Two instruments with one reach

The mark that changes what is reachable ends with a list of things it does not develop, and the first is a whole instrument set: “the reach of conic-assisted construction”. It also names the arithmetic that governs it — the Pierpont primes — and leaves the two connected by a claim.

Here is the connection, and the surprise is that two unrelated permissions turn out to be the same permission.

What an instrument set is allowed to be #

The framing the marked ruler’s reach insists on is that an impossibility proof is relative to a stated set of operations, and that the classical pair is one set among several. Three sets are in play here.

Compass and straightedge. Each step meets a line with a line, a line with a circle, or two circles. The resulting coordinates satisfy equations of degree at most two over what is already built, so the reachable numbers form a tower of quadratic extensions and every one has degree a power of two.

Conics, drawn at will. Allow any conic through five given points, or with given foci and a given point, to be drawn and intersected with anything already present. Two conics meet in up to four points, so a step solves an equation of degree at most four — and degree four factors through degree two, so what the conics add is degree three.

An angle trisector. A single device that takes an angle and produces a third of it. That solves a cubic, by the identity $4\cos^3\theta - 3\cos\theta = \cos 3\theta$, and nothing more.

The claim is that the second and third reach exactly the same numbers. A conic looks like far more apparatus than a trisector, and it buys nothing extra.

Why the two agree #

The argument is in both directions and each is a construction.

A trisector is enough to draw any conic’s intersections. Two conics meet in the roots of a quartic, and a quartic is solved by a resolvent cubic followed by two square roots — which is Ferrari’s method. A trisector plus the classical pair solves any cubic, and square roots are classical, so every intersection of conics is reachable.

And conics are enough to trisect. Trisecting means solving $4y^3 - 3y = c$, and a cubic’s real roots are the intersections of a parabola with a circle: substitute $y$ for one coordinate and the cubic becomes a system of two conics. Menaechmus knew this in the fourth century BC, for doubling the cube, and Pappus’s classification of problems as plane, solid and linear is exactly the observation that the cubics and quartics form one class.

So the reach is the same, and the reason is not a coincidence of the two devices. Each buys the ability to solve a cubic and nothing else buys anything more, because degree four reduces to degree three and degree three is what both supply.

The cubic, from two conics #

The construction that turns a cubic into two conics is short enough to have in full, because it is what the whole equality rests on.

Take $y^3 + py + q = 0$, which any cubic reduces to by a substitution. Introduce a second unknown $z = y^2$ — that is a parabola. Then $y^3 = yz$, so the equation becomes $yz + py + q = 0$, which is a hyperbola in $y$ and $z$. The two conics meet where both hold, so their intersections are the cubic’s roots.

Both conics are drawable from the coefficients: the parabola $z = y^2$ is a fixed curve once the axes are chosen, and the hyperbola $yz + py + q = 0$ has its asymptotes at $y = 0$ and $z = -p$, which are lines the classical pair constructs. So one parabola and one hyperbola solve any cubic, and the parabola can even be reused for every cubic — which is Menaechmus’s observation and the reason the subject calls these problems solid.

The other direction, that a trisector solves any quartic and hence any pair of conics, is Ferrari’s method: a quartic’s resolvent is a cubic, the trisector handles it, and two square roots finish. Neither direction is deep and both are constructions rather than existence arguments, which is what makes the equality of the two reaches a fact rather than an abstraction.

What the reach is, as arithmetic #

The reachable numbers are those lying in a tower whose steps have degree two or three, so their degrees over the rationals are products of twos and threes. Which regular polygons that permits is a condition on $n$.

Constructing the regular $n$-gon means constructing $\cos(2\pi/n)$, whose degree is $\varphi(n)/2$ for $n$ above two. So the polygon is reachable exactly when $\varphi(n)/2$ is a product of twos and threes — which, unwound, says the odd part of $n$ is a product of distinct primes $p$ with $p - 1$ of the form $2^u 3^v$, and a power of three is permitted.

Those primes are the Pierpont primes, and the next section unwinds why. Gauss’s condition allows only the Fermat primes, where $p - 1$ is a power of two — 3, 5, 17, 257, 65537 — and no factor of three beyond the first.

The figures compute both conditions from $n$’s factorisation and check them against the degree independently, so the two routes to each verdict are compared rather than one being quoted.

Where the sets part, and by how much #

Up to twenty-four sides, compass and straightedge reach eleven polygons and a conic reaches seventeen. The six that separate them are the ones whose cosine has degree divisible by three: the heptagon, the nonagon, the thirteen-gon, the fourteen-gon, the eighteen-gon and the nineteen-gon.

