All Questions
21 questions
1
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$F(x,y)$ absolutely irreducible over the rationals, but reducible modulo infinitely many primes? [duplicate]
In this two page note we give efficient probabilistic algorithm for factoring bivariate
polynomials in composite characteristic assuming the solution is unique
and we would like to test the algorithm ...
- 25.4k
2
votes
1
answer
134
views
On a efficient algorithm for factoring bivariate polynomials modulo composite modulus assuming the solution is unique
We found and implemented in sage efficient algorithm for factoring
bivariate polynomials modulo composite modulus assuming the solution is unique up to a constant factor.
More formally let $K=\mathbb{...
- 25.4k
0
votes
1
answer
177
views
Coefficients of 0,1-polynomials factorization
Let $n\in\mathbb{N}$ and $P_{n}$ is $0,1$-polynomial whose coefficients are binary digits of $n$.
Let $Q_{1}(x) \cdot Q_{2}(x) \cdots Q_{m}(x)$ - polynomial factorization (over integers) of $P_{n}$.
...
- 512
0
votes
1
answer
219
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Simple question about 0,1-polynomials
Being interested in these polynomials, would like to clarify one small observation.
Let $n\in\mathbb{N}$ and $P_{n}$ is $0,1$-polynomial whose coefficients are binary digits of $n$.
Let $n$ has prime ...
- 512
0
votes
0
answers
79
views
On the multiplicative group of quotients of polynomial rings
Related to this.
The $p+1$ factorization algorithm works over $\mathbb{Z}/n\mathbb{Z}[x]/f(x)$
and hopes $p+1$ to be smooth.
We are trying to generalize this to multivariate case
and also try to find ...
- 25.4k
3
votes
1
answer
82
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Evaluating the generalized continued fraction obtained from the factorization of a bivariate polynomial equation
Happy New Year, MO community!
We need someone expert in Generalized Continued Fractions (GCFs), with a deep knowledge of the GCFs’ convergence properties, to solve the following problem.
PROBLEM
...
- 125
0
votes
2
answers
328
views
A doubt regarding the extended form of the Weierstrass factorization theorem
I want to represent $\sin(x)-\dfrac{1}{\sqrt{2}}$ as a product of it's zeroes
According to the Weierstrass factorization theorem, the sine function can be represented as a product of its factors:
$$\...
- 103
1
vote
0
answers
138
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What are the limitations for calculating the inverse of a polynomial with the Lagrange inversion theorem?
I have been attempting to produce a series expression for the roots of high degree polynomial using the Lagrange Inversion theorem. I am curious about the statement from the Wikipedia page on Bring ...
- 547
1
vote
0
answers
107
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Polynomial divisible by unbounded primes with exponent one
Let $f(x)$ be squarefree polynomial with integer coefficients and
degree at least $3$.
Is it true that for all sufficiently large $n$, $f(n)$ is divisible
by prime $p$ with exponent one and $p$ is ...
- 25.4k
5
votes
0
answers
160
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Reducibility of $f(x)^{2^n}+1$ and $f(x)^{2^n}+g(x)^{2^n}$
Related to generalized Fermat numbers.
Let $f(x),g(x)$ be coprime polynomials with integer coefficients.
Assume that if $f(x)$ or $g(x)$ are of the form $h(x)^k$ then $k$ is power
of two.
Q1 Is it ...
- 25.4k
2
votes
1
answer
360
views
Positive divisors of $P(x,n)=1+x+x^2+ \cdots + x^n$ that are congruent to $1$ modulo $x$
This is a follow-up question to Positive integer solutions to the diophantine equation $(xz+1)(yz+1)=z^4+z^3 +z^2 +z+1$
Let \begin{equation}
P(x,n)= 1+x+x^2+ \cdots + x^n, \end{equation}
\begin{...
- 319
5
votes
1
answer
377
views
When $p(x)^2 \mid f(g(x))$?
Let $f(x),g(x),p(x)$ be non-constant polynomials with rational coefficients.
Is it true that for all $f$ exist $g,p$ such that $p(x)^2 \mid f(g(x))$?
Partial results:
$f(g(x))$ is divisible by square ...
- 25.4k
1
vote
0
answers
133
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Primes which do not divide certain homogeneous polynomials
It is known that if $x^2 + y^2 = z^2$ is a primitive Pythagorean triplet then $z$ is not divisible by any prime of the form $4k-1$. The following is a generalization of this classical result which ...
- 5,470
0
votes
1
answer
431
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Reason Coppersmith fails here?
Take classic problem of finding $P,Q$ in balanced semi-prime $N=PQ$.
$P$ has a binary expansion and so does $Q$. We can set the binary $0/1$ variables to be $x_1$ through $x_{\lceil\log P\rceil}$ and $...
- 13.9k
5
votes
0
answers
153
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On factorization algorithms for $\mathcal{O}[x]$
We know that $\mathsf{LLL}$ algorithm provides factorization procedure that runs in poly time for polynomials in $\Bbb Z[x]$ that are primitive.
What other rings $\mathcal{O}$ can we use instead of $\...
user76479
7
votes
2
answers
541
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Theorems of the Galois groups of quintics appears not to work for the ${F}_{20}$ group determination
I am computing the Galois groups of quintics using the theorems from Ryan Kavanagh paper "On Irreducible Rational Quintics" using the decic resolvent ${P}_{10} \left({x}\right) = \prod\limits_{1 \le i ...
- 441
4
votes
2
answers
411
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Find all possible rational values of a parametric quartic such that it is reducible
Description: Given the following parametric quartic polynomial
$y^4 - 28 z y^3 - 14 (656 - 328 z + 83 z^2) y^2 +
4 z (-20464 + 10232 z + 3409 z^2) y +
91 (62208 - 62208 z + 41504 z^2 - 12976 z^3 +...
- 441
0
votes
1
answer
263
views
Find all possible rational values of the parameter of a parametric cubic such that it is reducible
Description: Given the following parametric cubic polynomials ${E}^{3}
- 15\, {\beta}_{\pm}\, {E}^{2} - 3 \left({71\, {\beta}_{\pm}^{2} + 352\, {\beta}_{\mp}}\right) E
+ 135\, {\beta}_{\pm} \left({5\...
- 441
2
votes
1
answer
382
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Irreducibility of Faulhaber-like Polynomials over $\mathbb Q[x]$
Motivation: Inspired by the famous Faulhaber polynomials $F_k(N)=\displaystyle\sum_{n=0}^Nn^k,$ I decided to study their alternating versions, $\Phi_k(M)=\displaystyle\sum_{n=0}^M(-1)^nn^k$.
For $k\...
- 655
4
votes
0
answers
279
views
Analog of Euler's factoring technique
Is there an analog of Euler's Two Squares factoring theorem over polynomial rings $\Bbb Z[x]$ by considering a version for non-negative polynomials?
Euler's two squares factoring states that numbers ...
- 13.9k
6
votes
2
answers
312
views
Collision polynomials
Consider $P_n(x)$ polynomials defined through the recurrence relations
$$P_n(x)=2(1-x)P_{n-1}(x)-(1+x)^2P_{n-2}(x),$$ with $P_0(x)=1$ and $P_1(x)=1-3x$.
In fact, the explicit solution of these ...
- 16.5k