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Results tagged with nt.number-theory
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user 1384
Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
7
votes
An inequality for the Euler totient function
No it's not true. An equivalent question is whether the product of $(1-1/p)$, where $p$ ranges over all the primes for which 2 has odd order mod $p$, is greater than or equal to $1/2$. However an expl …
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3
votes
How to compute modular forms of weight one on Shimura curves?
I wonder whether you mean something else, but if you want to know what the answer is (dimensions of spaces, Hecke action etc) then all of this can be read off from Jacquet-Langlands and the analogous …
- 40.3k
5
votes
Accepted
Conditions of solution
As Gerhard says, $m=1$ works giving $p=5$ and $k=3$. I claim that this is the only time it happens.
Let $m\geq2$ be an integer, and set $p=2^{2m+1}-2^{m+1}+1$. I don't care if $p$ is prime or not, bu …
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7
votes
A very very good approximation to Ramanujan constant. Why?
This answer pertains to Michael's comment above.
Let $c$ be $e^{\pi\sqrt{163}}$. It is well-known that $c$ is close to an integer. More precisely, $c$ is about $10^{-12}$ from an integer of size abou …
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5
votes
Accepted
Proof in Schemmel's Paper
The case $n=4$ and $m=35$ given in both the question and the original German paper (link now removed from question) is a little misleading. In this case the sets are disjoint. The case $n=2$ and $m=35 …
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2
votes
Accepted
A question about numbers
No. How about $a=9$ and $b=16$? Then $c=25$ so $\Omega(a)=\Omega(c)=2\leq\Omega(b)$, the gcd's are $3,16,5$ so the left hand side is 6 and the right hand side only 5.
Edit: if $a=316$ and $b=27$ then …
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13
votes
Katz Modular Functions and Emerton's Completed Cohomology
I don't think I even know a natural non-zero map from the space $S_2(\Gamma_1(N);\mathbf{Q}_p)$ ("classical" modular forms with $p$-adic coefficients, defined for example as global sections of an appr …
- 40.3k
9
votes
Accepted
Density results for equality of Galois/automorphic representations
Firstly, at the beginning of the question you are missing irreducibility/cuspidality assumptions. If $\rho_1$ and $\rho_2$ are $\ell$-adic Galois reps with the same char poly in a set of primes of den …
- 40.3k
4
votes
Accepted
How can I evaluate the sum $\sum_{a+b=1;\ a,b \in \mathbb F_p} \left(\frac{a}{p}\right) \chi...
If I've understood the question correctly, this is just a Jacobi sum. The question only seems to make sense for $p$ a prime congruent to 1 mod 3. In this case, the order 3 character and the quadratic …
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13
votes
Accepted
Positive polynomial having roots modulo any integer
$\newcommand{\Q}{\mathbf{Q}}\newcommand{\Z}{\mathbf{Z}}$
I now suspect the answer might be no! This isn't a complete answer but it might be an idea that turns into one.
So let me assume that such $p$ …
- 40.3k
6
votes
Mid-Square with all bits set
Not an answer, but too long for a comment.
I can't see any tricks to do this other than a brute force computational approach. The naive approach would be to loop from $2^{96}$ to $2^{128}$ squaring e …
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14
votes
What is the limit of gcd(1! + 2! + ... + (n-1)! , n!) ?
Here's my guess: it might be out of reach to prove that $g_n$ tends to infinity, but it probably does, because $1!+2!+\ldots+(p-1)!$ is a "random" number mod $p$, so the chances that it's divisible by …
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10
votes
Reference for Deligne's construction of Galois representations attached to modular forms
I absolutely agree that Deligne is very terse. One thing that I ultimately found very helpful is Carayol's two papers where he proves the analogous theorem for Hilbert modular forms. I say "ultimately …
- 40.3k
54
votes
Accepted
Fermat's Last Theorem for Gaussian Integers ( excluding $\mathbb{Z}$ or $i\mathbb{Z}$ )
This is still way open, I should think. "Elementary" methods won't even solve the analogous problem over $\mathbf{Z}$, so you need to use "modular form" methods. The problem is that even if the result …
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15
votes
Ratio of triangular numbers
Any non-square $d$ can be written in this way. We need to solve $n(n+1)=dm(m+1)$, and multiplying by 4 and setting $a=2n+1$, $b=2m+1$ we get
$$a^2-db^2=1-d$$
and we need to solve this in odd integer …
- 40.3k
29
votes
Accepted
Complete discrete valuation rings with residue field ℤ/p
Greg, I want to say some basic things, but people are giving quite "high-brow" answers and what I want to say is a bit too big to fit into a comment. So let me leave an "answer" which is not really an …
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1
vote
Complete discrete valuation rings with residue field ℤ/p
Greg notes that $(Z/pZ)[[x]]$ is a limit of the integers of a sequence of totally ramified extensions of $Q_p$. At the back of my mind I wonder whether he'd be interested by the Fontaine-Wintenberger …
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12
votes
The product of n radii in an ellipse
OK so I bet my bottom dollar that this question is Fermat's Last Theorem in disguise. Very clever by the OP! Take for example $n=10$. Qiaochu's formula above, when expanded out and simplified, becomes …
- 40.3k
1
vote
Question on determining the minimal polynomial for an algebraic quotient
Here's a suggestion. Use polcompositum(FA,FC) (with FA the min poly of A, FC the min poly of C) to find a number field K=Q(alpha) containing roots of both your polynomials, and then use lindep() to fi …
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11
votes
Accepted
The closure of the set of rational points in the Adeles
Here's an example where $X(\mathbf{Q})$ is Zariski-dense but the first inequality is not an equality.
