183

Let me give an informal explanation using what little I know about complex analysis.
Suppose that $p(z)=a_{0}+...+a_{n}z^{n}$ is a polynomial with random complex coefficients and suppose that $p(z)=a_{n}(z-c_{1})\cdots(z-c_{n})$. Then take note that
$$\frac{p'(z)}{p(z)}=\frac{d}{dz}\log(p(z))=\frac{d}{dz}\log(z-c_{1})+...+\log(z-c_{n})=
\frac{1}{z-c_{1}}+.....

111

A complete derivation can be found in the classical paper of Shepp and Vanderbei:
Larry A. Shepp and Robert J. Vanderbei: The complex zeros of random polynomials, Trans. Amer. Math. Soc. 347 (1995), 4365-4384
But the heuristic explanation is that for small modulus the higher order terms contribute very little to the polynomials, and so can be thrown away ...

answered Oct 2 '14 at 21:04

Igor Rivin

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65

I think the following geometric argument is interesting and maybe sufficient to answer "why" at an intuitive level (?).
When we take the powers of $x$ in the complex plane, the absolute value scales geometrically ($|x^n| =|x|^n$) and the argument (angle with the x-axis) scales linearly ($\arg x^n = n \arg x$). So the powers of $x$ look like this:
If $x$ is ...

53

Think I will post this, the question remains popular and there are differing views on what it means; I wrote to Eric Kostlan, classmate, who has published on this sort of thing; he supported the answer by Phantom Hoover, saying " I was going to answer, but someone beat me to it" and sent an informal version later:
Eric Kostlan:
Its even easier, if ...

51

This is a special case of the Lee-Yang theorem. Let $G$ be a finite graph with vertex set $V$ and edge set $E$. Let $\beta \in (0,1)$ be a real number. Let $\sigma$ denote a ``spin function" $\sigma: V \to \{ -1, 1\}$. Put $m(\sigma)$ to be the number of vertices with positive spin, and $d(\sigma)$ to be the number of edges with vertices of opposite spin....

50

The minimal polynomial of $\cos(2\pi/n)$ (by William Watkins and Joel Zeitlin, The American Mathematical Monthly
Vol. 100, No. 5 (May, 1993), pp. 471-474) has full clarity on this matter (just take their result for even $n$ to resolve your case).

answered Nov 27 '17 at 18:29

Vladimir Dotsenko

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48

There is the following result of Davenport, Lewis, and Schinzel [DLS64, Cor to Thm 2]:
Theorem. Let $p \in \mathbf Z[x]$. Then the following are equivalent:
$p$ is a sum of two squares in $\mathbf Z[x]$;
$p(n)$ is a sum of two squares in $\mathbf Z$ for all $n \in \mathbf Z$;
Every arithmetic progression contains an $n$ such that $p(n)$ is a sum of two ...

ag.algebraic-geometry nt.number-theory ac.commutative-algebra polynomials oc.optimization-and-control

answered May 29 '20 at 4:56

R. van Dobben de Bruyn

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46

The following papers might be helpful:
Shmerling, E and Hochberg, K.J., Asymptotic Behavior Of Roots Of Random Polynomial Equations, Proceedings Of The American
Mathematical Society, Volume 130 (2002), Number 9, Pages 2761-2770.
Erdős, P. and Turan, P., On the distribution of roots of polynomials,
Ann. Math. 51 (1950), 105-119.
In particular, the Erdős-...

answered Oct 3 '14 at 1:22

Joe Silverman

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45

The answer to the question is yes (so the answer to the title is no) and I will give an example later.
Let me first recall a couple of results. The first one is the following, that can be found in [Cox, Galois Theory, Theorem 8.6.5].
Theorem 1. Let $F$ be a subfield of $\mathbb{R}$ and let $f \in F[x]$ be an irreducible polynomial with splitting field $F \...

answered Sep 30 '14 at 18:41

Francesco Polizzi

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43

Here is a proof, based on a trick that can be used to prove that
$x^n + x + 1$ is irreducible when $n \not\equiv 2 \bmod 3$.
We work with Laurent polynomials in $R = \mathbb Z[x,x^{-1}]$; note that
$R$ has unit group $R^\times = \pm x^{\mathbb Z}$.
We observe that for $f \in R$, the sum of the squares of the coefficients
is given by
$$\|f\| = \int_0^1 |f(e^{...

