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Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
3
votes
Accepted
Bounded gaps between powers
According to the Wikipedia page on Catalan's conjecture, the problem you pose is open. (Look under the "generalization" heading.) A more general problem is Pillai's conjecture.
4
votes
On the prime $k$-tuple problem
It is currently believed that the second conjecture is likely false, but it hasn't been proven quite yet. There is an interval of size 3159 which is not prevented from having more primes than the ini …
3
votes
Accepted
Estimate for the $2n$-th consecutive prime number
Let's use some standard bounds on the $n$th prime, as found in this paper by Pierre Dusart.
We have
$$
p_{2n} \leq 2n[\ln(2n)+\ln(\ln(2n))-0.9484] = 2n\left[\ln(n)+\ln\left(\frac{2}{e^{0.9484}}\ln(2n) …
5
votes
1
answer
320
views
A modification of the Ljunggren-Nagell equation
[Thanks to Gerhard Paseman for helping me reformulate my original question.]
The equation
$$
\frac{a^m-1}{a-1}=b^2
$$
was solved by Ljunggren, building on work of Nagell, who showed that if $a>1$, $b …
2
votes
Best known bounds for a product over primes in an interval
The best source for explicit, unconditional bounds on such products that I'm aware of is in the work of Pierre Dusart. See his paper Explicit estimates of some functions over primes.
In Section 5.4, …
12
votes
Accepted
Consecutive numbers with mutually distinct exponents in their canonical prime factorization
The answer to this question is almost certainly no.
It is well known that the ABC Conjecture implies that there are only finitely many triples $(n,n+1,n+2)$ which are all powerful. A similar argumen …
3
votes
Character sums over prime arguments
An update on this problem:
I found out how to compute effective (and asymptotically accurate) bounds for $\sum_{p\leq x,\, p\equiv a\pmod{k}}\log(p)/p$. Basically it boils down to the usual analytic …
3
votes
Factorizing polynomials in $\mathbf{Z}[[x]]$
Edited due to mistakes pointed out in the comments:
I think the answer to the problem might be yes. Here are some preliminary thoughts.
First, you know $f_{\nu}(x)$ divides $f(x)$, since $f(x)$ is …
10
votes
How do we recognize an integer inside the rationals?
I just ran across the following paper: Defining Z in Q
It uses another characterization of the integers inside the rationals that none of us listed, perhaps because it is so trivial. Namely, the int …
8
votes
Accepted
Ruling out an extremely specific class of Wieferich-like primes
This is impossible from standard results about cyclotomic polynomials and their factors. Note that $\frac{q^{p-1}-1}{p-1}$ is just
$$
\prod_{d|(p-1), d>1}\Phi_d(q),
$$
where $\Phi_d(x)$ is the $d$th …
4
votes
1
answer
671
views
Correct growth rate of logarithmic derivative of zeta, outside critical strip
Let $\zeta$ be the Riemann zeta-function, and let $t> 0$. I'm interested in the growth rate of
$$
\left|\frac{\zeta'}{\zeta}\left(-\frac{1}{2}+it\right)\right|
$$
as $t\to\infty$. It is easy to find …
14
votes
1
answer
2k
views
Character sums over prime arguments
Let $f$ be a monotone decreasing, continuously differentiable function with $\lim_{x\rightarrow \infty}f(x)=0$. Let $\chi$ be a non-principal Dirichlet character. It is standard to show that $\sum_{ …
12
votes
2
answers
1k
views
Density in van der Waerden's theorem
Color the positive integers using just two colors. By van der Waerden's theorem, we can find a $k$-term arithmetic progression as long as we consider a long interval.
I imagine it is possible to fin …
10
votes
2
answers
544
views
A back and forth Euclidean algorithm over the integers--does it have bounded length?
cLet $a,b,c,d\in \mathbb{Z}$ and suppose we have the equation $ac+bd=1$. One way of thinking about this equation is it expresses the fact $\gcd(c,d)=1$. It is well-known that all other similar equat …
3
votes
Does there exist a polynomial 𝑃(𝑥,𝑦) which detects all non-squares?
This is just a long comment that might be helpful:
Treat $a$ as a parameter, and treat $x$ as a variable. The Diophantine expression
$$
\exists x\ ((x^2<a) \land (a<(x+1)^2))
$$
defines $a$ as a nons …