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Results tagged with nt.number-theory
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user 48142
Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
7
votes
Divisibility condition implies $a_1=\dotsb=a_k$?
Here's a a tweak of Seva's idea that gives a counterexample. Note that if $r$ is odd, then $2^{n}+1$ divides $2^{rn} + 1$.
Let $k = 6$, $a_{1} = 1$, $a_{2} = a_{3} = a_{4} = 2$, $a_{5} = a_{6} = 4$. T …
- 19.9k
7
votes
Accepted
Isomorphism between genus 1 modular curves and elliptic curves
The maps $j : X_{0}(N) \to \mathbb{P}^{1}$ are given in Magma's ''Small modular curves'' database. In each case, they construct functions on the modular curves $X_{0}(N)$ out of eta products, modular …
- 19.9k
4
votes
Accepted
Generating DataSet of Strong PseudoPrimes?
I'm guessing you will want to be working with numbers larger than 64-bits, and so you probably want GMP (see this page). This library is used by much of the software that number theorists use. (Magma …
- 19.9k
12
votes
Accepted
Etale cohomology approach on $\tau(n)$
One of the goals of the development etale cohomology was to generate a cohomology theory that could successfully count points on varieties over finite fields, with one of the main goals of proving the …
- 19.9k
6
votes
Approximations to $\pi$
This is not exactly an answer to the stated question, but it's too long for a comment. Rather than the form given in the question, one could represent a number in the form $\frac{a + b \sqrt{d}}{c}$, …
- 19.9k
4
votes
Accepted
Best error terms for functions related to square free numbers
As I say in the comment, the asymptotics for $M_{+}$ and $M_{-}$ follow directly from those for $M$ and $\hat{M}$. Therefore $M_{+}(x) = \frac{1}{2 \zeta(2)} x + \frac{1}{2} M(x) + O(x^{1/2})$ and $M_ …
- 19.9k
7
votes
Accepted
Question about iterations not divisible by infinitely many prime numbers
Yes. This follows from a result of Corrales-Rodrigáñez and Schoof (see the paper here) solving the support problem of Erdős.
In particular, suppose that there are only finitely many primes $p$ that do …
- 19.9k
22
votes
Accepted
On Fibonacci numbers that are also highly composite
The largest highly composite Fibonacci number is $F_{3} = 2$.
If $p$ is a prime number, then either $p \mid F_{p-1}$ (if $p \equiv \pm 1 \pmod{5}$), $p \mid F_{p}$ (if $p = 5$), or $p \mid F_{p+1}$ (i …
- 19.9k
34
votes
Accepted
Question on a generalisation of a theorem by Euler
I suspect that $k = 4$ is good, but am not sure how to prove it. However, every positive integer $k \geq 5$ is good. This follows from the fact (see the proof of Theorem 1 from this preprint) that for …
- 19.9k
10
votes
Accepted
Prime numbers in a sparse set
Yes, there is a $c > 1$ for which infinitely many numbers of the form $\lfloor k^{c} \rfloor$ are prime. The first result of this type was proven in Ilya Piatetski-Shapiro's Ph.D. thesis (written in 1 …
- 19.9k
17
votes
Accepted
A divisor sum congruence for 8n+6
The congruence you state is true for all $m \equiv 6 \pmod{8}$. The proof I give below relies on the theory of modular forms. First, observe that
$$
\sum_{k=1}^{m-1} d(k) d(m-k) = 2 \sum_{k=1}^{\frac{ …
- 19.9k
9
votes
Accepted
Complexity of computing the number of visible points
There is an algorithm for computing $F(N) = \# \{ (a,b) : 1 \leq a, b \leq N, \gcd(a,b) = 1 \}$ in time $O(N^{5/6 + \epsilon})$. This relies on the algorithm of Deleglise and Rivat (see their paper he …
- 19.9k
7
votes
The Chebotarev Density Theorem and the representation of infinitely many numbers by forms
The example of Heath-Brown's article is a good one. For a bit more elementary examples, you can fix a number field $K$ with $[K : \mathbb{Q}] = n$ and ring of integers $\mathcal{O}_{K}$ and pick a bas …
- 19.9k
3
votes
Minimum number of unit fractions to sum up a given positive rational
No such polynomial-time algorithm exists because in some instances it would take too long to write down (or store in memory) the answer. In particular, if $n$ is a positive integer, then we have $\sum …
- 19.9k
10
votes
Accepted
Extension of a formula for the quadratic Gauss sums
No, the relation is not as simple in this case. For example, if $k = 3$ and $p = 7$, the three different cubic Gauss sums are roots of $y^{3} - 21y - 7$, and the three roots of this polynomial do not …
- 19.9k
8
votes
$2^n$-1 consisting only of small factors
It is true that if $N > 60$, then $2^{N} - 1$ has a prime factor $> 2500$.
