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Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
7
votes
On the prime factorisation of g(k) = (k^2)! * prod (j=0, k-1, j!/(j+k)!)
The trick is to simplify as much as possible. Have you noticed that the $n$ never appears alone in your question, but only in the context of $p(n)$, so why don't you just write $p$ for $p(n)$ ? Then, …
4
votes
Accepted
irrationality of square roots of all non perfect square naturals
The point is that you use this fact from number theory:
$p$ and $q$ are relatively prime $\Longrightarrow$ $p^2$ and $q^2$ are relatively prime.
While this is not hard, this is not trivial either. Y …
6
votes
Accepted
Implications of a relation on algebraic numbers
Proved by grobber (Alexandru Chirvasitu) on AoPS: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&t=157732
EDIT: Also, http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=335001 …
3
votes
Accepted
Proving inequation with ceilings in Finite Field of characteristic $p$
We have $1\equiv u\left( p-r\right) \equiv u\left( -r\right)
=-ur\operatorname{mod}p$, so that $p\mid1+ur=ur+1$. Hence, $\left\lceil
\dfrac{ur}{p}\right\rceil =\dfrac{ur+1}{p}$.
We need to prove t …
5
votes
Are there results in "Digit Theory"?
Persi Diaconis gave a talk about probabilities of carries at MIT about half a year ago. Unfortunately I was so busy trying to spot symmetric functions in the talk that most of it drifted past my mind, …
0
votes
Subfields joining an algebraic element to another
EDIT: Okay, wrong. Sorry for spamming.
This looks just too easy, so I guess I'm doing something stupid.
I'll prove that
(a) whenever a subfield $K$ of $\mathbb{C}$ satisfies $\beta\in K\left[\alpha …
15
votes
Accepted
A sum involving sines
No, such $c_x$ don't exist. Even if you replace $\lbrace 0,\pm 1,\pm 2\rbrace$ by $\mathbb Q$, this won't change. In fact, if they would exist, then, using the relation $\displaystyle \sin\frac{2\pi x …
1
vote
Accepted
adjoint map and number field
The answer is Yes. The only reason for the appearance of formally real fields and $\mathbb R$ in the question is to rule out roots of unity; in fact, we have the following (cleaner and more general) s …
12
votes
Accepted
a type of numbers resulting from the partitions of an integer
First of all, let's formalize the definition: Let $\lambda$ be a partition. For every integer $i \geq 1$, let $m_i\left(\lambda\right)$ be the number of appearances of $i$ in $\lambda$. Then your fanc …
24
votes
Accepted
A curious valuation of this sequence
To your first question: No, it is false. For $n = 256 - 13$, the number $a_n$ has $\nu_2\left(a_n\right) = 7 < 8 = \nu_2 \left(n+13\right)$.
HOWEVER, it is almost correct: namely, it is correct whene …
9
votes
0
answers
702
views
Ghost-Witt sequences vs. ghost-Polya-Burnside sequences?
If you're in a hurry scroll down until the questions:
First the known part:
A sequence $\left(b_1,b_2,b_3,...\right)$ of integers will be called a ghost-Witt sequence if there exists a sequence $\le …
5
votes
Accepted
Do n-th Witt polynomials generate {P | P' is divisible by n} ?
Solved. The key is that the set of polynomials $P$ for which there exist polynomials $P_d\in\mathbb{Z}\left[\Xi\right]$ for all divisors $d$ of $n$ such that $\displaystyle P=\sum_{d\mid n}dP_d^{n/d}$ …
3
votes
How to factorize X^n - 1 in Z/pZ?
This seems very much like homework to me, so I'll be brief. I assume that your $Z_p$ denotes the field with $p$ elements; I will call it $\mathbb{F}_p$ henceforth (lest it be confused with the ring $\ …
12
votes
Accepted
Reciprocal sum of binomials and divisibility by $3$
Here is the answer I hinted at in a comment, in real detail. Took me a while,
but I had no idea how tiresome such arguments are to expose...
Yes, it is true: see Corollary 6 (b) below. The proof reli …
22
votes
A gamma function identity
This is merely a variation of your own proof, Matt, but I believe it makes things clearer.
The first step is to define $c:=1-a-b$. Then, your identity takes the form
$\dfrac{\Gamma\left(b\right)\Gam …