41

Yes, $u$ must be a polynomial of degree $2$.
I had to draw on a few unexpected ingredients to prove this;
perhaps there's a simpler proof.
[EDIT Or maybe not: Peter Mueller's answer reports that
this was "an open problem on planar functions for many years",
and gives links to three independent papers c.1990 that independently
solved it. Two of them give ...

answered Jul 8 '13 at 5:02

Noam D. Elkies

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38

There is no such polynomial.
It is clear that $f$ cannot be a cyclotomic polynomial (your condition $\deg{f} > 1$ excludes $x-1$). So suppose $f$ is non-cyclotomic and irreducible, of degree $d$, and consider a prime $p$ for which $f(x^p)$ is reducible. For $\alpha$ a root of $f(x^p)$, reducibility means that $[\mathbb{Q}(\alpha):\mathbb{Q}] < pd$. ...

answered Aug 25 '14 at 20:05

Vesselin Dimitrov

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35

$$P_m(x)=\sum_{i=0}^{m}\sum_{j=0}^{m}\binom{x+j}{ j}\binom{x-1}{ j}\binom{j}{ i}\binom{m}{ i}\binom{i}{ m-j}\frac{3}{(2i-1)(2j+1)(2m-2i-1)}.$$
Our task is to show it takes integer values on integers.
Folowing Wadim Zudilin we put
$$B_k(x)=\binom{x+k}{2k}+\binom{-x+k}{2k}.$$
For $k\geq0$ the $B_k$ are even polynomials of degree $2k$ that take integer ...

answered Sep 18 '15 at 7:38

Wilberd van der Kallen

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33

Is there anything wrong with the following argument?
First of all, by scaling all $x_i$ and all $y_i$ by a positive constant, we may safely assume that $\prod_i x_i = \prod_i y_i =1$.
The result would now follow from the following more general conjecture.
$\bf Conjecture:$ For ${\bf a}=(a_1,\ldots,a_{n-1})\in (\mathbb R_{>0})^{n-1}$ consider
$h_{\bf a}(...

33

[edited to explain a few steps and connect with the Hahn polynomials]
The answer is
$$
\frac{(2n+1)(2n+3) n!^4}{(2n)!}
$$
assuming that I did the algebra right, which seems likely because this formula
agrees with the previously computed values $3,5,14,324/5$ for $n=0,1,2,3$.
Consider first the minimum of $\sum_{i=1}^{n+1} p(i)^2$ over monic $p$
of degree $...

answered Jul 6 '15 at 2:29

Noam D. Elkies

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33

It is known (though I don't have a reference except for the statement at http://ocw.mit.edu/courses/mathematics/18-s34-problem-solving-seminar-fall-2007/assignments/roots.pdf, #2) that the only real polynomials with nonzero constant term whose roots equal their coefficients are $x^2+x-2$, $x^3+x^2-x-1$, and (approximately)
$$ x^3 + .56519772x^2 - 1....

answered Sep 3 '15 at 19:32

Richard Stanley

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33

Although not the exactly the same due to $2^{n-1}$ instead of $2^n$ terms, the OP's formula seems to be essentially the well-known polarization formula for homogeneous polynomials, which is stated as following:
Any polynomial $f$, homogeneous of degree $n$ can be written as $f(x)=H(x,\ldots,x)$ for a specific multilinear form $H$. One has the following ...

33

Every real polynomial $f$ is the sum of TWO polynomials $g,h$ with all zeros real (of the same degree).
Indeed, take any $g_1$ of the same degree as $f$, whose roots are real and simple. Small perturbation does
not destroy this property. Therefore $h_1=g_1+\epsilon f$ also has real zeros
when $\epsilon$ is small and real.
Now take $g=-g_1/\epsilon, \; h=h_1/\...

answered Mar 10 '18 at 13:35

Alexandre Eremenko

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33

You should refer to Lemma 10 (page-233) in this paper by Schinzel where he proves that for any polynomial $F(x)$ of degree $d$ we have a polynomial $G(x)$ of degree $d-1$ such that their composition is reducible.

