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Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
2
votes
Accepted
Wiener-Ikehara tauberian theorem and order of pole at s=1
Here's a more detailed answer - fleshing out Daniel Loughran's comment. For
$\lim_{X \to \infty} \frac{1}{X} \sum_{n < X} \lambda(p,\pi,\rho)$, one applies the Wiener-Ikehara theorem to $L(s,\pi,\rho) …
6
votes
Accepted
Example of two p-Ordinary Elliptic Curves congruent to each other
There are many such curves. For $p = 2$, one can just take two curves $E$ and $F$ with $E : y^{2} = f(x)$ and $F : y^{2} = g(x)$ where $f(x)$ and $g(x)$ define the same number field. Also, if $E$ and …
7
votes
Accepted
Class number for binary quadratic forms discriminant $\Delta$ to class number $\mathbb Q(\sq...
It seems to me that what Buell says about the narrow class group is not quite right (it's hard for me to say, as I don't have a copy of it). Magma tells me that in $\mathbb{Q}(\sqrt{210})$, the narrow …
7
votes
Accepted
Density of prime divisors of $a^n + b$
This is only known conditional on the generalized Riemann hypothesis. A prime $p$ dividing the numerator of $a^{n} + b$ is more or less equivalent to the statement that the subgroup of $\mathbb{F}_{p} …
5
votes
Accepted
Exact bin packing the harmonic series: references?
The next $n$ for which there is an exact packing are $24 \leq n \leq 30$. This is because
$$
1 = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{1}{4} + \frac{1}{5} + \frac{1}{8} + \frac{1}{9} + \frac …
18
votes
Accepted
Would such polynomial identity exist? (related to sum of four squares)
To see that $f_{1}$, $f_{2}$, $\ldots$, $f_{k}$ must be scalar multiples of $f_{k+1}$, plug in the root of $f_{k+1}$ into both sides of the equation.
5
votes
Accepted
Adelic open image for modular forms?
As I mentioned in my comment, an adelic open image theorem should follow in a purely group-theoretic way from the knowledge that the $\ell$-adic representations are surjective (for an appropriately sp …
5
votes
Accepted
On the elliptic curve $x(x+a^2)(x+b^2) = y^2$
Yes, if $E : y^{2} = x(x+a^{2})(x+b^{2})$ has a point $P$ whose $x$-coordinate is $au^{2}$ for some nonzero rational number $w$, it also has a point $Q$ whose $x$-coordinate is $bv^{2}$ for some nonze …
10
votes
Number of points of elliptic curve over $\mathbb{F}_p$ with CM by $\sqrt{-2}$ when $p\equiv ...
Your conjecture is true and follows from a theorem in Cox's book ''Primes of the form $x^{2} + ny^{2}$.'' (Theorem 14.16 on page 317, although this is from the first edition.) In particular, if $\math …
15
votes
Existence of rational points on some genus 3 curves
There are no rational solutions to curve (b). This curve has the automorphism $(x,y) \mapsto (x,-y)$ and the quotient is the genus one curve
$$ -yz + w^{2} = 0 \quad x^{2} + y^{2} + xz - z^{2} + yw = …
17
votes
Accepted
Existence of rational points on a generalized Fermat quartic
This question is amenable to the use of the Mordell-Weil sieve. (For a good introduction to this technique, see the paper here.) In this situation, there is a simple version of it (using a single prim …
8
votes
Accepted
Trace-free basis for $\mathcal{O}_K$, $K$ a cubic field
No. There do not always exist such $\alpha$ and $\beta$. If $K$ is a cubic field and such $\alpha$ and $\beta$ exist, then for all $x \in \mathcal{O}_{K}$, ${\rm Tr}\left(\frac{1}{3} \cdot x\right) \i …
1
vote
Half integral weight modular forms that reduce to a nonzero constant modulo a given prime
If such a form does exist, then its level must be a multiple of $p$.
If $f = \sum a_{n} q^{n}$ is a half-integer weight modular form with integer coefficients with $a_{i} \equiv 0 \pmod{p}$ for all $ …
10
votes
Accepted
Integral points near elliptic curves
You can take $\theta = 0$, even $\theta = -1/6$ works.
Fix an integer $A \ne 0$ and an integer $B$. If $r$ is an integer, the elliptic curve $E : y^{2} = x^{3} + Ax + r^{2} A^{2}$ has the obvious poin …
3
votes
Accepted
How different can characters be for a sum of modular forms to still be in Gamma_0?
The bad news: Yes, you are forced to go to $\Gamma_{1}(N)$, and not $\Gamma_{0}(N)$. In particular, let's say that $E = \sum \sigma_{1}(2n+1) q^{2n+1}$ is the usual weight $2$ level $4$ Eisenstein ser …