Let $X$ be an elliptic curve over the rationals, such that the group $X(\mathbf{Q})$ is isomorph …
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6
votes
Conductor of monomial forms with trivial nebentypus
Thinking about Idoneal's question to Emerton about what is being used about the Steinberg, it's not just standard facts about the Steinberg one needs via this approach, but also local-global compatibi …
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9
votes
Accepted
If $f$ is an $p$-nonordinary eigenform of weight $k\leqslant p+1$ are there always two eigen...
It's deeper than theta cycles, I think.
I am going to assume that $N$ is prime to $p$ -- you don't say this in your question but most of my answer assumes this in a very serious way.
If $f$ is ordin …
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8
votes
What is the interpretation of complex multiplication in terms of Langlands?
Your assumption as written is not quite correct. The irreducible $n$-dimensional irreducible representations of the Langlands group should correspond I think to all cuspidal automorphic representation …
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9
votes
Accepted
Equality of the sum of powers
An even harder problem than $t>2$ and $n=m$ is the Prouhet–Tarry–Escott problem. Now I leave it to you and google to find lots of examples ;-)
http://en.wikipedia.org/wiki/Prouhet-Tarry-Escott_proble …
- 40.3k
18
votes
Accepted
Find $x,y,z \in \mathbb{Z}$ with $|x^2 + y^2 - \sqrt{3} z^2| < 10^{-6}$
$x=6627,y=314048,z=238678$ is an answer to the question in the title but I'm sure that the real question is something I don't really understand.
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8
votes
Properties shared by number fields with the same normal closure?
Fields with the same Galois closure can be completely different. For example take a random irreducible polynomial of huge degree, let $F$ be the field obtained by adjoining one root of the polynomial, …
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13
votes
Class number measuring the failure of unique factorization
For Dedekind domains, like the integers of a number field, PID iff UFD. There's definitely a quantitative statement relating the class number to failure of PIDness: the higher the class number, the sm …
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11
votes
Accepted
representability of consecutive integers by a binary quadratic form
Such a $C$ does not exist, if I got it right.
My example is the function $f(x,y)=(2x+1)(5y+1)$. An integer is represented by $f$ iff it can be written as the product of an odd integer and a number wh …
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6
votes
Are there primes p, q such that p^4+1 = 2q^2 ?
I don't know if there is a simple proof, but I know one which is easy to do because it lets a computer do all the work (but the work is perhaps complicated): you simply ask a computer to solve Y^2=2X^ …
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13
votes
Accepted
Can a the q-expansion of a p-adic modular form be a non-constant polynomial?
It is. I want to argue the following way: if the polynomial is non-constant then after scaling it has integral coefficients and so the reduction of the p-adic form mod p^n will be a classical form who …
- 40.3k
17
votes
Accepted
The first odd degree-2 Artin representation for which the Artin conjecture was proved
$R$ is a 2-dimensional conductor 133 representation of the absolute Galois group of the rationals into $GL(2,\mathbf{C})$, whose associated representation to $PGL(2,\mathbf{C})$ cuts out the $A_4$ ext …
- 40.3k
5
votes
Accepted
Fiddling with p-adics
A finite extension of a complete field is complete. This is standard and proved in lots of places e.g. one of the first two chapters of Cassels-Froehlich, or Bosch-Guentzer-Remmert, or lots of other p …
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3
votes
Is there a schemetical construction for modular curves over the rationals?
Passing comment: Mumford's GIT constructs modular curves as quotients---not of the upper half plane, but of some parameter space of subspaces of projective space, by an algebraic group. As for the las …
- 40.3k
9
votes
The resultant and the ideal generated by two polynomials in $\mathbb{Z}[x]$
$D=D(f,g)$ is not particularly well-behaved, is it. For example it's not multiplicative in the variables: if $g=x^2+1$ (nothing special about this example, I'm just fixing ideas) then $D(f,g)$ is the …
- 40.3k
19
votes
2
answers
576
views
Can something finite over $\mathbb{C}(q)$ be a modular form?