The count is worth reading against the classical criterion’s own list. Which polygons can be drawn decides Gauss’s condition for every $n$ up to a bound, and the polygons it refuses are exactly the ones the conic reaches and the compass does not — so the two essays’ verdicts are complementary halves of one table.

And the sets are nested rather than merely different. Everything the classical pair reaches, a conic reaches too, because a power of two is a product of twos and threes. The figures check that containment for every polygon they draw, which is the check that the two conditions have not been implemented as unrelated tests.

The nesting is what makes Pappus’s classification a hierarchy rather than a list. Problems soluble by line and circle are plane; those needing a conic are solid; those needing something else are linear. And the insistence that essay records — “a problem should be solved by the least powerful method that suffices” — is a statement about that hierarchy, which needs the levels to be nested to make sense.

Two sets whose reaches coincide, twice #

Two instrument sets coinciding in reach has now happened twice, and the reasons are different.

The straightedge buys nothing shows that the compass alone reaches everything compass and straightedge reach — the Mohr–Mascheroni theorem. There the two sets have the same reach because one operation is simulable by the other: every straightedge step can be replaced by a finite sequence of compass steps, and the proof is a construction.

Here the two sets have the same reach because both extend the field by the same degrees. There is no simulation in either direction that is short or natural; the conics and the trisector are genuinely different operations, and the agreement is an equality of closures rather than of procedures.

So “the same reach” comes in two flavours. One set can contain another’s operations in disguise, or two sets can happen to generate the same closure by different routes. The first is a theorem about constructions and the second about field degrees, and only the second is available here.

What the marked ruler adds beyond this #

That essay reports Baragar’s result: a marked ruler’s steps reach degrees 2, 3, 5 and 6, so it goes beyond the conics — and the polygon that shows it is the hendecagon, which a quintic a sliding mark reaches is about.

That is worth stating here because it says the hierarchy does not stop at the conics. Three levels have now been named — quadratics, cubics and quartics, and whatever a sliding mark reaches — and the third is not characterised. A hierarchy with an uncharacterised level is the honest state of the subject, and that essay’s closing observation applies: the mark moves the boundary rather than removing it, and where it moves it to is not fully known.

Doubling the cube, in the solid class #

The three classical problems sort themselves into the hierarchy neatly, and it is worth doing because the sorting is what the classification was invented for.

Doubling the cube needs $\sqrt[3]{2}$, of degree three, so it is solid: out of reach of circles and inside the reach of conics. Menaechmus’s solution is two parabolas, and Nicomedes’ is the conchoid the marked-ruler essay names.

Trisecting an angle needs a root of $4y^3 - 3y = c$, degree three for most angles, so it is solid as well. That is why an angle trisector and a conic have the same reach: each solves exactly the problems of this class.

Squaring the circle needs $\pi$, which has no degree. So it is not in the hierarchy at all — not solid, not linear in Pappus’s sense of needing a further curve, but outside every set whose steps solve polynomial equations. The circle that will not square is the essay for that, and it needs an entirely different theorem.

Two of the three are in the same class and the third is nowhere in the classification. That is a sharper statement than the three classical problems are all impossible with compass and straightedge, and it is the statement the hierarchy exists to make. That essay puts it as the three never having been the same problem; the class structure is why.

What the arithmetic does not settle #

It says nothing about how many steps. The degree condition says a number is reachable and gives no bound on the construction’s length. The regular 65537-gon is classically constructible and the shortest known construction is enormous; the condition is an existence statement.

It says nothing about the drawing. A conic drawn at will is an idealisation exactly as a circle is, and an earlier essay makes the point that the marked ruler is worse in one specific way: it requires a search rather than a drawing. A conic through five points is determined and drawable in principle, which puts it closer to the classical instruments than the mark is.

And it does not cover $\pi$. The circle that will not square fails for a reason no instrument set of this kind touches: $\pi$ satisfies no polynomial equation at all, so it has no degree to be a product of anything. Every set considered here reaches only algebraic numbers, and the transcendental case is outside all of them.

Where the Pierpont primes come from #

The condition on $n$ was stated above and its arithmetic is worth unwinding, because the primes in it are defined by a condition nobody would guess from the geometry.

