Yes - the standard proof of Vinogradov's result by means of the circle method gives this result. You just need to examine an integral $$\int_{\mathbb{R}/\mathbb{Z}} (\widehat{f}(\alpha))^2 \widehat{f}(... View answer 21 votes The aim of this answer is to sketch a proof of the fact that there are at most \epsilon p solutions to 2^{2^{2^x}} = x \mod p. The original question -- namely, to show the same for 2^{2^{2^{2^x}}}... View answer 11 votes Ke Gong kindly sent me a link to Vinogradov's work online: http://www.mathnet.ru/php/person.phtml?&personid=26537&option_lang=rus In summary, it seems clear that the 1937 proof was ... View answer 8 votes Gil Kalai writes: 2) Is it the case that people largely or even entirely lost their interest in the prime numbers for about fifteen centuries until Fermat? What are the facts of the matter and ... View answer 7 votes Instead of using van der Corput, why don't you express the sum as a complex integral? To simplify matters, consider a smooth sum, i.e.,$$S(x,t) = \sum_n f(n/x) n^{-i t},$$where f is fixed C^\... View answer 7 votes If nobody has a better idea, I will simply get a (real-variable) Taylor series for \zeta(\sigma+it) up to second-order with remainder. This is just (real) calculus - one can easily get the ... View answer 6 votes The "main part" of a conjecture such as Goldbach's is the statement that the number of counterexamples is finite (or even: that the number of ways of expressing a number as a sum of two primes is ... View answer 6 votes Not to toot my own tooting of other people's horns, but you'll find an explanation of k-ary Weisfeiler-Leman (complete with pseudocode) in section 2.5 of my Bourbaki talk on Babai's work (and Luks's, ... View answer 6 votes There is something the above posts missed (perhaps because the wording didn't make it explicit): the condition on the function f(x) is one sided (i.e., it assumes something on the decay as x\to \... View answer Accepted answer 6 votes Following up on Boris's suggestion, let me tell of my mostly happy experience with QEPCAD. First of all - QEPCAD seems to crash on three variables (at least for the slightly hairy expressions we are ... View answer 5 votes Self-answer: yes, the answer is Proposition 1.12 in Bennett-Martin-O'Bryant-Rechnitzer: https://projecteuclid.org/euclid.ijm/1552442669 They show that, for \chi a Dirichlet character modulo q\geq ... View answer 5 votes Here's a list of people in more than one section (in order of appearance...): a. Jaroslav Nesetril [1. Logic and 14. Combinatorics] b. Dmitry Kaledin [2. Algebra and 4. Algebraic Geometry] c. ... View answer 4 votes While @Lucia's answer is my favorite, I thought it might be worthwhile to sketch the Plancherel-based argument I alluded to in the above. First of all, let \phi:[0,\infty)\to \mathbb{R} be such that ... View answer 4 votes Wait, this isn't that hard. Let a positive integer \lambda\ll 1 be given. Let A=[0,1/2n]\subset \mathbb{R}/\mathbb{Z}. Let \phi be the multiplication-by-\lambda  map; then \phi^{-k}(A) is a ... View answer 3 votes One way to improve the explicit bound mentioned in the question is simply to compute L'(1,\chi)/L(1,\chi) for whatever characters \chi are needed. The bound in the question depends on a GRH ... View answer 3 votes The following is an answer that is also an attempt at interpreting Fedor Petrov's remark above. He may have had a somewhat different solution in mind, but the following procedure should be valid ... View answer 3 votes Thanks, GH! Let me have another go. I think the following is the right way to go about things, at least if one wants something self-contained and with good, explicit constants. (The latter more or ... View answer 3 votes Let me carry out matters using a complex-analytical approach, as Lucia suggests, and then say where the difficulty lies. Let 0<\beta<\alpha\leq 1. First of all, as Lucia says,$$\sum_{m\leq x}...

Actually, isn't question 2 brutally trivial? The following argument seems to show that the space $W$ spanned by the eigenspaces with eigenvalue $\geq \delta$ has dimension $\geq k$. Suppose this ...

Why this proposition is called "Flattening Lemma"? Is it because of the L2-norm of v∗v is smaller than the L2-norm of v? Yes, I'd think so. If you have two probability distributions $w$ and $v$, ...

There's a very nice answer to the first question (in the negative!) in https://arxiv.org/pdf/1910.01611.pdf The probability that $g$ be good actually goes to $0$ as $n\to \infty$. If one relaxes the ...

Unless I am very mistaken, there is an easy way to establish a bound of the "much better" kind mentioned in the comments above. (I don't doubt one can and should give a more precise answer.) Write $\... View answer 2 votes Mainly for purposes of comparison, let me flesh out what I called "a cheap version of Laplace". Write$\sigma = 2 r$. Choose$\rho\in (0,1)$. Let$g(y) = 1/(x_0^2+y^2)^{\sigma/2}$. Then$$g''(y) = ... View answer 2 votes This turns out to be an easy problem. For any$\epsilon>0$, we can cover all of$\mathbb{H}$except for a domain U of area$\epsilon$and its (horizontal) translates by$\mathbb{Z}$using words on$...

There are two approaches I can think of. (a) Analytic. We need only information about $\zeta(s)$, not about other $L$-functions. Hence we can use the fact that RH has been verified up to a very large ...

Ah, I get it - the factor of $\log x$ is really there because the truncation is sharp. If the truncation is continuous (and of bounded variation), then, not unexpectedly, the factor disappears. (See, ...

It seems to me that one can bound $|S|\leq (n-1) \deg(V)$. First, note that we can work projectively, that is, we will be able to work with the projective closure $\overline{V}\subset \mathbb{P}^n$. ...

Here is my self-answer to (c), based on my self-answer to (b). For any $f:V\to \mathbb{C}$ with $|f|_2^2=1$ and $|\langle f,\Delta f\rangle|\geq \alpha>0$, we obtain, proceeding as in my self-...

Let me show how to do (b), in a more general context than I set out in (b). Let $f:V\to \mathbb{C}$ with $|f|_2=1$ and $|\langle f, \Delta f\rangle|\geq \alpha>0$. Consider a partition of $V$ ...
Here's a very naive but arguably non-trivial bound. (Please feel free to do better!) Just choose a set $S_0$ in $\mathbf{P}$. It is clear that, for $\mathbf{S}\subset \mathbf{P}$ not containing $S_0$, ...