All Questions
58
questions
2
votes
1
answer
70
views
If $f(x)\in \mathbb{Z}[x]$ is irreducible (over $\mathbb{Q}$), is it always possible to find $a$ and $b$ in $\mathbb{Q}$ with $f(ax+b)$ Eisenstein? [duplicate]
My initial thought is no, simply because it seems too easy if it is true.
The simplest example of a nontrivial irreducible polynomial I could think of was $f(x)=x^2+1$. Unfortunately, $f(x+1)$ is ...
- 520
7
votes
1
answer
191
views
Factorization and irreducibilty for $x^n-2x^m+1$ trinomials.
I have encountered a weird phenomenon while trying to solve a problem on Reddit. Here is the phenomenon.
Let $a>b \in \mathbb{N}$ and $p_{(a,b)} = x^a - 2x^b + 1$
It seems that if $gcd(a,b,c,d) = 1,...
- 309
0
votes
1
answer
43
views
Reducibility of constrained polynomial
Let $f \in \mathbb{Z}[x, y]$ be a polynomial. Suppose that the list of terms of $f$ do not involve the $y$ variable except for a single $y^2$ term with some arbitrary coefficient. When is $f$ ...
- 932
2
votes
0
answers
60
views
Product Formula for Real Cyclotomic Polynomials
Let $n$ be a natural number, $\zeta_{n}$ be a primitive $n^{th}$ root of unity and $\Phi_{n}(x)$ be the $n^{th}$ cyclotomic polynomial. Let $\Psi_{n}(x)$ be the $n^{th}$ real cyclotomic polynomial (...
- 1,625
5
votes
7
answers
607
views
Showing $x^4 + 1$ is irreducible in $\mathbb{Q}[x]$.
Clearly, none of the roots are in $\mathbb{Q}$ so $f(x) = x^4 + 1$ does not have any linear factors. Thus, the only thing left to check is to show that $f(x)$ cannot reduce to two quadratic factors.
...
- 619
15
votes
3
answers
642
views
Under the which condition, factorisation of $a_1^n+a_2^n+\cdots+a_n^n-na_1a_2a_3...a_n ?$ is possible?
Under the which condition, factorisation of the polynomial
$$a_1^n+a_2^n+\cdots+a_n^n-na_1a_2a_3...a_n ?$$
is possible?
I know possible cases:
$$a^2+b^2-2ab=(a-b)^2$$
and
$$a^3+b^3+c^3-3abc=(a+b+c)(a^...
- 1,659
-1
votes
1
answer
157
views
Irreducibility of $p^{n-1}X^n+pX+1$ over $\mathbb{Q}$ [duplicate]
I am attempting to show that $f(X)=p^{n-1}X^n+pX+1$ are irreducible over $\mathbb{Q}$ for any positive integer $n$ and any prime $p$. At the behest of my teacher, and their hint, I would like to do so ...
- 71
-1
votes
1
answer
336
views
An irreducible polynomial in Z[x]
Let $p$ be a prime and let $n$ be an integer greater than $4$. Prove that if $a$ is an integer that is not divisible by $p$, then the polynomial
$$f(x)=ax^n-px^2+px+p^2$$is irreducible over $\mathbb{Z}...
14
votes
1
answer
420
views
Is $x^n-\sum_{i=0}^{n-1}x^i$ irreducible in $\mathbb{Z}[x]$, for all $n$?
Let the sequence of polynomials $p_n$ from $\mathbb{Z}[x]$ be defined recursively as $$p_n(x)= xp_{n-1}(x)-1$$
with initial term $p_0(x)=1$.
Then $$p_n(x)= x^n-\sum_{i=0}^{n-1}x^i $$
Question 1: is it ...
- 3,716
3
votes
0
answers
87
views
Showing the irreducibility over $\mathbb{Z}[X]$ of polynomials similar to the cyclotomic polynomials
This question follows this other question.
Let $y$ be a natural number, $x$ a variable and $$ f(x,y):= \frac{x^{2y}-1}{x+1}$$ and
$$ g(x,y):= \frac{f(x,y)^{2y+1}-1}{(f(x,y)-1)(xf(x,y)+1)}.$$
For a ...
- 3,716
0
votes
0
answers
51
views
Can an irreducible polynomial over $\mathbb{F}_{q}[T]$ have multiple roots?
Let $\mathbb{F}_{q}$ be a finite field of order $q = p^{l}$ for some prime $p$ and $l \geq 1$ and consider the ring of polynomials $R = (\mathbb{F}_{q}[T])[x] $. Can an irreducible element $g(x)$ in $...
- 1,625
2
votes
1
answer
159
views
Irreducible polynomial divisible by all primes
Does there exist an irreducible non-linear polynomial $P(x)\in\mathbb{Z}[x]$ such that for any prime number $q$ there exists $t\in\mathbb{N}$ such that $q|P(t)$ ?
Also (dis)proving whether there ...
- 1,528
0
votes
1
answer
260
views
construct irreducible polynomials of degree 32 over $Z_2[x]$ [duplicate]
I'm learning finite fields behind Advance Encryption Standard. As far as I know, the irreducible polynomial used in AES is $x^8+x^4+x^3+x+1$. This is because AES s-box is based on bytes(8bits). Now I ...
- 11
0
votes
0
answers
40
views
Irreducible polynomial cannot have only perfect power values
Does there exist a non-constant irreducible monic polynomial $P(x)$ with integer coefficients such that for every integer $n$ there are integers $a,b \geq 2$ such that $P(n) = a^b$?
No idea how to ...
- 2,733
2
votes
0
answers
31
views
Controlling the discriminant size of a polynomial using roots
I'm trying to generate random monic irreducible polynomials in $\mathbb{C}[x]$ whose absolute discriminant is within a certain size range (say $10^t$, where $t$ is a positive integer).
I also need to ...
- 197