. That is the real algebraic centre of the classical theory. From here one may take two different routes. The Bring–Jerrard route makes the residual quintic difficulty as sparse as possible. The Brioschi route makes its eventual icosahedral and modular structure more visible.

Quartic transformation

Suppose for the moment that the principal quintic is written as y^5+cy^2+dy+e=0. Its roots satisfy R_1=R_2=0, R_3=-3c, R_4=-4d,R_5=-5e, where R_m=\sum_{i=1}^5y_i^m. The defining equation gives the recurrence R_{m+5}+cR_{m+2}+dR_{m+1}+eR_m=0, and hence R_6=3c^2, R_7=7cd, R_8=8ce+4d^2. We now want one more coefficient to disappear. The final transformed roots should have their first three power sums equal to zero. Z_1=Z_2=Z_3=0. A naive attempt would use a cubic transformation z_i=y_i^3+u y_i^2+v y_i+w. There are three parameters and three conditions, so at first this seems ideal. The first condition is simple. Z_1=-3c+5w, so one must take w=\frac{3c}{5}. The second condition then becomes Z_2=-\frac25(-3c^2+15cuv+10du^2+20dv+25eu). Thus, provided 3cu+4d\ne0, it determines v in terms of u; v= \frac{3c^2-10du^2-25eu}{5(3cu+4d)}. The problem appears only when this expression is substituted into the final condition Z_3=0. Generically, the denominator clears and one obtains a sextic equation for u. A general sextic is not solvable by radicals. Thus the cubic transformation has not really reduced the generic quintic; it has simply shifted the unsolvable difficulty into an auxiliary equation.

Bring’s decisive idea is to use a quartic transformation,

\displaystyle z_i=y_i^4+\lambda y_i^3+\mu y_i^2+\nu y_i+\rho.

There are now four parameters but still only three power-sum conditions. The spare parameter is not used to eliminate a fourth coefficient. Instead, it is used to arrange the equations so that they can be solved in the order. That is the whole point of the quartic transformation.

The first condition gives \rho immediately. Since R_1=R_2=0,

\displaystyle Z_1=R_4+\lambda R_3+5\rho=-4d-3\lambda c+5\rho.

Therefore \displaystyle  \rho=\frac{3\lambda c+4d}{5}. Now expand Z_2. Before substituting for \rho, the calculation gives

\displaystyle Z_2= 3\lambda^2c^2-10\lambda\mu e+14\lambda cd-6\lambda c\rho -8\lambda d\nu -4\mu^2d+6\mu c^2-6\mu c\nu \\ ~~~~~~~~~~~\qquad +8ce+4d^2-8d\rho-10e\nu+5\rho^2.

After substituting \rho=\frac{3\lambda c+4d}{5}, this becomes

\displaystyle  Z_2=-10\lambda\mu e-4\mu^2d+\frac65\lambda^2c^2 +\frac{46}{5}\lambda cd+6\mu c^2+\frac45d^2+8ce \\ ~~~~~~\qquad-2\nu(5e+4\lambda d+3\mu c).

Now Bring uses the spare parameter. Assume first that c\ne0 and choose \mu=-\frac{5e+4\lambda d}{3c}. This makes the entire coefficient of \nu vanish. Thus the equation Z_2=0 becomes an equation only in \lambda, and it is only quadratic:

\displaystyle (27c^4-160d^3+300cde)\lambda^2 +(27c^3d-400cd^2e+375ce^2)\lambda + 18c^2d^2-45c^3e-250de^2=0.

Choose one solution \lambda. Then \mu=-\frac{5e+4\lambda d}{3c}, \rho=\frac{3\lambda c+4d}{5} are known. Only \nu remains. The third condition Z_3=0 is a cubic equation in \nu. which is solved by Cardano method. Thus every parameter in the quartic transformation is found by radicals.

Once the transformation has been fixed, the roots z_i have Z_1=Z_2=Z_3=0. Hence they satisfy a Bring–Jerrard equation z^5+Pz+Q=0.

The exceptional case c=0 is actually simpler: the principal quintic was already y^5+dy+e=0, so it was already in Bring–Jerrard form. Other vanishing denominators indicate that the chosen formulas have left their generic coordinate chart; they do not mean that the reduction theorem has failed.

A Tschirnhaus transformation need not be one-to-one. The final step is always to substitute the candidates solution obtained in the new form back into the original equation and compute the actual roots.

