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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
comment Nonlinear equations in integers
Quid: what about $3^2 + 4^2 = 5^2$?
May
22
comment 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
comment 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
comment 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
comment About unpublished lecture notes of Philip Hall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
comment 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
comment 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
comment 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
comment 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 $p \rightarrow \infty$.