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Results tagged with nt.number-theory
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user 127660
Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
9
votes
Accepted
Covering all but finitely many integers via some given polynomials
With the right adaptations, the answer should be yes. In particular, the trivially necessary assumption is that the gcd of the values (not just of the coefficients) is 1. Then it is shown in
K. F. Rot …
- 2,513
7
votes
Irreducibility measure of integer polynomials
For $f=x^6 - 3x^5 - 2x^4 + 10x^3 + x^2 - 8x - 5$, one has
$$f=(x^3 - 2x^2 - 2x + 5)(x^3 - x^2 - 2x - 1),$$
$$f+1=(x-2)(x^5 - x^4 - 4x^3 + 2x^2 + 5x + 2),$$
$$f+2=(x^2-x-1)(x^4 - 2x^3 - 3x^2 + 5x + 3), …
- 2,513
9
votes
Accepted
Galois groups of specific classes of polynomials with one coefficient fixed
This is equivalent (by Hilbert) to asking whether the partially specialized polynomial still has symmetric Galois group (over the respective function field). This holds unless you specialized the cons …
- 2,513
11
votes
Accepted
Integer solutions of an algebraic equation
There are no points other than the trivial points $(n, \pm n, 0)$, $(n,0,\pm n)$ and $(0,n,\pm n)$, and in particular, no integer points with all coordinates positive. For example, taking the affine p …
- 2,513
10
votes
Does there exist a number field, unramified over a predetermined finite set of primes of Q, ...
Not sure whether this is of help after such a long time, but anyway:
The answer is yes, and this even works over some number field in which all the primes in $P$ are completely split. Namely, denote …
- 2,513
4
votes
On integral points of $f(x,y)=z g(x,y)$
Regarding Q2, one has $F(x, x^{2D-1}+x^{D-1}) = (x^D+1)\cdot G(x, x^{2D-1}+x^{D-1})$, yielding infinitely many integral points.
- 2,513
2
votes
Accepted
Solving solvable septics using only cubics?
Regarding question 1), of course the obvious (sufficient) answer is "When the Galois group is contained in $C_7\rtimes C_3$". That's not quite the case here, but "almost". To be precise, your septic h …
- 2,513
5
votes
Accepted
On the refined minimal ramification problem for $p$-groups
The answer is affirmative at least for $p\ge 11$ by the following construction (for the smaller primes, the extension constructed is not of the demanded degree $>p^9$, but surely there will be some al …
- 2,513
4
votes
Radicands of square roots of the 2020s, written in simplest radical form
A simultaneous solution of $ax^2-c_1=b_1y^2$ and $ax^2-c_2=b_2z^2$ as demanded gives rise to an integral point on $b_1b_2T^2 = (ax^2-c_1)(ax^2-c_2)$, and since $a\ne 0$ and $c_1\ne c_2$, the degree-$4 …
- 2,513
2
votes
Accepted
Unramified composition for every extension
The Abhyankar construction can be generalized naturally as follows: For each $p\in S$, take the (finite) set of Galois extensions $F_p/K_p$ occurring as completions of the Galois closure of a degree-$ …
- 2,513
3
votes
Accepted
The distribution of certain Galois groups
Denote by $\Omega$ the splitting field of $f(x)-y$ over $\mathbb{Q}$, by $k/\mathbb{Q}$ the maximal constant extension inside $\Omega$, and set $G:=Gal(\Omega/\mathbb{Q}(y))$, $G^+ =Gal(\Omega/k(y))$. …
- 2,513
5
votes
Accepted
A method to generate solvable equations of degrees $p = 7, 13, 19, 31, 37,\dots$ using only ...
The following can explain the shape of the expression and also why you always get pairs (why the minimal polynomials for the two elements in a pair differ by few coefficients, I don't know).
Let $p$ b …
- 2,513
2
votes
Density of extended Mersenne numbers?
Not an answer, but since the graphic may not fit a comment:
The below is a plot of the value $M(n)/n$, where $M(n)$ is the $n$-th value of the sequence (in ascending order, computed such that it's gua …
- 2,513
9
votes
A cyclic Galois extension over $ \mathbb{Q}(\omega)$
The answer is ``yes".
Note first that in the question we may without loss restrict to cyclic extension of $2$-power degree, since the odd part of the order would split off as a direct factor, hence wo …
- 2,513
5
votes
Splitting fields of degree 4 irreducible polynomials containing a fixed quadratic extension
This works even for any symmetric group and any quadratic number field, and can be done very explicitly.
When $n$ is even, let $f(t,X) = X^{n-1}((n-1)X-n) + t$, and when $n$ is odd, let $f(t,X) = X^n- …
- 2,513