Bring Radical

Suppose now that we have reached z^5+Pz+Q=0. If P\ne0, choose a fourth root \sigma^4=P, and put z=\sigma w. Since P\sigma=\sigma^5, the equation becomes

\displaystyle w^5+w+a=0, \quad a=\frac{Q}{\sigma^5}.

Thus every nondegenerate Bring–Jerrard quintic is, over \mathbb C, equivalent to the normalized Bring equation w^5+w+a=0 .

For the equation w^5+w+a=0, define the principal Bring branch \text{Br}(a)  near a=0 by \text{Br}(a)^5+\text{Br}(a)+a=0, \quad \text{Br}(0)=0. We have a local power series definign the branch

\displaystyle  {\text{Br}}(a)= \sum_{m=0}^{\infty} \frac{(-1)^{m+1}}{4m+1} \binom{5m}{m}a^{4m+1}.

Hence

\displaystyle {\text{Br}}(a) =-a+a^5-5a^9+35a^{13}-285a^{17}+2530a^{21}-\cdots.

The Bring radical is not transcendental in the strict algebraic sense: it satisfies a polynomial equation of degree five over \mathbb C(a). What fails is radical expressibility. Generically, no finite tower of additions, multiplications, divisions, and ordinary root extractions can express this branch.

Once a single root w=r of w^5+w+a=0 has been found, the remaining four roots satisfy a quartic equation. Thus the true difficulty is the extraction of one Bring root. The other four roots are then accessible by Ferrari’s method.

The Brioschi normal form

The principal quintic also leads to another one-parameter normal form,

\displaystyle u^5-10\kappa u^3+45\kappa^2u-\kappa^2=0.

This is the Brioschi quintic. It is not obtained from the Bring equation by a mere scaling. It is a different normal form, related to the principal quintic by a rational Tschirnhaus transformation. Its importance is that it is especially well suited to the elliptic-modular and icosahedral solution of the quintic.

The discriminant of the Brioschi polynomial is exceptionally suggestive:

\displaystyle 3125\kappa^8(1728\kappa-1)^2.

The special values \kappa=0,\quad \kappa=\frac1{1728},\quad \kappa=\infty already point toward the three distinguished elliptic values \displaystyle j=\infty,\quad j=0,\quad j=1728.

After a change of variables Brioschi form becomes the modularly normalized quintic

\displaystyle v^5+10v^3+45v-\sqrt{j-1728}=0.

This is the form in which the elliptic parameter is visible in the coefficient itself.

Modular forms

At this last stage one does not need to know the detailed theory of modular forms in order to understand the shape of the answer. One introduces certain special functions of a complex variable \tau, called elliptic or modular functions. They are analogous, in spirit, to trigonometric functions: they are highly structured special functions, but they are adapted not to ordinary angles and circles, but to elliptic curves and their periods. One such function is the classical j -function. Given the invariant j attached to the quintic, one chooses a value of \tau for which j(\tau)=j.

There are then further special functions of \tau associated with the number 5. These level-five modular functions have exactly the symmetry needed for the general quintic: their transformations reflect the same icosahedral symmetry group A_5 which remains after the elementary reductions. Evaluating one of these functions at \tau gives an auxiliary quantity from which the five roots of the original quintic can be recovered by ordinary algebraic formulas. Thus the modular functions do not replace the roots by something unrelated; rather, they supply the one new special-function value which ordinary radicals cannot provide. Hermite’s classical construction carries this out most directly for the Bring–Jerrard equation, using special elliptic theta functions at five related arguments. Brioschi’s form packages the same final information more economically. In the Brioschi equation

\displaystyle u^5-10\kappa u^3+45\kappa^2u-\kappa^2=0,

the parameter \kappa is naturally related to the elliptic invariant by j=1728-\frac1\kappa. Thus, after the original quintic has been reduced to Brioschi form, one may think of the remaining task as follows: determine the appropriate elliptic parameter \tau from the number j, evaluate the relevant level-five special function at \tau, and then recover the roots by algebraic operations.

The algebraic reduction of the quintic is a masterclass in isolating complexity. By systematically removing coefficients through Tschirnhaus transformations, Bring’s quartic construction, and Brioschi’s rational normal form, one does not solve the quintic at once; one concentrates its irreducible difficulty into a single, highly structured residual problem. The failure of ordinary radicals is therefore not a defeat of algebra, but a diagnosis of the precise point at which a richer kind of function is required. In this sense, the quintic is not an exception to solvability, but the first great example of a broader principle: when elementary algebraic operations reach their limit, the geometry and symmetry of the equation indicate the special functions needed to continue.

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