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May
22 |
comment |
Nonlinear equations in integers
One of the references here, I bet: algo.inria.fr/csolve/triple |
May
22 |
comment |
Nonlinear equations in integers
I'm pretty sure, actually, that all sets of $99\%$ of $[1,...,N]$ contain a triple $(3k, 4k, 5k)$. I'm also pretty sure I can find a reference for this fact given a minute or two. |
May
22 |
comment |
Nonlinear equations in integers
Mark, for the Pell equation there is a huge set $A$, of size $N - O(\log N)$, just by deleting all $x$ and $y$ that are solutions to the Pell equation? |
May
22 |
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Nonlinear equations in integers
Quid: what about $3^2 + 4^2 = 5^2$? |
May
22 |
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Nonlinear equations in integers
Siming, I think that taking the squares $x^2$ with $x$ either odd or $\equiv 2 \mod{4}$ gives you a set consisting of $\frac{3}{4}$ of the squares with no solution to $x^2 + y^2 = z^2$ (look mod $8$). Probably it's true that if you take $99 \%$ of the squares then there's a solution to $x^2 + y^2 = z^2$. I don't immediately have a feel for whether this is doable or not; I'll get back to you. Certainly use of the circle method will be problematic if one proceeds naively. |
May
22 |
answered | Nonlinear equations in integers |
May
17 |
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Gauss sums over multiplicative subgroups
Igor, Thanks for this mention. I did try pretty hard with that particular set of notes, though I might change one or two things if I were doing them again. I had some help from Bourgain and Lindenstrauss. |
Apr
23 |
awarded | Yearling |
Apr
12 |
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Why groups that admit Folner Sequences are amenable
Jo, you can use the Folner sequence to define an almost invariant mean, then take a limit of these along an ultrafilter to get a genuinely invariant one. See for example these notes, starting page 26, for a discussion in the case of Z. dpmms.cam.ac.uk/~bjg23/ATG/Chapter3.pdf I'm speaking here of the case when G is a discrete group; in the locally compact case matters are a little more complicated. There are many better sources in the literature - recent blog notes of Tao, to give just one example. |
Apr
8 |
awarded | Necromancer |
Apr
2 |
revised |
Perron, Fourier
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Apr
1 |
awarded | Nice Answer |
Mar
28 |
answered | Perron, Fourier |
Mar
19 |
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About unpublished lecture notes of Philip Hall
There's a copy of the Edmonton notes in the Cambridge maths library; I consulted it a couple of years ago. |
Feb
3 |
revised |
minimum number of subsets?
added 392 characters in body; added 131 characters in body |
Feb
3 |
answered | minimum number of subsets? |
Dec
27 |
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Every prime number > 19 divides one plus the product of two smaller primes?
Denis, I don't understand your comment. I only know two things that count as good reasons in the theory of prime numbers. First, a proof. Second, a decent heuristic or "statistical" argument. Here, the set of $1 + rs$ ought to look like a fairly random (give or take some irregularities mod small primes) subset of $[1,p^2]$ of density $1/\log^2 p$. The probability of $x \leq p^2$ being divisible by $p$ is $1/p$. I'd expect subsets of $[1,X]$ of densities $\alpha$ and $\beta$ to intersect as soon as $\alpha \beta \gg 1/X$ unless there is some good reason why not. |
Dec
27 |
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Every prime number > 19 divides one plus the product of two smaller primes?
Geoff- there certainly is work on this. It's known that the smallest prime congruent to $a$ mod $q$ is $\ll q^{5.4}$ (or so). It's conjectured that it's $\ll q^{1 + \epsilon}$. The GRH would give $\ll q^{2 + \epsilon}$. |
Dec
27 |
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Every prime number > 19 divides one plus the product of two smaller primes?
Mark - yes, there are "sum-product" results of this type. The one most applicable here is that due to Bourgain-Katz-Tao: it says that if $A \subseteq Z/pZ$ and $|A + A|, |A \cdot A| \leq K|A|$ then either $|A| \leq K^C$ or $|A| \geq K^{-C} p$. When $A$ has size about $p/\log p$, it tells you rather little (namely either $|A + A|$ or $|A \cdot A|$ has size at least $p/(\log p)^{1 - c}$) and I think exponential sum techniques are likely to be better. |
Dec
27 |
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Every prime number > 19 divides one plus the product of two smaller primes?
I might add that I would expect trigonometric sums to allow one to show that $A \cdot A$ is almost all of $Z/pZ$ as $p \rightarrow \infty$. |