The degree of $\cos(2\pi/n)$ is $\varphi(n)/2$, and $\varphi$ is multiplicative over coprime factors: for $n = 2^a p_1^{e_1}\cdots$, the totient is $2^{a-1}\prod p_i^{e_i - 1}(p_i - 1)$. For that to be a product of twos and threes, two things must hold. No odd prime may appear squared, since $p^{e-1}$ would contribute $p$ itself. And each $p_i - 1$ must be a product of twos and threes.

A prime $p$ with $p - 1 = 2^u 3^v$ is a Pierpont prime, and the smallest are 2, 3, 5, 7, 13, 17, 19, 37, 73, 97. The exception is three, which may appear to any power in $n$ — because $\varphi(3^k) = 2\cdot 3^{k-1}$ is already a product of twos and threes.

So the condition is: $n$ is a power of two, times a power of three, times distinct Pierpont primes. Gauss’s condition is the same sentence with Fermat in place of Pierpont and no power of three — and a Fermat prime is a Pierpont prime with $v = 0$, so the nesting of the two conditions is visible in the definitions rather than needing an argument.

Whether there are infinitely many Pierpont primes is unknown, exactly as for the Fermat primes — where only five are known and the expectation is that there are no more. The Pierpont ones are believed to be infinite, which is the opposite expectation about a very similar condition, and the reason is that $2^u3^v$ takes many more values than $2^u$ does.

What the pictures cannot show #

The strip stops at twenty-four sides because a strip of a hundred is unreadable, and the interesting arithmetic continues — the next polygon separating the two sets after nineteen is twenty-one, and the pattern of which $n$ are Pierpont-reachable thins out slowly.

Nothing here draws a conic or a trisection. The verdicts are computed from factorisations, and the constructions the verdicts are about are the subject of an earlier essay; a figure of a parabola meeting a circle would illustrate one cubic and not the claim about all of them.

And the degree is computed as $\varphi(n)/2$ rather than by factoring a polynomial. That formula is a theorem about cyclotomic fields, the figures use it and check that the totient is even, and a reader wanting the degree established from the polynomial itself will find it in the hendecagon’s quintic, where one case is built and tested — the same method the tower law makes available.

Still open: what the third level is #

The conics’ level is completely characterised: a number is reachable exactly when its degree is a product of twos and threes, which is a clean statement with a clean proof.

The marked ruler’s level is not. Baragar showed every step reaches degree 2, 3, 5 or 6, which bounds what one step does and does not characterise what a sequence of them reaches — a tower of such steps has degree a product of those numbers, and whether every such degree is actually attained is open. So the reachable set is known to lie between two descriptions and is not pinned down.

The related question is whether the hierarchy is a hierarchy at all past the conics. Adding a device that solves every quintic gives a level above the marked ruler’s; adding one that solves every equation of degree at most $d$ gives a tower of levels; and whether the marked ruler’s reach is one of those or lies across several is exactly what is not known.

The least powerful method, as a rule #

Pappus’s insistence that a problem be solved by the least powerful method that suffices is usually read as aesthetic advice. Read against the hierarchy it is a mathematical instruction with content, and the content is worth stating.

To solve a problem by the least powerful sufficient method is to identify the degree of the number wanted and use an instrument set reaching exactly that degree. Using a conic where a circle suffices is not inelegant; it is a failure to notice that the degree was a power of two, which is a computation somebody skipped.

So the rule is a diagnostic rather than a preference. A solution using more than it needs is evidence that the solver did not know the degree — and the classical literature’s care about it is the care of people who understood that the class of a problem was the interesting fact about it, even before there was a theory of degrees to say what a class was.

The rule also explains why the impossibility proofs took until 1837. As long as the question was can this be done, the answer came from trying; once the question became what degree does this need, the answer came from arithmetic and the impossibility was a corollary. An earlier essay records that the folklore restriction to compass and straightedge hardened only when the impossibility proofs made the restricted question interesting, and that is the same observation from the other end.

What a degree was buying all along #

Three instrument sets, two of them with identical reach, and the whole account is arithmetic. That is the point that essay makes about one scratch on a ruler — “the answer is nearly always a change in the degree of some equation, and the geometry is downstream of it” — and this one is the same observation applied to two sets rather than one.

What an operation buys is a degree, and two operations buying the same degree are the same operation as far as reach is concerned. So the question to ask of a new instrument is not what it draws but what it solves, and the answer is a number.

The corollary is worth having as well. If two instrument sets have different reaches, some polygon or some length separates them, and the arithmetic says which — the six polygons in the figure are that separation made explicit. A hierarchy of instruments is a hierarchy of degrees with witnesses, and the witnesses are small enough to draw.