32

Consider $Q(x)=x(2x-1)(3x-1)$. This gives an injective map $\mathbb Z\to \mathbb Z$, because $n<m \implies Q(n)<Q(m)$. However, this $Q$ is not injective over $\mathbb Z/p\mathbb Z$ for any $p$ because $Q(x)=0$ has three solutions when $p\geq 5$ and two solutions when $p\in \{2,3\}$.

answered Aug 9 '20 at 17:49

Gjergji Zaimi

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31

The claim is true for $\lambda = 0$, when $f(x) = x^n + 1$,
and for $\lambda = 1$, when $f(x) = (x+1)^n$. We show that this
implies the claim for all intermediate $\lambda$, by proving that
the number of zeros on the unit circle is a non-decreasing
function of $\lambda$.
Let $\lambda = e^t$ with $-\infty < t < 0$, and $x = e^{iu}$ with
$u \in {\bf R} ...

answered May 7 '18 at 6:16

Noam D. Elkies

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30

This is true. Pass to an extension field where the polynomial has a root $r$, notice that the other roots are of the form $r+1$, $r+2$, ..., $r+p-1$. Suppose that $x^p - x +1 = f(x) g(x)$, with $f, g \in \mathbb{F}_p\left[x\right]$ and $\deg f = d$. Then $f(x) = (x-r-c_1) (x-r-c_2) \cdots (x-r-c_d)$ for some subset $\{ c_1, c_2, \ldots, c_d \}$ of $\mathbb{F}...

answered May 19 '14 at 15:45

David E Speyer

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29

You have two questions: 1) How Ramanujan discovered it? 2) Is an accidental, isolated result? The second one is easier to answer and may shed light on the first.
I. Define $F_n = x_1^n+x_2^n+x_3^n-(y_1^n+y_2^n+y_3^n),\;$ where $\,\small x_1+x_2+x_3=y_1+y_2+y_3 = 0$.
Theorem 1: If $F_1 = F_3 = 0$, then,
$$9x_1x_2x_3 F_6 = 2F_9 = 9y_1y_2y_3 F_6$$
Theorem ...

29

It is clear that $F_n(1)=0$. On the other hand, the derivative can be expressed in terms of $F_{n-2}$. Thus
$$
\begin{aligned}
F_n'(X) &=\sum_{k=0}^n\binom{n}{k}(-1)^k(k(n-k))X^{k(n-k)-1} \\
&= \sum_{k=1}^{n-1} \frac{n!}{(k-1)!(n-k-1)!} (-1)^k X^{k(n-k)-1} \\
&= n(n-1) \sum_{k=1}^{n-1} \frac{(n-2)!}{(k-1)!(n-k-1)!} (-1)^k X^{k(n-k)-1} \\
&= n(...

answered Mar 29 '18 at 0:26

Joe Silverman

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28

We treat all four problems in turn. In all that follows $n>1$.
Surjectivity over $\mathbb{Q}$:
If there is an algorithm to test whether an arbitrary polynomial with rational coefficients is surjective as a map from $\mathbb{Q}^n$ into $\mathbb{Q}$ then Hilbert's Tenth Problem for $\mathbb{Q}$ is effectively decidable.
Proof: Let $g(x_1,\ldots,x_n)$ be ...

28

This is true; it is due to Selmer. Ljunggren (On the irreducibility of certain trinomials and quadrinomials, Math. Scand. 1960) has obtained the complete list of reducible trinomials with $\pm 1$ coefficients. For more details see Gerry Myerson's answer to this question, which contains a review of Ljunggren's paper: About irreducible trinomials . To this I ...

answered Aug 4 '14 at 18:19

Vesselin Dimitrov

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28

Occasionally I wish somebody could give me a good whack on the head to keep my brains running and the older I get, the more frequently I need it. The problem is actually trivial.
I will prefer to think that $P$ is non-negative on some disjoint with integers arithmetic progression $\Lambda$ with step $1$ . Then we need to show that
$$
\sum_{k=0}^n {n\choose ...

28

For each positive integer $n$ and any rational $x$, we have
$$f(x)\geq (x-b_1)^2(x-b_2)^2\dots(x-b_n)^2.$$
For large $x$, we then have $f(x)\gg x^{2n}$, which implies that if $f$ is a polynomial, it must have degree $≥2n$.

answered Dec 11 '17 at 19:45

Laurent Moret-Bailly

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27

I am working on a book-length manusript, Around the Chevalley-Warning Theorem. A complete answer to your question is estimated at about 150 pages!
In terms of what exists at the moment, here are two papers. Both of them make connections between the classical results of Chevalley and Warning and modern polynomial methods. The first concerns a ...

answered Aug 22 '14 at 21:37

Pete L. Clark

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27

I was a bit skeptical of some of the explanations, so I ran my own experiments to see how varying the parameters affected the distribution of zeros. Note that I was only interested in the case where coefficients are independent and identically distributed. Computations were done using PARI/GP instead of Mathematica.
I first essentially repeated Andrej's ...

answered Oct 6 '14 at 23:35

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