The quintics that have a formula

The group that will not come apart proved that there is no formula for the general equation of degree five. The argument was about the symmetries of five roots: a formula built from radicals forces a chain of subgroups whose steps are commutative, and the group of all 120 permutations of five things cannot be climbed down that way, because the 60 even permutations inside it have no normal subgroup at all.

That theorem is about the general quintic, the one whose coefficients are letters. It says nothing against particular ones, and some particular ones are plainly solvable. The roots of $x^5 - 2$ are $\sqrt[5]{2}$ times the five fifth roots of unity, and the fifth roots of unity are themselves radicals — $\cos 72^\circ = (\sqrt5 - 1)/4$. So the honest form of the question is not “is there a formula for the quintic” but which quintics have one. The answer turns out to be as clean as the impossibility. A quintic with rational coefficients that cannot be factored is solvable by radicals exactly when the symmetries of its roots fit inside a single group of twenty elements, and that can be decided from the coefficients in finitely many exact steps.

The five shapes a quintic’s symmetry can take #

Start with what can happen at all. Label the five roots of an irreducible quintic $r_0, \dots, r_4$, and consider the Galois group: the permutations of the roots that respect every polynomial relation among them with rational coefficients. The lattice that runs the other way built this group and showed that its subgroups mirror the fields between the rationals and the field of all five roots.

Irreducibility constrains the group strongly. If some root could not be carried to some other root by a symmetry, the roots reachable from it would form a smaller set closed under every symmetry, and the polynomial whose roots are that set would have rational coefficients — a factor. So an irreducible quintic’s group is transitive: it can carry any root to any other. A transitive group on five points has a size divisible by five, by the count of the blocks a subgroup cuts out: the five roots are the orbit, and the size of an orbit divides the size of the group.

Up to relabelling, exactly five groups of permutations of five things are transitive. They are listed below with the shapes of their elements, written as the lengths of the cycles each permutation breaks into.

The smallest is the five rotations of a pentagon, $C_5$. Add the five reflections and the group is the ten symmetries of a pentagon, $D_5$. Add ten more permutations, each fixing one corner and cycling the other four, and the group has twenty elements; it has a name, $F_{20}$, and a description that is the key to everything below. The remaining two are the sixty even permutations, $A_5$, and all one hundred and twenty, $S_5$.

The table is worth reading column by column, because it already contains the theorem. The three-cycle column is empty for the first three groups and full for the last two. A group of permutations of five things that contains a three-cycle and is transitive must contain every even permutation — which is the fact the group that will not come apart used to show that $A_5$ is simple. So the five groups split cleanly into two families, and the dividing line is whether the roots are ever permuted in a cycle of three.

Twenty maps that are arithmetic #

Number the roots $0$ to $4$ and the twenty permutations in the third group are exactly the maps

$$x \;\mapsto\; a x + b \pmod 5, \qquad a \in \{1, 2, 3, 4\},\ b \in \{0, 1, 2, 3, 4\}.$$

The opening figure draws all of them. With slope $a = 1$ the map is a shift: the identity when $b = 0$, a turn of the pentagon otherwise. With slope $a = 4$, which is $-1$ modulo 5, the map is a reflection: it fixes the one corner solving $x = -x + b$ and swaps the other four in two pairs. With slope $2$ or $3$ the map fixes one corner and runs the other four round a four-cycle. That accounts for the table’s row: one identity, four five-cycles, five double swaps and ten four-cycles.

Composing two such maps gives a third, by arithmetic rather than by inspection: following $x \mapsto a'x + b'$ by $x \mapsto ax + b$ gives $x \mapsto (aa')x + (ab' + b)$. The slopes multiply and the shifts combine. That single line is the whole reason the group can be climbed down. The shifts, $a = 1$, form a subgroup of five, the rotations; it is normal, because conjugating a shift by any affine map gives another shift; and dividing it out leaves the slopes alone, which multiply among the four nonzero numbers modulo 5 — a cyclic group of four, since $2$ has powers that run through every nonzero remainder. Two steps, of sizes four and five, both commutative. The chain is

$$F_{20} \;\supset\; C_5 \;\supset\; \{1\},$$

and $D_5$ and $C_5$ inherit their chains from it by intersection.

The affine maps are not only a convenient group; for $x^5 - 2$ they are literally the symmetries. Write the roots as $\zeta^k \alpha$, where $\alpha = \sqrt[5]{2}$ and $\zeta = e^{2\pi i/5}$, and label the root $\zeta^k\alpha$ by $k$. A symmetry must send $\alpha$ to some root $\zeta^b\alpha$ and $\zeta$ to some primitive fifth root of unity $\zeta^a$, and then it sends $\zeta^k\alpha$ to $\zeta^{ak+b}\alpha$ — the root labelled $k$ goes to the root labelled $ak + b$. Twenty choices of $(a, b)$, twenty symmetries, and the radical tower that solves the equation is the chain read in the other direction: first adjoin $\zeta$, which is a degree-four step whose symmetries are the slopes, then $\alpha$, a degree-five step whose symmetries are the shifts.