Here's another approach. First, observe that every prime factor of $2^{p} - 1$ is $\equiv 1 \pmod{p}$. Combining this with t …
- 19.9k
3
votes
Accepted
When are the powers of 2 sum-free mod n?
This question is very similar to the one here, and the heuristic should apply equally well. In particular, $A$ is sum-free if and only if there does not exist a $k$ with $k \ne \frac{n+1}{2}$ so that …
- 19.9k
7
votes
Accepted
primitive prime divisor of $2^{8n+4} - 1 $
No. We have that $p = 709$ is a primitive prime divisor of $2^{708} - 1$. However, $\frac{2^{708} - 1}{2^{177} + 1}$ is a multiple of the prime $q = 5521693$ and therefore $q-1 | \gamma\left(\frac{2^{ …
- 19.9k
5
votes
Accepted
$p$-th root of non-torsion points on elliptic curves
No. Lemma 3.7 on page 11 from my paper here implies that if $\mathcal{T}_{1} = {\rm Gal}(K(E[p])/K)$ and there is a normal subgroup $H \unlhd \mathcal{T}_{1}$ with order coprime to $p$ for which $E[p] …
- 19.9k
16
votes
Diophantine equation $3(a^4+a^2b^2+b^4)+(c^4+c^2d^2+d^4)=3(a^2+b^2)(c^2+d^2)$
The equation you specify defines a surface $X$ in $\mathbb{P}^{3}$, and this surface is a K3 surface. It is conjectured that if $X$ is a K3 surface, there is a field extension $K/\mathbb{Q}$ over whic …
- 19.9k
9
votes
Accepted
Congruences among primes modulo which a given polynomial has roots
Here's a survey of the possible things that can happen. In regards to your first question, given any polynomial $f(x)$, there is a positive integer $M$ so that if $\gcd(b,M) = 1$, then there are infin …
- 19.9k
9
votes
Accepted
Sets of squares representing all squares up to $n^2$
We can have the size of $S$ as small as $c \ln(n)$ for some constant $c$, and we can do this in such a way that every element of $\{1, 2, \ldots, n^2 \}$ can be represented by adding or subtracting at …
- 19.9k
4
votes
Relations of eisenstein series with eta quotient
The answer is probably some fairly basic linear algebra. For the first one, each term on the right hand side is a modular form of weight $4$ and level $2$. The space $M_{4}(\Gamma_{0}(2))$ has dimensi …
- 19.9k
8
votes
Accepted
Is $p$ is square modulo $F_p$ when $p=4k+1 > 5$?
The answer to the first question is yes, although the argument I give below is not along the lines that you were originally thinking. I will show that $p$ is a square modulo $q$ for every prime factor …
- 19.9k
40
votes
Accepted
When $\frac{a^2}{b+c}＋\frac{b^2}{a+c}＋\frac{c^2}{a+b}$ is integer and $a,b,c$ are coprime na...
Yes, there is another solution. The next one I found is a bit big, namely
$$ a = 15349474555424019, b = 35633837601183731, c = 105699057106239769. $$
This solution also satisfies the property that
$$ …
- 19.9k
12
votes
Accepted
Extension of $\mathbb Q$ which splits only at primes in $S$
For many choices of $R$ and $S$ the answer is obviously no. For example, if $R$ is empty, then the answer is no, because there are no unramified extensions of $\mathbb{Q}$.
For a more interesting exa …
- 19.9k
4
votes
Accepted
Solutions to diophantine equation
I probably put a little bit too much effort into this. The only rational point on this curve is $(0,0)$ (as well as the points at infinity $(1 : 5 : 0)$ and $(1 : -5 : 0)$).
There's a slightly non-ob …
- 19.9k
10
votes
Accepted
Cusp forms with integer Fourier-coefficients
No. Take $k = 24$ and $N = 1$. Then $\Delta^{2} = q^{2} - 48q^{3} + 1080q^{4} + \cdots \in S_{K}(\Gamma_{1}(N),\mathbb{Z})$. However, if we write $\Delta^{2} = c_{1} f_{1} + c_{2} f_{2}$, where $f_{1} …
- 19.9k
5
votes
Accepted
Representation of integers by positive definite ternary quadratic polynomials with linear terms
Yes, it is possible to extend these methods, although the picture is somewhat
less clear than in the quadratic form case. When one discusses representations by a quadratic polynomial, this is equival …
- 19.9k
6
votes
Accepted
How does this sequence grow
The answer is yes, and the number of solutions with a prime $p$ is $\lfloor \frac{p+5}{8} \rfloor$ when $p \not\equiv 1 \pmod{8}$ and is $\lfloor \frac{p+5}{8} \rfloor + 1$ when $p \equiv 1 \pmod{8}$. …
- 19.9k