If $f\in\mathbf{C}[[q]]$ is non-constant, and algebraic over $\mathbf{C}[q]$ (in the sense that it is a root of a polynomial with coefficients in in $\mathbf{C}[q]$) then can $f$ be the $q$-expansion …
- 40.3k
14
votes
2
answers
2k
views
Lifting the p-torsion of a supersingular elliptic curve.
Let $K$ be a finite extension of $\mathbf{Q}_p$, with integer ring $R$ and residue field $k$. Say $G/R$ is a finite flat (commutative) group scheme of order $p^2$, killed by $p$. Say the special fibre …
- 40.3k
18
votes
What is the difference between a zeta function and an L-function?
Although no-one else seems to have suggested this, my personal take on this is that there's no difference whatsoever. Sure, if people start talking about "Dedekind zeta functions" or "Artin $L$-functi …
- 40.3k
17
votes
Uniform Faltings
On the contrary, some conjectures suggest that the answer is NO! It follows from the Bombieri-Lang conjecture (sometimes known as Lang's conjectures) that a uniform bound should exist.
More precisel …
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25
votes
2
answers
2k
views
Proving non-existence of solutions to $3^n-2^m=t$ without using congruences
I made a passing comment under Max Alekseyev's cute answer to this question and Pete Clark suggested I raise it explicitly as a different question. I cannot give any motivation for it however---it was …
- 40.3k
11
votes
1
answer
554
views
Non-trivial class number at some finite level in the cyclotomic $\mathbf{Z}_p$-extension of ...
An MSc student asked me if I knew an example of a prime $p$ and some finite layer $K_n$ in the cyclotomic $\mathbf{Z}_p$-extension of $\mathbf{Q}$ (so $[K_n:\mathbf{Q}]=p^n$) which had non-trivial cla …
- 40.3k
18
votes
Accepted
Is there a canonical notion of "mod-l automorphic representation"?
This is in my mind a central open problem.
Here is an explicit example which I believe is still wide open. Serre's conjecture (the Khare-Wintenberger theorem) says that if I have a continuous odd irre …
- 40.3k
29
votes
6
answers
4k
views
Infinitely many primes of the form $2^n+c$ as $n$ varies?
At the time of writing, question 5191 is closed with the accusation of homework. But I don't have a clue about what is going on in that question (other than part 3) [Edit: Anton's comments at 5191 cla …
- 40.3k
68
votes
Accepted
Smooth proper scheme over Z
Hey Bjorn. Let me try for a counterexample. Consider a hypersurface in projective $N$-space, defined by one degree 2 equation with integral coefficients. When is such a gadget smooth? Well the partial …
- 40.3k
19
votes
1
answer
1k
views
Are Q-curves now known to be modular?
I really should know the answer to this, but I don't, so I'll ask here.
A Q-curve is an elliptic curve E over Q-bar which is isogenous to all its Galois conjugates. A Q-curve is modular if it's isoge …
- 40.3k
6
votes
Proof of no rational point on Selmer's Curve $3x^3+4y^3+5z^3=0$
There's a proof in Cassels' little blue book on elliptic curves which the OP might find more to his taste than some others mentioned here.
- 40.3k
11
votes
Accepted
Hilbert Modular Newforms
If I've understood your question correctly, you're right that $C(q,f)\not=0$ always and there is a natural representation-theoretic proof of this result (before I start let me say that I don't know ho …
- 40.3k
11
votes
0
answers
412
views
Growth of $n=n(k)$ for which there's a non-trivial solution to $x_1^k+\cdots+x_n^k=y^k$?
Walter Hayman just asked me the following question. What, if anything, is known about the growth of the function $n(k)$, where $k\geq1$ is an integer, and $n=n(k)\geq2$ is the smallest integer for whi …
- 40.3k
8
votes
Can the failure of the multiplicativity of Euler factors at bad primes be corrected?
James: given that no-one else answered this yet, let me just make some naive comments that you probably know already.
Of course the problem is that if $I_p$ isn't acting trivially, then "taking $I_p …
- 40.3k
40
votes
Accepted
How did Birch and Swinnerton Dyer arrive at their conjecture?
For what it's worth, here are some historical comments.
Both Birch and S-D spoke in Cambridge a few weeks ago, on the history of their conjecture. To my surprise, both of them emphasized the role not …
- 40.3k
1
vote
How badly can strong multiplicity one fail in the theory of automorphic representations?
For what it's worth I can now answer Q0. I believe it is not true in general that $\pi_v$ and $\pi'_v$ will have to have the same central character. We can let $G$ be a torus $T$. If $S$ is a finite s …
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