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<-------- Over there, you'll find my website.


11h
comment Approximating the number of primes
Minor correction, according to polymathprojects.files.wordpress.com/2010/07/polymath.pdf there exists a $O(n^{2/3})$ algorithm to compute $\pi(n)$ due to Meissel-Lehmer, ($O(n^{1/2})$ might need RH) but that's not the $O(n^{1/3})$ that you want.
12h
comment Approximating the number of primes
"easy" and "nice" are not well-defined. If you want an algorithm that, in deterministic polynomial time (in $\log n$) computes $\pi(n)$ with an error better than $\sqrt{n}$, this is an open problem related to that considered by the polymath4 project.
12h
comment Approximating the number of primes
Depends on what you mean by nice. Riemann found an explicit formula for $\pi(x)$, which I find very nice.
1d
comment Computing a projection of a $p$-adic plane curve
It's a union of disks. To get a description of the disks, you need to know the description of the set of solutions to $f(x,y) \equiv 0 \pmod{p}$. What do mean by a description of that?
1d
comment Strong divisibility of Lucas sequences
I think I was optimistic in my eventual injectivity. It's clear when the roots of the characteristic equation are real but not when they are complex. But to answer the question it enough to establish that if $k|m$ and $u_m=u_k$, then $m=k$. For that, even in the complex case, the growth is fast enough so that this implication holds for $m$ large enough.
2d
comment Old Math books, will research and sell most
This will be closed as off-topic but, back in 2009, when this website started, it wasn't. This question was asked and answered: mathoverflow.net/questions/9793/…
Sep
19
comment Strong divisibility of Lucas sequences
@Fry If $|\alpha| = |\beta| = 1$, then $|b| = |\alpha||\beta| = 1$ and for the roots to be complex $a^2 +4b <0$ so $b=-1, |a| < 2$.
Sep
19
comment Strong divisibility of Lucas sequences
$u_n$ grows exponentially (except in trivial cases) so you get eventual injectivity. That's the best you can expect, e.g. $a=b=1$ (Fibonacci) has $u_2=u_1$.
Sep
19
comment Proof that at least one of the nontrivial zeta zeroes has an irrational height (assuming RH)
An infinite product of rational numbers is not necessarily rational.
Sep
19
comment Endomorphism Ring of Simple Abelian Varieties
See here: mathoverflow.net/questions/119956/…
Sep
19
comment Algebraic approach to showing trigonometric equations have no solution
After your step 1., you are better off using techniques from real algebraic geometry. The Tarski-Seidenberg theorem should provide an algorithmic way of deciding if your system has a solution.
Sep
19
comment Is there a von Koch-type theorem for the generalized Riemann hypothesis?
Yes. If the estimate you wrote holds for fixed $d$ and all $a, (a,d)=1$, then RH is true for the zeta function of the cyclotomic field of $d$-th roots of unity by the explicit formula. This should be in standard books (e.g. Lang Algebraic Numbers, Davenport, Iwaniec-Kowalski,...)
Sep
18
comment Is the support of two odd theta characteristics on a generic curve disjoint?
@Torquemada I don't understand your final objection. I can't rule out lots of points, yes. But as you show in genus four, ruling out lots of points should be easier. You should reconsider your username, BTW.
Sep
18
comment Is the support of two odd theta characteristics on a generic curve disjoint?
@abx If the point is unique, it's defined over $K$. If there is more than one, then I guess you are right but then it's even more unlikely. Thanks for the correct attribution.
Sep
17
answered Is the support of two odd theta characteristics on a generic curve disjoint?
Sep
16
comment Go I Know Not Whither and Fetch I Know Not What
I am honored to be confused with GH.
Sep
16
comment Go I Know Not Whither and Fetch I Know Not What
That's what I meant by congruences and it's not automatic. By pure thought, if the value of polynomial is divisible by a suitably high power of three, then the variables are divisible by three. To figure out what suitably high means, requires a calculation which I won't do.
Sep
16
comment Go I Know Not Whither and Fetch I Know Not What
This is the norm of $a+b\sqrt{3}+c\sqrt{5}+d\sqrt{15}$ from the quartic field to the rationals, so it's a norm form. I don't think $1,\sqrt{3},\sqrt{5},\sqrt{15}$ form a basis of the ring of integers over the usual integers, so the congruences don't automatically follow from facts about norms, but it should not be hard to prove them that way. You can also use that to get information on numbers (integers?, rationals?) represented by it. I don't know how hard it would be to answer your last question.
Sep
15
comment Is it easy to prove that $\sum_n |X(\mathbb{F}_{q^n})| t^n$ is rational?
For curves, the rationality of the zeta function follows easily from the expression of the zeta function in terms of divisors and Riemann-Roch. But I don't see how to get the rationality of $P$ directly for curves.
Sep
12
comment Subgroup cliques in the Paley graph
You should double check, but I think you get $c\sqrt{p}$ for some $c<1, c=1/\sqrt{2}$ maybe.