Why a group inside twenty is enough, and a larger one is not #

The theorem has two halves, and the pieces for both are now in hand.

If the group of an irreducible quintic fits inside $F_{20}$ — after relabelling the roots, which is always allowed — it is $C_5$, $D_5$ or $F_{20}$ itself, and it has a chain with commutative steps. The lattice that runs the other way turns that chain into a tower of fields, each step an extension with a commutative group of symmetries. Once enough roots of unity are adjoined — the fourth and fifth roots suffice here, and both are themselves radicals — every such step is reached by adjoining a single root of the form $\sqrt[n]{c}$. That is the classical half of Galois’s theorem, and it is what makes “fits inside twenty” sufficient.

If the group does not fit inside $F_{20}$, it is $A_5$ or $S_5$, and both contain $A_5$, which has no chain at all. No tower of radicals reaches the roots, by the argument of the essay before this one.

So the question “which quintics can be solved” is the question “which quintics have a group inside the twenty affine maps”. Put that way, it is a question about a finite object attached to the polynomial, and the remaining work is to read that object off the coefficients without knowing the roots exactly. There is a natural way to do it, and it begins with a picture of what the twenty maps leave alone.

Six pictures, and the twenty maps that keep one of them #

Any group of permutations is the set of symmetries of something; the problem is to find a something that can be computed. For $F_{20}$ the something is a drawing.

Put the five roots at the corners of a regular pentagon and consider the ten pairs of roots. Five of the pairs are the sides of a closed loop through all five corners; the other five pairs then form the five-pointed star inscribed in it — and a five-pointed star is itself a closed loop through all five corners, visiting them in a different order. So the ten pairs of five labelled points split into a loop and a star, and since there are twelve loops through five labelled points, there are six such splittings.

The count of twenty is checked in the figure by trying all 120 permutations, but it can be seen directly. A shift turns the pentagon and keeps both the loop and the star. A reflection flips it and keeps both. A slope of $2$ sends consecutive labels $k, k+1$ to labels $2k, 2k+2$, which differ by two — it turns the loop into the star, and the star, whose pairs differ by two, into pairs differing by four, which is the loop again. So the affine maps keep the splitting as a whole while sometimes exchanging its two halves. And since $120 / 6 = 20$, nothing else can keep it: the six splittings are permuted transitively, so each is kept by a group of exactly twenty.

That is the object the test is built on. A quintic’s group fits inside a conjugate of $F_{20}$ exactly when it leaves one of the six splittings alone.

A number that only the twenty can leave alone #

Turn the picture into a number. For one splitting, add up the products of the roots joined along the loop and subtract the products along the star:

$$t = (r_0 r_1 + r_1 r_2 + r_2 r_3 + r_3 r_4 + r_4 r_0) - (r_0 r_2 + r_2 r_4 + r_4 r_1 + r_1 r_3 + r_3 r_0).$$

A shift or a reflection leaves both sums as they are, so it leaves $t$ alone. A slope of $2$ exchanges the sums and so changes the sign of $t$. The square $\theta = t^2$ is therefore unchanged by all twenty maps, and generically by nothing else. Doing the same for each of the six splittings gives six numbers $\theta_1, \dots, \theta_6$, and every permutation of the roots permutes these six among themselves.

That last fact is the engine of the method. Any symmetric function of $\theta_1, \dots, \theta_6$ is unchanged by every permutation of the roots, so — by the theorem what the coefficients already know proved — it is a polynomial in the quintic’s coefficients. The polynomial with roots $\theta_1, \dots, \theta_6$, a sextic resolvent, therefore has rational coefficients, and for a monic quintic with whole-number coefficients they are whole numbers.

Now suppose the quintic’s group fits inside the stabiliser of one splitting. Then the corresponding $\theta$ is left alone by every symmetry of the roots, which by the correspondence means it lies in the field fixed by the whole group: it is rational. Conversely, if one of the six values is rational and differs from the other five, the group can only permute the splittings in a way that keeps that one, and so fits inside its stabiliser. The test, then, is whether the sextic has a rational root — and a rational root of a monic polynomial with whole-number coefficients is a whole number, so the test is finite. It is the same move three ways to pair four roots made for the quartic, where three pairings gave a cubic whose rational roots decided the group, carried up one degree.

