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Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
23
votes
Accepted
Algorithmic (un-)solvability of diophantine equations of given degree with given number of v...
I'm going to take a stab at this. First (as mentioned in Andres Caicedo's answer to this question), Siegel proved in 1972 that there is an algorithm to determine whether a quadratic equation in any nu …
11
votes
Accepted
Primes dividing $2^a+2^b-1$
This is a heuristic which suggests that the problem is probably quite hard. We have that $p | 2^{a} + 2^{b} - 1$ if and only if there is some integer $k$, $1 \leq k \leq p-1$ with $k \ne \frac{p+1}{2} …
5
votes
What are the $j$-invariants of the genus 1 modular curves?
First, there is no comprehensive list of all models of modular curves of genus $1$. (There is a list of congruence subgroups of $SL_{2}(\mathbb{Z})$ here.) Many cases have been computed, including the …
10
votes
Accepted
A strengthening of base 2 Fermat pseudoprime
Such an $n$ must be prime. If $\frac{1}{k} \binom{n-1}{2k-1}$ is an integer for all $1 \leq k \leq \lfloor \frac{n}{2} \rfloor$, then $n$ divides $\binom{n}{2k}$
for all $1 \leq k \leq \lfloor \frac{n …
7
votes
Accepted
Can the Petersson inner product $\langle f(z), f(2z) \rangle$ be zero?
Yes, the Petersson inner product can be zero. In my paper "Explicit bounds for sums of squares (see Lemma 5) I show that if $f$ is a newform of level $N$ and $p$ is a prime that does not divide $N$, t …
6
votes
Accepted
Intersecting Sets of Pythagorean Triples with Common Hypotenuses
Yes, given any $N \in \mathbb{N}$ there do exist $r$ and $s$ so that $A_{r} \cap A_{s}$ contains at least $N$ elements. In fact, I will show that one can take $s = 6r$. (The approach I suggest below i …
6
votes
Accepted
Can you find squares in this class?
Here's a bit more detail about my comment. We're searching for integer solutions to $y^{2} = p(l^{4} + 6l^{2}m^{2} - 3m^{4})$. Assume for simplicity that $\gcd(l,m) = 1$ and $p \geq 5$. If $p \equiv 2 …
10
votes
Accepted
Confusion regarding the definition of semistable reduction of an elliptic curve at a prime $p$
The criteria given by Silverman and Husemoller are equivalent, and so determining the "correct definition" is a bit tricky. In particular, on page 361 of Arithmetic of Elliptic Curves, Silverman (the …
3
votes
Finite generation of module of modular forms
Edit: This is an answer to the wrong question. This explains the finite generation of $M_{k}(\Gamma_{1}(n), R)$ as an algebra, not as a module over the graded ring of level $1$ modular forms.
This re …
7
votes
Euclid-style proof of Dirichlet’s theorem on primes in certain arithmetic progression
Yes. According to Paul Pollack's paper Hypothesis H and an impossibility theorem of Ram Murty, Murty gave an argument that an elementary Euclid-style proof is impossible when $a^{2} \not\equiv 1 \pmod …
3
votes
An upper bound of number of some fractional ideals
For any $n > 2$, there is no such upper bound. Consider for simplicity the case that $\mathcal{O} = \mathcal{O}_{K}$. If you take $J$ to be the identity class, then $f^{-1}(J)$ is the set of elements …
9
votes
Accepted
For which $n$ is it true that all surjections $SL_2(\mathbb{Z})\rightarrow SL_2(\mathbb{Z}/n...
The following mostly answers the question:
Claim 1: If every surjection $SL_{2}(\mathbb{Z}) \to SL_{2}(\mathbb{Z}/n\mathbb{Z})$ has kernel $\Gamma(n)$, then all prime factors $p | n$ have $p \leq 11$ …
7
votes
Accepted
Characterizing the newforms s.t. the associated symmetric square $L$-function has a pole
Yes, your understanding is correct. Here's a little bit more detail. If $f$ is a CM form, $f$ is associated to a Hecke Grössencharacter $\xi$. If $k \geq 2$, then $\xi$ is associated to an imaginary q …
3
votes
Accepted
Number of representations as sums of squares in rings of integers of number fields
There are some results known in the $n = 4$ case. In particular, in 1928 Gotzky (see Mathematische Annalen volume 100 pages 411-437) proved a formula for the number of representations of a totally pos …
1
vote
Finding integer representation as difference of two triangular numbers
There is a quick way to find solutions even without (completely) factoring. In particular, write $2n = 2^{\alpha} m$, where $m$ is odd and use the factorization $2n = m \cdot 2^{\alpha}$ to find $a$ a …