H A Helfgott
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A variant of the Goldbach Conjecture
Accepted answer
37 votes

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}(...

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Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$
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}}}...

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Was Vinogradov's 1937 proof of the three-prime theorem effective?
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 ...

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At what times were people interested in prime numbers
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 ...

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Sum over characters
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^\...

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Non-vanishing of zeta(s), Re(s)=1, without complex analysis?
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 ...

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Can anything deep be said uniformly about conjectures like Goldbach's?
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 ...

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On the use of Weisfeiler-Leman refinement in Babai's GI proof
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, ...

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Uncertainty principle for Mellin transform
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 \...

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Bounding Euler products (or almost) by products of zeta functions
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 ...

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Effective, explicit bound on $L'(1,\chi)/L(1,\chi)$
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 ...

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Work of ICM 2010 plenary speakers (and other humans)
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. ...

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Bound on $L^2$ norm of $1/\zeta(1+i t)$?
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 ...

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Showing non-expansion for $x\rightarrow x+1, x\rightarrow 2x.$
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 ...

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$|L'(1,\chi)/L(1,\chi)|$
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 ...

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Diameter for permutations of bounded support
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 ...

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$L_2$ bounds for tails of $\zeta(s)$ on a vertical line
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 ...

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Double sum of negative powers of integers: a direct approach?
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}...

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If many orthogonal vectors are respected (somewhat), are there many eigenvectors with large eigenvalues?
3 votes

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 ...

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L^2-Flattening Lemma
3 votes

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 $v$, ...

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Escaping from a centralizer
2 votes

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 ...

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How rare are unholey permutations?
2 votes

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 $\...

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$\int_{-\infty}^\infty \frac{e^{-y^2/2}}{((y+y_0)^2+x_0^2)^r} dy$
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) = ...

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Images of the fundamental domain of $\text{SL}_2(\mathbb{Z})\backslash \mathbb{H}$ whose Euclidean area is large
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 $...

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Counting square-free numbers smoothly
2 votes

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 ...

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Truncated Perron - logarithm-free error term?
1 votes

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, ...

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How many points can the projection of a variety to a line omit?
Accepted answer
1 votes

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 $\overline{V}\subset \mathbb{P}^n$. ...

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From one eigenvector to many, in a very local graph?
1 votes

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-...

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From one eigenvector to many, in a very local graph?
1 votes

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 $V$ ...

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Alternating sum over collections of sets
1 votes

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 $S_0$, ...

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