6
votes
How weird can the level set of a polynomial be?
Let me explain an upper bound on the number of components, which is asymptotically consistent with your conjecture.
I will be using your notation where the dimension is $d$ and the degree is $n$ even ...
- 105k
5
votes
Is every element of this sequence of polynomials irreducible?
As suggested in the comments by Mr. Sassatelli, we may alternatively define the polynomials $f_n$ by $f_1=X$, $f_2=X+1$, and
$$
f_{n} = f_{n-1}^2 - f_{n-1} + 1
$$
for all $n \geq 3$.
Writing $g$ for ...
- 1,517
5
votes
Solve the equation $x^3-3x=\frac{a^6+1}{a^3}$
Let $A=a+\frac{1}{a}$. Notice that $$x^3-3x=A^3-3A\tag{1}$$ so it is clear that $x=A$ satisfies $(1)$. We then have $$x^3-3x-A^3+3A=(x-A)(x^2+Ax+A^2-3)$$ The quadratic polynomial has roots: $$x=\frac{...
- 13.6k
3
votes
Question about polynomials, their degrees and their real roots.
Since $A^2=(C-B)(C+B)$ and $A^2$ has degree $4$ it follows that one of $C-B, C+B$ has degree $1$ and one has degree $3$ and since the problem is invariant when dividing by a nonzero constant we can ...
- 29.5k
2
votes
Find $m \in \mathbb{Z} $ so that all roots of $x^2-mx+m+2=0$ are also integers.
To use your approach OP: The determinant of $x^2 - mx +(m+2)$ is in fact $m^2-4m-8 = (m-2)^2-12$. So for the polynomial $x^2-mx +(m+12)$ to have integral roots, $(m-2)^2-12$ has to be an integral ...
- 22k
2
votes
Accepted
Solution of coupled trigonometric equations
Using the tangent half-angle substitutions
$u=\tan\frac12x$ and $v=\tan\frac12y$, we have $$\sin x=\frac{2u}{1+u^2}\quad \cos x=\frac{1-u^2}{1+u^2}\quad \sin y=\frac{2v}{1+v^2}\quad \cos y=\frac{1-v^2}...
- 80.3k
1
vote
Solve the equation $x^3-3x=\frac{a^6+1}{a^3}$
$\textbf{The strategic approach:}$
Our equation is $x^3 - 3x = a^3 + \frac{1}{a^3}$. Bringing the $3x$ over, we see
$$ x^3 = a^3 + 3x + \frac{1}{a^3} $$
The right side of the equation looks an awful ...
- 624
1
vote
Accepted
Unit in $(\mathbb{Z}/5)[x]/\langle x^4-x^2\rangle$
In $\mathbf Z/m\mathbf Z$, $a$ is a unit if and only if $a$ and $m$ are relatively prime.
Similarly, when $K$ is a field and $f(x) \in K[x]$, $f(x)$ is a unit in $K[x]/(g(x))$ if and only if $f(x)$ ...
- 50.8k
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