Three quintics put to the test #

The figure carries out the computation for three quintics. In each case the five roots are found numerically, the six $\theta$’s formed from them, and the sextic built by multiplying out $(y - \theta_1)\cdots(y - \theta_6)$; its coefficients come out within rounding of whole numbers, and are rounded. Then — and this is the part that makes the verdict a fact rather than an estimate — each value near a whole number is substituted into the rounded sextic in exact integer arithmetic. Either the result is exactly zero or it is not.

For $x^5 - 5x + 12$ the sextic is

$$y^6 + 200y^5 + 22000y^4 + 1120000y^3 + 28000000y^2 - 66016000000y + 1600000000,$$

and $y = 100$ is a root: the sextic factors as $(y - 100)$ times a quintic. So $x^5 - 5x + 12$ is solvable by radicals, although nothing in its appearance suggests it — it looks exactly like $x^5 - x - 1$ with different numbers. For $x^5 - 2$ the sextic is $y(y^5 - 51{,}200{,}000)$, with the root $0$, which says that for this polynomial the loop sum and the star sum are equal; that is what one would expect, since every product of two roots is a power of $\alpha$ times a root of unity, and the two sums are the same multiple of the sum of all the fifth roots of unity. For $x^5 - x - 1$ the sextic is $y^6 + 40y^5 + 880y^4 + 8960y^3 + 44800y^2 - 3091456y + 102400$, and no whole number is a root. Its group is not inside twenty, so it is $A_5$ or $S_5$, and either way no formula in radicals gives its roots.

There is one hypothesis to keep in sight. The converse direction of the test needs the six $\theta$’s to be different numbers, since a repeated value could be rational without being fixed by the group for the right reason. When two coincide there is a standard repair: transform the quintic first by a substitution that preserves its group and separates the values. Every computation in this essay checks that the six values are distinct, and for every polynomial drawn here they are.

The discriminant finishes the sorting #

The sextic separates the solvable three groups from the unsolvable two. A second exact test separates within each family, and it is older: the discriminant, the product of the squared differences of the roots. Its square root, $\prod_{i<j}(r_i - r_j)$, is changed in sign by any swap of two roots and so by any odd permutation, and kept by any even one. So the discriminant is a perfect square of a rational exactly when the group contains only even permutations.

The reflections of a pentagon are double swaps, which are even, so $C_5$ and $D_5$ lie inside $A_5$ and give square discriminants; $F_{20}$ contains the four-cycles, which are odd, and gives a discriminant that is not a square. The same test separates $A_5$ from $S_5$ on the unsolvable side. In the table, $x^5 + 20x + 16$ has discriminant $1{,}024{,}000{,}000 = 32{,}000^2$ and no whole root of its sextic, so its group is exactly the sixty even permutations: unsolvable, but with a symmetry group only half as large as the generic one. And $x^5 + 15x + 12$, whose discriminant $259{,}200{,}000$ is not a square and whose sextic has the root $180$, has the full group of twenty.

The last line of the sorting needs something else, because the five rotations and the ten symmetries of the pentagon pass both tests identically. One way is to count real roots. The rotations fix no root, so complex conjugation — which is always a symmetry, and swaps the non-real roots in pairs — must be the identity in a group of rotations; a quintic with group $C_5$ has all five roots real. $x^5 - 5x + 12$ has one real root and four complex ones, so its group is the ten-element $D_5$. The polynomial $x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1$, whose roots are the numbers $2\cos(2\pi k/11)$, has all five roots real and the group $C_5$ — it is the quintic that the construction of polygons meets at the eleven-sided polygon, where the group is commutative and the radicals are as simple as radicals can be.

The primes see the same group #

There is an independent check that uses no roots at all, and it turns the abstract group into something observable. Reduce a quintic modulo a prime $p$ and factor it over the integers modulo $p$. How a polynomial breaks modulo the primes explained what the pattern of factor degrees means: for all but finitely many primes, the pattern is the cycle shape of a symmetry of the roots — the one the prime singles out, Frobenius’s element — and Chebotarev’s theorem says each shape occurs for a share of primes equal to its share of the group.

The agreement is good to a percentage point or two in every cell, and the zeros are exact. That gives a practical and rather striking test in one direction. If a quintic factors modulo some prime that does not divide its discriminant as a linear factor times a linear factor times a cubic, or as a quadratic times a cubic, it is not solvable by radicals — because the shape $1 + 1 + 3$ is a three-cycle and $2 + 3$ is a swap times a three-cycle, and the solvable groups contain neither. $x^5 - x - 1$ modulo 7 is such a case. A single factorisation modulo a small prime can prove that no formula exists. The other direction is only statistical: a quintic that has shown no forbidden pattern in a thousand primes is probably solvable, and the sextic is what makes “probably” certain.

