24

The answer is 'no', because polynomial mappings with polynomial inverses preserve volumes up to a constant multiple.
To see why this property holds, suppose that $p:\mathbb{R}^d\to\mathbb{R}^d$ is a polynomial mapping with polynomial inverse $q:\mathbb{R}^d\to\mathbb{R}^d$. Then $p$ and $q$ extend to $\mathbb{C}^d$ as polynomial maps with polynomial ...

5

Francesco Polizzi's idea is enough to solve the problem:
Lemma. Let $f = ax^3 + bx^2 + cx + d \in \mathbf Z_{\geq 0}[x]$ nonconstant with $d > 0$, such that $f(1)$ is a prime $p > 2$. Then $f$ is irreducible in $\mathbf Z[x]$.
Proof. Suppose $f = gh$ for $g,h \in \mathbf Z[x]$; we must show that $g \in \{\pm 1\}$ or $h \in \{\pm 1\}$.
First assume $\...

answered Jul 8 at 1:48

R. van Dobben de Bruyn

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4

I will concentrate on the case of odd $q$ and $\xi,-\eta$ being squares, but the solution should be extendable to the remaining cases.
It is convenient to use the language of characters sums in order to compute $N$. To reduce your point counting problem to a character sum problem, the following observation is useful:
$$\#\{ x \in \mathbb{F}_q : ax^2+bx+c=0\} ...

3

Integral points on elliptic can often be computed routinely. In Question 1, the curve can be rewritten as
$$Y^2 = 5184 + 432 X -12X^2 + X^3,$$
where $X:=6n$ and $Y:=72y$. SageMath computes:
sage: EllipticCurve([0,-12,0,432,5184]).integral_points()
[(0 : 72 : 1), (21 : 135 : 1)]
So, the only integer solution is $(n,y) = (0,1)$.
For Question 2 with $N=4$, we ...

2

You can get all the way up to $\binom{n-1}{2}+1 = g+1$, where $g$ is the genus. This is the maximal number of connected components a real curve of genus $g$ can have, so this is optimal.
To do this, I will use Viro's patchworking method. Itenberg and Viro already give an example of how to use patchworking to build a plane curve with $g+1$ connected ...

answered Jun 22 at 14:37

David E Speyer

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2

An illustration for one of the examples in the answer by Robert Bryant.
It is supposed to convey the feeling of something extremely rigid, unyielding and inflexible.
Image of the square $[-1,1]\times[-1,1]$ under the map $(x,y)\mapsto(x-y^2-2x^2y-x^4,y+x^2)$ (composite of $(x,y)\mapsto(x-y^2,y)$ with $(x,y)\mapsto(x,y+x^2)$).

1

Equivalently, you want to interpolate the points $(i, p_i)$, $i = m \ldots n$ where $p_i$ is the $i$'th prime.
The prime $k$-tuples conjecture implies that for each integer $k > 2$ and each $d$ from $1$ to $k-1$, there are infinitely many $m$ such that with $n=m+k$ the minimum degree of the interpolating polynomial is $d$.
The conjecture says that $(p_{m+...

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