^{n}+ a

_{n-1}z

^{n-1}+ ... + a

_{2}z

^{2}+ a

_{1}z + a

_{0}= 0, where the leading coefficient is a

_{n}=1, and the other coefficients a

_{k}are ordinary integers (sometimes called rational integers to distinguish from the more general notion). This implies that the minimal polynomial of z also has that form.

For example, φ = (1+√5)/2 is an algebraic integer because φ

^{2}- φ - 1 = 0, and 6cos(π/7) is an algebraic integer because it satisfies the equation x

^{3}- 3x

^{2}- 18x + 27 = 0, but 1/√2 is not an algebraic integer because its minimal polynomial is either 2x

^{2}- 1 (which doesn't have leading coefficient 1) or x

^{2}- 1/2 (which doesn't have integer coefficients).

It can be proven that the sum, difference, or product of any two algints is another algint. (That is, the set of algebraic integers is a ring.) Also, if a complex number z satisfies a polynomial equation as above, where the leading coefficient is a

_{n}=1, and the other coefficients a

_{k}are algints, then z is an algint (generally of much higher degree than n).

Here Wendy suggested that polytopes of certain types have algints as chord lengths. I would like to know which types.

I was able to prove that, for any (finite) Wythoffian n-polytope, if the edge lengths are algints, then the chord lengths are algints. Here's the proof.

The polytope's vertices are generated by reflections from a single vertex v. A reflection sends v to an adjacent vertex v - (2v•u)u, where u is the unit vector perpendicular to the mirror; so the length of the edge connecting these two vertices is just 2v•u.

Any vertex has the form v + sum

_{1≤k≤n}a

_{k}u

_{k}, involving a combination of the mirror vectors u

_{1}, u

_{2}, ... , u

_{n}, where the coefficients a

_{k}are algints. We prove this by induction: Suppose some vertex v' has that form. Then any adjacent vertex v'' is a reflection of v', say, across the j'th mirror. (Some would say "reflection in the mirror", which suggests a mere visual effect; I prefer "reflection across the mirror", which suggests an actual transformation, moving the point from one side to the other.)

v'' = v' - (2v'•u

_{j})u

_{j}

= (v + sum

_{1≤k≤n}a

_{k}u

_{k}) - (2(v + sum

_{1≤k≤n}a

_{k}u

_{k})•u

_{j})u

_{j}

= v + sum

_{1≤k≤n}a

_{k}u

_{k}- (2v•u

_{j})u

_{j}- sum

_{1≤k≤n}a

_{k}(2u

_{k}•u

_{j})u

_{j}

We know that 2v•u

_{j}(being an edge length) is an algint, and 2u

_{k}•u

_{j}= -2cos(θ

_{k,j}) is an algint for any k, because the angle between the k'th and j'th mirrors is a rational angle (otherwise the symmetry group would be infinite). Therefore v'' has the same form as v'.

Now the chord vector between any two vertices has the form

(v + sum

_{1≤k≤n}a

_{k}u

_{k}) - (v + sum

_{1≤k≤n}b

_{k}u

_{k})

= sum

_{1≤k≤n}(a

_{k}- b

_{k})u

_{k}

= sum

_{1≤k≤n}c

_{k}u

_{k}

where the coefficients c

_{k}are algints, so the chordsquare is

(sum

_{1≤k≤n}c

_{k}u

_{k})

^{2}

= sum

_{1≤k≤n}sum

_{1≤j≤n}c

_{k}c

_{j}u

_{k}•u

_{j}

= sum

_{1≤k≤n}c

_{k}

^{2}+ 2 sum

_{1≤k<j≤n}c

_{k}c

_{j}u

_{k}•u

_{j}

= sum

_{1≤k≤n}c

_{k}

^{2}+ sum

_{1≤k<j≤n}c

_{k}c

_{j}(2u

_{k}•u

_{j})

which is clearly a combination of algints and thus itself an algint.

(In fact this shows that any chordsquare is in the ring generated by the edge lengths and doubled cosines of angles between mirrors.)

But this doesn't work for infinite polytopes, even regular polygons. Consider a hyperbolic apeirogon, or a zigzag apeirogon ( /\/\/\/\/ ), or a Wythoffian polygon with an irrational angle between mirrors (so the vertices are dense on the circumcircle); given edge length 1, the chord lengths may vary in a continuum, so they're not necessarily algebraic numbers, much less algebraic integers.

And, as I noted in the other thread, J37 (elongated square gyrobicupola) has chordsquares involving 1/√2, so a finite unit-edged polyhedron with congruent vertices doesn't necessarily have algint chords.

I suspect that any finite isogonal polytope with algint edges (and maybe some non-degeneracy conditions) has algint chords. Can anyone prove this?