A census of the trinomials #

Every example so far has had a short form, $x^5 + ax + b$, and that family can be examined whole. Every reduced quintic can be brought to the form $x^5 + px^3 + qx^2 + rx + s$, and a classical sequence of substitutions due to Bring and Jerrard reduces it further to $x^5 + ax + b$ — though the coefficients of the new polynomial lie in a larger field, so solvability over the rationals is not simply transferred. The question for the trinomials themselves is concrete: among all $x^5 + ax + b$ with whole-number coefficients in a box, how many are solvable?

The answer the figure gives is extreme. Of the 6,561 polynomials in the square, 337 are reducible — the faint lines are the families with a root $r$, and the row $b = 0$ — and of the 6,224 irreducible ones, 6,140 have all 120 permutations as symmetries. Two have only the even ones, $x^5 + 20x \pm 16$. The whole column $a = 0$, the pure quintics $x^5 + b$, has the group of twenty, as $x^5 - 2$ did. And off that column exactly six trinomials are solvable: $x^5 - 5x \pm 12$, $x^5 + 15x \pm 12$ and $x^5 + 20x \pm 32$. Four of the six have the ten-element group of the pentagon, and the two with $15$ have the full twenty; none has the five rotations alone, since every one of them has a single real root.

Their rarity is not an accident of the box. B. K. Spearman and K. S. Williams showed in 1994 that the solvable irreducible trinomials $x^5 + ax + b$ with rational coefficients are exactly those given by one explicit family of rational expressions in two parameters, so in the plane of coefficients they lie along a thin set of curves rather than filling any region, and only scattered points of those curves have whole-number coordinates. The figure is the same fact seen from below: growing the box adds solvable trinomials far more slowly than it adds polynomials.

What a census of primes cannot certify, and what a verdict does not supply #

Three distinctions are easy to lose in pictures as clean as these.

The census figure is evidence, not proof. Agreement between factor patterns and cycle shapes over five hundred primes is what Chebotarev’s theorem predicts, and a missing pattern after five hundred primes is very strong evidence of a missing shape — but only the sextic and the discriminant, both exact, decide the group. The census is drawn as a check on the exact tests, never as a substitute for them.

The verdict “solvable” says that a formula exists, not that it is short, and the pictures do not show one. The radical expression for a root of $x^5 - 5x + 12$ is a sum of four fifth roots of numbers that themselves involve square roots, and it fills several lines; the group says how many radicals of each kind are needed and in what order, but writing them out is a separate and laborious computation that this essay does not perform.

And the trinomial census is a box. The six solvable trinomials it finds are all of them in that box, checked exhaustively, but the box says nothing about larger coefficients except through the theorem that describes the family — and that theorem, not the figure, is what says the six are typical of how rare such polynomials are.

Still open: how rare the smaller groups are #

Most polynomials have the largest possible group. Bartel van der Waerden conjectured in 1936 a precise form of this: among monic polynomials of degree $n$ with whole-number coefficients up to $H$ in size, the number whose group is not the full symmetric group is of order at most $H^{n-1}$, against roughly $(2H)^n$ polynomials in all — the same order as the reducible ones, which come for free along lines like the faint ones in the census. Manjul Bhargava proved the conjecture in 2021 after a long series of partial results.

What it does not give is the count for each smaller group separately. How many quintics with coefficients up to $H$ have the twenty-element group, or the ten, or the five, is a sharper question about thinner sets, and the exponents and constants for each are not settled in general; the counts for particular groups are known in some families, and the solvable trinomials are one such family, completely described by Spearman and Williams’ parametrisation. In the other direction, building polynomials with a given small group is easy for these five groups and hard in general: that is the inverse question the quartic essay ended on, and for most finite groups it is open.

What the group was deciding all along #

The general quintic has no formula, and the reason is one finite group. The particular quintics that do have a formula have one for exactly the opposite reason: their roots are permuted only by maps of the form $x \mapsto ax + b$, arithmetic on five numbers, and that arithmetic can be undone one commutative step at a time. A solvable quintic and an unsolvable one can look identical on the page — $x^5 - 5x + 12$ and $x^5 - 4x + 12$ differ by a single unit — and what separates them is whether one of six numbers built from their roots is a whole number.

That is the Galois correspondence doing exactly what it was built to do. A question about formulas, which seems to ask about the endless variety of expressions one could write down, is turned into a question about a group, which has finitely many possibilities and can be settled by computation. Five shapes, two families, one test for each: and of all the trinomials anyone is likely to write down, almost none fall on the side where the formula exists.