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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.
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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 …
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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) …
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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 …
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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, …
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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 …
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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 …
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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 …
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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 …
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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 …
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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 …
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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_{ …
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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 …
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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 …
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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 …
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