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It is well known from this famous question that the roots of a random polynomial tend to be close to the unit circle. So I was wondering in a somewhat converse sense: for an irreducible polynomial, is it possible that many roots are collinear (on a line other than the real or the imaginary axis)?

Now, if we take for instance a Chebyshev polynomial $T_k(x^n-1)$, we get trivially $k$ roots on each of the $n$ rays from $O$ through each $n$th root of unity, see e.g. here for $k=5,n=7$ (left picture). But what if we want roots on a line which are not collinear with the origin, i.e. not proportional over $\mathbb R$? (And of course mere linear shifts should be excluded, like replacing in such a polynomial $x$ by $x-1$.)
We can easily get twice four of them on vertical lines, e.g. for the minimal polynomial of $ \dfrac{1+i}{\sqrt[4]{2}}+(\sqrt{2}+1)i $, which is $x^8+12 x^6+16 x^5+42 x^4+32 x^3+68 x^2+112 x+73$ (right picture).
enter image description here

This raises several questions (remember that it is all about irreducible polynomials, so we could talk as well of the Galois conjugates of a given algebraic number):
Assuming that the pattern has no rotational symmetry and taking the word 'line' to exclude both axis,

1. Is it possible to have more than 4 roots on a line?
2. Can a line with 4 or more roots be other than horizontal or vertical? If so, which slopes can occur?
3. Is it even possible to have exactly 3 of the roots on a line?

EDIT: Q1 and Q3 => Yes by Ilya Bogdanov's comments. So Q2 remains, and I've added Q4:

4. Is it possible that the roots make up a $p\times q$ grid where $p$, the number of rows, is not a power of $2$?

Note that there can be patterns with only one line of 4 collinear roots, e-g. the minimal polynomial of $\sqrt{2}+i+ \dfrac{i}{\sqrt[4]{2}} $ , viz. ${ {4 x^8-16 x^6+32 x^5+84 x^4-192 x^3-40 x^2-112 x+353}}$: enter image description here

For Q1, it might be possible to find constructions by some sort of "bifurcation" in a sense similar to what happens with the logistic map, but I am not very sure about that. If at all, I'd conjecture that for any [edit: vertical] line, the number of roots it contains always is a power of 2.

EDIT: Q4 has a positive answer. (And so has Q2.) Based on the construction of Jeremy Rouse in the related question, we have e.g. $$x^{18}+6x^{15}+54x^{14}-54x^{13}+24x^{12}+378x^{11}+189x^{10}\\ -1672x^9+5184x^8+162x^7-11649x^6+13824x^5\\ +28944x^4-80904x^3+138240x^2-86400x+44992$$ whose roots come in a nice $3\times 6$ pack as displayed here: enter image description here

Wolfgang
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  • If you have some $\xi$ such that its minimal polynomial $P(x)$ has several collinear roots, you may choose some non-real $\eta$ whose minimal polynomial remains irreducible in the splitting field of $P$; then the minimal polynomial of $\eta\xi$ also has the same number of roots lying on the line rotated by $\arg\eta$ (and scaled by $|\eta|$). – Ilya Bogdanov Sep 26 '15 at 11:01
  • Also, you may see that the minimal polynomial of $\sqrt2+i/(\root4\of 2+\root4\of 3+\root4\of5+\root4\of7)$ has quite more collinear roots... – Ilya Bogdanov Sep 26 '15 at 11:06
  • Do you mean polynomial with integer coefficients? (Otherwise just choose the roots)? – joro Sep 26 '15 at 11:16
  • As for Q3, why not take three real roots of an irreducible degree 3 polynomial and shif them by $i\sqrt n$? You may then rotate them if you wish. – Ilya Bogdanov Sep 26 '15 at 11:19
  • @joro yes, of course I mean irreducible over the integers. – Wolfgang Sep 26 '15 at 11:50
  • @IlyaBogdanov For your last comment: Oh sure, how could I miss such a simple idea? So e.g. $x^6+2 x^3+21 x^2-18 x+51$. – Wolfgang Sep 26 '15 at 12:09
  • Even better when shifting by $i\cos\frac{\pi}{5}$ which yields a 4×3 grid! Or by $i\cos\frac{\pi}{20}$ which yields a 8×3 grid.There may be some of that bifurcation here. – Wolfgang Sep 26 '15 at 14:21
  • Are you aware of Theodore Motzkin's paper Sur l'equation irreductible $z^n + a_{n-1}z^{n-1} + \cdots + a_0 = 0$, $n > 1$, a coefficients complexes entiers, dont toutes les racines sont sur une droite. Les $11$ classes de droites admissibles (Comptes Rendues hebdomadaires..., 1945)? – Vesselin Dimitrov Oct 02 '15 at 22:39
  • @VesselinDimitrov No, I'm not. Can you access it? Even though the subject seems quite different (especially considering complex coefficients), maybe some of the methods are applicable here? – Wolfgang Oct 03 '15 at 12:58
  • @Wolfgang: I have not looked at it closely, but it describes all classes of lines $L \subset \mathbb{C}$ that contain all complex roots of a monic irreducible polynomial of arbitrarily large degree over the Gaussian ring $\mathbb{Z}[i]$. This could be relevant since the $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$-orbit of an algebraic number partitions into (at most) two $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q}(i))$-orbits, giving irreducible polynomials over $\mathbb{Q}$ whose roots are covered by the union of two lines: $L$ and its complex conjugate. – Vesselin Dimitrov Oct 03 '15 at 18:16
  • I took this reference from the endnotes in Chapter 2.4.3 of Narkiewicz's Elementary and Analytic Theory of Algebraic Numbers. This contains some further references (Robinson, Ennola, Smyth,...) to similar questions which could be of interest to you. – Vesselin Dimitrov Oct 03 '15 at 18:19

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